3.10.39 \(\int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx\) [939]

Optimal. Leaf size=134 \[ -\frac {8 a^5 x}{c^3}+\frac {8 i a^5 \log (\cos (e+f x))}{c^3 f}+\frac {a^5 \tan (e+f x)}{c^3 f}-\frac {16 i a^5}{3 f (c-i c \tan (e+f x))^3}-\frac {24 i a^5}{f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {16 i a^5 c^5}{f \left (c^4-i c^4 \tan (e+f x)\right )^2} \]

[Out]

-8*a^5*x/c^3+8*I*a^5*ln(cos(f*x+e))/c^3/f+a^5*tan(f*x+e)/c^3/f-16/3*I*a^5/f/(c-I*c*tan(f*x+e))^3-24*I*a^5/f/(c
^3-I*c^3*tan(f*x+e))+16*I*a^5*c^5/f/(c^4-I*c^4*tan(f*x+e))^2

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Rubi [A]
time = 0.10, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 45} \begin {gather*} \frac {a^5 \tan (e+f x)}{c^3 f}-\frac {24 i a^5}{f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {8 i a^5 \log (\cos (e+f x))}{c^3 f}-\frac {8 a^5 x}{c^3}+\frac {16 i a^5 c^5}{f \left (c^4-i c^4 \tan (e+f x)\right )^2}-\frac {16 i a^5}{3 f (c-i c \tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^5/(c - I*c*Tan[e + f*x])^3,x]

[Out]

(-8*a^5*x)/c^3 + ((8*I)*a^5*Log[Cos[e + f*x]])/(c^3*f) + (a^5*Tan[e + f*x])/(c^3*f) - (((16*I)/3)*a^5)/(f*(c -
I*c*Tan[e + f*x])^3) - ((24*I)*a^5)/(f*(c^3 - I*c^3*Tan[e + f*x])) + ((16*I)*a^5*c^5)/(f*(c^4 - I*c^4*Tan[e +
f*x])^2)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

