3.815 \(\int \frac{(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx\)

Optimal. Leaf size=261 \[ -\frac{2 (-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{15015 c^4 f (c-i c \tan (e+f x))^{5/2}}-\frac{2 (-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{3003 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac{(-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{429 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac{(-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}} \]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(13*f*(c - I*c*Tan[e + f*x])^(13/2)) - (((4*I)*A - 9*B)*(a + I*a*Tan
[e + f*x])^(5/2))/(143*c*f*(c - I*c*Tan[e + f*x])^(11/2)) - (((4*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(5/2))/(42
9*c^2*f*(c - I*c*Tan[e + f*x])^(9/2)) - (2*((4*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(5/2))/(3003*c^3*f*(c - I*c*
Tan[e + f*x])^(7/2)) - (2*((4*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(5/2))/(15015*c^4*f*(c - I*c*Tan[e + f*x])^(5
/2))

________________________________________________________________________________________

Rubi [A]  time = 0.321313, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ -\frac{2 (-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{15015 c^4 f (c-i c \tan (e+f x))^{5/2}}-\frac{2 (-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{3003 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac{(-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{429 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac{(-9 B+4 i A) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(13/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(13*f*(c - I*c*Tan[e + f*x])^(13/2)) - (((4*I)*A - 9*B)*(a + I*a*Tan
[e + f*x])^(5/2))/(143*c*f*(c - I*c*Tan[e + f*x])^(11/2)) - (((4*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(5/2))/(42
9*c^2*f*(c - I*c*Tan[e + f*x])^(9/2)) - (2*((4*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(5/2))/(3003*c^3*f*(c - I*c*
Tan[e + f*x])^(7/2)) - (2*((4*I)*A - 9*B)*(a + I*a*Tan[e + f*x])^(5/2))/(15015*c^4*f*(c - I*c*Tan[e + f*x])^(5
/2))

Rule 3588

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2} (A+B x)}{(c-i c x)^{15/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}+\frac{(a (4 A+9 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(c-i c x)^{13/2}} \, dx,x,\tan (e+f x)\right )}{13 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac{(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}+\frac{(3 a (4 A+9 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{143 c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac{(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{429 c^2 f (c-i c \tan (e+f x))^{9/2}}+\frac{(2 a (4 A+9 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{429 c^2 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac{(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{429 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac{2 (4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{3003 c^3 f (c-i c \tan (e+f x))^{7/2}}+\frac{(2 a (4 A+9 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{3/2}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{3003 c^3 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac{(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac{(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{429 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac{2 (4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{3003 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac{2 (4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{15015 c^4 f (c-i c \tan (e+f x))^{5/2}}\\ \end{align*}

Mathematica [B]  time = 17.0542, size = 577, normalized size = 2.21 \[ \frac{\cos ^3(e+f x) (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((B-i A) \cos (4 f x) \left (\frac{\cos (2 e)}{160 c^7}+\frac{i \sin (2 e)}{160 c^7}\right )+(A+i B) \sin (4 f x) \left (\frac{\cos (2 e)}{160 c^7}+\frac{i \sin (2 e)}{160 c^7}\right )+(17 B-27 i A) \cos (6 f x) \left (\frac{\cos (4 e)}{1120 c^7}+\frac{i \sin (4 e)}{1120 c^7}\right )+(3 B-13 i A) \cos (8 f x) \left (\frac{\cos (6 e)}{336 c^7}+\frac{i \sin (6 e)}{336 c^7}\right )+(17 A-3 i B) \cos (10 f x) \left (\frac{\sin (8 e)}{528 c^7}-\frac{i \cos (8 e)}{528 c^7}\right )+(63 A-37 i B) \cos (12 f x) \left (\frac{\sin (10 e)}{4576 c^7}-\frac{i \cos (10 e)}{4576 c^7}\right )+(A-i B) \cos (14 f x) \left (\frac{\sin (12 e)}{416 c^7}-\frac{i \cos (12 e)}{416 c^7}\right )+(27 A+17 i B) \sin (6 f x) \left (\frac{\cos (4 e)}{1120 c^7}+\frac{i \sin (4 e)}{1120 c^7}\right )+(13 A+3 i B) \sin (8 f x) \left (\frac{\cos (6 e)}{336 c^7}+\frac{i \sin (6 e)}{336 c^7}\right )+(17 A-3 i B) \sin (10 f x) \left (\frac{\cos (8 e)}{528 c^7}+\frac{i \sin (8 e)}{528 c^7}\right )+(63 A-37 i B) \sin (12 f x) \left (\frac{\cos (10 e)}{4576 c^7}+\frac{i \sin (10 e)}{4576 c^7}\right )+(A-i B) \sin (14 f x) \left (\frac{\cos (12 e)}{416 c^7}+\frac{i \sin (12 e)}{416 c^7}\right )\right )}{f (\cos (f x)+i \sin (f x))^2 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(13/2),x]

