\(\int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx\) [948]

Optimal result

Integrand size = 31, antiderivative size = 139 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx=\frac {a^5 x}{c^4}-\frac {i a^5 \log (\cos (e+f x))}{c^4 f}-\frac {4 i a^5}{c^4 f (1-i \tan (e+f x))^4}+\frac {32 i a^5}{3 c^4 f (1-i \tan (e+f x))^3}-\frac {12 i a^5}{c^4 f (1-i \tan (e+f x))^2}+\frac {8 i a^5}{c^4 f (1-i \tan (e+f x))} \] Output:

a^5*x/c^4-I*a^5*ln(cos(f*x+e))/c^4/f-4*I*a^5/c^4/f/(1-I*tan(f*x+e))^4+32/3 
*I*a^5/c^4/f/(1-I*tan(f*x+e))^3-12*I*a^5/c^4/f/(1-I*tan(f*x+e))^2+8*I*a^5/ 
c^4/f/(1-I*tan(f*x+e))
 

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.54 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx=\frac {i a^5 \left (\log (i+\tan (e+f x))+\frac {4 \left (2-8 i \tan (e+f x)-9 \tan ^2(e+f x)+6 i \tan ^3(e+f x)\right )}{3 (i+\tan (e+f x))^4}\right )}{c^4 f} \] Input:

Integrate[(a + I*a*Tan[e + f*x])^5/(c - I*c*Tan[e + f*x])^4,x]
 

Output:

(I*a^5*(Log[I + Tan[e + f*x]] + (4*(2 - (8*I)*Tan[e + f*x] - 9*Tan[e + f*x 
]^2 + (6*I)*Tan[e + f*x]^3))/(3*(I + Tan[e + f*x])^4)))/(c^4*f)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 4005, 3042, 3968, 49, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4}dx\)

\(\Big \downarrow \) 4005

\(\displaystyle a^5 c^5 \int \frac {\sec ^{10}(e+f x)}{(c-i c \tan (e+f x))^9}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^5 c^5 \int \frac {\sec (e+f x)^{10}}{(c-i c \tan (e+f x))^9}dx\)

\(\Big \downarrow \) 3968

\(\displaystyle \frac {i a^5 \int \frac {(i \tan (e+f x) c+c)^4}{(c-i c \tan (e+f x))^5}d(-i c \tan (e+f x))}{c^4 f}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {i a^5 \int \left (\frac {16 c^4}{(c-i c \tan (e+f x))^5}-\frac {32 c^3}{(c-i c \tan (e+f x))^4}+\frac {24 c^2}{(c-i c \tan (e+f x))^3}-\frac {8 c}{(c-i c \tan (e+f x))^2}+\frac {1}{c-i c \tan (e+f x)}\right )d(-i c \tan (e+f x))}{c^4 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i a^5 \left (-\frac {4 c^4}{(c-i c \tan (e+f x))^4}+\frac {32 c^3}{3 (c-i c \tan (e+f x))^3}-\frac {12 c^2}{(c-i c \tan (e+f x))^2}+\frac {8 c}{c-i c \tan (e+f x)}+\log (c-i c \tan (e+f x))\right )}{c^4 f}\)

Input:

Int[(a + I*a*Tan[e + f*x])^5/(c - I*c*Tan[e + f*x])^4,x]
 

Output:

(I*a^5*(Log[c - I*c*Tan[e + f*x]] - (4*c^4)/(c - I*c*Tan[e + f*x])^4 + (32 
*c^3)/(3*(c - I*c*Tan[e + f*x])^3) - (12*c^2)/(c - I*c*Tan[e + f*x])^2 + ( 
8*c)/(c - I*c*Tan[e + f*x])))/(c^4*f)
 

Defintions of rubi rules used

rule 49

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}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3968

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[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 4005

