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

Optimal result

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

-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
 

Mathematica [A] (verified)

Time = 2.40 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.60 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=\frac {i a^5 \left (-8 c \log (i+\tan (e+f x))-i c \tan (e+f x)+\frac {8 c \left (5 i+12 \tan (e+f x)-9 i \tan ^2(e+f x)\right )}{3 (i+\tan (e+f x))^3}\right )}{c^4 f} \] Input:

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

Output:

(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)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 104, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4005

\(\displaystyle a^5 c^5 \int \frac {\sec ^{10}(e+f x)}{(c-i c \tan (e+f x))^8}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^5 c^5 \int \frac {\sec (e+f x)^{10}}{(c-i c \tan (e+f x))^8}dx\)

\(\Big \downarrow \) 3968

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(I*a^5*(-8*c*Log[c - I*c*Tan[e + f*x]] - I*c*Tan[e + f*x] - (16*c^4)/(3*(c 
 - I*c*Tan[e + f*x])^3) + (16*c^3)/(c - I*c*Tan[e + f*x])^2 - (24*c^2)/(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 = 0.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.94

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.90 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=-\frac {2 \, {\left (i \, a^{5} e^{\left (8 i \, f x + 8 i \, e\right )} - 2 i \, a^{5} e^{\left (6 i \, f x + 6 i \, e\right )} + 6 i \, a^{5} e^{\left (4 i \, f x + 4 i \, e\right )} + 9 i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, a^{5} + 12 \, {\left (-i \, a^{5} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{5}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{3 \, {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{3} f\right )}} \] Input:

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

Output:

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

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

Output:

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.63 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=-\frac {8 i \, a^{5} \log \left (\tan \left (f x + e\right ) + i\right )}{c^{3} f} + \frac {a^{5} \tan \left (f x + e\right )}{c^{3} f} + \frac {8 \, {\left (9 \, a^{5} \tan \left (f x + e\right )^{2} + 12 i \, a^{5} \tan \left (f x + e\right ) - 5 \, a^{5}\right )}}{3 \, c^{3} f {\left (\tan \left (f x + e\right ) + i\right )}^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 2.81 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.03 \[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=\frac {a^5\,\left (\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,8{}\mathrm {i}+31\,\mathrm {tan}\left (e+f\,x\right )+24\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )-\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2\,24{}\mathrm {i}-8\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,21{}\mathrm {i}+3\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}+\frac {40}{3}{}\mathrm {i}\right )}{c^3\,f\,{\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3} \] Input:

int((a + a*tan(e + f*x)*1i)^5/(c - c*tan(e + f*x)*1i)^3,x)
 

Output:

(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 - tan 
(e + f*x)^4*1i + 40i/3))/(c^3*f*(tan(e + f*x)*1i - 1)^3)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^5}{(c-i c \tan (e+f x))^3} \, dx=\frac {a^{5} \left (80 \left (\int \frac {\tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3}+3 \tan \left (f x +e \right )^{2} i -3 \tan \left (f x +e \right )-i}d x \right ) f i -80 \left (\int \frac {\tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3}+3 \tan \left (f x +e \right )^{2} i -3 \tan \left (f x +e \right )-i}d x \right ) f -32 \left (\int \frac {1}{\tan \left (f x +e \right )^{3}+3 \tan \left (f x +e \right )^{2} i -3 \tan \left (f x +e \right )-i}d x \right ) f i -4 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) i +\tan \left (f x +e \right )-32 f x \right )}{c^{3} f} \] Input:

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

Output:

(a**5*(80*int(tan(e + f*x)**2/(tan(e + f*x)**3 + 3*tan(e + f*x)**2*i - 3*t 
an(e + f*x) - i),x)*f*i - 80*int(tan(e + f*x)/(tan(e + f*x)**3 + 3*tan(e + 
 f*x)**2*i - 3*tan(e + f*x) - i),x)*f - 32*int(1/(tan(e + f*x)**3 + 3*tan( 
e + f*x)**2*i - 3*tan(e + f*x) - i),x)*f*i - 4*log(tan(e + f*x)**2 + 1)*i 
+ tan(e + f*x) - 32*f*x))/(c**3*f)