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

Optimal result

Integrand size = 31, antiderivative size = 134 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=-\frac {8 a^5 x}{c^3}+\frac {8 i a^5 \log (\cos (e+f x))}{c^3 f}+\frac {a^5 \tan (e+f x)}{c^3 f}-\frac {16 i a^5}{3 f (c-i c \tan (e+f x))^3}-\frac {24 i a^5}{f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {16 i a^5 c^5}{f \left (c^4-i c^4 \tan (e+f x)\right )^2} \]

[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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 45} \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=\frac {a^5 \tan (e+f x)}{c^3 f}-\frac {24 i a^5}{f \left (c^3-i c^3 \tan (e+f x)\right )}+\frac {8 i a^5 \log (\cos (e+f x))}{c^3 f}-\frac {8 a^5 x}{c^3}+\frac {16 i a^5 c^5}{f \left (c^4-i c^4 \tan (e+f x)\right )^2}-\frac {16 i a^5}{3 f (c-i c \tan (e+f x))^3} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 3.82 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.60 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=\frac {i a^5 \left (-8 c \log (i+\tan (e+f x))-i c \tan (e+f x)+\frac {8 c \left (5 i+12 \tan (e+f x)-9 i \tan ^2(e+f x)\right )}{3 (i+\tan (e+f x))^3}\right )}{c^4 f} \]

[In]

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

[Out]

(I*a^5*(-8*c*Log[I + Tan[e + f*x]] - I*c*Tan[e + f*x] + (8*c*(5*I + 12*Tan[e + f*x] - (9*I)*Tan[e + f*x]^2))/(
3*(I + Tan[e + f*x])^3)))/(c^4*f)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.94

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

[In]

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

[Out]

a^5*tan(f*x+e)/c^3/f+24/f*a^5/c^3/(tan(f*x+e)+I)-16/3/f*a^5/c^3/(tan(f*x+e)+I)^3-16*I/f*a^5/c^3/(tan(f*x+e)+I)
^2-8/f*a^5/c^3*arctan(tan(f*x+e))-4*I/f*a^5/c^3*ln(1+tan(f*x+e)^2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.90 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=-\frac {2 \, {\left (i \, a^{5} e^{\left (8 i \, f x + 8 i \, e\right )} - 2 i \, a^{5} e^{\left (6 i \, f x + 6 i \, e\right )} + 6 i \, a^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + 9 i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, a^{5} + 12 \, {\left (-i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{5}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{3 \, {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.51 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=\frac {2 i a^{5}}{c^{3} f e^{2 i e} e^{2 i f x} + c^{3} f} + \frac {8 i a^{5} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{3} f} + \begin {cases} \frac {- 2 i a^{5} c^{6} f^{2} e^{6 i e} e^{6 i f x} + 6 i a^{5} c^{6} f^{2} e^{4 i e} e^{4 i f x} - 18 i a^{5} c^{6} f^{2} e^{2 i e} e^{2 i f x}}{3 c^{9} f^{3}} & \text {for}\: c^{9} f^{3} \neq 0 \\\frac {x \left (4 a^{5} e^{6 i e} - 8 a^{5} e^{4 i e} + 12 a^{5} e^{2 i e}\right )}{c^{3}} & \text {otherwise} \end {cases} \]

[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))

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]

[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: expt: undefined: 0 to a negative exponent.

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (118) = 236\).

Time = 0.94 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.79 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=-\frac {2 \, {\left (-\frac {60 i \, a^{5} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{3}} + \frac {120 i \, a^{5} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c^{3}} - \frac {60 i \, a^{5} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c^{3}} - \frac {15 \, {\left (-4 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 i \, a^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{3}} + \frac {2 \, {\left (-147 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 942 \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 2445 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3460 \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2445 i \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 942 \, a^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 147 i \, a^{5}\right )}}{c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{6}}\right )}}{15 \, f} \]

[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

Mupad [B] (verification not implemented)

Time = 6.58 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.03 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=\frac {a^5\,\left (\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,8{}\mathrm {i}+31\,\mathrm {tan}\left (e+f\,x\right )+24\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )-\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2\,24{}\mathrm {i}-8\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,21{}\mathrm {i}+3\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}+\frac {40}{3}{}\mathrm {i}\right )}{c^3\,f\,{\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3} \]

[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)