3.133 \(\int \frac{(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=169 \[ -\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (7 B+11 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 a^3 (B+i A)}{d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac{7}{2}}(c+d x)} \]

[Out]

(-8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (8*a^3*(23*A - (21*I)*B))/(105*d*Tan[c
+ d*x]^(3/2)) + (8*a^3*(I*A + B))/(d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^2)/(7*d*Tan[c + d*x]
^(7/2)) - (2*((11*I)*A + 7*B)*(a^3 + I*a^3*Tan[c + d*x]))/(35*d*Tan[c + d*x]^(5/2))

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Rubi [A]  time = 0.424736, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {3593, 3591, 3529, 3533, 205} \[ -\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (7 B+11 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 a^3 (B+i A)}{d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(9/2),x]

[Out]

(-8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d + (8*a^3*(23*A - (21*I)*B))/(105*d*Tan[c
+ d*x]^(3/2)) + (8*a^3*(I*A + B))/(d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^2)/(7*d*Tan[c + d*x]
^(7/2)) - (2*((11*I)*A + 7*B)*(a^3 + I*a^3*Tan[c + d*x]))/(35*d*Tan[c + d*x]^(5/2))

Rule 3593

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

Rule 3591

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

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac{9}{2}}(c+d x)} \, dx &=-\frac{2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2}{7} \int \frac{(a+i a \tan (c+d x))^2 \left (\frac{1}{2} a (11 i A+7 B)-\frac{1}{2} a (3 A-7 i B) \tan (c+d x)\right )}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 (11 i A+7 B) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4}{35} \int \frac{(a+i a \tan (c+d x)) \left (-a^2 (23 A-21 i B)-2 a^2 (6 i A+7 B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 (11 i A+7 B) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4}{35} \int \frac{-35 a^3 (i A+B)+35 a^3 (A-i B) \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{8 a^3 (i A+B)}{d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 (11 i A+7 B) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4}{35} \int \frac{35 a^3 (A-i B)+35 a^3 (i A+B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{8 a^3 (i A+B)}{d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 (11 i A+7 B) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{\left (280 a^6 (A-i B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{35 a^3 (A-i B)-35 a^3 (i A+B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{8 a^3 (23 A-21 i B)}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{8 a^3 (i A+B)}{d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^2}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 (11 i A+7 B) \left (a^3+i a^3 \tan (c+d x)\right )}{35 d \tan ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [B]  time = 12.0529, size = 495, normalized size = 2.93 \[ \frac{8 e^{-3 i c} (A-i B) \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{d \sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{\cos ^4(c+d x) \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \left (\csc (c) \left (\frac{2}{5} \cos (3 c)-\frac{2}{5} i \sin (3 c)\right ) \csc ^3(c+d x) (B \sin (d x)+3 i A \sin (d x))+\csc (c) \left (\frac{2}{105} \cos (3 c)-\frac{2}{105} i \sin (3 c)\right ) \csc ^2(c+d x) (170 A \sin (c)-63 i A \cos (c)-105 i B \sin (c)-21 B \cos (c))+\csc (c) \left (\frac{2}{5} \cos (3 c)-\frac{2}{5} i \sin (3 c)\right ) \csc (c+d x) (-21 B \sin (d x)-23 i A \sin (d x))+\csc (c) \left (\frac{2}{105} \cos (3 c)-\frac{2}{105} i \sin (3 c)\right ) (-155 A \sin (c)+483 i A \cos (c)+105 i B \sin (c)+441 B \cos (c))+\left (-\frac{2}{7} A \cos (3 c)+\frac{2}{7} i A \sin (3 c)\right ) \csc ^4(c+d x)\right )}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(9/2),x]

[Out]

(8*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c
+ d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*E^((3*I)
*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*S
in[c + d*x])) + (Cos[c + d*x]^4*(Csc[c]*Csc[c + d*x]^2*((-63*I)*A*Cos[c] - 21*B*Cos[c] + 170*A*Sin[c] - (105*I
)*B*Sin[c])*((2*Cos[3*c])/105 - ((2*I)/105)*Sin[3*c]) + Csc[c]*((483*I)*A*Cos[c] + 441*B*Cos[c] - 155*A*Sin[c]
+ (105*I)*B*Sin[c])*((2*Cos[3*c])/105 - ((2*I)/105)*Sin[3*c]) + Csc[c + d*x]^4*((-2*A*Cos[3*c])/7 + ((2*I)/7)
*A*Sin[3*c]) + Csc[c]*Csc[c + d*x]*((2*Cos[3*c])/5 - ((2*I)/5)*Sin[3*c])*((-23*I)*A*Sin[d*x] - 21*B*Sin[d*x])
+ Csc[c]*Csc[c + d*x]^3*((2*Cos[3*c])/5 - ((2*I)/5)*Sin[3*c])*((3*I)*A*Sin[d*x] + B*Sin[d*x]))*Sqrt[Tan[c + d*
x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*
x]))

