3.828 \(\int \frac{(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{15/2}} \, dx\)

Optimal. Leaf size=261 \[ -\frac{2 (-11 B+4 i A) (a+i a \tan (e+f x))^{7/2}}{45045 c^4 f (c-i c \tan (e+f x))^{7/2}}-\frac{2 (-11 B+4 i A) (a+i a \tan (e+f x))^{7/2}}{6435 c^3 f (c-i c \tan (e+f x))^{9/2}}-\frac{(-11 B+4 i A) (a+i a \tan (e+f x))^{7/2}}{715 c^2 f (c-i c \tan (e+f x))^{11/2}}-\frac{(-11 B+4 i A) (a+i a \tan (e+f x))^{7/2}}{195 c f (c-i c \tan (e+f x))^{13/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{15 f (c-i c \tan (e+f x))^{15/2}} \]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(15*f*(c - I*c*Tan[e + f*x])^(15/2)) - (((4*I)*A - 11*B)*(a + I*a*Ta
n[e + f*x])^(7/2))/(195*c*f*(c - I*c*Tan[e + f*x])^(13/2)) - (((4*I)*A - 11*B)*(a + I*a*Tan[e + f*x])^(7/2))/(
715*c^2*f*(c - I*c*Tan[e + f*x])^(11/2)) - (2*((4*I)*A - 11*B)*(a + I*a*Tan[e + f*x])^(7/2))/(6435*c^3*f*(c -
I*c*Tan[e + f*x])^(9/2)) - (2*((4*I)*A - 11*B)*(a + I*a*Tan[e + f*x])^(7/2))/(45045*c^4*f*(c - I*c*Tan[e + f*x
])^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.315509, 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} \[ -\frac{2 (-11 B+4 i A) (a+i a \tan (e+f x))^{7/2}}{45045 c^4 f (c-i c \tan (e+f x))^{7/2}}-\frac{2 (-11 B+4 i A) (a+i a \tan (e+f x))^{7/2}}{6435 c^3 f (c-i c \tan (e+f x))^{9/2}}-\frac{(-11 B+4 i A) (a+i a \tan (e+f x))^{7/2}}{715 c^2 f (c-i c \tan (e+f x))^{11/2}}-\frac{(-11 B+4 i A) (a+i a \tan (e+f x))^{7/2}}{195 c f (c-i c \tan (e+f x))^{13/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{7/2}}{15 f (c-i c \tan (e+f x))^{15/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(15/2),x]

[Out]

-((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(15*f*(c - I*c*Tan[e + f*x])^(15/2)) - (((4*I)*A - 11*B)*(a + I*a*Ta
n[e + f*x])^(7/2))/(195*c*f*(c - I*c*Tan[e + f*x])^(13/2)) - (((4*I)*A - 11*B)*(a + I*a*Tan[e + f*x])^(7/2))/(
715*c^2*f*(c - I*c*Tan[e + f*x])^(11/2)) - (2*((4*I)*A - 11*B)*(a + I*a*Tan[e + f*x])^(7/2))/(6435*c^3*f*(c -
I*c*Tan[e + f*x])^(9/2)) - (2*((4*I)*A - 11*B)*(a + I*a*Tan[e + f*x])^(7/2))/(45045*c^4*f*(c - I*c*Tan[e + f*x
])^(7/2))

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))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{15/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2} (A+B x)}{(c-i c x)^{17/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{15 f (c-i c \tan (e+f x))^{15/2}}+\frac{(a (4 A+11 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{15/2}} \, dx,x,\tan (e+f x)\right )}{15 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{15 f (c-i c \tan (e+f x))^{15/2}}-\frac{(4 i A-11 B) (a+i a \tan (e+f x))^{7/2}}{195 c f (c-i c \tan (e+f x))^{13/2}}+\frac{(a (4 A+11 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{13/2}} \, dx,x,\tan (e+f x)\right )}{65 c f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{15 f (c-i c \tan (e+f x))^{15/2}}-\frac{(4 i A-11 B) (a+i a \tan (e+f x))^{7/2}}{195 c f (c-i c \tan (e+f x))^{13/2}}-\frac{(4 i A-11 B) (a+i a \tan (e+f x))^{7/2}}{715 c^2 f (c-i c \tan (e+f x))^{11/2}}+\frac{(2 a (4 A+11 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{11/2}} \, dx,x,\tan (e+f x)\right )}{715 c^2 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{15 f (c-i c \tan (e+f x))^{15/2}}-\frac{(4 i A-11 B) (a+i a \tan (e+f x))^{7/2}}{195 c f (c-i c \tan (e+f x))^{13/2}}-\frac{(4 i A-11 B) (a+i a \tan (e+f x))^{7/2}}{715 c^2 f (c-i c \tan (e+f x))^{11/2}}-\frac{2 (4 i A-11 B) (a+i a \tan (e+f x))^{7/2}}{6435 c^3 f (c-i c \tan (e+f x))^{9/2}}+\frac{(2 a (4 A+11 i B)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{5/2}}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{6435 c^3 f}\\ &=-\frac{(i A+B) (a+i a \tan (e+f x))^{7/2}}{15 f (c-i c \tan (e+f x))^{15/2}}-\frac{(4 i A-11 B) (a+i a \tan (e+f x))^{7/2}}{195 c f (c-i c \tan (e+f x))^{13/2}}-\frac{(4 i A-11 B) (a+i a \tan (e+f x))^{7/2}}{715 c^2 f (c-i c \tan (e+f x))^{11/2}}-\frac{2 (4 i A-11 B) (a+i a \tan (e+f x))^{7/2}}{6435 c^3 f (c-i c \tan (e+f x))^{9/2}}-\frac{2 (4 i A-11 B) (a+i a \tan (e+f x))^{7/2}}{45045 c^4 f (c-i c \tan (e+f x))^{7/2}}\\ \end{align*}