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[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]))
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.88

method	result	size
risch	\(-\frac {i a^{5} {\mathrm e}^{8 i \left (f x +e \right )}}{4 c^{4} f}+\frac {i a^{5} {\mathrm e}^{6 i \left (f x +e \right )}}{3 c^{4} f}-\frac {i a^{5} {\mathrm e}^{4 i \left (f x +e \right )}}{2 c^{4} f}+\frac {i a^{5} {\mathrm e}^{2 i \left (f x +e \right )}}{c^{4} f}-\frac {2 a^{5} e}{c^{4} f}-\frac {i a^{5} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{c^{4} f}\)	\(122\)
derivativedivides	\(\frac {32 a^{5}}{3 f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {4 i a^{5}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}+\frac {i a^{5} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{4}}+\frac {a^{5} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{4}}+\frac {12 i a^{5}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {8 a^{5}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )}\)	\(132\)
default	\(\frac {32 a^{5}}{3 f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {4 i a^{5}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}+\frac {i a^{5} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{4}}+\frac {a^{5} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{4}}+\frac {12 i a^{5}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {8 a^{5}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )}\)	\(132\)
norman	\(\frac {\frac {a^{5} x}{c}+\frac {a^{5} x \tan \left (f x +e \right )^{8}}{c}+\frac {8 i a^{5}}{3 c f}+\frac {4 a^{5} x \tan \left (f x +e \right )^{2}}{c}+\frac {6 a^{5} x \tan \left (f x +e \right )^{4}}{c}+\frac {4 a^{5} x \tan \left (f x +e \right )^{6}}{c}-\frac {40 a^{5} \tan \left (f x +e \right )^{3}}{3 c f}+\frac {32 a^{5} \tan \left (f x +e \right )^{5}}{3 c f}-\frac {8 a^{5} \tan \left (f x +e \right )^{7}}{c f}+\frac {20 i a^{5} \tan \left (f x +e \right )^{6}}{c f}+\frac {44 i a^{5} \tan \left (f x +e \right )^{2}}{3 c f}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{4} c^{3}}+\frac {i a^{5} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{4}}\)	\(226\)

Input:

int((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.59 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx=\frac {-3 i \, a^{5} e^{\left (8 i \, f x + 8 i \, e\right )} + 4 i \, a^{5} e^{\left (6 i \, f x + 6 i \, e\right )} - 6 i \, a^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - 12 i \, a^{5} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{12 \, c^{4} f} \] Input:

integrate((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.50 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx=- \frac {i a^{5} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{4} f} + \begin {cases} \frac {- 6 i a^{5} c^{12} f^{3} e^{8 i e} e^{8 i f x} + 8 i a^{5} c^{12} f^{3} e^{6 i e} e^{6 i f x} - 12 i a^{5} c^{12} f^{3} e^{4 i e} e^{4 i f x} + 24 i a^{5} c^{12} f^{3} e^{2 i e} e^{2 i f x}}{24 c^{16} f^{4}} & \text {for}\: c^{16} f^{4} \neq 0 \\\frac {x \left (2 a^{5} e^{8 i e} - 2 a^{5} e^{6 i e} + 2 a^{5} e^{4 i e} - 2 a^{5} e^{2 i e}\right )}{c^{4}} & \text {otherwise} \end {cases} \] Input:

integrate((a+I*a*tan(f*x+e))**5/(c-I*c*tan(f*x+e))**4,x)
 

Output:

-I*a**5*log(exp(2*I*f*x) + exp(-2*I*e))/(c**4*f) + Piecewise(((-6*I*a**5*c 
**12*f**3*exp(8*I*e)*exp(8*I*f*x) + 8*I*a**5*c**12*f**3*exp(6*I*e)*exp(6*I 
*f*x) - 12*I*a**5*c**12*f**3*exp(4*I*e)*exp(4*I*f*x) + 24*I*a**5*c**12*f** 
3*exp(2*I*e)*exp(2*I*f*x))/(24*c**16*f**4), Ne(c**16*f**4, 0)), (x*(2*a**5 
*exp(8*I*e) - 2*a**5*exp(6*I*e) + 2*a**5*exp(4*I*e) - 2*a**5*exp(2*I*e))/c 
**4, True))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.59 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx=\frac {i \, a^{5} \log \left (\tan \left (f x + e\right ) + i\right )}{c^{4} f} - \frac {4 \, {\left (6 \, a^{5} \tan \left (f x + e\right )^{3} + 9 i \, a^{5} \tan \left (f x + e\right )^{2} - 8 \, a^{5} \tan \left (f x + e\right ) - 2 i \, a^{5}\right )}}{3 \, c^{4} f {\left (\tan \left (f x + e\right ) + i\right )}^{4}} \] Input:

integrate((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.20 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.05 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx=\frac {a^5\,\left (\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )-6\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^4-12\,{\mathrm {tan}\left (e+f\,x\right )}^2+\frac {8}{3}-\frac {\mathrm {tan}\left (e+f\,x\right )\,32{}\mathrm {i}}{3}-\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}+\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3\,4{}\mathrm {i}+{\mathrm {tan}\left (e+f\,x\right )}^3\,8{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^4\,f\,{\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4} \] Input:

int((a + a*tan(e + f*x)*1i)^5/(c - c*tan(e + f*x)*1i)^4,x)
 

Output:

(a^5*(log(tan(e + f*x) + 1i) - (tan(e + f*x)*32i)/3 - log(tan(e + f*x) + 1 
i)*tan(e + f*x)*4i - 6*log(tan(e + f*x) + 1i)*tan(e + f*x)^2 + log(tan(e + 
 f*x) + 1i)*tan(e + f*x)^3*4i + log(tan(e + f*x) + 1i)*tan(e + f*x)^4 - 12 
*tan(e + f*x)^2 + tan(e + f*x)^3*8i + 8/3)*1i)/(c^4*f*(tan(e + f*x)*1i - 1 
)^4)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^4} \, dx=\frac {a^{5} \left (160 \left (\int \frac {\tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{4} i -4 \tan \left (f x +e \right )^{3}-6 \tan \left (f x +e \right )^{2} i +4 \tan \left (f x +e \right )+i}d x \right ) f +160 \left (\int \frac {\tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{4} i -4 \tan \left (f x +e \right )^{3}-6 \tan \left (f x +e \right )^{2} i +4 \tan \left (f x +e \right )+i}d x \right ) f i -160 \left (\int \frac {\tan \left (f x +e \right )}{\tan \left (f x +e \right )^{4} i -4 \tan \left (f x +e \right )^{3}-6 \tan \left (f x +e \right )^{2} i +4 \tan \left (f x +e \right )+i}d x \right ) f -32 \left (\int \frac {1}{\tan \left (f x +e \right )^{4} i -4 \tan \left (f x +e \right )^{3}-6 \tan \left (f x +e \right )^{2} i +4 \tan \left (f x +e \right )+i}d x \right ) f i +9 \,\mathrm {log}\left (\tan \left (f x +e \right )^{4}+4 \tan \left (f x +e \right )^{3} i -6 \tan \left (f x +e \right )^{2}-4 \tan \left (f x +e \right ) i +1\right ) i -16 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) i \right )}{4 c^{4} f} \] Input:

int((a+I*a*tan(f*x+e))^5/(c-I*c*tan(f*x+e))^4,x)
 

Output:

(a**5*(160*int(tan(e + f*x)**3/(tan(e + f*x)**4*i - 4*tan(e + f*x)**3 - 6* 
tan(e + f*x)**2*i + 4*tan(e + f*x) + i),x)*f + 160*int(tan(e + f*x)**2/(ta 
n(e + f*x)**4*i - 4*tan(e + f*x)**3 - 6*tan(e + f*x)**2*i + 4*tan(e + f*x) 
 + i),x)*f*i - 160*int(tan(e + f*x)/(tan(e + f*x)**4*i - 4*tan(e + f*x)**3 
 - 6*tan(e + f*x)**2*i + 4*tan(e + f*x) + i),x)*f - 32*int(1/(tan(e + f*x) 
**4*i - 4*tan(e + f*x)**3 - 6*tan(e + f*x)**2*i + 4*tan(e + f*x) + i),x)*f 
*i + 9*log(tan(e + f*x)**4 + 4*tan(e + f*x)**3*i - 6*tan(e + f*x)**2 - 4*t 
an(e + f*x)*i + 1)*i - 16*log(tan(e + f*x)**2 + 1)*i))/(4*c**4*f)