[Out]

(Cos[e + f*x]^3*(((-I)*A + B)*Cos[4*f*x]*(Cos[2*e]/(160*c^7) + ((I/160)*Sin[2*e])/c^7) + ((-27*I)*A + 17*B)*Co
s[6*f*x]*(Cos[4*e]/(1120*c^7) + ((I/1120)*Sin[4*e])/c^7) + ((-13*I)*A + 3*B)*Cos[8*f*x]*(Cos[6*e]/(336*c^7) +
((I/336)*Sin[6*e])/c^7) + (17*A - (3*I)*B)*Cos[10*f*x]*(((-I/528)*Cos[8*e])/c^7 + Sin[8*e]/(528*c^7)) + (63*A
- (37*I)*B)*Cos[12*f*x]*(((-I/4576)*Cos[10*e])/c^7 + Sin[10*e]/(4576*c^7)) + (A - I*B)*Cos[14*f*x]*(((-I/416)*
Cos[12*e])/c^7 + Sin[12*e]/(416*c^7)) + (A + I*B)*(Cos[2*e]/(160*c^7) + ((I/160)*Sin[2*e])/c^7)*Sin[4*f*x] + (
27*A + (17*I)*B)*(Cos[4*e]/(1120*c^7) + ((I/1120)*Sin[4*e])/c^7)*Sin[6*f*x] + (13*A + (3*I)*B)*(Cos[6*e]/(336*
c^7) + ((I/336)*Sin[6*e])/c^7)*Sin[8*f*x] + (17*A - (3*I)*B)*(Cos[8*e]/(528*c^7) + ((I/528)*Sin[8*e])/c^7)*Sin
[10*f*x] + (63*A - (37*I)*B)*(Cos[10*e]/(4576*c^7) + ((I/4576)*Sin[10*e])/c^7)*Sin[12*f*x] + (A - I*B)*(Cos[12
*e]/(416*c^7) + ((I/416)*Sin[12*e])/c^7)*Sin[14*f*x])*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(
a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^2*(A*Cos[e + f*x] + B*Sin[e + f*x
]))

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Maple [A]  time = 0.11, size = 183, normalized size = 0.7 \begin{align*}{\frac{{a}^{2} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 18\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{5}+64\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{4}+8\,A \left ( \tan \left ( fx+e \right ) \right ) ^{5}-531\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{3}-144\,B \left ( \tan \left ( fx+e \right ) \right ) ^{4}-544\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{2}-236\,A \left ( \tan \left ( fx+e \right ) \right ) ^{3}-1704\,iB\tan \left ( fx+e \right ) +1224\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}-1763\,iA+911\,A\tan \left ( fx+e \right ) +213\,B \right ) }{15015\,f{c}^{7} \left ( \tan \left ( fx+e \right ) +i \right ) ^{8}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x)

[Out]