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

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx &=\left (a^5 c^5\right ) \int \frac {\sec ^{10}(e+f x)}{(c-i c \tan (e+f x))^8} \, dx\\ &=\frac {\left (i a^5\right ) \text {Subst}\left (\int \frac {(c-x)^4}{(c+x)^4} \, dx,x,-i c \tan (e+f x)\right )}{c^4 f}\\ &=\frac {\left (i a^5\right ) \text {Subst}\left (\int \left (1+\frac {16 c^4}{(c+x)^4}-\frac {32 c^3}{(c+x)^3}+\frac {24 c^2}{(c+x)^2}-\frac {8 c}{c+x}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^4 f}\\ &=-\frac {8 a^5 x}{c^3}+\frac {8 i a^5 \log (\cos (e+f x))}{c^3 f}+\frac {a^5 \tan (e+f x)}{c^3 f}-\frac {16 i a^5}{3 f (c-i c \tan (e+f x))^3}+\frac {16 i a^5}{c f (c-i c \tan (e+f x))^2}-\frac {24 i a^5}{f \left (c^3-i c^3 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(923\) vs. \(2(134)=268\).
time = 6.91, size = 923, normalized size = 6.89 \begin {gather*} -\frac {8 x \cos (5 e) \cos ^5(e+f x) (a+i a \tan (e+f x))^5}{c^3 (\cos (f x)+i \sin (f x))^5}+\frac {4 i \cos (5 e) \cos ^5(e+f x) \log \left (\cos ^2(e+f x)\right ) (a+i a \tan (e+f x))^5}{c^3 f (\cos (f x)+i \sin (f x))^5}+\frac {\cos (6 f x) \cos ^5(e+f x) \left (-\frac {2 i \cos (e)}{3 c^3}+\frac {2 \sin (e)}{3 c^3}\right ) (a+i a \tan (e+f x))^5}{f (\cos (f x)+i \sin (f x))^5}+\frac {\cos (4 f x) \cos ^5(e+f x) \left (\frac {2 i \cos (e)}{c^3}+\frac {2 \sin (e)}{c^3}\right ) (a+i a \tan (e+f x))^5}{f (\cos (f x)+i \sin (f x))^5}+\frac {\cos (2 f x) \cos ^5(e+f x) \left (-\frac {6 i \cos (3 e)}{c^3}-\frac {6 \sin (3 e)}{c^3}\right ) (a+i a \tan (e+f x))^5}{f (\cos (f x)+i \sin (f x))^5}+\frac {8 i x \cos ^5(e+f x) \sin (5 e) (a+i a \tan (e+f x))^5}{c^3 (\cos (f x)+i \sin (f x))^5}+\frac {4 \cos ^5(e+f x) \log \left (\cos ^2(e+f x)\right ) \sin (5 e) (a+i a \tan (e+f x))^5}{c^3 f (\cos (f x)+i \sin (f x))^5}+\frac {\cos ^4(e+f x) \left (\frac {\cos (5 e)}{c^3}-\frac {i \sin (5 e)}{c^3}\right ) \sin (f x) (a+i a \tan (e+f x))^5}{f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) (\cos (f x)+i \sin (f x))^5}+\frac {\cos ^5(e+f x) \left (\frac {6 \cos (3 e)}{c^3}-\frac {6 i \sin (3 e)}{c^3}\right ) \sin (2 f x) (a+i a \tan (e+f x))^5}{f (\cos (f x)+i \sin (f x))^5}+\frac {\cos ^5(e+f x) \left (-\frac {2 \cos (e)}{c^3}+\frac {2 i \sin (e)}{c^3}\right ) \sin (4 f x) (a+i a \tan (e+f x))^5}{f (\cos (f x)+i \sin (f x))^5}+\frac {\cos ^5(e+f x) \left (\frac {2 \cos (e)}{3 c^3}+\frac {2 i \sin (e)}{3 c^3}\right ) \sin (6 f x) (a+i a \tan (e+f x))^5}{f (\cos (f x)+i \sin (f x))^5}+\frac {x \cos ^5(e+f x) \left (\frac {4 \cos ^3(e)}{c^3}-\frac {4 \cos ^5(e)}{c^3}-\frac {16 i \cos ^2(e) \sin (e)}{c^3}+\frac {24 i \cos ^4(e) \sin (e)}{c^3}-\frac {24 \cos (e) \sin ^2(e)}{c^3}+\frac {60 \cos ^3(e) \sin ^2(e)}{c^3}+\frac {16 i \sin ^3(e)}{c^3}-\frac {80 i \cos ^2(e) \sin ^3(e)}{c^3}-\frac {60 \cos (e) \sin ^4(e)}{c^3}+\frac {24 i \sin ^5(e)}{c^3}+\frac {4 \sin ^3(e) \tan (e)}{c^3}+\frac {4 \sin ^5(e) \tan (e)}{c^3}-i \left (\frac {8 \cos (5 e)}{c^3}-\frac {8 i \sin (5 e)}{c^3}\right ) \tan (e)\right ) (a+i a \tan (e+f x))^5}{(\cos (f x)+i \sin (f x))^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^5/(c - I*c*Tan[e + f*x])^3,x]

[Out]

