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

Optimal. Leaf size=194 \[ \frac{c^5 (-7 B+i A) \tan ^2(e+f x)}{2 a^2 f}-\frac{c^5 (7 A+24 i B) \tan (e+f x)}{a^2 f}+\frac{16 c^5 (2 A+3 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac{8 c^5 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}+\frac{8 c^5 (-7 B+3 i A) \log (\cos (e+f x))}{a^2 f}+\frac{8 c^5 x (3 A+7 i B)}{a^2}+\frac{i B c^5 \tan ^3(e+f x)}{3 a^2 f} \]

[Out]

(8*(3*A + (7*I)*B)*c^5*x)/a^2 + (8*((3*I)*A - 7*B)*c^5*Log[Cos[e + f*x]])/(a^2*f) - (8*(I*A - B)*c^5)/(a^2*f*(
I - Tan[e + f*x])^2) + (16*(2*A + (3*I)*B)*c^5)/(a^2*f*(I - Tan[e + f*x])) - ((7*A + (24*I)*B)*c^5*Tan[e + f*x
])/(a^2*f) + ((I*A - 7*B)*c^5*Tan[e + f*x]^2)/(2*a^2*f) + ((I/3)*B*c^5*Tan[e + f*x]^3)/(a^2*f)

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Rubi [A]  time = 0.248915, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.049, Rules used = {3588, 77} \[ \frac{c^5 (-7 B+i A) \tan ^2(e+f x)}{2 a^2 f}-\frac{c^5 (7 A+24 i B) \tan (e+f x)}{a^2 f}+\frac{16 c^5 (2 A+3 i B)}{a^2 f (-\tan (e+f x)+i)}-\frac{8 c^5 (-B+i A)}{a^2 f (-\tan (e+f x)+i)^2}+\frac{8 c^5 (-7 B+3 i A) \log (\cos (e+f x))}{a^2 f}+\frac{8 c^5 x (3 A+7 i B)}{a^2}+\frac{i B c^5 \tan ^3(e+f x)}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(3*A + (7*I)*B)*c^5*x)/a^2 + (8*((3*I)*A - 7*B)*c^5*Log[Cos[e + f*x]])/(a^2*f) - (8*(I*A - B)*c^5)/(a^2*f*(
I - Tan[e + f*x])^2) + (16*(2*A + (3*I)*B)*c^5)/(a^2*f*(I - Tan[e + f*x])) - ((7*A + (24*I)*B)*c^5*Tan[e + f*x
])/(a^2*f) + ((I*A - 7*B)*c^5*Tan[e + f*x]^2)/(2*a^2*f) + ((I/3)*B*c^5*Tan[e + f*x]^3)/(a^2*f)

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^5}{(a+i a \tan (e+f x))^2} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^4}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(a c) \operatorname{Subst}\left (\int \left (-\frac{(7 A+24 i B) c^4}{a^3}+\frac{i (A+7 i B) c^4 x}{a^3}+\frac{i B c^4 x^2}{a^3}+\frac{16 i (A+i B) c^4}{a^3 (-i+x)^3}+\frac{16 (2 A+3 i B) c^4}{a^3 (-i+x)^2}+\frac{8 (-3 i A+7 B) c^4}{a^3 (-i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{8 (3 A+7 i B) c^5 x}{a^2}+\frac{8 (3 i A-7 B) c^5 \log (\cos (e+f x))}{a^2 f}-\frac{8 (i A-B) c^5}{a^2 f (i-\tan (e+f x))^2}+\frac{16 (2 A+3 i B) c^5}{a^2 f (i-\tan (e+f x))}-\frac{(7 A+24 i B) c^5 \tan (e+f x)}{a^2 f}+\frac{(i A-7 B) c^5 \tan ^2(e+f x)}{2 a^2 f}+\frac{i B c^5 \tan ^3(e+f x)}{3 a^2 f}\\ \end{align*}

