3.8.29 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^4}{(a+i a \tan (e+f x))^3} \, dx\) [729]

Optimal. Leaf size=164 \[ -\frac {(A+7 i B) c^4 x}{a^3}-\frac {(i A-7 B) c^4 \log (\cos (e+f x))}{a^3 f}+\frac {8 (A+i B) c^4}{3 a^3 f (i-\tan (e+f x))^3}+\frac {2 (3 i A-5 B) c^4}{a^3 f (i-\tan (e+f x))^2}-\frac {6 (A+3 i B) c^4}{a^3 f (i-\tan (e+f x))}+\frac {i B c^4 \tan (e+f x)}{a^3 f} \]

[Out]

-(A+7*I*B)*c^4*x/a^3-(I*A-7*B)*c^4*ln(cos(f*x+e))/a^3/f+8/3*(A+I*B)*c^4/a^3/f/(I-tan(f*x+e))^3+2*(3*I*A-5*B)*c
^4/a^3/f/(I-tan(f*x+e))^2-6*(A+3*I*B)*c^4/a^3/f/(I-tan(f*x+e))+I*B*c^4*tan(f*x+e)/a^3/f

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Rubi [A]
time = 0.15, antiderivative size = 164, 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 {6 c^4 (A+3 i B)}{a^3 f (-\tan (e+f x)+i)}+\frac {2 c^4 (-5 B+3 i A)}{a^3 f (-\tan (e+f x)+i)^2}+\frac {8 c^4 (A+i B)}{3 a^3 f (-\tan (e+f x)+i)^3}-\frac {c^4 (-7 B+i A) \log (\cos (e+f x))}{a^3 f}-\frac {c^4 x (A+7 i B)}{a^3}+\frac {i B c^4 \tan (e+f x)}{a^3 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])^3,x]

[Out]

-(((A + (7*I)*B)*c^4*x)/a^3) - ((I*A - 7*B)*c^4*Log[Cos[e + f*x]])/(a^3*f) + (8*(A + I*B)*c^4)/(3*a^3*f*(I - T
an[e + f*x])^3) + (2*((3*I)*A - 5*B)*c^4)/(a^3*f*(I - Tan[e + f*x])^2) - (6*(A + (3*I)*B)*c^4)/(a^3*f*(I - Tan
[e + f*x])) + (I*B*c^4*Tan[e + f*x])/(a^3*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))^3} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {(A+B x) (c-i c x)^3}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \text {Subst}\left (\int \left (\frac {i B c^3}{a^4}+\frac {8 (A+i B) c^3}{a^4 (-i+x)^4}+\frac {4 (-3 i A+5 B) c^3}{a^4 (-i+x)^3}-\frac {6 (A+3 i B) c^3}{a^4 (-i+x)^2}+\frac {i (A+7 i B) c^3}{a^4 (-i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(A+7 i B) c^4 x}{a^3}-\frac {(i A-7 B) c^4 \log (\cos (e+f x))}{a^3 f}+\frac {8 (A+i B) c^4}{3 a^3 f (i-\tan (e+f x))^3}+\frac {2 (3 i A-5 B) c^4}{a^3 f (i-\tan (e+f x))^2}-\frac {6 (A+3 i B) c^4}{a^3 f (i-\tan (e+f x))}+\frac {i B c^4 \tan (e+f x)}{a^3 f}\\ \end {align*}

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Mathematica [A]
time = 7.07, size = 319, normalized size = 1.95 \begin {gather*} \frac {c^4 \sec ^2(e+f x) (\cos (f x)+i \sin (f x))^3 \left ((A+5 i B) \cos (2 f x) (6 i \cos (e)-6 \sin (e))+3 (-i A+3 B) \cos (4 f x) (\cos (e)-i \sin (e))+6 (-i A+7 B) \log (\cos (e+f x)) \left (\cos \left (\frac {3 e}{2}\right )+i \sin \left (\frac {3 e}{2}\right )\right )^2-6 (A+7 i B) f x (\cos (3 e)+i \sin (3 e))+2 (A+i B) \cos (6 f x) (i \cos (3 e)+\sin (3 e))+6 i B \sec (e) \sec (e+f x) (\cos (3 e)+i \sin (3 e)) \sin (f x)+6 (A+5 i B) (\cos (e)+i \sin (e)) \sin (2 f x)+(A+3 i B) (-3 \cos (e)+3 i \sin (e)) \sin (4 f x)+2 (A+i B) (\cos (3 e)-i \sin (3 e)) \sin (6 f x)\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))^3} \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])^3,x]

[Out]