(-8*x*Cos[5*e]*Cos[e + f*x]^5*(a + I*a*Tan[e + f*x])^5)/(c^3*(Cos[f*x] + I*Sin[f*x])^5) + ((4*I)*Cos[5*e]*Cos[
e + f*x]^5*Log[Cos[e + f*x]^2]*(a + I*a*Tan[e + f*x])^5)/(c^3*f*(Cos[f*x] + I*Sin[f*x])^5) + (Cos[6*f*x]*Cos[e
+ f*x]^5*((((-2*I)/3)*Cos[e])/c^3 + (2*Sin[e])/(3*c^3))*(a + I*a*Tan[e + f*x])^5)/(f*(Cos[f*x] + I*Sin[f*x])^
5) + (Cos[4*f*x]*Cos[e + f*x]^5*(((2*I)*Cos[e])/c^3 + (2*Sin[e])/c^3)*(a + I*a*Tan[e + f*x])^5)/(f*(Cos[f*x] +
I*Sin[f*x])^5) + (Cos[2*f*x]*Cos[e + f*x]^5*(((-6*I)*Cos[3*e])/c^3 - (6*Sin[3*e])/c^3)*(a + I*a*Tan[e + f*x])
^5)/(f*(Cos[f*x] + I*Sin[f*x])^5) + ((8*I)*x*Cos[e + f*x]^5*Sin[5*e]*(a + I*a*Tan[e + f*x])^5)/(c^3*(Cos[f*x]
+ I*Sin[f*x])^5) + (4*Cos[e + f*x]^5*Log[Cos[e + f*x]^2]*Sin[5*e]*(a + I*a*Tan[e + f*x])^5)/(c^3*f*(Cos[f*x] +
I*Sin[f*x])^5) + (Cos[e + f*x]^4*(Cos[5*e]/c^3 - (I*Sin[5*e])/c^3)*Sin[f*x]*(a + I*a*Tan[e + f*x])^5)/(f*(Cos
[e/2] - Sin[e/2])*(Cos[e/2] + Sin[e/2])*(Cos[f*x] + I*Sin[f*x])^5) + (Cos[e + f*x]^5*((6*Cos[3*e])/c^3 - ((6*I
)*Sin[3*e])/c^3)*Sin[2*f*x]*(a + I*a*Tan[e + f*x])^5)/(f*(Cos[f*x] + I*Sin[f*x])^5) + (Cos[e + f*x]^5*((-2*Cos
[e])/c^3 + ((2*I)*Sin[e])/c^3)*Sin[4*f*x]*(a + I*a*Tan[e + f*x])^5)/(f*(Cos[f*x] + I*Sin[f*x])^5) + (Cos[e + f
*x]^5*((2*Cos[e])/(3*c^3) + (((2*I)/3)*Sin[e])/c^3)*Sin[6*f*x]*(a + I*a*Tan[e + f*x])^5)/(f*(Cos[f*x] + I*Sin[
f*x])^5) + (x*Cos[e + f*x]^5*((4*Cos[e]^3)/c^3 - (4*Cos[e]^5)/c^3 - ((16*I)*Cos[e]^2*Sin[e])/c^3 + ((24*I)*Cos
[e]^4*Sin[e])/c^3 - (24*Cos[e]*Sin[e]^2)/c^3 + (60*Cos[e]^3*Sin[e]^2)/c^3 + ((16*I)*Sin[e]^3)/c^3 - ((80*I)*Co
s[e]^2*Sin[e]^3)/c^3 - (60*Cos[e]*Sin[e]^4)/c^3 + ((24*I)*Sin[e]^5)/c^3 + (4*Sin[e]^3*Tan[e])/c^3 + (4*Sin[e]^
5*Tan[e])/c^3 - I*((8*Cos[5*e])/c^3 - ((8*I)*Sin[5*e])/c^3)*Tan[e])*(a + I*a*Tan[e + f*x])^5)/(Cos[f*x] + I*Si
n[f*x])^5

