\(\int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4} \, dx\) [947]

Optimal result

Integrand size = 31, antiderivative size = 160 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4} \, dx=\frac {10 a^6 x}{c^4}-\frac {10 i a^6 \log (\cos (e+f x))}{c^4 f}-\frac {a^6 \tan (e+f x)}{c^4 f}-\frac {8 i a^6}{f (c-i c \tan (e+f x))^4}+\frac {80 i a^6}{3 c f (c-i c \tan (e+f x))^3}-\frac {40 i a^6}{f \left (c^2-i c^2 \tan (e+f x)\right )^2}+\frac {40 i a^6}{f \left (c^4-i c^4 \tan (e+f x)\right )} \] Output:

10*a^6*x/c^4-10*I*a^6*ln(cos(f*x+e))/c^4/f-a^6*tan(f*x+e)/c^4/f-8*I*a^6/f/ 
(c-I*c*tan(f*x+e))^4+80/3*I*a^6/c/f/(c-I*c*tan(f*x+e))^3-40*I*a^6/f/(c^2-I 
*c^2*tan(f*x+e))^2+40*I*a^6/f/(c^4-I*c^4*tan(f*x+e))
 

Mathematica [A] (verified)

Time = 3.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.56 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4} \, dx=\frac {i a^6 \left (10 c \log (i+\tan (e+f x))+i c \tan (e+f x)+\frac {8 c \left (7-25 i \tan (e+f x)-30 \tan ^2(e+f x)+15 i \tan ^3(e+f x)\right )}{3 (i+\tan (e+f x))^4}\right )}{c^5 f} \] Input:

Integrate[(a + I*a*Tan[e + f*x])^6/(c - I*c*Tan[e + f*x])^4,x]
 

Output:

(I*a^6*(10*c*Log[I + Tan[e + f*x]] + I*c*Tan[e + f*x] + (8*c*(7 - (25*I)*T 
an[e + f*x] - 30*Tan[e + f*x]^2 + (15*I)*Tan[e + f*x]^3))/(3*(I + Tan[e + 
f*x])^4)))/(c^5*f)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 124, 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.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4}dx\)

\(\Big \downarrow \) 4005

\(\displaystyle a^6 c^6 \int \frac {\sec ^{12}(e+f x)}{(c-i c \tan (e+f x))^{10}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^6 c^6 \int \frac {\sec (e+f x)^{12}}{(c-i c \tan (e+f x))^{10}}dx\)

\(\Big \downarrow \) 3968

\(\displaystyle \frac {i a^6 \int \frac {(i \tan (e+f x) c+c)^5}{(c-i c \tan (e+f x))^5}d(-i c \tan (e+f x))}{c^5 f}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {i a^6 \int \left (\frac {32 c^5}{(c-i c \tan (e+f x))^5}-\frac {80 c^4}{(c-i c \tan (e+f x))^4}+\frac {80 c^3}{(c-i c \tan (e+f x))^3}-\frac {40 c^2}{(c-i c \tan (e+f x))^2}+\frac {10 c}{c-i c \tan (e+f x)}-1\right )d(-i c \tan (e+f x))}{c^5 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i a^6 \left (-\frac {8 c^5}{(c-i c \tan (e+f x))^4}+\frac {80 c^4}{3 (c-i c \tan (e+f x))^3}-\frac {40 c^3}{(c-i c \tan (e+f x))^2}+\frac {40 c^2}{c-i c \tan (e+f x)}+i c \tan (e+f x)+10 c \log (c-i c \tan (e+f x))\right )}{c^5 f}\)

Input:

Int[(a + I*a*Tan[e + f*x])^6/(c - I*c*Tan[e + f*x])^4,x]
 

Output:

(I*a^6*(10*c*Log[c - I*c*Tan[e + f*x]] + I*c*Tan[e + f*x] - (8*c^5)/(c - I 
*c*Tan[e + f*x])^4 + (80*c^4)/(3*(c - I*c*Tan[e + f*x])^3) - (40*c^3)/(c - 
 I*c*Tan[e + f*x])^2 + (40*c^2)/(c - I*c*Tan[e + f*x])))/(c^5*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 = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.92

