\(\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [393]
Optimal result
Integrand size = 33, antiderivative size = 421 \[
\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}
\]
[Out]
1/2*(a^3*(A-B)-3*a*b^2*(A-B)-3*a^2*b*(A+B)+b^3*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/2*(a^3*(
A-B)-3*a*b^2*(A-B)-3*a^2*b*(A+B)+b^3*(A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/4*(3*a^2*b*(A-B)-b^
3*(A-B)+a^3*(A+B)-3*a*b^2*(A+B))*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-1/4*(3*a^2*b*(A-B)-b^3*(A
-B)+a^3*(A+B)-3*a*b^2*(A+B))*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)+2*(3*A*a^2*b-A*b^3+B*a^3-3*B*
a*b^2)*tan(d*x+c)^(1/2)/d+2/21*b*(21*A*a*b+18*B*a^2-7*B*b^2)*tan(d*x+c)^(3/2)/d+2/35*b^2*(7*A*b+11*B*a)*tan(d*
x+c)^(5/2)/d+2/7*b*B*tan(d*x+c)^(3/2)*(a+b*tan(d*x+c))^2/d
Rubi [A] (verified)
Time = 0.98 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.00, number of
steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3688, 3718, 3711, 3609,
3615, 1182, 1176, 631, 210, 1179, 642} \[
\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 b \left (18 a^2 B+21 a A b-7 b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)}{21 d}-\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \sqrt {\tan (c+d x)}}{d}+\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 b^2 (11 a B+7 A b) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}
\]
[In]
Int[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]
[Out]
-(((a^3*(A - B) - 3*a*b^2*(A - B) - 3*a^2*b*(A + B) + b^3*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sq
rt[2]*d)) + ((a^3*(A - B) - 3*a*b^2*(A - B) - 3*a^2*b*(A + B) + b^3*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d
*x]]])/(Sqrt[2]*d) + ((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan
[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - ((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*L
og[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (2*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*
Sqrt[Tan[c + d*x]])/d + (2*b*(21*a*A*b + 18*a^2*B - 7*b^2*B)*Tan[c + d*x]^(3/2))/(21*d) + (2*b^2*(7*A*b + 11*a
*B)*Tan[c + d*x]^(5/2))/(35*d) + (2*b*B*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^2)/(7*d)
Rule 210
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 631
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 642
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Rule 1176
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Rule 1179
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Rule 1182
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]
Rule 3609
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Rule 3615
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]
Rule 3688
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f
*(m + n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[a^2*A*d*(m +
n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m
- 1) - b*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&
!(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Rule 3711
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&& !LeQ[m, -1]
Rule 3718
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])
^(n + 1)/(d*f*(n + 2))), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && !LtQ[n, -1]
Rubi steps \begin{align*}
\text {integral}& = \frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}+\frac {2}{7} \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x)) \left (\frac {1}{2} a (7 a A-3 b B)+\frac {7}{2} \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {1}{2} b (7 A b+11 a B) \tan ^2(c+d x)\right ) \, dx \\ & = \frac {2 b^2 (7 A b+11 a B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}-\frac {4}{35} \int \sqrt {\tan (c+d x)} \left (-\frac {5}{4} a^2 (7 a A-3 b B)-\frac {35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-\frac {5}{4} b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^2(c+d x)\right ) \, dx \\ & = \frac {2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}-\frac {4}{35} \int \sqrt {\tan (c+d x)} \left (-\frac {35}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )-\frac {35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)\right ) \, dx \\ & = \frac {2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}-\frac {4}{35} \int \frac {\frac {35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac {35}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}-\frac {8 \text {Subst}\left (\int \frac {\frac {35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac {35}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{35 d} \\ & = \frac {2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d} \\ & = \frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d} \\ & = -\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {2 b^2 (7 A b+11 a B) \tan ^{\frac {5}{2}}(c+d x)}{35 d}+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2}{7 d} \\
\end{align*}
Mathematica [C] (verified)
Result contains complex when optimal does not.
