3.8.6 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^4}{a+i a \tan (e+f x)} \, dx\) [706]

Optimal. Leaf size=157 \[ -\frac {4 (3 A+5 i B) c^4 x}{a}-\frac {4 (3 i A-5 B) c^4 \log (\cos (e+f x))}{a f}-\frac {8 (A+i B) c^4}{a f (i-\tan (e+f x))}+\frac {(5 A+12 i B) c^4 \tan (e+f x)}{a f}-\frac {(i A-5 B) c^4 \tan ^2(e+f x)}{2 a f}-\frac {i B c^4 \tan ^3(e+f x)}{3 a f} \]

[Out]

-4*(3*A+5*I*B)*c^4*x/a-4*(3*I*A-5*B)*c^4*ln(cos(f*x+e))/a/f-8*(A+I*B)*c^4/a/f/(I-tan(f*x+e))+(5*A+12*I*B)*c^4*
tan(f*x+e)/a/f-1/2*(I*A-5*B)*c^4*tan(f*x+e)^2/a/f-1/3*I*B*c^4*tan(f*x+e)^3/a/f

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Rubi [A]
time = 0.15, antiderivative size = 157, normalized size of antiderivative = 1.00, 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 = {3669, 78} \begin {gather*} -\frac {c^4 (-5 B+i A) \tan ^2(e+f x)}{2 a f}+\frac {c^4 (5 A+12 i B) \tan (e+f x)}{a f}-\frac {8 c^4 (A+i B)}{a f (-\tan (e+f x)+i)}-\frac {4 c^4 (-5 B+3 i A) \log (\cos (e+f x))}{a f}-\frac {4 c^4 x (3 A+5 i B)}{a}-\frac {i B c^4 \tan ^3(e+f x)}{3 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4)/(a + I*a*Tan[e + f*x]),x]

[Out]

(-4*(3*A + (5*I)*B)*c^4*x)/a - (4*((3*I)*A - 5*B)*c^4*Log[Cos[e + f*x]])/(a*f) - (8*(A + I*B)*c^4)/(a*f*(I - T
an[e + f*x])) + ((5*A + (12*I)*B)*c^4*Tan[e + f*x])/(a*f) - ((I*A - 5*B)*c^4*Tan[e + f*x]^2)/(2*a*f) - ((I/3)*
B*c^4*Tan[e + f*x]^3)/(a*f)

Rule 78

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]))))

Rule 3669

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]

Rubi steps

\begin {align*} \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^4}{a+i a \tan (e+f x)} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(A+B x) (c-i c x)^3}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \text {Subst}\left (\int \left (\frac {(5 A+12 i B) c^3}{a^2}+\frac {(-i A+5 B) c^3 x}{a^2}-\frac {i B c^3 x^2}{a^2}-\frac {8 (A+i B) c^3}{a^2 (-i+x)^2}+\frac {4 i (3 A+5 i B) c^3}{a^2 (-i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {4 (3 A+5 i B) c^4 x}{a}-\frac {4 (3 i A-5 B) c^4 \log (\cos (e+f x))}{a f}-\frac {8 (A+i B) c^4}{a f (i-\tan (e+f x))}+\frac {(5 A+12 i B) c^4 \tan (e+f x)}{a f}-\frac {(i A-5 B) c^4 \tan ^2(e+f x)}{2 a f}-\frac {i B c^4 \tan ^3(e+f x)}{3 a f}\\ \end {align*}

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Mathematica [A]
time = 1.60, size = 260, normalized size = 1.66 \begin {gather*} \frac {c^4 (\cos (f x)+i \sin (f x)) \left (12 (-3 i A+5 B) \log \left (\cos ^2(e+f x)\right ) \left (\cos \left (\frac {e}{2}\right )+i \sin \left (\frac {e}{2}\right )\right )^2-24 (3 A+5 i B) \text {ArcTan}(\tan (f x)) (\cos (e)+i \sin (e))+24 (A+i B) \cos (2 f x) (i \cos (e)+\sin (e))+24 (A+i B) (\cos (e)-i \sin (e)) \sin (2 f x)+2 (15 A+37 i B) \sec (e+f x) \sin (f x) (1+i \tan (e))+2 B \sec ^3(e+f x) \sin (f x) (-i+\tan (e))+\cos (e) \sec ^2(e+f x) (-i+\tan (e)) (3 (A+5 i B)+2 B \tan (e))\right ) (A+B \tan (e+f x))}{6 f (A \cos (e+f x)+B \sin (e+f x)) (a+i a \tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^4)/(a + I*a*Tan[e + f*x]),x]

[Out]