Mathematica [B]  time = 11.1381, size = 1357, normalized size = 6.99 \[ \frac{4 (5 B-3 i A) \cos (2 f x) \sec (e+f x) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) c^5}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}-\frac{4 (3 A+5 i B) \sec (e+f x) (\cos (f x)+i \sin (f x))^2 \sin (2 f x) (A+B \tan (e+f x)) c^5}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{\sec (e) \sec ^4(e+f x) (\cos (f x)+i \sin (f x))^2 \left (-\frac{1}{2} B \cos (2 e-f x) c^5+\frac{1}{2} B \cos (2 e+f x) c^5-\frac{1}{2} i B \sin (2 e-f x) c^5+\frac{1}{2} i B \sin (2 e+f x) c^5\right ) (A+B \tan (e+f x))}{3 f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{\sec (e) \sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left (-\frac{21}{2} i A \cos (2 e-f x) c^5+\frac{73}{2} B \cos (2 e-f x) c^5+\frac{21}{2} i A \cos (2 e+f x) c^5-\frac{73}{2} B \cos (2 e+f x) c^5+\frac{21}{2} A \sin (2 e-f x) c^5+\frac{73}{2} i B \sin (2 e-f x) c^5-\frac{21}{2} A \sin (2 e+f x) c^5-\frac{73}{2} i B \sin (2 e+f x) c^5\right ) (A+B \tan (e+f x))}{3 f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{x \sec (e+f x) (\cos (f x)+i \sin (f x))^2 \left (-24 A c^5-56 i B c^5-24 i A \tan (e) c^5+56 B \tan (e) c^5+(7 B-3 i A) \left (8 \cos (2 e) c^5+8 i \sin (2 e) c^5\right ) \tan (e)\right ) (A+B \tan (e+f x))}{(A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{\sec (e+f x) \left (3 A \cos (e) c^5+7 i B \cos (e) c^5+3 i A \sin (e) c^5-7 B \sin (e) c^5\right ) \left (8 \tan ^{-1}(\tan (f x)) \cos (e)+8 i \tan ^{-1}(\tan (f x)) \sin (e)\right ) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x))}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{\sec (e+f x) \left (3 A \cos (e) c^5+7 i B \cos (e) c^5+3 i A \sin (e) c^5-7 B \sin (e) c^5\right ) \left (4 i \cos (e) \log \left (\cos ^2(e+f x)\right )-4 \log \left (\cos ^2(e+f x)\right ) \sin (e)\right ) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x))}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{\sec (e) \sec ^3(e+f x) (3 A \cos (e)+21 i B \cos (e)+2 B \sin (e)) \left (\frac{1}{6} i c^5 \cos (2 e)-\frac{1}{6} c^5 \sin (2 e)\right ) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x))}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{(A+i B) \cos (4 f x) \sec (e+f x) \left (2 i \cos (2 e) c^5+2 \sin (2 e) c^5\right ) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x))}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{(3 A+7 i B) \sec (e+f x) \left (8 f x \cos (2 e) c^5+8 i f x \sin (2 e) c^5\right ) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x))}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2}+\frac{(A+i B) \sec (e+f x) \left (2 c^5 \cos (2 e)-2 i c^5 \sin (2 e)\right ) (\cos (f x)+i \sin (f x))^2 \sin (4 f x) (A+B \tan (e+f x))}{f (A \cos (e+f x)+B \sin (e+f x)) (i \tan (e+f x) a+a)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*((-3*I)*A + 5*B)*c^5*Cos[2*f*x]*Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])^2*(A + B*Tan[e + f*x]))/(f*(A*Cos[e +
f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + (Sec[e + f*x]*(3*A*c^5*Cos[e] + (7*I)*B*c^5*Cos[e] + (3*I)*
A*c^5*Sin[e] - 7*B*c^5*Sin[e])*(8*ArcTan[Tan[f*x]]*Cos[e] + (8*I)*ArcTan[Tan[f*x]]*Sin[e])*(Cos[f*x] + I*Sin[f
*x])^2*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + (Sec[e + f*x]*(3
*A*c^5*Cos[e] + (7*I)*B*c^5*Cos[e] + (3*I)*A*c^5*Sin[e] - 7*B*c^5*Sin[e])*((4*I)*Cos[e]*Log[Cos[e + f*x]^2] -
4*Log[Cos[e + f*x]^2]*Sin[e])*(Cos[f*x] + I*Sin[f*x])^2*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f
*x])*(a + I*a*Tan[e + f*x])^2) + (Sec[e]*Sec[e + f*x]^3*(3*A*Cos[e] + (21*I)*B*Cos[e] + 2*B*Sin[e])*((I/6)*c^5
*Cos[2*e] - (c^5*Sin[2*e])/6)*(Cos[f*x] + I*Sin[f*x])^2*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f
*x])*(a + I*a*Tan[e + f*x])^2) + ((A + I*B)*Cos[4*f*x]*Sec[e + f*x]*((2*I)*c^5*Cos[2*e] + 2*c^5*Sin[2*e])*(Cos
[f*x] + I*Sin[f*x])^2*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + (
(3*A + (7*I)*B)*Sec[e + f*x]*(8*c^5*f*x*Cos[2*e] + (8*I)*c^5*f*x*Sin[2*e])*(Cos[f*x] + I*Sin[f*x])^2*(A + B*Ta
n[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) - (4*(3*A + (5*I)*B)*c^5*Sec[e + f
*x]*(Cos[f*x] + I*Sin[f*x])^2*Sin[2*f*x]*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*T
an[e + f*x])^2) + ((A + I*B)*Sec[e + f*x]*(2*c^5*Cos[2*e] - (2*I)*c^5*Sin[2*e])*(Cos[f*x] + I*Sin[f*x])^2*Sin[
4*f*x]*(A + B*Tan[e + f*x]))/(f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + (Sec[e]*Sec[e +
f*x]^4*(Cos[f*x] + I*Sin[f*x])^2*(-(B*c^5*Cos[2*e - f*x])/2 + (B*c^5*Cos[2*e + f*x])/2 - (I/2)*B*c^5*Sin[2*e -
f*x] + (I/2)*B*c^5*Sin[2*e + f*x])*(A + B*Tan[e + f*x]))/(3*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[
e + f*x])^2) + (Sec[e]*Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*(((-21*I)/2)*A*c^5*Cos[2*e - f*x] + (73*B*c^5*
Cos[2*e - f*x])/2 + ((21*I)/2)*A*c^5*Cos[2*e + f*x] - (73*B*c^5*Cos[2*e + f*x])/2 + (21*A*c^5*Sin[2*e - f*x])/
2 + ((73*I)/2)*B*c^5*Sin[2*e - f*x] - (21*A*c^5*Sin[2*e + f*x])/2 - ((73*I)/2)*B*c^5*Sin[2*e + f*x])*(A + B*Ta
n[e + f*x]))/(3*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2) + (x*Sec[e + f*x]*(Cos[f*x] + I*
Sin[f*x])^2*(-24*A*c^5 - (56*I)*B*c^5 - (24*I)*A*c^5*Tan[e] + 56*B*c^5*Tan[e] + ((-3*I)*A + 7*B)*(8*c^5*Cos[2*
e] + (8*I)*c^5*Sin[2*e])*Tan[e])*(A + B*Tan[e + f*x]))/((A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x
])^2)