Time = 2.18 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.47
\[
\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 \left (\frac {105}{2} (a-i b)^3 (i A+B) \left (\sqrt [4]{-1} \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\sqrt {\tan (c+d x)}\right )+\frac {105}{2} (a+i b)^3 (-i A+B) \left (\sqrt [4]{-1} \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\sqrt {\tan (c+d x)}\right )+5 b \left (21 a A b+18 a^2 B-7 b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)+3 b^2 (7 A b+11 a B) \tan ^{\frac {5}{2}}(c+d x)+15 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^2\right )}{105 d}
\]
[In]
Integrate[Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]
[Out]
(2*((105*(a - I*b)^3*(I*A + B)*((-1)^(1/4)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + Sqrt[Tan[c + d*x]]))/2 + (1
05*(a + I*b)^3*((-I)*A + B)*((-1)^(1/4)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + Sqrt[Tan[c + d*x]]))/2 + 5*b*
(21*a*A*b + 18*a^2*B - 7*b^2*B)*Tan[c + d*x]^(3/2) + 3*b^2*(7*A*b + 11*a*B)*Tan[c + d*x]^(5/2) + 15*b*B*Tan[c
+ d*x]^(3/2)*(a + b*Tan[c + d*x])^2))/(105*d)
Maple [A] (verified)
Time = 0.04 (sec) , antiderivative size = 372, normalized size of antiderivative = 0.88
| | |
| method | result | size |
| | |
| derivativedivides |
\(\frac {\frac {2 B \,b^{3} \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}+\frac {2 A \,b^{3} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {6 B a \,b^{2} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+2 A a \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+2 B \,a^{2} b \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )-\frac {2 B \,b^{3} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+6 A \,a^{2} b \left (\sqrt {\tan }\left (d x +c \right )\right )-2 \left (\sqrt {\tan }\left (d x +c \right )\right ) A \,b^{3}+2 B \,a^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )-6 \left (\sqrt {\tan }\left (d x +c \right )\right ) B a \,b^{2}+\frac {\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) |
\(372\) |
| default |
\(\frac {\frac {2 B \,b^{3} \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}+\frac {2 A \,b^{3} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {6 B a \,b^{2} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+2 A a \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+2 B \,a^{2} b \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )-\frac {2 B \,b^{3} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+6 A \,a^{2} b \left (\sqrt {\tan }\left (d x +c \right )\right )-2 \left (\sqrt {\tan }\left (d x +c \right )\right ) A \,b^{3}+2 B \,a^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )-6 \left (\sqrt {\tan }\left (d x +c \right )\right ) B a \,b^{2}+\frac {\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) |
\(372\) |
| parts |
\(\frac {\left (A \,b^{3}+3 B a \,b^{2}\right ) \left (\frac {2 \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-2 \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {\left (3 A a \,b^{2}+3 B \,a^{2} b \right ) \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {A \,a^{3} \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {B \,b^{3} \left (\frac {2 \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}-\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\) |
\(559\) |
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[In]
int(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)
[Out]
1/d*(2/7*B*b^3*tan(d*x+c)^(7/2)+2/5*A*b^3*tan(d*x+c)^(5/2)+6/5*B*a*b^2*tan(d*x+c)^(5/2)+2*A*a*b^2*tan(d*x+c)^(
3/2)+2*B*a^2*b*tan(d*x+c)^(3/2)-2/3*B*b^3*tan(d*x+c)^(3/2)+6*A*a^2*b*tan(d*x+c)^(1/2)-2*tan(d*x+c)^(1/2)*A*b^3
+2*B*a^3*tan(d*x+c)^(1/2)-6*tan(d*x+c)^(1/2)*B*a*b^2+1/4*(-3*A*a^2*b+A*b^3-B*a^3+3*B*a*b^2)*2^(1/2)*(ln((1+2^(
1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2)
)+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(A*a^3-3*A*a*b^2-3*B*a^2*b+B*b^3)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+
c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1
+2^(1/2)*tan(d*x+c)^(1/2))))
Fricas [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 6157 vs. \(2 (379) = 758\).
