3.177 \(\int \frac{(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac{13}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=323 \[ -\frac{8 a^2 (655 A-649 i B) \sqrt{a+i a \tan (c+d x)}}{3465 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (253 B+250 i A) \sqrt{a+i a \tan (c+d x)}}{1155 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 a^2 (212 A-209 i B) \sqrt{a+i a \tan (c+d x)}}{693 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 a^2 (11 B+14 i A) \sqrt{a+i a \tan (c+d x)}}{99 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{8 a^2 (2167 B+2155 i A) \sqrt{a+i a \tan (c+d x)}}{3465 d \sqrt{\tan (c+d x)}}+\frac{(4+4 i) a^{5/2} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac{11}{2}}(c+d x)} \]

[Out]

((4 + 4*I)*a^(5/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a^2*((14*I)*A + 11*B)*Sqrt[a + I*a*Tan[c + d*x]])/(99*d*Tan[c + d*x]^(9/2)) + (2*a^2*(212*A - (209*I)*B)*Sqrt[
a + I*a*Tan[c + d*x]])/(693*d*Tan[c + d*x]^(7/2)) + (4*a^2*((250*I)*A + 253*B)*Sqrt[a + I*a*Tan[c + d*x]])/(11
55*d*Tan[c + d*x]^(5/2)) - (8*a^2*(655*A - (649*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3465*d*Tan[c + d*x]^(3/2))
- (8*a^2*((2155*I)*A + 2167*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3465*d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c
+ d*x])^(3/2))/(11*d*Tan[c + d*x]^(11/2))

________________________________________________________________________________________

Rubi [A]  time = 1.16084, antiderivative size = 323, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{8 a^2 (655 A-649 i B) \sqrt{a+i a \tan (c+d x)}}{3465 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (253 B+250 i A) \sqrt{a+i a \tan (c+d x)}}{1155 d \tan ^{\frac{5}{2}}(c+d x)}+\frac{2 a^2 (212 A-209 i B) \sqrt{a+i a \tan (c+d x)}}{693 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 a^2 (11 B+14 i A) \sqrt{a+i a \tan (c+d x)}}{99 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{8 a^2 (2167 B+2155 i A) \sqrt{a+i a \tan (c+d x)}}{3465 d \sqrt{\tan (c+d x)}}+\frac{(4+4 i) a^{5/2} (B+i A) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac{11}{2}}(c+d x)} \]

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]^(13/2),x]

[Out]

((4 + 4*I)*a^(5/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*
a^2*((14*I)*A + 11*B)*Sqrt[a + I*a*Tan[c + d*x]])/(99*d*Tan[c + d*x]^(9/2)) + (2*a^2*(212*A - (209*I)*B)*Sqrt[
a + I*a*Tan[c + d*x]])/(693*d*Tan[c + d*x]^(7/2)) + (4*a^2*((250*I)*A + 253*B)*Sqrt[a + I*a*Tan[c + d*x]])/(11
55*d*Tan[c + d*x]^(5/2)) - (8*a^2*(655*A - (649*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3465*d*Tan[c + d*x]^(3/2))
- (8*a^2*((2155*I)*A + 2167*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3465*d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c
+ d*x])^(3/2))/(11*d*Tan[c + d*x]^(11/2))