(c^4*Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^3*((A + (5*I)*B)*Cos[2*f*x]*((6*I)*Cos[e] - 6*Sin[e]) + 3*((-I)*A
+ 3*B)*Cos[4*f*x]*(Cos[e] - I*Sin[e]) + 6*((-I)*A + 7*B)*Log[Cos[e + f*x]]*(Cos[(3*e)/2] + I*Sin[(3*e)/2])^2 -
6*(A + (7*I)*B)*f*x*(Cos[3*e] + I*Sin[3*e]) + 2*(A + I*B)*Cos[6*f*x]*(I*Cos[3*e] + Sin[3*e]) + (6*I)*B*Sec[e]
*Sec[e + f*x]*(Cos[3*e] + I*Sin[3*e])*Sin[f*x] + 6*(A + (5*I)*B)*(Cos[e] + I*Sin[e])*Sin[2*f*x] + (A + (3*I)*B
)*(-3*Cos[e] + (3*I)*Sin[e])*Sin[4*f*x] + 2*(A + I*B)*(Cos[3*e] - I*Sin[3*e])*Sin[6*f*x])*(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])^3)

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Maple [A]
time = 0.32, size = 104, normalized size = 0.63

method	result	size
derivativedivides	\(\frac {c^{4} \left (i B \tan \left (f x +e \right )-\frac {8 i B +8 A}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {-18 i B -6 A}{-i+\tan \left (f x +e \right )}+\left (i A -7 B \right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-12 i A +20 B}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}\right )}{f \,a^{3}}\)	\(104\)
default	\(\frac {c^{4} \left (i B \tan \left (f x +e \right )-\frac {8 i B +8 A}{3 \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {-18 i B -6 A}{-i+\tan \left (f x +e \right )}+\left (i A -7 B \right ) \ln \left (-i+\tan \left (f x +e \right )\right )-\frac {-12 i A +20 B}{2 \left (-i+\tan \left (f x +e \right )\right )^{2}}\right )}{f \,a^{3}}\)	\(104\)
risch	\(-\frac {5 c^{4} {\mathrm e}^{-2 i \left (f x +e \right )} B}{a^{3} f}+\frac {i c^{4} {\mathrm e}^{-2 i \left (f x +e \right )} A}{a^{3} f}+\frac {3 c^{4} {\mathrm e}^{-4 i \left (f x +e \right )} B}{2 a^{3} f}-\frac {i c^{4} {\mathrm e}^{-4 i \left (f x +e \right )} A}{2 a^{3} f}-\frac {c^{4} {\mathrm e}^{-6 i \left (f x +e \right )} B}{3 a^{3} f}+\frac {i c^{4} {\mathrm e}^{-6 i \left (f x +e \right )} A}{3 a^{3} f}-\frac {14 i c^{4} B x}{a^{3}}-\frac {2 c^{4} A x}{a^{3}}-\frac {14 i c^{4} B e}{a^{3} f}-\frac {2 c^{4} A e}{a^{3} f}-\frac {2 c^{4} B}{f \,a^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {7 c^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) B}{a^{3} f}-\frac {i c^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) A}{a^{3} f}\)	\(253\)
norman	\(\frac {\frac {\left (7 i c^{4} B +2 A \,c^{4}\right ) \tan \left (f x +e \right )}{a f}+\frac {i c^{4} B \left (\tan ^{7}\left (f x +e \right )\right )}{a f}-\frac {\left (7 i c^{4} B +A \,c^{4}\right ) x}{a}-\frac {-8 i c^{4} A +32 B \,c^{4}}{3 a f}-\frac {3 \left (7 i c^{4} B +A \,c^{4}\right ) x \left (\tan ^{2}\left (f x +e \right )\right )}{a}-\frac {3 \left (7 i c^{4} B +A \,c^{4}\right ) x \left (\tan ^{4}\left (f x +e \right )\right )}{a}-\frac {\left (7 i c^{4} B +A \,c^{4}\right ) x \left (\tan ^{6}\left (f x +e \right )\right )}{a}-\frac {\left (-49 i c^{4} B +8 A \,c^{4}\right ) \left (\tan ^{3}\left (f x +e \right )\right )}{3 a f}+\frac {3 \left (7 i c^{4} B +2 A \,c^{4}\right ) \left (\tan ^{5}\left (f x +e \right )\right )}{a f}-\frac {\left (-4 i c^{4} A +28 B \,c^{4}\right ) \left (\tan ^{2}\left (f x +e \right )\right )}{a f}-\frac {\left (-12 i c^{4} A +28 B \,c^{4}\right ) \left (\tan ^{4}\left (f x +e \right )\right )}{a f}}{a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {\left (-i c^{4} A +7 B \,c^{4}\right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 a^{3} f}\)	\(342\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