(c^4*(Cos[f*x] + I*Sin[f*x])*(12*((-3*I)*A + 5*B)*Log[Cos[e + f*x]^2]*(Cos[e/2] + I*Sin[e/2])^2 - 24*(3*A + (5
*I)*B)*ArcTan[Tan[f*x]]*(Cos[e] + I*Sin[e]) + 24*(A + I*B)*Cos[2*f*x]*(I*Cos[e] + Sin[e]) + 24*(A + I*B)*(Cos[
e] - I*Sin[e])*Sin[2*f*x] + 2*(15*A + (37*I)*B)*Sec[e + f*x]*Sin[f*x]*(1 + I*Tan[e]) + 2*B*Sec[e + f*x]^3*Sin[
f*x]*(-I + Tan[e]) + Cos[e]*Sec[e + f*x]^2*(-I + Tan[e])*(3*(A + (5*I)*B) + 2*B*Tan[e]))*(A + B*Tan[e + f*x]))
/(6*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x]))

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Maple [A]
time = 0.22, size = 106, normalized size = 0.68

method	result	size
derivativedivides	\(\frac {c^{4} \left (\frac {5 B \left (\tan ^{2}\left (f x +e \right )\right )}{2}-\frac {i B \left (\tan ^{3}\left (f x +e \right )\right )}{3}+5 A \tan \left (f x +e \right )-\frac {i A \left (\tan ^{2}\left (f x +e \right )\right )}{2}+12 i B \tan \left (f x +e \right )-\frac {-8 i B -8 A}{-i+\tan \left (f x +e \right )}+\left (12 i A -20 B \right ) \ln \left (-i+\tan \left (f x +e \right )\right )\right )}{f a}\)	\(106\)
default	\(\frac {c^{4} \left (\frac {5 B \left (\tan ^{2}\left (f x +e \right )\right )}{2}-\frac {i B \left (\tan ^{3}\left (f x +e \right )\right )}{3}+5 A \tan \left (f x +e \right )-\frac {i A \left (\tan ^{2}\left (f x +e \right )\right )}{2}+12 i B \tan \left (f x +e \right )-\frac {-8 i B -8 A}{-i+\tan \left (f x +e \right )}+\left (12 i A -20 B \right ) \ln \left (-i+\tan \left (f x +e \right )\right )\right )}{f a}\)	\(106\)
risch	\(-\frac {4 c^{4} {\mathrm e}^{-2 i \left (f x +e \right )} B}{a f}+\frac {4 i c^{4} {\mathrm e}^{-2 i \left (f x +e \right )} A}{a f}-\frac {40 i c^{4} B x}{a}-\frac {24 c^{4} A x}{a}-\frac {40 i c^{4} B e}{f a}-\frac {24 c^{4} A e}{f a}-\frac {2 c^{4} \left (-12 i A \,{\mathrm e}^{4 i \left (f x +e \right )}+24 B \,{\mathrm e}^{4 i \left (f x +e \right )}-27 i A \,{\mathrm e}^{2 i \left (f x +e \right )}+57 B \,{\mathrm e}^{2 i \left (f x +e \right )}-15 i A +37 B \right )}{3 f a \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}+\frac {20 c^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) B}{f a}-\frac {12 i c^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) A}{f a}\)	\(224\)
norman	\(\frac {\frac {\left (20 i c^{4} B +13 A \,c^{4}\right ) \tan \left (f x +e \right )}{a f}-\frac {4 \left (5 i c^{4} B +3 A \,c^{4}\right ) x}{a}-\frac {-17 i c^{4} A +21 B \,c^{4}}{2 a f}+\frac {5 \left (7 i c^{4} B +3 A \,c^{4}\right ) \left (\tan ^{3}\left (f x +e \right )\right )}{3 a f}-\frac {4 \left (5 i c^{4} B +3 A \,c^{4}\right ) x \left (\tan ^{2}\left (f x +e \right )\right )}{a}+\frac {\left (-i c^{4} A +5 B \,c^{4}\right ) \left (\tan ^{4}\left (f x +e \right )\right )}{2 a f}-\frac {i c^{4} B \left (\tan ^{5}\left (f x +e \right )\right )}{3 a f}}{1+\tan ^{2}\left (f x +e \right )}-\frac {2 \left (-3 i c^{4} A +5 B \,c^{4}\right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{a f}\)	\(227\)