Time = 1.44 (sec) , antiderivative size = 6157, normalized size of antiderivative = 14.62
\[
\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\text {Too large to display}
\]
[In]
integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")
[Out]
Too large to include
Sympy [F]
\[
\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sqrt {\tan {\left (c + d x \right )}}\, dx
\]
[In]
integrate(tan(d*x+c)**(1/2)*(a+b*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)
[Out]
Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**3*sqrt(tan(c + d*x)), x)
Maxima [A] (verification not implemented)
none
Time = 0.33 (sec) , antiderivative size = 363, normalized size of antiderivative = 0.86
\[
\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {120 \, B b^{3} \tan \left (d x + c\right )^{\frac {7}{2}} + 168 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \tan \left (d x + c\right )^{\frac {5}{2}} + 210 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 210 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 105 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 105 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 280 \, {\left (3 \, B a^{2} b + 3 \, A a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + 840 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{420 \, d}
\]
[In]
integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
1/420*(120*B*b^3*tan(d*x + c)^(7/2) + 168*(3*B*a*b^2 + A*b^3)*tan(d*x + c)^(5/2) + 210*sqrt(2)*((A - B)*a^3 -
3*(A + B)*a^2*b - 3*(A - B)*a*b^2 + (A + B)*b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 210*sq
rt(2)*((A - B)*a^3 - 3*(A + B)*a^2*b - 3*(A - B)*a*b^2 + (A + B)*b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(ta
n(d*x + c)))) - 105*sqrt(2)*((A + B)*a^3 + 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 - (A - B)*b^3)*log(sqrt(2)*sqrt(t
an(d*x + c)) + tan(d*x + c) + 1) + 105*sqrt(2)*((A + B)*a^3 + 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 - (A - B)*b^3)
*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 280*(3*B*a^2*b + 3*A*a*b^2 - B*b^3)*tan(d*x + c)^(3/2)
+ 840*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*sqrt(tan(d*x + c)))/d
Giac [F(-1)]
Timed out. \[
\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\text {Timed out}
\]
[In]
integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")
[Out]
Timed out
Mupad [B] (verification not implemented)
Time = 25.34 (sec) , antiderivative size = 6716, normalized size of antiderivative = 15.95
\[
\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\text {Too large to display}
\]
[In]
int(tan(c + d*x)^(1/2)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3,x)
[Out]
tan(c + d*x)^(1/2)*((2*B*a^3)/d - (6*B*a*b^2)/d) - tan(c + d*x)^(1/2)*((2*A*b^3)/d - (6*A*a^2*b)/d) - tan(c +
d*x)^(3/2)*((2*B*b^3)/(3*d) - (2*B*a^2*b)/d) - atan((((8*(4*A*b^3*d^2 - 12*A*a^2*b*d^2)*((3*A^2*a*b^5)/(2*d^2)
- (5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^
6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*A^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16
*tan(c + d*x)^(1/2)*(A^2*a^6 - A^2*b^6 + 15*A^2*a^2*b^4 - 15*A^2*a^4*b^2))/d^2)*((3*A^2*a*b^5)/(2*d^2) - (5*A^
2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^
4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*A^2*a^5*b)/(2*d^2))^(1/2)*1i - ((8*(4*A*b^3*
d^2 - 12*A*a^2*b*d^2)*((3*A^2*a*b^5)/(2*d^2) - (5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4
*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^
4) + (3*A^2*a^5*b)/(2*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(A^2*a^6 - A^2*b^6 + 15*A^2*a^2*b^4 - 15*A^2*a
^4*b^2))/d^2)*((3*A^2*a*b^5)/(2*d^2) - (5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^
4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*
A^2*a^5*b)/(2*d^2))^(1/2)*1i)/(((8*(4*A*b^3*d^2 - 12*A*a^2*b*d^2)*((3*A^2*a*b^5)/(2*d^2) - (5*A^2*a^3*b^3)/d^2
- (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^
8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*A^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(A
^2*a^6 - A^2*b^6 + 15*A^2*a^2*b^4 - 15*A^2*a^4*b^2))/d^2)*((3*A^2*a*b^5)/(2*d^2) - (5*A^2*a^3*b^3)/d^2 - (30*A
^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^
4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*A^2*a^5*b)/(2*d^2))^(1/2) - (16*(3*A^3*a*b^8 - A^3*a^9 + 8*A^3*a^3
*b^6 + 6*A^3*a^5*b^4))/d^3 + ((8*(4*A*b^3*d^2 - 12*A*a^2*b*d^2)*((3*A^2*a*b^5)/(2*d^2) - (5*A^2*a^3*b^3)/d^2 -
(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*
b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*A^2*a^5*b)/(2*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(A^2
*a^6 - A^2*b^6 + 15*A^2*a^2*b^4 - 15*A^2*a^4*b^2))/d^2)*((3*A^2*a*b^5)/(2*d^2) - (5*A^2*a^3*b^3)/d^2 - (30*A^4
*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4
+ 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*A^2*a^5*b)/(2*d^2))^(1/2)))*((3*A^2*a*b^5)/(2*d^2) - (5*A^2*a^3*b^3)
/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^
4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*A^2*a^5*b)/(2*d^2))^(1/2)*2i - atan((((8*(4*A*b^3*d^2
- 12*A*a^2*b*d^2)*((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*
d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) +
(3*A^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(A^2*a^6 - A^2*b^6 + 15*A^2*a^2*b^4 - 15*A^2*a^4*b
^2))/d^2)*((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 25
5*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*