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Maple [B]  time = 0.019, size = 572, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x)

[Out]

-2/7/d*a^3*A/tan(d*x+c)^(7/2)-6/5*I/d*a^3/tan(d*x+c)^(5/2)*A+8/d*a^3/tan(d*x+c)^(1/2)*B+I/d*a^3*A*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^(1/2)-2/5/d*a^3/tan(d*x+c)^(5/2)*B+2
*I/d*a^3*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+8/3/d*a^3*A/tan(d*x+c)^(3/2)-2*I/d*a^3*B*arctan(1+2^(1/2
)*tan(d*x+c)^(1/2))*2^(1/2)-I/d*a^3*B*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*I/d*a^3*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+2/d*a^3*A*arctan(-1+2^(1/2)*tan(d*
x+c)^(1/2))*2^(1/2)+1/d*a^3*A*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)+t
an(d*x+c)))+2/d*a^3*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-2*I/d*a^3/tan(d*x+c)^(3/2)*B+8*I/d*a^3/tan(d*
x+c)^(1/2)*A+2*I/d*a^3*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+1/d*a^3*B*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^(1/2)+2/d*a^3*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(
1/2)+2/d*a^3*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)

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Maxima [A]  time = 2.10347, size = 286, normalized size = 1.69 \begin{align*} -\frac{105 \,{\left (\sqrt{2}{\left (-\left (2 i + 2\right ) \, A + \left (2 i - 2\right ) \, B\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + \sqrt{2}{\left (-\left (2 i + 2\right ) \, A + \left (2 i - 2\right ) \, B\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{3} - \frac{2 \,{\left (420 \,{\left (i \, A + B\right )} a^{3} \tan \left (d x + c\right )^{3} +{\left (140 \, A - 105 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} + 21 \,{\left (-3 i \, A - B\right )} a^{3} \tan \left (d x + c\right ) - 15 \, A a^{3}\right )}}{\tan \left (d x + c\right )^{\frac{7}{2}}}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

-1/105*(105*(sqrt(2)*(-(2*I + 2)*A + (2*I - 2)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + sqrt(
2)*(-(2*I + 2)*A + (2*I - 2)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*((I - 1)*A + (
I + 1)*B)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*((I - 1)*A + (I + 1)*B)*log(-sqrt(2)*sq
rt(tan(d*x + c)) + tan(d*x + c) + 1))*a^3 - 2*(420*(I*A + B)*a^3*tan(d*x + c)^3 + (140*A - 105*I*B)*a^3*tan(d*
x + c)^2 + 21*(-3*I*A - B)*a^3*tan(d*x + c) - 15*A*a^3)/tan(d*x + c)^(7/2))/d

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Fricas [B]  time = 2.07294, size = 1558, normalized size = 9.22 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

1/420*(105*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6
*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)*log((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I*A
^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*
x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^3)) - 105*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^
2)*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)*l
og((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(-I*d*e^(2*I*d*x + 2*
I*c) - I*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*
a^3)) - 16*((319*A - 273*I*B)*a^3*e^(8*I*d*x + 8*I*c) - 3*(109*A - 133*I*B)*a^3*e^(6*I*d*x + 6*I*c) - 5*(19*A
- 21*I*B)*a^3*e^(4*I*d*x + 4*I*c) + 3*(129*A - 133*I*B)*a^3*e^(2*I*d*x + 2*I*c) - 4*(41*A - 42*I*B)*a^3)*sqrt(
(-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*
d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.37301, size = 184, normalized size = 1.09 \begin{align*} \frac{\left (4 i - 4\right ) \, \sqrt{2}{\left (-i \, A a^{3} - B a^{3}\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{d} - \frac{-840 i \, A a^{3} \tan \left (d x + c\right )^{3} - 840 \, B a^{3} \tan \left (d x + c\right )^{3} - 280 \, A a^{3} \tan \left (d x + c\right )^{2} + 210 i \, B a^{3} \tan \left (d x + c\right )^{2} + 126 i \, A a^{3} \tan \left (d x + c\right ) + 42 \, B a^{3} \tan \left (d x + c\right ) + 30 \, A a^{3}}{105 \, d \tan \left (d x + c\right )^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="giac")

[Out]

(4*I - 4)*sqrt(2)*(-I*A*a^3 - B*a^3)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 1/105*(-840*I*A*a^3
*tan(d*x + c)^3 - 840*B*a^3*tan(d*x + c)^3 - 280*A*a^3*tan(d*x + c)^2 + 210*I*B*a^3*tan(d*x + c)^2 + 126*I*A*a
^3*tan(d*x + c) + 42*B*a^3*tan(d*x + c) + 30*A*a^3)/(d*tan(d*x + c)^(7/2))