1/15015/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^2/c^7*(1+tan(f*x+e)^2)*(18*I*B*tan(f*x+e)^
5+64*I*A*tan(f*x+e)^4+8*A*tan(f*x+e)^5-531*I*B*tan(f*x+e)^3-144*B*tan(f*x+e)^4-544*I*A*tan(f*x+e)^2-236*A*tan(
f*x+e)^3-1704*I*B*tan(f*x+e)+1224*B*tan(f*x+e)^2-1763*I*A+911*A*tan(f*x+e)+213*B)/(tan(f*x+e)+I)^8

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Maxima [A]  time = 2.53554, size = 448, normalized size = 1.72 \begin{align*} \frac{{\left (1155 \,{\left (-i \, A - B\right )} a^{2} \cos \left (\frac{13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 2730 \,{\left (-2 i \, A - B\right )} a^{2} \cos \left (\frac{11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 10010 i \, A a^{2} \cos \left (\frac{9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 4290 \,{\left (-2 i \, A + B\right )} a^{2} \cos \left (\frac{7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3003 \,{\left (-i \, A + B\right )} a^{2} \cos \left (\frac{5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (1155 \, A - 1155 i \, B\right )} a^{2} \sin \left (\frac{13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (5460 \, A - 2730 i \, B\right )} a^{2} \sin \left (\frac{11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 10010 \, A a^{2} \sin \left (\frac{9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (8580 \, A + 4290 i \, B\right )} a^{2} \sin \left (\frac{7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (3003 \, A + 3003 i \, B\right )} a^{2} \sin \left (\frac{5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt{a}}{240240 \, c^{\frac{13}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x, algorithm="maxima")

[Out]

1/240240*(1155*(-I*A - B)*a^2*cos(13/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2730*(-2*I*A - B)*a^2*co
s(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 10010*I*A*a^2*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*
x + 2*e))) + 4290*(-2*I*A + B)*a^2*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 3003*(-I*A + B)*a^2*
cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (1155*A - 1155*I*B)*a^2*sin(13/2*arctan2(sin(2*f*x + 2*
e), cos(2*f*x + 2*e))) + (5460*A - 2730*I*B)*a^2*sin(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 10010
*A*a^2*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (8580*A + 4290*I*B)*a^2*sin(7/2*arctan2(sin(2*f*
x + 2*e), cos(2*f*x + 2*e))) + (3003*A + 3003*I*B)*a^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*s
qrt(a)/(c^(13/2)*f)

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Fricas [A]  time = 1.41336, size = 531, normalized size = 2.03 \begin{align*} \frac{{\left ({\left (-1155 i \, A - 1155 \, B\right )} a^{2} e^{\left (14 i \, f x + 14 i \, e\right )} +{\left (-6615 i \, A - 3885 \, B\right )} a^{2} e^{\left (12 i \, f x + 12 i \, e\right )} +{\left (-15470 i \, A - 2730 \, B\right )} a^{2} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-18590 i \, A + 4290 \, B\right )} a^{2} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-11583 i \, A + 7293 \, B\right )} a^{2} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-3003 i \, A + 3003 \, B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt{\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )}}{240240 \, c^{7} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x, algorithm="fricas")

[Out]

1/240240*((-1155*I*A - 1155*B)*a^2*e^(14*I*f*x + 14*I*e) + (-6615*I*A - 3885*B)*a^2*e^(12*I*f*x + 12*I*e) + (-
15470*I*A - 2730*B)*a^2*e^(10*I*f*x + 10*I*e) + (-18590*I*A + 4290*B)*a^2*e^(8*I*f*x + 8*I*e) + (-11583*I*A +
7293*B)*a^2*e^(6*I*f*x + 6*I*e) + (-3003*I*A + 3003*B)*a^2*e^(4*I*f*x + 4*I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) +
1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e)/(c^7*f)

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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(f*x+e))**(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(13/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (f x + e\right ) + A\right )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{13}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(5/2)/(-I*c*tan(f*x + e) + c)^(13/2), x)