1/f*c^4/a^3*(I*B*tan(f*x+e)-1/3*(8*A+8*I*B)/(-I+tan(f*x+e))^3-(-6*A-18*I*B)/(-I+tan(f*x+e))+(I*A-7*B)*ln(-I+ta
n(f*x+e))-1/2*(-12*I*A+20*B)/(-I+tan(f*x+e))^2)

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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))^3,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 [A]
time = 5.03, size = 206, normalized size = 1.26 \begin {gather*} -\frac {12 \, {\left (A + 7 i \, B\right )} c^{4} f x e^{\left (8 i \, f x + 8 i \, e\right )} + 3 \, {\left (-i \, A + 7 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-i \, A + 7 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (-i \, A + B\right )} c^{4} + 6 \, {\left (2 \, {\left (A + 7 i \, B\right )} c^{4} f x + {\left (-i \, A + 7 \, B\right )} c^{4}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, {\left ({\left (i \, A - 7 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (i \, A - 7 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{6 \, {\left (a^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} + a^{3} f e^{\left (6 i \, f x + 6 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))^3,x, algorithm="fricas")

[Out]

-1/6*(12*(A + 7*I*B)*c^4*f*x*e^(8*I*f*x + 8*I*e) + 3*(-I*A + 7*B)*c^4*e^(4*I*f*x + 4*I*e) - (-I*A + 7*B)*c^4*e
^(2*I*f*x + 2*I*e) + 2*(-I*A + B)*c^4 + 6*(2*(A + 7*I*B)*c^4*f*x + (-I*A + 7*B)*c^4)*e^(6*I*f*x + 6*I*e) + 6*(
(I*A - 7*B)*c^4*e^(8*I*f*x + 8*I*e) + (I*A - 7*B)*c^4*e^(6*I*f*x + 6*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(a^3*
f*e^(8*I*f*x + 8*I*e) + a^3*f*e^(6*I*f*x + 6*I*e))

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Sympy [A]
time = 0.75, size = 403, normalized size = 2.46 \begin {gather*} - \frac {2 B c^{4}}{a^{3} f e^{2 i e} e^{2 i f x} + a^{3} f} + \begin {cases} \frac {\left (\left (2 i A a^{6} c^{4} f^{2} e^{6 i e} - 2 B a^{6} c^{4} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (- 3 i A a^{6} c^{4} f^{2} e^{8 i e} + 9 B a^{6} c^{4} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (6 i A a^{6} c^{4} f^{2} e^{10 i e} - 30 B a^{6} c^{4} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{6 a^{9} f^{3}} & \text {for}\: a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac {- 2 A c^{4} - 14 i B c^{4}}{a^{3}} + \frac {\left (- 2 A c^{4} e^{6 i e} + 2 A c^{4} e^{4 i e} - 2 A c^{4} e^{2 i e} + 2 A c^{4} - 14 i B c^{4} e^{6 i e} + 10 i B c^{4} e^{4 i e} - 6 i B c^{4} e^{2 i e} + 2 i B c^{4}\right ) e^{- 6 i e}}{a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {i c^{4} \left (A + 7 i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{3} f} + \frac {x \left (- 2 A c^{4} - 14 i B c^{4}\right )}{a^{3}} \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))**3,x)

[Out]

-2*B*c**4/(a**3*f*exp(2*I*e)*exp(2*I*f*x) + a**3*f) + Piecewise((((2*I*A*a**6*c**4*f**2*exp(6*I*e) - 2*B*a**6*
c**4*f**2*exp(6*I*e))*exp(-6*I*f*x) + (-3*I*A*a**6*c**4*f**2*exp(8*I*e) + 9*B*a**6*c**4*f**2*exp(8*I*e))*exp(-
4*I*f*x) + (6*I*A*a**6*c**4*f**2*exp(10*I*e) - 30*B*a**6*c**4*f**2*exp(10*I*e))*exp(-2*I*f*x))*exp(-12*I*e)/(6
*a**9*f**3), Ne(a**9*f**3*exp(12*I*e), 0)), (x*(-(-2*A*c**4 - 14*I*B*c**4)/a**3 + (-2*A*c**4*exp(6*I*e) + 2*A*
c**4*exp(4*I*e) - 2*A*c**4*exp(2*I*e) + 2*A*c**4 - 14*I*B*c**4*exp(6*I*e) + 10*I*B*c**4*exp(4*I*e) - 6*I*B*c**
4*exp(2*I*e) + 2*I*B*c**4)*exp(-6*I*e)/a**3), True)) - I*c**4*(A + 7*I*B)*log(exp(2*I*f*x) + exp(-2*I*e))/(a**
3*f) + x*(-2*A*c**4 - 14*I*B*c**4)/a**3

