3.175 \(\int \frac{(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac{9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=231 \[ \frac{2 a^2 (80 A-77 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a^2 (7 B+10 i A) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2 (133 B+130 i A) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}+\frac{(4-4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)} \]

[Out]

((4 - 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a^2*((10*I)*A + 7*B)*Sqrt[a + I*a*Tan[c + d*x]])/(35*d*Tan[c + d*x]^(5/2)) + (2*a^2*(80*A - (77*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(3/2)) + (4*a^2*((130*I)*A + 133*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d
*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^(3/2))/(7*d*Tan[c + d*x]^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.75676, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {3593, 3598, 12, 3544, 205} \[ \frac{2 a^2 (80 A-77 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a^2 (7 B+10 i A) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{4 a^2 (133 B+130 i A) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}+\frac{(4-4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(9/2),x]

[Out]

((4 - 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a^2*((10*I)*A + 7*B)*Sqrt[a + I*a*Tan[c + d*x]])/(35*d*Tan[c + d*x]^(5/2)) + (2*a^2*(80*A - (77*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(3/2)) + (4*a^2*((130*I)*A + 133*B)*Sqrt[a + I*a*Tan[c + d*x]])/(105*d
*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^(3/2))/(7*d*Tan[c + d*x]^(7/2))

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 3598

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[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))^{5/2} (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))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2}{7} \int \frac{(a+i a \tan (c+d x))^{3/2} \left (\frac{1}{2} a (10 i A+7 B)-\frac{1}{2} a (4 A-7 i B) \tan (c+d x)\right )}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (10 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4}{35} \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 (80 A-77 i B)-\frac{3}{4} a^2 (20 i A+21 B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (10 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 a^2 (80 A-77 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{8 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^3 (130 i A+133 B)+\frac{1}{4} a^3 (80 A-77 i B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{105 a}\\ &=-\frac{2 a^2 (10 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 a^2 (80 A-77 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (130 i A+133 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{16 \int \frac{105 a^4 (A-i B) \sqrt{a+i a \tan (c+d x)}}{4 \sqrt{\tan (c+d x)}} \, dx}{105 a^2}\\ &=-\frac{2 a^2 (10 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 a^2 (80 A-77 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (130 i A+133 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}+\left (4 a^2 (A-i B)\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 (10 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 a^2 (80 A-77 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (130 i A+133 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{\left (8 a^4 (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac{(4+4 i) a^{5/2} (i A+B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a^2 (10 i A+7 B) \sqrt{a+i a \tan (c+d x)}}{35 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 a^2 (80 A-77 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (130 i A+133 B) \sqrt{a+i a \tan (c+d x)}}{105 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{7 d \tan ^{\frac{7}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 12.9675, size = 363, normalized size = 1.57 \[ \frac{4 \sqrt{2} e^{-2 i c} \sqrt{e^{i d x}} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} (a+i a \tan (c+d x))^{5/2} \left (e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \left (7 i B \left (50 e^{2 i (c+d x)}-61 e^{4 i (c+d x)}+26 e^{6 i (c+d x)}-15\right )-5 A \left (70 e^{2 i (c+d x)}-77 e^{4 i (c+d x)}+40 e^{6 i (c+d x)}-21\right )\right )+105 (A-i B) \left (-1+e^{2 i (c+d x)}\right )^4 \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )\right ) (A+B \tan (c+d x))}{105 d \left (-1+e^{2 i (c+d x)}\right )^{9/2} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec ^{\frac{7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(9/2),x]

[Out]

(4*Sqrt[2]*Sqrt[E^(I*d*x)]*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(E^(I*(c + d*x))*
Sqrt[-1 + E^((2*I)*(c + d*x))]*((7*I)*B*(-15 + 50*E^((2*I)*(c + d*x)) - 61*E^((4*I)*(c + d*x)) + 26*E^((6*I)*(
c + d*x))) - 5*A*(-21 + 70*E^((2*I)*(c + d*x)) - 77*E^((4*I)*(c + d*x)) + 40*E^((6*I)*(c + d*x)))) + 105*(A -
I*B)*(-1 + E^((2*I)*(c + d*x)))^4*Log[E^(I*(c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]])*(a + I*a*Tan[c + d*x]
)^(5/2)*(A + B*Tan[c + d*x]))/(105*d*E^((2*I)*c)*(-1 + E^((2*I)*(c + d*x)))^(9/2)*Sqrt[E^(I*(c + d*x))/(1 + E^
((2*I)*(c + d*x)))]*Sec[c + d*x]^(7/2)*(Cos[d*x] + I*Sin[d*x])^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

________________________________________________________________________________________

Maple [B]  time = 0.044, size = 798, normalized size = 3.5 \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))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x)

[Out]

1/105/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(7/2)*(532*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3*(a*tan(d*
x+c)*(1+I*tan(d*x+c)))^(1/2)-154*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/
2)+210*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(
1/2)*tan(d*x+c)^4*a+160*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+105*I*(I
*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(
d*x+c)+I))*tan(d*x+c)^4*a+420*A*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a
)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^4*a-90*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d
*x+c)))^(1/2)-105*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3
*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^4*a-420*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)
))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^4*a+210*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^4*a+520*I*A*(I*a)^(1/2)*(-I*a)^(1/
2)*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-42*B*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x
+c)))^(1/2)*tan(d*x+c)-30*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2))/(I*a)^(1/2)/(-I*a)
^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)

________________________________________________________________________________________

Maxima [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))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [B]  time = 1.82178, size = 1813, normalized size = 7.85 \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))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

-1/210*(8*sqrt(2)*(2*(100*A - 91*I*B)*a^2*e^(8*I*d*x + 8*I*c) - 5*(37*A - 49*I*B)*a^2*e^(6*I*d*x + 6*I*c) - 7*
(5*A - 11*I*B)*a^2*e^(4*I*d*x + 4*I*c) + 245*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - 105*(A - I*B)*a^2)*sqrt(a/(e^
(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 105*sqr
t((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/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((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a^2
)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*
c) + I*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B
)*a^2)) + 105*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/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((sqrt(2)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) +
(4*I*A + 4*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) +
1))*e^(I*d*x + I*c) - I*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I
*c)/((4*I*A + 4*B)*a^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)

________________________________________________________________________________________

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))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(9/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.63827, size = 312, normalized size = 1.35 \begin{align*} \frac{-\left (i - 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{6} +{\left (-\left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{5} + \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{6}\right )} \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}} \sqrt{i \, a \tan \left (d x + c\right ) + a} B}{{\left (2 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a - 14 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} + 40 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} - 60 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} + 50 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} - 22 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} + 4 i \, a^{7}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(9/2),x, algorithm="giac")

[Out]

(-(I - 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)^3*a^6 + (-(2*I + 2)*(I*a*tan(d*x +
c) + a)^3*a^5 + (2*I + 2)*(I*a*tan(d*x + c) + a)^2*a^6)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*tan
(d*x + c) + a)*B)/((2*I*(I*a*tan(d*x + c) + a)^6*a - 14*I*(I*a*tan(d*x + c) + a)^5*a^2 + 40*I*(I*a*tan(d*x + c
) + a)^4*a^3 - 60*I*(I*a*tan(d*x + c) + a)^3*a^4 + 50*I*(I*a*tan(d*x + c) + a)^2*a^5 - 22*I*(I*a*tan(d*x + c)
+ a)*a^6 + 4*I*a^7)*d)