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{13}{2}}(c+d x)} \, dx &=-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac{11}{2}}(c+d x)}+\frac{2}{11} \int \frac{(a+i a \tan (c+d x))^{3/2} \left (\frac{1}{2} a (14 i A+11 B)-\frac{1}{2} a (8 A-11 i B) \tan (c+d x)\right )}{\tan ^{\frac{11}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (14 i A+11 B) \sqrt{a+i a \tan (c+d x)}}{99 d \tan ^{\frac{9}{2}}(c+d x)}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac{11}{2}}(c+d x)}+\frac{4}{99} \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{1}{4} a^2 (212 A-209 i B)-\frac{1}{4} a^2 (184 i A+187 B) \tan (c+d x)\right )}{\tan ^{\frac{9}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (14 i A+11 B) \sqrt{a+i a \tan (c+d x)}}{99 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{2 a^2 (212 A-209 i B) \sqrt{a+i a \tan (c+d x)}}{693 d \tan ^{\frac{7}{2}}(c+d x)}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac{11}{2}}(c+d x)}+\frac{8 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{3}{4} a^3 (250 i A+253 B)+\frac{3}{4} a^3 (212 A-209 i B) \tan (c+d x)\right )}{\tan ^{\frac{7}{2}}(c+d x)} \, dx}{693 a}\\ &=-\frac{2 a^2 (14 i A+11 B) \sqrt{a+i a \tan (c+d x)}}{99 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{2 a^2 (212 A-209 i B) \sqrt{a+i a \tan (c+d x)}}{693 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a^2 (250 i A+253 B) \sqrt{a+i a \tan (c+d x)}}{1155 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac{11}{2}}(c+d x)}+\frac{16 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{3}{4} a^4 (655 A-649 i B)+\frac{3}{2} a^4 (250 i A+253 B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{3465 a^2}\\ &=-\frac{2 a^2 (14 i A+11 B) \sqrt{a+i a \tan (c+d x)}}{99 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{2 a^2 (212 A-209 i B) \sqrt{a+i a \tan (c+d x)}}{693 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a^2 (250 i A+253 B) \sqrt{a+i a \tan (c+d x)}}{1155 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{8 a^2 (655 A-649 i B) \sqrt{a+i a \tan (c+d x)}}{3465 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac{11}{2}}(c+d x)}+\frac{32 \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{3}{8} a^5 (2155 i A+2167 B)-\frac{3}{4} a^5 (655 A-649 i B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{10395 a^3}\\ &=-\frac{2 a^2 (14 i A+11 B) \sqrt{a+i a \tan (c+d x)}}{99 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{2 a^2 (212 A-209 i B) \sqrt{a+i a \tan (c+d x)}}{693 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a^2 (250 i A+253 B) \sqrt{a+i a \tan (c+d x)}}{1155 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{8 a^2 (655 A-649 i B) \sqrt{a+i a \tan (c+d x)}}{3465 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{8 a^2 (2155 i A+2167 B) \sqrt{a+i a \tan (c+d x)}}{3465 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac{11}{2}}(c+d x)}+\frac{64 \int -\frac{10395 a^6 (A-i B) \sqrt{a+i a \tan (c+d x)}}{16 \sqrt{\tan (c+d x)}} \, dx}{10395 a^4}\\ &=-\frac{2 a^2 (14 i A+11 B) \sqrt{a+i a \tan (c+d x)}}{99 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{2 a^2 (212 A-209 i B) \sqrt{a+i a \tan (c+d x)}}{693 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a^2 (250 i A+253 B) \sqrt{a+i a \tan (c+d x)}}{1155 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{8 a^2 (655 A-649 i B) \sqrt{a+i a \tan (c+d x)}}{3465 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{8 a^2 (2155 i A+2167 B) \sqrt{a+i a \tan (c+d x)}}{3465 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac{11}{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 (14 i A+11 B) \sqrt{a+i a \tan (c+d x)}}{99 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{2 a^2 (212 A-209 i B) \sqrt{a+i a \tan (c+d x)}}{693 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a^2 (250 i A+253 B) \sqrt{a+i a \tan (c+d x)}}{1155 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{8 a^2 (655 A-649 i B) \sqrt{a+i a \tan (c+d x)}}{3465 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{8 a^2 (2155 i A+2167 B) \sqrt{a+i a \tan (c+d x)}}{3465 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac{11}{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 (14 i A+11 B) \sqrt{a+i a \tan (c+d x)}}{99 d \tan ^{\frac{9}{2}}(c+d x)}+\frac{2 a^2 (212 A-209 i B) \sqrt{a+i a \tan (c+d x)}}{693 d \tan ^{\frac{7}{2}}(c+d x)}+\frac{4 a^2 (250 i A+253 B) \sqrt{a+i a \tan (c+d x)}}{1155 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{8 a^2 (655 A-649 i B) \sqrt{a+i a \tan (c+d x)}}{3465 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{8 a^2 (2155 i A+2167 B) \sqrt{a+i a \tan (c+d x)}}{3465 d \sqrt{\tan (c+d x)}}-\frac{2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac{11}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 19.3325, size = 328, normalized size = 1.02 \[ \frac{4 \sqrt{2} a^2 (B+i A) e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )}{d \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-\frac{a^2 \csc ^3(c+d x) \sec ^2(c+d x) \sqrt{a+i a \tan (c+d x)} (66 (95 A-47 i B) \cos (c+d x)+(-5225 A+6743 i B) \cos (3 (c+d x))+84810 i A \sin (c+d x)-42185 i A \sin (3 (c+d x))+10925 i A \sin (5 (c+d x))+3995 A \cos (5 (c+d x))+84414 B \sin (c+d x)-43703 B \sin (3 (c+d x))+10571 B \sin (5 (c+d x))-3641 i B \cos (5 (c+d x)))}{27720 d \tan ^{\frac{5}{2}}(c+d x)} \]

