\(\int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx\) [394]

Optimal result

Integrand size = 33, antiderivative size = 380 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{5 d}+\frac {2 b^2 (5 A b+9 a B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d} \]

[Out]

1/2*(3*a^2*b*(A-B)-b^3*(A-B)+a^3*(A+B)-3*a*b^2*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/2*(3*a^2
*b*(A-B)-b^3*(A-B)+a^3*(A+B)-3*a*b^2*(A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/4*(a^3*(A-B)-3*a*b^
2*(A-B)-3*a^2*b*(A+B)+b^3*(A+B))*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)+1/4*(a^3*(A-B)-3*a*b^2*(A
-B)-3*a^2*b*(A+B)+b^3*(A+B))*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)+2/5*b*(15*A*a*b+14*B*a^2-5*B*
b^2)*tan(d*x+c)^(1/2)/d+2/15*b^2*(5*A*b+9*B*a)*tan(d*x+c)^(3/2)/d+2/5*b*B*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^2/
d

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3688, 3718, 3711, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{5 d}-\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}-\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d} \]

[In]

Int[((a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Sqrt[Tan[c + d*x]],x]

[Out]

-(((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sq
rt[2]*d)) + ((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d
*x]]])/(Sqrt[2]*d) - ((a^3*(A - B) - 3*a*b^2*(A - B) - 3*a^2*b*(A + B) + b^3*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan
[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + ((a^3*(A - B) - 3*a*b^2*(A - B) - 3*a^2*b*(A + B) + b^3*(A + B))*L
og[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (2*b*(15*a*A*b + 14*a^2*B - 5*b^2*B)*Sqrt[T
an[c + d*x]])/(5*d) + (2*b^2*(5*A*b + 9*a*B)*Tan[c + d*x]^(3/2))/(15*d) + (2*b*B*Sqrt[Tan[c + d*x]]*(a + b*Tan
[c + d*x])^2)/(5*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 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 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}+\frac {2}{5} \int \frac {(a+b \tan (c+d x)) \left (\frac {1}{2} a (5 a A-b B)+\frac {5}{2} \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac {1}{2} b (5 A b+9 a B) \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 b^2 (5 A b+9 a B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}-\frac {4}{15} \int \frac {-\frac {3}{4} a^2 (5 a A-b B)-\frac {15}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-\frac {3}{4} b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{5 d}+\frac {2 b^2 (5 A b+9 a B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}-\frac {4}{15} \int \frac {-\frac {15}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )-\frac {15}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{5 d}+\frac {2 b^2 (5 A b+9 a B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}-\frac {8 \text {Subst}\left (\int \frac {-\frac {15}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )-\frac {15}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{15 d} \\ & = \frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{5 d}+\frac {2 b^2 (5 A b+9 a B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 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 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{5 d}+\frac {2 b^2 (5 A b+9 a B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 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}{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 ) \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 {\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 {\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 ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\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 ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{5 d}+\frac {2 b^2 (5 A b+9 a B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 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}{-1-x^2} \, dx,x,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 ) \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 (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (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 ) \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 ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\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 ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{5 d}+\frac {2 b^2 (5 A b+9 a B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.59 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.40 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {-15 \sqrt [4]{-1} (a-i b)^3 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-15 \sqrt [4]{-1} (a+i b)^3 (A+i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 b \sqrt {\tan (c+d x)} \left (15 \left (3 a A b+3 a^2 B-b^2 B\right )+5 b (A b+3 a B) \tan (c+d x)+3 b^2 B \tan ^2(c+d x)\right )}{15 d} \]

[In]

Integrate[((a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Sqrt[Tan[c + d*x]],x]

[Out]

(-15*(-1)^(1/4)*(a - I*b)^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - 15*(-1)^(1/4)*(a + I*b)^3*(A + I
*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 2*b*Sqrt[Tan[c + d*x]]*(15*(3*a*A*b + 3*a^2*B - b^2*B) + 5*b*(A*b
+ 3*a*B)*Tan[c + d*x] + 3*b^2*B*Tan[c + d*x]^2))/(15*d)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\frac {2 B \,b^{3} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 A \,b^{3} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 B a \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+6 \left (\sqrt {\tan }\left (d x +c \right )\right ) A a \,b^{2}+6 \left (\sqrt {\tan }\left (d x +c \right )\right ) B \,a^{2} b -2 \left (\sqrt {\tan }\left (d x +c \right )\right ) B \,b^{3}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(314\)
default	\(\frac {\frac {2 B \,b^{3} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 A \,b^{3} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+2 B a \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+6 \left (\sqrt {\tan }\left (d x +c \right )\right ) A a \,b^{2}+6 \left (\sqrt {\tan }\left (d x +c \right )\right ) B \,a^{2} b -2 \left (\sqrt {\tan }\left (d x +c \right )\right ) B \,b^{3}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(314\)
parts	\(\frac {\left (A \,b^{3}+3 B a \,b^{2}\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 \,b^{2}+3 B \,a^{2} b \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 {\left (3 A \,a^{2} b +B \,a^{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}+\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 {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}\)	\(537\)

[In]

int((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(2/5*B*b^3*tan(d*x+c)^(5/2)+2/3*A*b^3*tan(d*x+c)^(3/2)+2*B*a*b^2*tan(d*x+c)^(3/2)+6*tan(d*x+c)^(1/2)*A*a*b
^2+6*tan(d*x+c)^(1/2)*B*a^2*b-2*tan(d*x+c)^(1/2)*B*b^3+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)))+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))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6136 vs. \(2 (340) = 680\).