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Maple [A]  time = 0.06, size = 240, normalized size = 1.2 \begin{align*}{\frac{{\frac{i}{3}}B{c}^{5} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{{a}^{2}f}}+{\frac{{\frac{i}{2}}{c}^{5}A \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{{a}^{2}f}}-{\frac{24\,i{c}^{5}B\tan \left ( fx+e \right ) }{{a}^{2}f}}-{\frac{7\,B{c}^{5} \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{2\,{a}^{2}f}}-7\,{\frac{A{c}^{5}\tan \left ( fx+e \right ) }{{a}^{2}f}}-{\frac{48\,i{c}^{5}B}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) }}-32\,{\frac{A{c}^{5}}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{8\,i{c}^{5}A}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+8\,{\frac{B{c}^{5}}{{a}^{2}f \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{24\,i{c}^{5}A\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{{a}^{2}f}}+56\,{\frac{B{c}^{5}\ln \left ( \tan \left ( fx+e \right ) -i \right ) }{{a}^{2}f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*I*B*c^5*tan(f*x+e)^3/a^2/f+1/2*I/f*c^5/a^2*A*tan(f*x+e)^2-24*I/f*c^5/a^2*B*tan(f*x+e)-7/2/f*c^5/a^2*B*tan(
f*x+e)^2-7/f*c^5/a^2*A*tan(f*x+e)-48*I/f*c^5/a^2/(tan(f*x+e)-I)*B-32/f*c^5/a^2/(tan(f*x+e)-I)*A-8*I/f*c^5/a^2/
(tan(f*x+e)-I)^2*A+8/f*c^5/a^2/(tan(f*x+e)-I)^2*B-24*I/f*c^5/a^2*A*ln(tan(f*x+e)-I)+56/f*c^5/a^2*B*ln(tan(f*x+
e)-I)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.16834, size = 898, normalized size = 4.63 \begin{align*} \frac{48 \,{\left (3 \, A + 7 i \, B\right )} c^{5} f x e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (-18 i \, A + 42 \, B\right )} c^{5} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (6 i \, A - 6 \, B\right )} c^{5} +{\left (144 \,{\left (3 \, A + 7 i \, B\right )} c^{5} f x +{\left (-72 i \, A + 168 \, B\right )} c^{5}\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (144 \,{\left (3 \, A + 7 i \, B\right )} c^{5} f x +{\left (-180 i \, A + 420 \, B\right )} c^{5}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (48 \,{\left (3 \, A + 7 i \, B\right )} c^{5} f x +{\left (-132 i \, A + 308 \, B\right )} c^{5}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left ({\left (72 i \, A - 168 \, B\right )} c^{5} e^{\left (10 i \, f x + 10 i \, e\right )} +{\left (216 i \, A - 504 \, B\right )} c^{5} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (216 i \, A - 504 \, B\right )} c^{5} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (72 i \, A - 168 \, B\right )} c^{5} e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{3 \,{\left (a^{2} f e^{\left (10 i \, f x + 10 i \, e\right )} + 3 \, a^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 3 \, a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(48*(3*A + 7*I*B)*c^5*f*x*e^(10*I*f*x + 10*I*e) + (-18*I*A + 42*B)*c^5*e^(2*I*f*x + 2*I*e) + (6*I*A - 6*B)
*c^5 + (144*(3*A + 7*I*B)*c^5*f*x + (-72*I*A + 168*B)*c^5)*e^(8*I*f*x + 8*I*e) + (144*(3*A + 7*I*B)*c^5*f*x +
(-180*I*A + 420*B)*c^5)*e^(6*I*f*x + 6*I*e) + (48*(3*A + 7*I*B)*c^5*f*x + (-132*I*A + 308*B)*c^5)*e^(4*I*f*x +
4*I*e) + ((72*I*A - 168*B)*c^5*e^(10*I*f*x + 10*I*e) + (216*I*A - 504*B)*c^5*e^(8*I*f*x + 8*I*e) + (216*I*A -
504*B)*c^5*e^(6*I*f*x + 6*I*e) + (72*I*A - 168*B)*c^5*e^(4*I*f*x + 4*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(a^2
*f*e^(10*I*f*x + 10*I*e) + 3*a^2*f*e^(8*I*f*x + 8*I*e) + 3*a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*
e))