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))^4/(a+I*a*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/f*c^4/a*(5/2*B*tan(f*x+e)^2-1/3*I*B*tan(f*x+e)^3+5*A*tan(f*x+e)-1/2*I*A*tan(f*x+e)^2+12*I*B*tan(f*x+e)-(-8*I
*B-8*A)/(-I+tan(f*x+e))+(12*I*A-20*B)*ln(-I+tan(f*x+e)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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))^4/(a+I*a*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (142) = 284\).
time = 2.83, size = 311, normalized size = 1.98 \begin {gather*} -\frac {2 \, {\left (12 \, {\left (3 \, A + 5 i \, B\right )} c^{4} f x e^{\left (8 i \, f x + 8 i \, e\right )} + 6 \, {\left (-i \, A + B\right )} c^{4} + 6 \, {\left (6 \, {\left (3 \, A + 5 i \, B\right )} c^{4} f x + {\left (-3 i \, A + 5 \, B\right )} c^{4}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, {\left (12 \, {\left (3 \, A + 5 i \, B\right )} c^{4} f x + 5 \, {\left (-3 i \, A + 5 \, B\right )} c^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (12 \, {\left (3 \, A + 5 i \, B\right )} c^{4} f x + 11 \, {\left (-3 i \, A + 5 \, B\right )} c^{4}\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 \, {\left ({\left (3 i \, A - 5 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} + 3 \, {\left (3 i \, A - 5 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, {\left (3 i \, A - 5 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (3 i \, A - 5 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{3 \, {\left (a f e^{\left (8 i \, f x + 8 i \, e\right )} + 3 \, a f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \end {gather*}

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))^4/(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

-2/3*(12*(3*A + 5*I*B)*c^4*f*x*e^(8*I*f*x + 8*I*e) + 6*(-I*A + B)*c^4 + 6*(6*(3*A + 5*I*B)*c^4*f*x + (-3*I*A +
5*B)*c^4)*e^(6*I*f*x + 6*I*e) + 3*(12*(3*A + 5*I*B)*c^4*f*x + 5*(-3*I*A + 5*B)*c^4)*e^(4*I*f*x + 4*I*e) + (12
*(3*A + 5*I*B)*c^4*f*x + 11*(-3*I*A + 5*B)*c^4)*e^(2*I*f*x + 2*I*e) + 6*((3*I*A - 5*B)*c^4*e^(8*I*f*x + 8*I*e)
+ 3*(3*I*A - 5*B)*c^4*e^(6*I*f*x + 6*I*e) + 3*(3*I*A - 5*B)*c^4*e^(4*I*f*x + 4*I*e) + (3*I*A - 5*B)*c^4*e^(2*
I*f*x + 2*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(a*f*e^(8*I*f*x + 8*I*e) + 3*a*f*e^(6*I*f*x + 6*I*e) + 3*a*f*e^(
4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))

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Sympy [A]
time = 0.57, size = 328, normalized size = 2.09 \begin {gather*} \frac {30 i A c^{4} - 74 B c^{4} + \left (54 i A c^{4} e^{2 i e} - 114 B c^{4} e^{2 i e}\right ) e^{2 i f x} + \left (24 i A c^{4} e^{4 i e} - 48 B c^{4} e^{4 i e}\right ) e^{4 i f x}}{3 a f e^{6 i e} e^{6 i f x} + 9 a f e^{4 i e} e^{4 i f x} + 9 a f e^{2 i e} e^{2 i f x} + 3 a f} + \begin {cases} \frac {\left (4 i A c^{4} - 4 B c^{4}\right ) e^{- 2 i e} e^{- 2 i f x}}{a f} & \text {for}\: a f e^{2 i e} \neq 0 \\x \left (- \frac {- 24 A c^{4} - 40 i B c^{4}}{a} + \frac {\left (- 24 A c^{4} e^{2 i e} + 8 A c^{4} - 40 i B c^{4} e^{2 i e} + 8 i B c^{4}\right ) e^{- 2 i e}}{a}\right ) & \text {otherwise} \end {cases} - \frac {4 i c^{4} \cdot \left (3 A + 5 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a f} + \frac {x \left (- 24 A c^{4} - 40 i B c^{4}\right )}{a} \end {gather*}

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))**4/(a+I*a*tan(f*x+e)),x)

[Out]

(30*I*A*c**4 - 74*B*c**4 + (54*I*A*c**4*exp(2*I*e) - 114*B*c**4*exp(2*I*e))*exp(2*I*f*x) + (24*I*A*c**4*exp(4*
I*e) - 48*B*c**4*exp(4*I*e))*exp(4*I*f*x))/(3*a*f*exp(6*I*e)*exp(6*I*f*x) + 9*a*f*exp(4*I*e)*exp(4*I*f*x) + 9*
a*f*exp(2*I*e)*exp(2*I*f*x) + 3*a*f) + Piecewise(((4*I*A*c**4 - 4*B*c**4)*exp(-2*I*e)*exp(-2*I*f*x)/(a*f), Ne(
a*f*exp(2*I*e), 0)), (x*(-(-24*A*c**4 - 40*I*B*c**4)/a + (-24*A*c**4*exp(2*I*e) + 8*A*c**4 - 40*I*B*c**4*exp(2
*I*e) + 8*I*B*c**4)*exp(-2*I*e)/a), True)) - 4*I*c**4*(3*A + 5*I*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(a*f) + x*
(-24*A*c**4 - 40*I*B*c**4)/a