Mathematica [B]  time = 17.3174, size = 577, normalized size = 2.21 \[ \frac{\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt{\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((B-i A) \cos (6 f x) \left (\frac{\cos (3 e)}{224 c^8}+\frac{i \sin (3 e)}{224 c^8}\right )+(A+i B) \sin (6 f x) \left (\frac{\cos (3 e)}{224 c^8}+\frac{i \sin (3 e)}{224 c^8}\right )+(23 B-37 i A) \cos (8 f x) \left (\frac{\cos (5 e)}{2016 c^8}+\frac{i \sin (5 e)}{2016 c^8}\right )+(11 B-49 i A) \cos (10 f x) \left (\frac{\cos (7 e)}{1584 c^8}+\frac{i \sin (7 e)}{1584 c^8}\right )+(61 A-11 i B) \cos (12 f x) \left (\frac{\sin (9 e)}{2288 c^8}-\frac{i \cos (9 e)}{2288 c^8}\right )+(73 A-43 i B) \cos (14 f x) \left (\frac{\sin (11 e)}{6240 c^8}-\frac{i \cos (11 e)}{6240 c^8}\right )+(A-i B) \cos (16 f x) \left (\frac{\sin (13 e)}{480 c^8}-\frac{i \cos (13 e)}{480 c^8}\right )+(37 A+23 i B) \sin (8 f x) \left (\frac{\cos (5 e)}{2016 c^8}+\frac{i \sin (5 e)}{2016 c^8}\right )+(49 A+11 i B) \sin (10 f x) \left (\frac{\cos (7 e)}{1584 c^8}+\frac{i \sin (7 e)}{1584 c^8}\right )+(61 A-11 i B) \sin (12 f x) \left (\frac{\cos (9 e)}{2288 c^8}+\frac{i \sin (9 e)}{2288 c^8}\right )+(73 A-43 i B) \sin (14 f x) \left (\frac{\cos (11 e)}{6240 c^8}+\frac{i \sin (11 e)}{6240 c^8}\right )+(A-i B) \sin (16 f x) \left (\frac{\cos (13 e)}{480 c^8}+\frac{i \sin (13 e)}{480 c^8}\right )\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(15/2),x]

[Out]

(Cos[e + f*x]^4*(((-I)*A + B)*Cos[6*f*x]*(Cos[3*e]/(224*c^8) + ((I/224)*Sin[3*e])/c^8) + ((-37*I)*A + 23*B)*Co
s[8*f*x]*(Cos[5*e]/(2016*c^8) + ((I/2016)*Sin[5*e])/c^8) + ((-49*I)*A + 11*B)*Cos[10*f*x]*(Cos[7*e]/(1584*c^8)
+ ((I/1584)*Sin[7*e])/c^8) + (61*A - (11*I)*B)*Cos[12*f*x]*(((-I/2288)*Cos[9*e])/c^8 + Sin[9*e]/(2288*c^8)) +
(73*A - (43*I)*B)*Cos[14*f*x]*(((-I/6240)*Cos[11*e])/c^8 + Sin[11*e]/(6240*c^8)) + (A - I*B)*Cos[16*f*x]*(((-
I/480)*Cos[13*e])/c^8 + Sin[13*e]/(480*c^8)) + (A + I*B)*(Cos[3*e]/(224*c^8) + ((I/224)*Sin[3*e])/c^8)*Sin[6*f
*x] + (37*A + (23*I)*B)*(Cos[5*e]/(2016*c^8) + ((I/2016)*Sin[5*e])/c^8)*Sin[8*f*x] + (49*A + (11*I)*B)*(Cos[7*
e]/(1584*c^8) + ((I/1584)*Sin[7*e])/c^8)*Sin[10*f*x] + (61*A - (11*I)*B)*(Cos[9*e]/(2288*c^8) + ((I/2288)*Sin[
9*e])/c^8)*Sin[12*f*x] + (73*A - (43*I)*B)*(Cos[11*e]/(6240*c^8) + ((I/6240)*Sin[11*e])/c^8)*Sin[14*f*x] + (A
- I*B)*(Cos[13*e]/(480*c^8) + ((I/480)*Sin[13*e])/c^8)*Sin[16*f*x])*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] - I*c*Si
n[e + f*x])]*(a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] +
B*Sin[e + f*x]))

