Integrand size = 31, antiderivative size = 71 \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=-\frac {4 a^3 x}{c}+\frac {4 i a^3 \log (\cos (e+f x))}{c f}+\frac {a^3 \tan (e+f x)}{c f}-\frac {4 i a^3}{f (c-i c \tan (e+f x))} \] Output:
-4*a^3*x/c+4*I*a^3*ln(cos(f*x+e))/c/f+a^3*tan(f*x+e)/c/f-4*I*a^3/f/(c-I*c* tan(f*x+e))
Time = 2.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.75 \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=\frac {i a^3 \left (-4 \log (i+\tan (e+f x))-i \tan (e+f x)-\frac {4 i}{i+\tan (e+f x)}\right )}{c f} \] Input:
Integrate[(a + I*a*Tan[e + f*x])^3/(c - I*c*Tan[e + f*x]),x]
Output:
(I*a^3*(-4*Log[I + Tan[e + f*x]] - I*Tan[e + f*x] - (4*I)/(I + Tan[e + f*x ])))/(c*f)
Time = 0.36 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.87, 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.
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle a^3 c^3 \int \frac {\sec ^6(e+f x)}{(c-i c \tan (e+f x))^4}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \int \frac {\sec (e+f x)^6}{(c-i c \tan (e+f x))^4}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i a^3 \int \frac {(i \tan (e+f x) c+c)^2}{(c-i c \tan (e+f x))^2}d(-i c \tan (e+f x))}{c^2 f}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {i a^3 \int \left (\frac {4 c^2}{(c-i c \tan (e+f x))^2}-\frac {4 c}{c-i c \tan (e+f x)}+1\right )d(-i c \tan (e+f x))}{c^2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i a^3 \left (-\frac {4 c^2}{c-i c \tan (e+f x)}-i c \tan (e+f x)-4 c \log (c-i c \tan (e+f x))\right )}{c^2 f}\) |
Input:
Int[(a + I*a*Tan[e + f*x])^3/(c - I*c*Tan[e + f*x]),x]
Output:
(I*a^3*(-4*c*Log[c - I*c*Tan[e + f*x]] - I*c*Tan[e + f*x] - (4*c^2)/(c - I *c*Tan[e + f*x])))/(c^2*f)
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]
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]
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]))
Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {a^{3} \tan \left (f x +e \right )}{c f}+\frac {4 a^{3}}{f c \left (i+\tan \left (f x +e \right )\right )}-\frac {4 a^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f c}-\frac {2 i a^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f c}\) | \(81\) |
default | \(\frac {a^{3} \tan \left (f x +e \right )}{c f}+\frac {4 a^{3}}{f c \left (i+\tan \left (f x +e \right )\right )}-\frac {4 a^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f c}-\frac {2 i a^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f c}\) | \(81\) |
risch | \(-\frac {2 i a^{3} {\mathrm e}^{2 i \left (f x +e \right )}}{c f}+\frac {8 a^{3} e}{c f}+\frac {2 i a^{3}}{f c \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f c}\) | \(84\) |
norman | \(\frac {-\frac {4 i a^{3}}{c f}+\frac {a^{3} \tan \left (f x +e \right )^{3}}{c f}-\frac {4 a^{3} x}{c}-\frac {4 a^{3} x \tan \left (f x +e \right )^{2}}{c}+\frac {5 a^{3} \tan \left (f x +e \right )}{c f}}{1+\tan \left (f x +e \right )^{2}}-\frac {2 i a^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{f c}\) | \(112\) |
Input:
int((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e)),x,method=_RETURNVERBOSE)
Output:
a^3*tan(f*x+e)/c/f+4/f*a^3/c/(I+tan(f*x+e))-4/f*a^3/c*arctan(tan(f*x+e))-2 *I/f*a^3/c*ln(1+tan(f*x+e)^2)
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.24 \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=-\frac {2 \, {\left (i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} + 2 \, {\left (-i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e)),x, algorithm="fricas")
Output:
-2*(I*a^3*e^(4*I*f*x + 4*I*e) + I*a^3*e^(2*I*f*x + 2*I*e) - I*a^3 + 2*(-I* a^3*e^(2*I*f*x + 2*I*e) - I*a^3)*log(e^(2*I*f*x + 2*I*e) + 1))/(c*f*e^(2*I *f*x + 2*I*e) + c*f)
Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=\frac {2 i a^{3}}{c f e^{2 i e} e^{2 i f x} + c f} + \frac {4 i a^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c f} + \begin {cases} - \frac {2 i a^{3} e^{2 i e} e^{2 i f x}}{c f} & \text {for}\: c f \neq 0 \\\frac {4 a^{3} x e^{2 i e}}{c} & \text {otherwise} \end {cases} \] Input:
integrate((a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e)),x)
Output:
2*I*a**3/(c*f*exp(2*I*e)*exp(2*I*f*x) + c*f) + 4*I*a**3*log(exp(2*I*f*x) + exp(-2*I*e))/(c*f) + Piecewise((-2*I*a**3*exp(2*I*e)*exp(2*I*f*x)/(c*f), Ne(c*f, 0)), (4*a**3*x*exp(2*I*e)/c, True))
Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e)),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.52 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.82 \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=-\frac {4 i \, a^{3} \log \left (\tan \left (f x + e\right ) + i\right )}{c f} + \frac {a^{3} \tan \left (f x + e\right )}{c f} + \frac {4 \, a^{3}}{c f {\left (\tan \left (f x + e\right ) + i\right )}} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e)),x, algorithm="giac")
Output:
-4*I*a^3*log(tan(f*x + e) + I)/(c*f) + a^3*tan(f*x + e)/(c*f) + 4*a^3/(c*f *(tan(f*x + e) + I))
Time = 1.79 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=\frac {a^3\,\mathrm {tan}\left (e+f\,x\right )}{c\,f}+\frac {4\,a^3}{c\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {a^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c\,f} \] Input:
int((a + a*tan(e + f*x)*1i)^3/(c - c*tan(e + f*x)*1i),x)
Output:
(a^3*tan(e + f*x))/(c*f) + (4*a^3)/(c*f*(tan(e + f*x) + 1i)) - (a^3*log(ta n(e + f*x) + 1i)*4i)/(c*f)
\[ \int \frac {(a+i a \tan (e+f x))^3}{c-i c \tan (e+f x)} \, dx=\frac {a^{3} \left (-8 \left (\int \frac {1}{\tan \left (f x +e \right ) i -1}d x \right ) f -2 \,\mathrm {log}\left (\tan \left (f x +e \right )^{2}+1\right ) i +\tan \left (f x +e \right )-8 f x \right )}{c f} \] Input:
int((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e)),x)
Output:
(a**3*( - 8*int(1/(tan(e + f*x)*i - 1),x)*f - 2*log(tan(e + f*x)**2 + 1)*i + tan(e + f*x) - 8*f*x))/(c*f)