3.176 \(\int \frac{(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac{11}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=277 \[ \frac{8 a^2 (60 B+59 i A) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (46 A-45 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a^2 (3 B+4 i A) \sqrt{a+i a \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{8 a^2 (197 A-195 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}+\frac{(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)} \]

[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*((4*I)*A + 3*B)*Sqrt[a + I*a*Tan[c + d*x]])/(21*d*Tan[c + d*x]^(7/2)) + (2*a^2*(46*A - (45*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(5/2)) + (8*a^2*((59*I)*A + 60*B)*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*Ta
n[c + d*x]^(3/2)) - (8*a^2*(197*A - (195*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*Sqrt[Tan[c + d*x]]) - (2*a*A
*(a + I*a*Tan[c + d*x])^(3/2))/(9*d*Tan[c + d*x]^(9/2))

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Rubi [A]  time = 0.950501, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{8 a^2 (60 B+59 i A) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2 a^2 (46 A-45 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a^2 (3 B+4 i A) \sqrt{a+i a \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{8 a^2 (197 A-195 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}+\frac{(4+4 i) a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)} \]

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]^(11/2),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*((4*I)*A + 3*B)*Sqrt[a + I*a*Tan[c + d*x]])/(21*d*Tan[c + d*x]^(7/2)) + (2*a^2*(46*A - (45*I)*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(5/2)) + (8*a^2*((59*I)*A + 60*B)*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*Ta
n[c + d*x]^(3/2)) - (8*a^2*(197*A - (195*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(315*d*Sqrt[Tan[c + d*x]]) - (2*a*A
*(a + I*a*Tan[c + d*x])^(3/2))/(9*d*Tan[c + d*x]^(9/2))

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{11}{2}}(c+d x)} \, dx &=-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{2}{9} \int \frac{(a+i a \tan (c+d x))^{3/2} \left (\frac{3}{2} a (4 i A+3 B)-\frac{3}{2} a (2 A-3 i B) \tan (c+d x)\right )}{\tan ^{\frac{9}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (4 i A+3 B) \sqrt{a+i a \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{4}{63} \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{4} a^2 (46 A-45 i B)-\frac{3}{4} a^2 (38 i A+39 B) \tan (c+d x)\right )}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (4 i A+3 B) \sqrt{a+i a \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 a^2 (46 A-45 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{8 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{2} a^3 (59 i A+60 B)+\frac{3}{2} a^3 (46 A-45 i B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{315 a}\\ &=-\frac{2 a^2 (4 i A+3 B) \sqrt{a+i a \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 a^2 (46 A-45 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 (59 i A+60 B) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{16 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{3}{4} a^4 (197 A-195 i B)+\frac{3}{2} a^4 (59 i A+60 B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{945 a^2}\\ &=-\frac{2 a^2 (4 i A+3 B) \sqrt{a+i a \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 a^2 (46 A-45 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 (59 i A+60 B) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{8 a^2 (197 A-195 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{32 \int \frac{945 a^5 (i A+B) \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{945 a^3}\\ &=-\frac{2 a^2 (4 i A+3 B) \sqrt{a+i a \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 a^2 (46 A-45 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 (59 i A+60 B) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{8 a^2 (197 A-195 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\left (4 a^2 (i A+B)\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a^2 (4 i A+3 B) \sqrt{a+i a \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 a^2 (46 A-45 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 (59 i A+60 B) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{8 a^2 (197 A-195 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{\left (8 a^4 (A-i 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} (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^2 (4 i A+3 B) \sqrt{a+i a \tan (c+d x)}}{21 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{2 a^2 (46 A-45 i B) \sqrt{a+i a \tan (c+d x)}}{105 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{8 a^2 (59 i A+60 B) \sqrt{a+i a \tan (c+d x)}}{315 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{8 a^2 (197 A-195 i B) \sqrt{a+i a \tan (c+d x)}}{315 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{9 d \tan ^{\frac{9}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 14.9964, size = 246, normalized size = 0.89 \[ \frac{a^2 \sqrt{a+i a \tan (c+d x)} \left (\frac{1260 (A-i B) e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )}{\sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+\frac{\csc ^2(2 (c+d x)) (12 (251 A-260 i B) \cos (2 (c+d x))+(-961 A+915 i B) \cos (4 (c+d x))+282 i A \sin (2 (c+d x))-331 i A \sin (4 (c+d x))-2331 A+390 B \sin (2 (c+d x))-285 B \sin (4 (c+d x))+2205 i B)}{\tan ^{\frac{5}{2}}(c+d x)}\right )}{315 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(11/2),x]