method	result	size
risch	\(-\frac {i a^{6} {\mathrm e}^{8 i \left (f x +e \right )}}{2 c^{4} f}+\frac {4 i a^{6} {\mathrm e}^{6 i \left (f x +e \right )}}{3 c^{4} f}-\frac {3 i a^{6} {\mathrm e}^{4 i \left (f x +e \right )}}{c^{4} f}+\frac {8 i a^{6} {\mathrm e}^{2 i \left (f x +e \right )}}{c^{4} f}-\frac {20 a^{6} e}{f \,c^{4}}-\frac {2 i a^{6}}{f \,c^{4} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {10 i a^{6} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f \,c^{4}}\)	\(147\)
derivativedivides	\(-\frac {a^{6} \tan \left (f x +e \right )}{c^{4} f}+\frac {40 i a^{6}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {8 i a^{6}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {40 a^{6}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )}+\frac {80 a^{6}}{3 f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {10 a^{6} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{4}}+\frac {5 i a^{6} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f \,c^{4}}\)	\(150\)
default	\(-\frac {a^{6} \tan \left (f x +e \right )}{c^{4} f}+\frac {40 i a^{6}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {8 i a^{6}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {40 a^{6}}{f \,c^{4} \left (i+\tan \left (f x +e \right )\right )}+\frac {80 a^{6}}{3 f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {10 a^{6} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{4}}+\frac {5 i a^{6} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f \,c^{4}}\)	\(150\)
norman	\(\frac {\frac {10 a^{6} x}{c}+\frac {56 i a^{6}}{3 c f}+\frac {40 a^{6} x \tan \left (f x +e \right )^{2}}{c}+\frac {60 a^{6} x \tan \left (f x +e \right )^{4}}{c}+\frac {40 a^{6} x \tan \left (f x +e \right )^{6}}{c}+\frac {10 a^{6} x \tan \left (f x +e \right )^{8}}{c}-\frac {9 a^{6} \tan \left (f x +e \right )}{c f}-\frac {148 a^{6} \tan \left (f x +e \right )^{3}}{3 c f}-\frac {58 a^{6} \tan \left (f x +e \right )^{5}}{3 c f}-\frac {44 a^{6} \tan \left (f x +e \right )^{7}}{c f}-\frac {a^{6} \tan \left (f x +e \right )^{9}}{c f}+\frac {72 i a^{6} \tan \left (f x +e \right )^{4}}{c f}+\frac {80 i a^{6} \tan \left (f x +e \right )^{6}}{c f}+\frac {224 i a^{6} \tan \left (f x +e \right )^{2}}{3 c f}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{4} c^{3}}+\frac {5 i a^{6} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f \,c^{4}}\)	\(284\)

Input:

int((a+I*a*tan(f*x+e))^6/(c-I*c*tan(f*x+e))^4,x,method=_RETURNVERBOSE)
 

Output:

-1/2*I/c^4/f*a^6*exp(8*I*(f*x+e))+4/3*I/c^4/f*a^6*exp(6*I*(f*x+e))-3*I/c^4 
/f*a^6*exp(4*I*(f*x+e))+8*I/c^4/f*a^6*exp(2*I*(f*x+e))-20/f*a^6/c^4*e-2*I* 
a^6/f/c^4/(exp(2*I*(f*x+e))+1)-10*I/f*a^6/c^4*ln(exp(2*I*(f*x+e))+1)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.84 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4} \, dx=\frac {-3 i \, a^{6} e^{\left (10 i \, f x + 10 i \, e\right )} + 5 i \, a^{6} e^{\left (8 i \, f x + 8 i \, e\right )} - 10 i \, a^{6} e^{\left (6 i \, f x + 6 i \, e\right )} + 30 i \, a^{6} e^{\left (4 i \, f x + 4 i \, e\right )} + 48 i \, a^{6} e^{\left (2 i \, f x + 2 i \, e\right )} - 12 i \, a^{6} - 60 \, {\left (i \, a^{6} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{6}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{6 \, {\left (c^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} + c^{4} f\right )}} \] Input:

integrate((a+I*a*tan(f*x+e))^6/(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")
 

Output:

1/6*(-3*I*a^6*e^(10*I*f*x + 10*I*e) + 5*I*a^6*e^(8*I*f*x + 8*I*e) - 10*I*a 
^6*e^(6*I*f*x + 6*I*e) + 30*I*a^6*e^(4*I*f*x + 4*I*e) + 48*I*a^6*e^(2*I*f* 
x + 2*I*e) - 12*I*a^6 - 60*(I*a^6*e^(2*I*f*x + 2*I*e) + I*a^6)*log(e^(2*I* 
f*x + 2*I*e) + 1))/(c^4*f*e^(2*I*f*x + 2*I*e) + c^4*f)
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.52 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4} \, dx=- \frac {2 i a^{6}}{c^{4} f e^{2 i e} e^{2 i f x} + c^{4} f} - \frac {10 i a^{6} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{4} f} + \begin {cases} \frac {- 3 i a^{6} c^{12} f^{3} e^{8 i e} e^{8 i f x} + 8 i a^{6} c^{12} f^{3} e^{6 i e} e^{6 i f x} - 18 i a^{6} c^{12} f^{3} e^{4 i e} e^{4 i f x} + 48 i a^{6} c^{12} f^{3} e^{2 i e} e^{2 i f x}}{6 c^{16} f^{4}} & \text {for}\: c^{16} f^{4} \neq 0 \\\frac {x \left (4 a^{6} e^{8 i e} - 8 a^{6} e^{6 i e} + 12 a^{6} e^{4 i e} - 16 a^{6} e^{2 i e}\right )}{c^{4}} & \text {otherwise} \end {cases} \] Input:

integrate((a+I*a*tan(f*x+e))**6/(c-I*c*tan(f*x+e))**4,x)
 

Output:

-2*I*a**6/(c**4*f*exp(2*I*e)*exp(2*I*f*x) + c**4*f) - 10*I*a**6*log(exp(2* 
I*f*x) + exp(-2*I*e))/(c**4*f) + Piecewise(((-3*I*a**6*c**12*f**3*exp(8*I* 
e)*exp(8*I*f*x) + 8*I*a**6*c**12*f**3*exp(6*I*e)*exp(6*I*f*x) - 18*I*a**6* 
c**12*f**3*exp(4*I*e)*exp(4*I*f*x) + 48*I*a**6*c**12*f**3*exp(2*I*e)*exp(2 
*I*f*x))/(6*c**16*f**4), Ne(c**16*f**4, 0)), (x*(4*a**6*exp(8*I*e) - 8*a** 
6*exp(6*I*e) + 12*a**6*exp(4*I*e) - 16*a**6*exp(2*I*e))/c**4, True))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((a+I*a*tan(f*x+e))^6/(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 = 0.77 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.62 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4} \, dx=\frac {10 i \, a^{6} \log \left (\tan \left (f x + e\right ) + i\right )}{c^{4} f} - \frac {a^{6} \tan \left (f x + e\right )}{c^{4} f} - \frac {8 \, {\left (15 \, a^{6} \tan \left (f x + e\right )^{3} + 30 i \, a^{6} \tan \left (f x + e\right )^{2} - 25 \, a^{6} \tan \left (f x + e\right ) - 7 i \, a^{6}\right )}}{3 \, c^{4} f {\left (\tan \left (f x + e\right ) + i\right )}^{4}} \] Input:

integrate((a+I*a*tan(f*x+e))^6/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")
 

Output:

10*I*a^6*log(tan(f*x + e) + I)/(c^4*f) - a^6*tan(f*x + e)/(c^4*f) - 8/3*(1 
5*a^6*tan(f*x + e)^3 + 30*I*a^6*tan(f*x + e)^2 - 25*a^6*tan(f*x + e) - 7*I 
*a^6)/(c^4*f*(tan(f*x + e) + I)^4)
 

Mupad [B] (verification not implemented)

Time = 4.08 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.06 \[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4} \, dx=\frac {a^6\,\left (10\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )-60\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+10\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^4-76\,{\mathrm {tan}\left (e+f\,x\right )}^2-4\,{\mathrm {tan}\left (e+f\,x\right )}^4+\frac {56}{3}-\frac {\mathrm {tan}\left (e+f\,x\right )\,197{}\mathrm {i}}{3}-\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\mathrm {tan}\left (e+f\,x\right )\,40{}\mathrm {i}+\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3\,40{}\mathrm {i}+{\mathrm {tan}\left (e+f\,x\right )}^3\,34{}\mathrm {i}+{\mathrm {tan}\left (e+f\,x\right )}^5\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^4\,f\,{\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4} \] Input:

int((a + a*tan(e + f*x)*1i)^6/(c - c*tan(e + f*x)*1i)^4,x)
 

Output:

(a^6*(10*log(tan(e + f*x) + 1i) - (tan(e + f*x)*197i)/3 - log(tan(e + f*x) 
 + 1i)*tan(e + f*x)*40i - 60*log(tan(e + f*x) + 1i)*tan(e + f*x)^2 + log(t 
an(e + f*x) + 1i)*tan(e + f*x)^3*40i + 10*log(tan(e + f*x) + 1i)*tan(e + f 
*x)^4 - 76*tan(e + f*x)^2 + tan(e + f*x)^3*34i - 4*tan(e + f*x)^4 + tan(e 
+ f*x)^5*1i + 56/3)*1i)/(c^4*f*(tan(e + f*x)*1i - 1)^4)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^6}{(c-i c \tan (e+f x))^4} \, dx=\frac {a^{6} \left (160 \left (\int \frac {\tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{4} i -4 \tan \left (f x +e \right )^{3}-6 \tan \left (f x +e \right )^{2} i +4 \tan \left (f x +e \right )+i}d x \right ) f +240 \left (\int \frac {\tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{4} i -4 \tan \left (f x +e \right )^{3}-6 \tan \left (f x +e \right )^{2} i +4 \tan \left (f x +e \right )+i}d x \right ) f i -192 \left (\int \frac {\tan \left (f x +e \right )}{\tan \left (f x +e \right )^{4} i -4 \tan \left (f x +e \right )^{3}-6 \tan \left (f x +e \right )^{2} i +4 \tan \left (f x +e \right )+i}d x \right ) f -48 \left (\int \frac {1}{\tan \left (f x +e \right )^{4} i -4 \tan \left (f x +e \right )^{3}-6 \tan \left (f x +e \right )^{2} i +4 \tan \left (f x +e \right )+i}d x \right ) f i +5 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) i -\tan \left (f x +e \right )+50 f x \right )}{c^{4} f} \] Input:

int((a+I*a*tan(f*x+e))^6/(c-I*c*tan(f*x+e))^4,x)
 

Output:

(a**6*(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 + 240*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 - 192*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 - 48*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 + 5*log(tan(e + f*x)**2 + 1)*i - tan(e + f*x) + 50*f*x))/(c**4*f)