a^5*b)/(2*d^2))^(1/2)*1i - ((8*(4*A*b^3*d^2 - 12*A*a^2*b*d^2)*((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*
d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (
5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*a^5*b)/(2*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(A^2*a
^6 - A^2*b^6 + 15*A^2*a^2*b^4 - 15*A^2*a^4*b^2))/d^2)*((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 25
5*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*A^2*a^
3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*a^5*b)/(2*d^2))^(1/2)*1i)/(((8*(4*A*b^3*d^2 - 12*A*a^2*b*d^2)*((30
*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*
d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*a^5*b)/(2*d^2)
)^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(A^2*a^6 - A^2*b^6 + 15*A^2*a^2*b^4 - 15*A^2*a^4*b^2))/d^2)*((30*A^4*a^2
*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30
*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*a^5*b)/(2*d^2))^(1/2)
- (16*(3*A^3*a*b^8 - A^3*a^9 + 8*A^3*a^3*b^6 + 6*A^3*a^5*b^4))/d^3 + ((8*(4*A*b^3*d^2 - 12*A*a^2*b*d^2)*((30*A
^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^
4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*a^5*b)/(2*d^2))^
(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(A^2*a^6 - A^2*b^6 + 15*A^2*a^2*b^4 - 15*A^2*a^4*b^2))/d^2)*((30*A^4*a^2*b
^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A
^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*a^5*b)/(2*d^2))^(1/2)))*
((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*
b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*A^2*a^3*b^3)/d^2 + (3*A^2*a*b^5)/(2*d^2) + (3*A^2*a^5*b)/(2*
d^2))^(1/2)*2i - atan((((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^
12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)
^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(B^2*a^6 -
B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 -
B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(
4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*1i - ((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((5*B^2*
a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4
- 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1
/2))/d^3 + (16*tan(c + d*x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*((5*B^2*a^3*b^3)
/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^
4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*1i)/
(((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4
- 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^
2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16*tan(c + d*x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2*a^2*
b^4 - 15*B^2*a^4*b^2))/d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^
4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)
/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2) + ((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^
2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 3
0*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 + (16*tan(c + d*
x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d
^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^
10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2) + (16*(3*B^3*a^8*b - B^3*b^9
+ 6*B^3*a^4*b^5 + 8*B^3*a^6*b^3))/d^3))*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*
d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (
3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*2i - atan((((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((30*B^4*a^2
*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30
*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))
/d^3 - (16*tan(c + d*x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*((30*B^4*a^2*b^10*d^
4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^1
0*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*1i - ((8
*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 45
2*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*
b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2*a^2*b^4
- 15*B^2*a^4*b^2))/d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^
6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*
d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*1i)/(((8*(4*B*a^3*d^2 - 12*B*a*b^2*d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d
^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/
2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16*tan(c + d*x
)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4
*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^
4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2) + ((8*(4*B*a^3*d^2 - 12*B*a*b^
2*d^2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B
^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5
*b)/(2*d^2))^(1/2))/d^3 + (16*tan(c + d*x)^(1/2)*(B^2*a^6 - B^2*b^6 + 15*B^2*a^2*b^4 - 15*B^2*a^4*b^2))/d^2)*(
(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b
^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d
^2))^(1/2) + (16*(3*B^3*a^8*b - B^3*b^9 + 6*B^3*a^4*b^5 + 8*B^3*a^6*b^3))/d^3))*((30*B^4*a^2*b^10*d^4 - B^4*b^
12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)
^(1/2)/(4*d^4) + (5*B^2*a^3*b^3)/d^2 - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2)*2i + (2*A*b^3*tan(
c + d*x)^(5/2))/(5*d) + (2*B*b^3*tan(c + d*x)^(7/2))/(7*d) + (2*A*a*b^2*tan(c + d*x)^(3/2))/d + (6*B*a*b^2*tan
(c + d*x)^(5/2))/(5*d)