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Sympy [A]  time = 14.6862, size = 389, normalized size = 2.01 \begin{align*} \frac{- \frac{\left (12 i A c^{5} - 36 B c^{5}\right ) e^{- 2 i e} e^{4 i f x}}{a^{2} f} - \frac{\left (26 i A c^{5} - 82 B c^{5}\right ) e^{- 4 i e} e^{2 i f x}}{a^{2} f} - \frac{\left (42 i A c^{5} - 146 B c^{5}\right ) e^{- 6 i e}}{3 a^{2} f}}{e^{6 i f x} + 3 e^{- 2 i e} e^{4 i f x} + 3 e^{- 4 i e} e^{2 i f x} + e^{- 6 i e}} + \frac{c^{5} \left (24 i A - 56 B\right ) \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} + \frac{\left (\begin{cases} 48 A c^{5} x e^{4 i e} - \frac{12 i A c^{5} e^{2 i e} e^{- 2 i f x}}{f} + \frac{2 i A c^{5} e^{- 4 i f x}}{f} + 112 i B c^{5} x e^{4 i e} + \frac{20 B c^{5} e^{2 i e} e^{- 2 i f x}}{f} - \frac{2 B c^{5} e^{- 4 i f x}}{f} & \text{for}\: f \neq 0 \\x \left (48 A c^{5} e^{4 i e} - 24 A c^{5} e^{2 i e} + 8 A c^{5} + 112 i B c^{5} e^{4 i e} - 40 i B c^{5} e^{2 i e} + 8 i B c^{5}\right ) & \text{otherwise} \end{cases}\right ) e^{- 4 i e}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(12*I*A*c**5 - 36*B*c**5)*exp(-2*I*e)*exp(4*I*f*x)/(a**2*f) - (26*I*A*c**5 - 82*B*c**5)*exp(-4*I*e)*exp(2*I*
f*x)/(a**2*f) - (42*I*A*c**5 - 146*B*c**5)*exp(-6*I*e)/(3*a**2*f))/(exp(6*I*f*x) + 3*exp(-2*I*e)*exp(4*I*f*x)
+ 3*exp(-4*I*e)*exp(2*I*f*x) + exp(-6*I*e)) + c**5*(24*I*A - 56*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(a**2*f) +
Piecewise((48*A*c**5*x*exp(4*I*e) - 12*I*A*c**5*exp(2*I*e)*exp(-2*I*f*x)/f + 2*I*A*c**5*exp(-4*I*f*x)/f + 112*
I*B*c**5*x*exp(4*I*e) + 20*B*c**5*exp(2*I*e)*exp(-2*I*f*x)/f - 2*B*c**5*exp(-4*I*f*x)/f, Ne(f, 0)), (x*(48*A*c
**5*exp(4*I*e) - 24*A*c**5*exp(2*I*e) + 8*A*c**5 + 112*I*B*c**5*exp(4*I*e) - 40*I*B*c**5*exp(2*I*e) + 8*I*B*c*
*5), True))*exp(-4*I*e)/a**2