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Maple [A]  time = 0.122, size = 206, normalized size = 0.8 \begin{align*} -{\frac{{a}^{3} \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 22\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{6}+72\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{5}+8\,A \left ( \tan \left ( fx+e \right ) \right ) ^{6}-825\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{4}-198\,B \left ( \tan \left ( fx+e \right ) \right ) ^{5}-780\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{3}-300\,A \left ( \tan \left ( fx+e \right ) \right ) ^{4}-7260\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{2}+2145\,B \left ( \tan \left ( fx+e \right ) \right ) ^{3}-6858\,iA\tan \left ( fx+e \right ) +1455\,A \left ( \tan \left ( fx+e \right ) \right ) ^{2}-407\,iB-3663\,B\tan \left ( fx+e \right ) -4243\,A \right ) }{45045\,f{c}^{8} \left ( \tan \left ( fx+e \right ) +i \right ) ^{9}}\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))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(15/2),x)

[Out]

-1/45045/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^8*(1+tan(f*x+e)^2)*(22*I*B*tan(f*x+e)
^6+72*I*A*tan(f*x+e)^5+8*A*tan(f*x+e)^6-825*I*B*tan(f*x+e)^4-198*B*tan(f*x+e)^5-780*I*A*tan(f*x+e)^3-300*A*tan
(f*x+e)^4-7260*I*B*tan(f*x+e)^2+2145*B*tan(f*x+e)^3-6858*I*A*tan(f*x+e)+1455*A*tan(f*x+e)^2-407*I*B-3663*B*tan
(f*x+e)-4243*A)/(tan(f*x+e)+I)^9

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Maxima [A]  time = 3.28053, size = 448, normalized size = 1.72 \begin{align*} \frac{{\left (3003 \,{\left (-i \, A - B\right )} a^{3} \cos \left (\frac{15}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 6930 \,{\left (-2 i \, A - B\right )} a^{3} \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 ) - 24570 i \, A a^{3} \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 \,{\left (-2 i \, A + B\right )} a^{3} \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 ) + 6435 \,{\left (-i \, A + B\right )} a^{3} \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 ) +{\left (3003 \, A - 3003 i \, B\right )} a^{3} \sin \left (\frac{15}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (13860 \, A - 6930 i \, B\right )} a^{3} \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 ) + 24570 \, A a^{3} \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 ) +{\left (20020 \, A + 10010 i \, B\right )} a^{3} \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 (6435 \, A + 6435 i \, B\right )} a^{3} \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 )\right )} \sqrt{a}}{720720 \, c^{\frac{15}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(15/2),x, algorithm="maxima")

[Out]

1/720720*(3003*(-I*A - B)*a^3*cos(15/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 6930*(-2*I*A - B)*a^3*co
s(13/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 24570*I*A*a^3*cos(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e))) + 10010*(-2*I*A + B)*a^3*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 6435*(-I*A + B)*a^
3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (3003*A - 3003*I*B)*a^3*sin(15/2*arctan2(sin(2*f*x +
2*e), cos(2*f*x + 2*e))) + (13860*A - 6930*I*B)*a^3*sin(13/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 24
570*A*a^3*sin(11/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (20020*A + 10010*I*B)*a^3*sin(9/2*arctan2(si
n(2*f*x + 2*e), cos(2*f*x + 2*e))) + (6435*A + 6435*I*B)*a^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e
))))*sqrt(a)/(c^(15/2)*f)

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Fricas [A]  time = 1.41013, size = 537, normalized size = 2.06 \begin{align*} \frac{{\left ({\left (-3003 i \, A - 3003 \, B\right )} a^{3} e^{\left (16 i \, f x + 16 i \, e\right )} +{\left (-16863 i \, A - 9933 \, B\right )} a^{3} e^{\left (14 i \, f x + 14 i \, e\right )} +{\left (-38430 i \, A - 6930 \, B\right )} a^{3} e^{\left (12 i \, f x + 12 i \, e\right )} +{\left (-44590 i \, A + 10010 \, B\right )} a^{3} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-26455 i \, A + 16445 \, B\right )} a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-6435 i \, A + 6435 \, B\right )} a^{3} e^{\left (6 i \, f x + 6 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 )}}{720720 \, c^{8} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(15/2),x, algorithm="fricas")

[Out]

1/720720*((-3003*I*A - 3003*B)*a^3*e^(16*I*f*x + 16*I*e) + (-16863*I*A - 9933*B)*a^3*e^(14*I*f*x + 14*I*e) + (
-38430*I*A - 6930*B)*a^3*e^(12*I*f*x + 12*I*e) + (-44590*I*A + 10010*B)*a^3*e^(10*I*f*x + 10*I*e) + (-26455*I*
A + 16445*B)*a^3*e^(8*I*f*x + 8*I*e) + (-6435*I*A + 6435*B)*a^3*e^(6*I*f*x + 6*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^8*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))**(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(15/2),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{7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac{15}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(15/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(7/2)/(-I*c*tan(f*x + e) + c)^(15/2), x)