\(\int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) [397]

Optimal result

Integrand size = 33, antiderivative size = 380 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

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

Rubi [A] (verified)

Time = 0.70 (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 = {3686, 3716, 3709, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{5 d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{15 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]

[In]

Int[((a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),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*a^2*(9*A*b + 5*a*B))/(15*d*Tan[c + d*x]^
(3/2)) + (2*a*(5*a^2*A - 14*A*b^2 - 15*a*b*B))/(5*d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + b*Tan[c + d*x])^2)/(5*d*
Tan[c + d*x]^(5/2))

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 3686

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*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e
+ f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

Rule 3709

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2)
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

Rule 3716

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 - a*d))*(c^2*C - B*c*d + A*d^2)
*((c + d*Tan[e + f*x])^(n + 1)/(d^2*f*(n + 1)*(c^2 + d^2))), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f
*x])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b*(c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d
+ a*C*d)*Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] &
& NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(a+b \tan (c+d x)) \left (\frac {1}{2} a (9 A b+5 a B)-\frac {5}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac {1}{2} b (a A-5 b B) \tan ^2(c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx \\ & = -\frac {2 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {-\frac {1}{2} a \left (5 a^2 A-14 A b^2-15 a b B\right )-\frac {5}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-\frac {1}{2} b^2 (a A-5 b B) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx \\ & = -\frac {2 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {-\frac {5}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+\frac {5}{2} \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 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {4 \text {Subst}\left (\int \frac {-\frac {5}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+\frac {5}{2} \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 )}{5 d} \\ & = -\frac {2 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\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 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\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 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\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 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.44 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.44 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \left (3 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},1,-\frac {1}{4},-\tan ^2(c+d x)\right )+5 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\tan ^2(c+d x)\right ) \tan (c+d x)+b \left (9 a A b+9 a^2 B-3 b^2 B+5 b (A b+3 a B) \tan (c+d x)+15 b^2 B \tan ^2(c+d x)\right )\right )}{15 d \tan ^{\frac {5}{2}}(c+d x)} \]

[In]

Integrate[((a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]

[Out]

(-2*(3*(a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*Hypergeometric2F1[-5/4, 1, -1/4, -Tan[c + d*x]^2] + 5*(3*a^2*A*
b - A*b^3 + a^3*B - 3*a*b^2*B)*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2]*Tan[c + d*x] + b*(9*a*A*b + 9*
a^2*B - 3*b^2*B + 5*b*(A*b + 3*a*B)*Tan[c + d*x] + 15*b^2*B*Tan[c + d*x]^2)))/(15*d*Tan[c + d*x]^(5/2))

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {-\frac {2 A \,a^{3}}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 a^{2} \left (3 A b +B a \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(290\)
default	\(\frac {-\frac {2 A \,a^{3}}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 a^{2} \left (3 A b +B a \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(290\)
parts	\(\frac {\left (A \,b^{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}+\frac {\left (3 A a \,b^{2}+3 B \,a^{2} b \right ) \left (-\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}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\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} \left (\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}-\frac {2}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {B \,b^{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}\)	\(537\)

[In]

int((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(-2/5*A*a^3/tan(d*x+c)^(5/2)-2/3*a^2*(3*A*b+B*a)/tan(d*x+c)^(3/2)+2*a*(A*a^2-3*A*b^2-3*B*a*b)/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)))+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. 6162 vs. \(2 (340) = 680\).

Time = 1.45 (sec) , antiderivative size = 6162, normalized size of antiderivative = 16.22 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

Too large to include

Sympy [F]

\[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{3}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a+b*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(7/2),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**3/tan(c + d*x)**(7/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {30 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 30 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \frac {8 \, {\left (3 \, A a^{3} - 15 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{60 \, d} \]

[In]

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/60*(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) - 8*(3*A*a^3 - 15*(A*a^3 - 3
*B*a^2*b - 3*A*a*b^2)*tan(d*x + c)^2 + 5*(B*a^3 + 3*A*a^2*b)*tan(d*x + c))/tan(d*x + c)^(5/2))/d

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 14.60 (sec) , antiderivative size = 7591, normalized size of antiderivative = 19.98 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3)/tan(c + d*x)^(7/2),x)

[Out]