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Giac [B]  time = 1.88084, size = 698, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(24*(3*I*A*c^5 - 7*B*c^5)*log(tan(1/2*f*x + 1/2*e) - I)/a^2 - 12*(3*I*A*c^5 - 7*B*c^5)*log(abs(tan(1/2*f*
x + 1/2*e) + 1))/a^2 + 12*(-3*I*A*c^5 + 7*B*c^5)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 + (66*I*A*c^5*tan(1/2*
f*x + 1/2*e)^6 - 154*B*c^5*tan(1/2*f*x + 1/2*e)^6 - 21*A*c^5*tan(1/2*f*x + 1/2*e)^5 - 72*I*B*c^5*tan(1/2*f*x +
1/2*e)^5 - 201*I*A*c^5*tan(1/2*f*x + 1/2*e)^4 + 483*B*c^5*tan(1/2*f*x + 1/2*e)^4 + 42*A*c^5*tan(1/2*f*x + 1/2
*e)^3 + 148*I*B*c^5*tan(1/2*f*x + 1/2*e)^3 + 201*I*A*c^5*tan(1/2*f*x + 1/2*e)^2 - 483*B*c^5*tan(1/2*f*x + 1/2*
e)^2 - 21*A*c^5*tan(1/2*f*x + 1/2*e) - 72*I*B*c^5*tan(1/2*f*x + 1/2*e) - 66*I*A*c^5 + 154*B*c^5)/((tan(1/2*f*x
+ 1/2*e)^2 - 1)^3*a^2) + (-150*I*A*c^5*tan(1/2*f*x + 1/2*e)^4 + 350*B*c^5*tan(1/2*f*x + 1/2*e)^4 - 648*A*c^5*
tan(1/2*f*x + 1/2*e)^3 - 1496*I*B*c^5*tan(1/2*f*x + 1/2*e)^3 + 1044*I*A*c^5*tan(1/2*f*x + 1/2*e)^2 - 2340*B*c^
5*tan(1/2*f*x + 1/2*e)^2 + 648*A*c^5*tan(1/2*f*x + 1/2*e) + 1496*I*B*c^5*tan(1/2*f*x + 1/2*e) - 150*I*A*c^5 +
350*B*c^5)/(a^2*(tan(1/2*f*x + 1/2*e) - I)^4))/f