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Maple [A]
time = 0.24, size = 71, normalized size = 0.53

method	result	size
derivativedivides	\(\frac {a^{5} \left (\tan \left (f x +e \right )+\frac {24}{\tan \left (f x +e \right )+i}-\frac {16}{3 \left (\tan \left (f x +e \right )+i\right )^{3}}-\frac {16 i}{\left (\tan \left (f x +e \right )+i\right )^{2}}-8 i \ln \left (\tan \left (f x +e \right )+i\right )\right )}{f \,c^{3}}\)	\(71\)
default	\(\frac {a^{5} \left (\tan \left (f x +e \right )+\frac {24}{\tan \left (f x +e \right )+i}-\frac {16}{3 \left (\tan \left (f x +e \right )+i\right )^{3}}-\frac {16 i}{\left (\tan \left (f x +e \right )+i\right )^{2}}-8 i \ln \left (\tan \left (f x +e \right )+i\right )\right )}{f \,c^{3}}\)	\(71\)
risch	\(-\frac {2 i a^{5} {\mathrm e}^{6 i \left (f x +e \right )}}{3 c^{3} f}+\frac {2 i a^{5} {\mathrm e}^{4 i \left (f x +e \right )}}{c^{3} f}-\frac {6 i a^{5} {\mathrm e}^{2 i \left (f x +e \right )}}{c^{3} f}+\frac {16 a^{5} e}{f \,c^{3}}+\frac {2 i a^{5}}{f \,c^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {8 i a^{5} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f \,c^{3}}\)	\(126\)
norman	\(\frac {\frac {a^{5} \left (\tan ^{7}\left (f x +e \right )\right )}{c f}-\frac {40 i a^{5} \left (\tan ^{4}\left (f x +e \right )\right )}{c f}-\frac {32 i a^{5} \left (\tan ^{2}\left (f x +e \right )\right )}{c f}-\frac {8 a^{5} x}{c}-\frac {40 i a^{5}}{3 c f}-\frac {24 a^{5} x \left (\tan ^{2}\left (f x +e \right )\right )}{c}-\frac {24 a^{5} x \left (\tan ^{4}\left (f x +e \right )\right )}{c}-\frac {8 a^{5} x \left (\tan ^{6}\left (f x +e \right )\right )}{c}+\frac {9 a^{5} \tan \left (f x +e \right )}{c f}+\frac {41 a^{5} \left (\tan ^{3}\left (f x +e \right )\right )}{3 c f}+\frac {27 a^{5} \left (\tan ^{5}\left (f x +e \right )\right )}{c f}}{c^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {4 i a^{5} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{c^{3} f}\)	\(227\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

1/f*a^5/c^3*(tan(f*x+e)+24/(tan(f*x+e)+I)-16/3/(tan(f*x+e)+I)^3-16*I/(tan(f*x+e)+I)^2-8*I*ln(tan(f*x+e)+I))

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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+I*a*tan(f*x+e))^5/(c-I*c*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 = 1.53, size = 127, normalized size = 0.95 \begin {gather*} -\frac {2 \, {\left (i \, a^{5} e^{\left (8 i \, f x + 8 i \, e\right )} - 2 i \, a^{5} e^{\left (6 i \, f x + 6 i \, e\right )} + 6 i \, a^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + 9 i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, a^{5} + 12 \, {\left (-i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{5}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{3 \, {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

-2/3*(I*a^5*e^(8*I*f*x + 8*I*e) - 2*I*a^5*e^(6*I*f*x + 6*I*e) + 6*I*a^5*e^(4*I*f*x + 4*I*e) + 9*I*a^5*e^(2*I*f
*x + 2*I*e) - 3*I*a^5 + 12*(-I*a^5*e^(2*I*f*x + 2*I*e) - I*a^5)*log(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f*e^(2*I*f*
x + 2*I*e) + c^3*f)

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Sympy [A]
time = 0.39, size = 202, normalized size = 1.51 \begin {gather*} \frac {2 i a^{5}}{c^{3} f e^{2 i e} e^{2 i f x} + c^{3} f} + \frac {8 i a^{5} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{3} f} + \begin {cases} \frac {- 2 i a^{5} c^{6} f^{2} e^{6 i e} e^{6 i f x} + 6 i a^{5} c^{6} f^{2} e^{4 i e} e^{4 i f x} - 18 i a^{5} c^{6} f^{2} e^{2 i e} e^{2 i f x}}{3 c^{9} f^{3}} & \text {for}\: c^{9} f^{3} \neq 0 \\\frac {x \left (4 a^{5} e^{6 i e} - 8 a^{5} e^{4 i e} + 12 a^{5} e^{2 i e}\right )}{c^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**5/(c-I*c*tan(f*x+e))**3,x)