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]^(13/2),x]

[Out]

(4*Sqrt[2]*a^2*(I*A + B)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))
]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])/(d*E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d
*x))))/(1 + E^((2*I)*(c + d*x)))]) - (a^2*Csc[c + d*x]^3*Sec[c + d*x]^2*(66*(95*A - (47*I)*B)*Cos[c + d*x] + (
-5225*A + (6743*I)*B)*Cos[3*(c + d*x)] + 3995*A*Cos[5*(c + d*x)] - (3641*I)*B*Cos[5*(c + d*x)] + (84810*I)*A*S
in[c + d*x] + 84414*B*Sin[c + d*x] - (42185*I)*A*Sin[3*(c + d*x)] - 43703*B*Sin[3*(c + d*x)] + (10925*I)*A*Sin
[5*(c + d*x)] + 10571*B*Sin[5*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(27720*d*Tan[c + d*x]^(5/2))

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Maple [B]  time = 0.046, size = 976, normalized size = 3. \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)^(13/2),x)

[Out]

-1/3465/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(11/2)*(17336*B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^5*(a*t
an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-5192*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)-13860*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)^6*a+5240*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)+2090*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)+13860*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)^6*a-300
0*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)-3465*(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)
^6*a-3036*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)+17240*I*A*(I*a)^(1/2)*
(-I*a)^(1/2)*tan(d*x+c)^5*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+6930*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)^6*a-2120*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)+1610*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)+3465*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)^6*a+6930*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)^6*a+770*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)+630*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)

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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)^(13/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.88638, size = 2203, normalized size = 6.82 \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)^(13/2),x, algorithm="fricas")

[Out]

1/6930*(8*sqrt(2)*(2*(3730*A - 3553*I*B)*a^2*e^(12*I*d*x + 12*I*c) - 9*(1805*A - 2013*I*B)*a^2*e^(10*I*d*x + 1
0*I*c) + 55*(397*A - 337*I*B)*a^2*e^(8*I*d*x + 8*I*c) + 66*(95*A - 47*I*B)*a^2*e^(6*I*d*x + 6*I*c) - 1386*(15*
A - 16*I*B)*a^2*e^(4*I*d*x + 4*I*c) + 15015*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - 3465*(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) - 3465*sq
rt((-32*I*A^2 - 64*A*B + 32*I*B^2)*a^5/d^2)*(d*e^(12*I*d*x + 12*I*c) - 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I
*d*x + 8*I*c) - 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) - 6*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)) + 3465*sqrt((-32*I*A^2 - 64*A*B + 32*I*B^2
)*a^5/d^2)*(d*e^(12*I*d*x + 12*I*c) - 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) - 20*d*e^(6*I*d*x +
6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) - 6*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^(12*I*d*x + 12*I*c) - 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8
*I*c) - 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) - 6*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)**(13/2),x)

[Out]

Timed out

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Giac [A]  time = 1.71255, size = 360, normalized size = 1.11 \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^{8} +{\left (\left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{7} - \left (2 i + 2\right ) \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{8}\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 )}^{8} a - 18 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{7} a^{2} + 70 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{6} a^{3} - 154 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{4} + 210 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{5} - 182 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{6} + 98 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{7} - 30 i \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{8} + 4 i \, a^{9}\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)^(13/2),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^8 + ((2*I + 2)*(I*a*tan(d*x + c)
+ a)^3*a^7 - (2*I + 2)*(I*a*tan(d*x + c) + a)^2*a^8)*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)^8*a - 18*I*(I*a*tan(d*x + c) + a)^7*a^2 + 70*I*(I*a*tan(d*x + c)
+ a)^6*a^3 - 154*I*(I*a*tan(d*x + c) + a)^5*a^4 + 210*I*(I*a*tan(d*x + c) + a)^4*a^5 - 182*I*(I*a*tan(d*x + c)
+ a)^3*a^6 + 98*I*(I*a*tan(d*x + c) + a)^2*a^7 - 30*I*(I*a*tan(d*x + c) + a)*a^8 + 4*I*a^9)*d)