[Out]

(a^2*((1260*(A - I*B)*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])/
(E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]) + (Csc[2*(c + d*x)]^2*(-23
31*A + (2205*I)*B + 12*(251*A - (260*I)*B)*Cos[2*(c + d*x)] + (-961*A + (915*I)*B)*Cos[4*(c + d*x)] + (282*I)*
A*Sin[2*(c + d*x)] + 390*B*Sin[2*(c + d*x)] - (331*I)*A*Sin[4*(c + d*x)] - 285*B*Sin[4*(c + d*x)]))/Tan[c + d*
x]^(5/2))*Sqrt[a + I*a*Tan[c + d*x]])/(315*d)

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Maple [B]  time = 0.048, size = 887, normalized size = 3.2 \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)^(11/2),x)

[Out]

1/315/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(9/2)*(-1576*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4*(a*tan(
d*x+c)*(1+I*tan(d*x+c)))^(1/2)+1560*I*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^
(1/2)-270*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)+480*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)-190*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)*(a*t
an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+1260*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)^5*a+1260*I*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)^5*a-315*(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)^5*a+276*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)-315*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)^5*a-
630*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)^5*a+630*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)^5*a+472*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)-90*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)-70*A*(a*tan(d*x+c)*(1+I*t
an(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)

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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)^(11/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.87754, size = 2014, normalized size = 7.27 \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)^(11/2),x, algorithm="fricas")

[Out]

1/630*(sqrt(2)*((-5168*I*A - 4800*B)*a^2*e^(10*I*d*x + 10*I*c) + (8008*I*A + 9240*B)*a^2*e^(8*I*d*x + 8*I*c) +
(-5472*I*A - 3600*B)*a^2*e^(6*I*d*x + 6*I*c) + (-7728*I*A - 6720*B)*a^2*e^(4*I*d*x + 4*I*c) + (8400*I*A + 840
0*B)*a^2*e^(2*I*d*x + 2*I*c) + (-2520*I*A - 2520*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) + 315*sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*(
d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*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) + 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)) - 315*sqr
t((32*I*A^2 + 64*A*B - 32*I*B^2)*a^5/d^2)*(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x
+ 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*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) - 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^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*
c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*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))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(11/2),x)

[Out]

Timed out

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Giac [A]  time = 1.71951, size = 336, normalized size = 1.21 \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^{7} +{\left (\left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{6} - \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{7}\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}{2 \,{\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{7} a - 8 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a^{2} + 27 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{3} - 50 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{4} + 55 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{5} - 36 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{6} + 13 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{7} - 2 \, a^{8}\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)^(11/2),x, algorithm="giac")

[Out]

-1/2*((I - 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)^3*a^7 + ((2*I + 2)*(I*a*tan(d*x
+ c) + a)^3*a^6 - (2*I + 2)*(I*a*tan(d*x + c) + a)^2*a^7)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*sqrt(I*a*
tan(d*x + c) + a)*B)/(((I*a*tan(d*x + c) + a)^7*a - 8*(I*a*tan(d*x + c) + a)^6*a^2 + 27*(I*a*tan(d*x + c) + a)
^5*a^3 - 50*(I*a*tan(d*x + c) + a)^4*a^4 + 55*(I*a*tan(d*x + c) + a)^3*a^5 - 36*(I*a*tan(d*x + c) + a)^2*a^6 +
13*(I*a*tan(d*x + c) + a)*a^7 - 2*a^8)*d)