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 445 vs. \(2 (142) = 284\).
time = 0.83, size = 445, normalized size = 2.83 \begin {gather*} \frac {2 \, {\left (\frac {6 \, {\left (-3 i \, A c^{4} + 5 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a} - \frac {12 \, {\left (-3 i \, A c^{4} + 5 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a} - \frac {6 \, {\left (3 i \, A c^{4} - 5 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a} - \frac {6 \, {\left (9 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 15 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 22 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 34 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 i \, A c^{4} + 15 \, B c^{4}\right )}}{a {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{2}} - \frac {-33 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 55 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 15 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 36 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 102 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 180 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 30 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 76 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 102 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 180 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 36 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 33 i \, A c^{4} - 55 \, B c^{4}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3} a}\right )}}{3 \, f} \end {gather*}

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))^4/(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

2/3*(6*(-3*I*A*c^4 + 5*B*c^4)*log(tan(1/2*f*x + 1/2*e) + 1)/a - 12*(-3*I*A*c^4 + 5*B*c^4)*log(tan(1/2*f*x + 1/
2*e) - I)/a - 6*(3*I*A*c^4 - 5*B*c^4)*log(tan(1/2*f*x + 1/2*e) - 1)/a - 6*(9*I*A*c^4*tan(1/2*f*x + 1/2*e)^2 -
15*B*c^4*tan(1/2*f*x + 1/2*e)^2 + 22*A*c^4*tan(1/2*f*x + 1/2*e) + 34*I*B*c^4*tan(1/2*f*x + 1/2*e) - 9*I*A*c^4
+ 15*B*c^4)/(a*(tan(1/2*f*x + 1/2*e) - I)^2) - (-33*I*A*c^4*tan(1/2*f*x + 1/2*e)^6 + 55*B*c^4*tan(1/2*f*x + 1/
2*e)^6 + 15*A*c^4*tan(1/2*f*x + 1/2*e)^5 + 36*I*B*c^4*tan(1/2*f*x + 1/2*e)^5 + 102*I*A*c^4*tan(1/2*f*x + 1/2*e
)^4 - 180*B*c^4*tan(1/2*f*x + 1/2*e)^4 - 30*A*c^4*tan(1/2*f*x + 1/2*e)^3 - 76*I*B*c^4*tan(1/2*f*x + 1/2*e)^3 -
102*I*A*c^4*tan(1/2*f*x + 1/2*e)^2 + 180*B*c^4*tan(1/2*f*x + 1/2*e)^2 + 15*A*c^4*tan(1/2*f*x + 1/2*e) + 36*I*
B*c^4*tan(1/2*f*x + 1/2*e) + 33*I*A*c^4 - 55*B*c^4)/((tan(1/2*f*x + 1/2*e)^2 - 1)^3*a))/f

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Mupad [B]
time = 8.80, size = 205, normalized size = 1.31 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (-\frac {20\,B\,c^4}{a}+\frac {A\,c^4\,12{}\mathrm {i}}{a}\right )}{f}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (-\frac {B\,c^4}{a}+\frac {c^4\,\left (A+B\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a}\right )}{f}-\frac {\frac {\left (4\,A\,c^4+B\,c^4\,12{}\mathrm {i}\right )\,1{}\mathrm {i}}{a}-\frac {\left (12\,A\,c^4+B\,c^4\,20{}\mathrm {i}\right )\,1{}\mathrm {i}}{a}}{f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {2\,c^4\,\left (A+B\,3{}\mathrm {i}\right )}{a}+\frac {B\,c^4\,3{}\mathrm {i}}{a}-\frac {c^4\,\left (-B+A\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{a}\right )}{f}-\frac {B\,c^4\,{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}}{3\,a\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^4)/(a + a*tan(e + f*x)*1i),x)

[Out]

(log(tan(e + f*x) - 1i)*((A*c^4*12i)/a - (20*B*c^4)/a))/f - (tan(e + f*x)^2*((c^4*(A + B*3i)*1i)/(2*a) - (B*c^
4)/a))/f - (((4*A*c^4 + B*c^4*12i)*1i)/a - ((12*A*c^4 + B*c^4*20i)*1i)/a)/(f*(tan(e + f*x)*1i + 1)) + (tan(e +
f*x)*((2*c^4*(A + B*3i))/a + (B*c^4*3i)/a - (c^4*(A*1i - B)*3i)/a))/f - (B*c^4*tan(e + f*x)^3*1i)/(3*a*f)

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