2*atanh((32*B^2*a^6*d^3*tan(c + d*x)^(1/2)*((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*B^3*b^9*d^2 + 16*B*a^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) - 288*B^3*a^2*b^7*d^2 + 960*B^3*a^4*b^5*d^2 - 736*B^3*a^6*b^3*d^2 + 48*B^3*a^8*b*d^2 - 48*B*a*b^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)) - (32*B^2*b^6*d^3*tan(c + d*x)^(1/2)*((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*B^3*b^9*d^2 + 16*B*a^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) - 288*B^3*a^2*b^7*d^2 + 960*B^3*a^4*b^5*d^2 - 736*B^3*a^6*b^3*d^2 + 48*B^
3*a^8*b*d^2 - 48*B*a*b^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)) + (480*B^2*a^2*b^4*d^3*tan(c + d*x)^(1/2)*((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*B^3*b^9*d^2 + 16*B*a^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 + 4
52*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) - 288*B^3*a^2*b^7*d^2 + 960*B^3*a^4*b^5*
d^2 - 736*B^3*a^6*b^3*d^2 + 48*B^3*a^8*b*d^2 - 48*B*a*b^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)) - (480*B^2*a^4*
b^2*d^3*tan(c + d*x)^(1/2)*((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*B^3*b^9*d^2 + 16*B*a^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) - 288*B
^3*a^2*b^7*d^2 + 960*B^3*a^4*b^5*d^2 - 736*B^3*a^6*b^3*d^2 + 48*B^3*a^8*b*d^2 - 48*B*a*b^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)))*((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) + 2*atanh((32*B^2*a^6*d^3*tan(c + d*x)^(1/2)*((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*B^3*b^9*d^2 - 16*
B*a^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) - 288*B^3*a^2*b^7*d^2 + 960*B^3*a^4*b^5*d^2 - 736*B^3*a^6*b^3*d^2 +
48*B^3*a^8*b*d^2 + 48*B*a*b^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)) - (32*B^2*b^6*d^3*tan(c + d*x)^(1/2)*((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*B^3*b^9*d^2 - 16*B*a^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 + 4
52*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) - 288*B^3*a^2*b^7*d^2 + 960*B^3*a^4*b^5*
d^2 - 736*B^3*a^6*b^3*d^2 + 48*B^3*a^8*b*d^2 + 48*B*a*b^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)) + (480*B^2*a^2*
b^4*d^3*tan(c + d*x)^(1/2)*((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*B^3*b^9*d^2 - 16*B*a^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) - 288*B
^3*a^2*b^7*d^2 + 960*B^3*a^4*b^5*d^2 - 736*B^3*a^6*b^3*d^2 + 48*B^3*a^8*b*d^2 + 48*B*a*b^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)) - (480*B^2*a^4*b^2*d^3*tan(c + d*x)^(1/2)*((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*B^3*b^9*d^2 - 16*B*a^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 + 3
0*B^4*a^10*b^2*d^4)^(1/2) - 288*B^3*a^2*b^7*d^2 + 960*B^3*a^4*b^5*d^2 - 736*B^3*a^6*b^3*d^2 + 48*B^3*a^8*b*d^2
+ 48*B*a*b^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)))*((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) - 2*atanh((32*A^2*a^6*d^3*tan(c + d*x)^(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))/(16*A^3*a^9*d^2 - 16*A*b^3*(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) - 736*A^3*a^3*b^6*d^2 + 960*A^3*a^5
*b^4*d^2 - 288*A^3*a^7*b^2*d^2 + 48*A*a^2*b*(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) + 48*A^3*a*b^8*d^2) - (32*A^2*
b^6*d^3*tan(c + d*x)^(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))/(16*A^3*a^9*d^2 - 16*A*b^3*(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) - 736*A
^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4*d^2 - 288*A^3*a^7*b^2*d^2 + 48*A*a^2*b*(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) +
48*A^3*a*b^8*d^2) + (480*A^2*a^2*b^4*d^3*tan(c + d*x)^(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))/(16*A^3*a^9*d^2 - 16*A*b^3*(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) - 736*A^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4*d^2 - 288*A^3*a^7*b^2*d^2 + 48*A*a^2*b*(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) + 48*A^3*a*b^8*d^2) - (480*A^2*a^4*b^2*d^3*tan(c + d*x)^(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))/(16*
A^3*a^9*d^2 - 16*A*b^3*(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) - 736*A^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4*d^2 - 288*A
^3*a^7*b^2*d^2 + 48*A*a^2*b*(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) + 48*A^3*a*b^8*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) - 2*atanh((32
*A^2*a^6*d^3*tan(c + d*x)^(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 + 45
2*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*A*b^3*(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) + 16*A^3*a^9*d^2 -
736*A^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4*d^2 - 288*A^3*a^7*b^2*d^2 - 48*A*a^2*b*(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) + 48*A^3*a*b^8*d^2) - (32*A^2*b^6*d^3*tan(c + d*x)^(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))/(16*A*b^3*(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) + 16*A^3*a^9*d^2 - 736*A^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4*d^2 - 288*A^3*a^7*b^2*d^2 - 48*A*a^2*b*(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) + 48*A^3*a*b^8*d^2) + (480*A^2*a^2*b^4*d^3*tan(c + d*x)^(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))/(16*
A*b^3*(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) + 16*A^3*a^9*d^2 - 736*A^3*a^3*b^6*d^2 + 960*A^3*a^5*b^4*d^2 - 288*A
^3*a^7*b^2*d^2 - 48*A*a^2*b*(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) + 48*A^3*a*b^8*d^2) - (480*A^2*a^4*b^2*d^3*tan
(c + d*x)^(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))/(16*A*b^3*(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) + 16*A^3*a^9*d^2 - 736*A^3*a^3*b^6*
d^2 + 960*A^3*a^5*b^4*d^2 - 288*A^3*a^7*b^2*d^2 - 48*A*a^2*b*(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) + 48*A^3*a*b^
8*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) - ((2*A*a^3)/5 - tan(c + d*x)^2*(2*A*a^3 - 6*A*a*b^2) + 2*A*a^2*b*tan(c + d*x))/(d*tan(c +
d*x)^(5/2)) - ((2*B*a^3)/3 + 6*B*a^2*b*tan(c + d*x))/(d*tan(c + d*x)^(3/2))