[Out]

2*I*a**5/(c**3*f*exp(2*I*e)*exp(2*I*f*x) + c**3*f) + 8*I*a**5*log(exp(2*I*f*x) + exp(-2*I*e))/(c**3*f) + Piece
wise(((-2*I*a**5*c**6*f**2*exp(6*I*e)*exp(6*I*f*x) + 6*I*a**5*c**6*f**2*exp(4*I*e)*exp(4*I*f*x) - 18*I*a**5*c*
*6*f**2*exp(2*I*e)*exp(2*I*f*x))/(3*c**9*f**3), Ne(c**9*f**3, 0)), (x*(4*a**5*exp(6*I*e) - 8*a**5*exp(4*I*e) +
12*a**5*exp(2*I*e))/c**3, True))

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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. 253 vs. \(2 (123) = 246\).
time = 1.03, size = 253, normalized size = 1.89 \begin {gather*} -\frac {2 \, {\left (-\frac {60 i \, a^{5} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{3}} + \frac {120 i \, a^{5} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c^{3}} - \frac {60 i \, a^{5} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c^{3}} - \frac {15 \, {\left (-4 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 i \, a^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{3}} + \frac {2 \, {\left (-147 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 942 \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 2445 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3460 \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2445 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 942 \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 147 i \, a^{5}\right )}}{c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{6}}\right )}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-2/15*(-60*I*a^5*log(tan(1/2*f*x + 1/2*e) + 1)/c^3 + 120*I*a^5*log(tan(1/2*f*x + 1/2*e) + I)/c^3 - 60*I*a^5*lo
g(tan(1/2*f*x + 1/2*e) - 1)/c^3 - 15*(-4*I*a^5*tan(1/2*f*x + 1/2*e)^2 - a^5*tan(1/2*f*x + 1/2*e) + 4*I*a^5)/((
tan(1/2*f*x + 1/2*e)^2 - 1)*c^3) + 2*(-147*I*a^5*tan(1/2*f*x + 1/2*e)^6 + 942*a^5*tan(1/2*f*x + 1/2*e)^5 + 244
5*I*a^5*tan(1/2*f*x + 1/2*e)^4 - 3460*a^5*tan(1/2*f*x + 1/2*e)^3 - 2445*I*a^5*tan(1/2*f*x + 1/2*e)^2 + 942*a^5
*tan(1/2*f*x + 1/2*e) + 147*I*a^5)/(c^3*(tan(1/2*f*x + 1/2*e) + I)^6))/f

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Mupad [B]
time = 5.69, size = 138, normalized size = 1.03 \begin {gather*} \frac {a^5\,\left (\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,8{}\mathrm {i}+31\,\mathrm {tan}\left (e+f\,x\right )+24\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )-\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2\,24{}\mathrm {i}-8\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,21{}\mathrm {i}+3\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}+\frac {40}{3}{}\mathrm {i}\right )}{c^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 + a*tan(e + f*x)*1i)^5/(c - c*tan(e + f*x)*1i)^3,x)

[Out]

(a^5*(log(tan(e + f*x) + 1i)*8i + 31*tan(e + f*x) + 24*log(tan(e + f*x) + 1i)*tan(e + f*x) - log(tan(e + f*x)
+ 1i)*tan(e + f*x)^2*24i - 8*log(tan(e + f*x) + 1i)*tan(e + f*x)^3 - tan(e + f*x)^2*21i + 3*tan(e + f*x)^3 - t
an(e + f*x)^4*1i + 40i/3))/(c^3*f*(tan(e + f*x)*1i - 1)^3)

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