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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. 429 vs. \(2 (143) = 286\).
time = 1.09, size = 429, normalized size = 2.62 \begin {gather*} \frac {\frac {30 \, {\left (-i \, A c^{4} + 7 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3}} + \frac {60 \, {\left (i \, A c^{4} - 7 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{3}} - \frac {30 \, {\left (i \, A c^{4} - 7 \, B c^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a^{3}} - \frac {30 \, {\left (-i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 7 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i \, A c^{4} - 7 \, B c^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{3}} - \frac {147 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 1029 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 1002 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 6534 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 2445 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 17115 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3820 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 23860 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2445 i \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 17115 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1002 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6534 i \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 147 i \, A c^{4} + 1029 \, B c^{4}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{6}}}{30 \, 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))^3,x, algorithm="giac")

[Out]

1/30*(30*(-I*A*c^4 + 7*B*c^4)*log(tan(1/2*f*x + 1/2*e) + 1)/a^3 + 60*(I*A*c^4 - 7*B*c^4)*log(tan(1/2*f*x + 1/2
*e) - I)/a^3 - 30*(I*A*c^4 - 7*B*c^4)*log(tan(1/2*f*x + 1/2*e) - 1)/a^3 - 30*(-I*A*c^4*tan(1/2*f*x + 1/2*e)^2
+ 7*B*c^4*tan(1/2*f*x + 1/2*e)^2 + 2*I*B*c^4*tan(1/2*f*x + 1/2*e) + I*A*c^4 - 7*B*c^4)/((tan(1/2*f*x + 1/2*e)^
2 - 1)*a^3) - (147*I*A*c^4*tan(1/2*f*x + 1/2*e)^6 - 1029*B*c^4*tan(1/2*f*x + 1/2*e)^6 + 1002*A*c^4*tan(1/2*f*x
+ 1/2*e)^5 + 6534*I*B*c^4*tan(1/2*f*x + 1/2*e)^5 - 2445*I*A*c^4*tan(1/2*f*x + 1/2*e)^4 + 17115*B*c^4*tan(1/2*
f*x + 1/2*e)^4 - 3820*A*c^4*tan(1/2*f*x + 1/2*e)^3 - 23860*I*B*c^4*tan(1/2*f*x + 1/2*e)^3 + 2445*I*A*c^4*tan(1
/2*f*x + 1/2*e)^2 - 17115*B*c^4*tan(1/2*f*x + 1/2*e)^2 + 1002*A*c^4*tan(1/2*f*x + 1/2*e) + 6534*I*B*c^4*tan(1/
2*f*x + 1/2*e) - 147*I*A*c^4 + 1029*B*c^4)/(a^3*(tan(1/2*f*x + 1/2*e) - I)^6))/f

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Mupad [B]
time = 11.01, size = 266, normalized size = 1.62 \begin {gather*} -\frac {c^4\,\left (25\,B\,\mathrm {tan}\left (e+f\,x\right )-\frac {B\,32{}\mathrm {i}}{3}-A\,\mathrm {tan}\left (e+f\,x\right )\,6{}\mathrm {i}-\frac {8\,A}{3}-A\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-B\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}+6\,A\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,{\mathrm {tan}\left (e+f\,x\right )}^2\,15{}\mathrm {i}+3\,B\,{\mathrm {tan}\left (e+f\,x\right )}^3+B\,{\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}+3\,A\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+A\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}+B\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,21{}\mathrm {i}-7\,B\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-A\,\mathrm {tan}\left (e+f\,x\right )\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}+21\,B\,\mathrm {tan}\left (e+f\,x\right )\,\ln \left (-1-\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{a^3\,f\,{\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3} \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)^3,x)

[Out]

-(c^4*(25*B*tan(e + f*x) - (B*32i)/3 - A*tan(e + f*x)*6i - (8*A)/3 - A*log(- tan(e + f*x)*1i - 1) - B*log(- ta
n(e + f*x)*1i - 1)*7i + 6*A*tan(e + f*x)^2 + B*tan(e + f*x)^2*15i + 3*B*tan(e + f*x)^3 + B*tan(e + f*x)^4*1i +
3*A*tan(e + f*x)^2*log(- tan(e + f*x)*1i - 1) + A*tan(e + f*x)^3*log(- tan(e + f*x)*1i - 1)*1i + B*tan(e + f*
x)^2*log(- tan(e + f*x)*1i - 1)*21i - 7*B*tan(e + f*x)^3*log(- tan(e + f*x)*1i - 1) - A*tan(e + f*x)*log(- tan
(e + f*x)*1i - 1)*3i + 21*B*tan(e + f*x)*log(- tan(e + f*x)*1i - 1))*1i)/(a^3*f*(tan(e + f*x)*1i + 1)^3)

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