Time = 1.49 (sec) , antiderivative size = 6136, normalized size of antiderivative = 16.15 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{3}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx \]

[In]

integrate((a+b*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(1/2),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 = 0.34 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {24 \, B b^{3} \tan \left (d x + c\right )^{\frac {5}{2}} + 30 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 30 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 15 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 15 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 40 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + 120 \, {\left (3 \, B a^{2} b + 3 \, A a b^{2} - B b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{60 \, d} \]

[In]

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/60*(24*B*b^3*tan(d*x + c)^(5/2) + 30*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)))) + 30*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)))) + 15*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) - 15*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
) + 40*(3*B*a*b^2 + A*b^3)*tan(d*x + c)^(3/2) + 120*(3*B*a^2*b + 3*A*a*b^2 - B*b^3)*sqrt(tan(d*x + c)))/d

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 14.53 (sec) , antiderivative size = 6657, normalized size of antiderivative = 17.52 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Too large to display} \]

[In]

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3)/tan(c + d*x)^(1/2),x)

[Out]

atan((((8*(4*A*a^3*d^2 - 12*A*a*b^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*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)*((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 - 2
55*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
*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2)*1i - ((8*(4*A*a^3*d^2 - 12*A*a*b^2*d^2)*((5*A^2*a^3*b^3)/d^2 - (3
0*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*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2))/d^3 + (16*ta
n(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)*((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 + 3
0*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*A^2*a*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2)*1i)/((16*(3*A^3*a^8*b
- A^3*b^9 + 6*A^3*a^4*b^5 + 8*A^3*a^6*b^3))/d^3 + ((8*(4*A*a^3*d^2 - 12*A*a*b^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*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16*t
an(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)*((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*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2) + ((8*(4*A*a^3*d^2 -
12*A*a*b^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*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)*((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*b^5)/(2*d^2) - (3*A^2*a
^5*b)/(2*d^2))^(1/2)))*((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*b^5)/(2*d
^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2)*2i - tan(c + d*x)^(1/2)*((2*B*b^3)/d - (6*B*a^2*b)/d) + atan((((8*(4*A*a^3*
d^2 - 12*A*a*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))/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)*1i - ((8*(4*A*a^3*d^2 - 12*A*a*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))/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)*1i)/((16*(3*A^3*a^8*b - A^3*b^9 + 6*A^3*
a^4*b^5 + 8*A^3*a^6*b^3))/d^3 + ((8*(4*A*a^3*d^2 - 12*A*a*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))/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) + ((8*(4*A*a^3*d^2 - 12*A*a*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))/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*b^3*d^2 - 12*B*a^2*b*d^2)*((3*B^2*a*b^5)/(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^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)*((3*B^2*a*b^5)/(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^5*b)/(2*d^2))^(1/2)*1i - ((8*(4*B*b^3*d^2 - 12*B*a^2*b*d^2)*((3*
B^2*a*b^5)/(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^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)*((3*B^2*a*b^
5)/(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^5*b)/(2*d^2))^(1/2)*
1i)/(((8*(4*B*b^3*d^2 - 12*B*a^2*b*d^2)*((3*B^2*a*b^5)/(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^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)*((3*B^2*a*b^5)/(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^5*b)/(2*d^2))^(1/2) + ((8*(4*B*b^3*d^2 - 12*B*a^2*b*d^2)*((3*B^2*a*b^5)/(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^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)*((3*B^2*a*b^5)/(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^5*b)/(2*d^2))^(1/2) - (16*(3*B^3*a*b^8 - B^3*
a^9 + 8*B^3*a^3*b^6 + 6*B^3*a^5*b^4))/d^3))*((3*B^2*a*b^5)/(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^1
0*b^2*d^4)^(1/2)/(4*d^4) + (3*B^2*a^5*b)/(2*d^2))^(1/2)*2i - atan((((8*(4*B*b^3*d^2 - 12*B*a^2*b*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^1
0*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*b^3*d^2 - 12*B*a^2*b*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)*1i)/(((8*(4*B*b^3*d^2 - 12*B*a^2*b*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*b^3*d^2 - 12*B*
a^2*b*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 - 2
55*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*b^8 - B^3*a^9 + 8*B^3*a^3*b^6 + 6*B^3*a^5*b^4))/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)^(3/2))/(3*d) + (2*B*b^3*tan(c + d*x)^(5/2))/(5*d) + (6*A*a*b^2*tan(c + d*x)^(1/2))/d + (2*B*a*b^2
*tan(c + d*x)^(3/2))/d