Integrand size = 31, antiderivative size = 83 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=\frac {a^3 x}{c^2}-\frac {i a^3 \log (\cos (e+f x))}{c^2 f}-\frac {2 i a^3}{c^2 f (1-i \tan (e+f x))^2}+\frac {4 i a^3}{c^2 f (1-i \tan (e+f x))} \] Output:
a^3*x/c^2-I*a^3*ln(cos(f*x+e))/c^2/f-2*I*a^3/c^2/f/(1-I*tan(f*x+e))^2+4*I* a^3/c^2/f/(1-I*tan(f*x+e))
Time = 2.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.60 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=\frac {i a^3 \left (\log (i+\tan (e+f x))+\frac {-2+4 i \tan (e+f x)}{(i+\tan (e+f x))^2}\right )}{c^2 f} \] Input:
Integrate[(a + I*a*Tan[e + f*x])^3/(c - I*c*Tan[e + f*x])^2,x]
Output:
(I*a^3*(Log[I + Tan[e + f*x]] + (-2 + (4*I)*Tan[e + f*x])/(I + Tan[e + f*x ])^2))/(c^2*f)
Time = 0.36 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80, 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.
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle a^3 c^3 \int \frac {\sec ^6(e+f x)}{(c-i c \tan (e+f x))^5}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \int \frac {\sec (e+f x)^6}{(c-i c \tan (e+f x))^5}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i a^3 \int \frac {(i \tan (e+f x) c+c)^2}{(c-i c \tan (e+f x))^3}d(-i c \tan (e+f x))}{c^2 f}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {i a^3 \int \left (\frac {4 c^2}{(c-i c \tan (e+f x))^3}-\frac {4 c}{(c-i c \tan (e+f x))^2}+\frac {1}{c-i c \tan (e+f x)}\right )d(-i c \tan (e+f x))}{c^2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i a^3 \left (-\frac {2 c^2}{(c-i c \tan (e+f x))^2}+\frac {4 c}{c-i c \tan (e+f x)}+\log (c-i c \tan (e+f x))\right )}{c^2 f}\) |
Input:
Int[(a + I*a*Tan[e + f*x])^3/(c - I*c*Tan[e + f*x])^2,x]
Output:
(I*a^3*(Log[c - I*c*Tan[e + f*x]] - (2*c^2)/(c - I*c*Tan[e + f*x])^2 + (4* c)/(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 = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96
method | result | size |
risch | \(-\frac {i a^{3} {\mathrm e}^{4 i \left (f x +e \right )}}{2 c^{2} f}+\frac {i a^{3} {\mathrm e}^{2 i \left (f x +e \right )}}{c^{2} f}-\frac {2 a^{3} e}{c^{2} f}-\frac {i a^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{c^{2} f}\) | \(80\) |
derivativedivides | \(\frac {2 i a^{3}}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {i a^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{2}}+\frac {a^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{2}}-\frac {4 a^{3}}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )}\) | \(87\) |
default | \(\frac {2 i a^{3}}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {i a^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{2}}+\frac {a^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{2}}-\frac {4 a^{3}}{f \,c^{2} \left (i+\tan \left (f x +e \right )\right )}\) | \(87\) |
norman | \(\frac {\frac {a^{3} x}{c}+\frac {2 i a^{3}}{c f}+\frac {a^{3} x \tan \left (f x +e \right )^{4}}{c}+\frac {2 a^{3} x \tan \left (f x +e \right )^{2}}{c}-\frac {4 a^{3} \tan \left (f x +e \right )^{3}}{c f}+\frac {6 i a^{3} \tan \left (f x +e \right )^{2}}{c f}}{c \left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {i a^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{2}}\) | \(134\) |
Input:
int((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
Output:
-1/2*I/c^2/f*a^3*exp(4*I*(f*x+e))+I/c^2/f*a^3*exp(2*I*(f*x+e))-2*a^3/c^2/f *e-I*a^3/c^2/f*ln(exp(2*I*(f*x+e))+1)
Time = 0.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.65 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=\frac {-i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, a^{3} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{2 \, c^{2} f} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="fricas")
Output:
1/2*(-I*a^3*e^(4*I*f*x + 4*I*e) + 2*I*a^3*e^(2*I*f*x + 2*I*e) - 2*I*a^3*lo g(e^(2*I*f*x + 2*I*e) + 1))/(c^2*f)
Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.47 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=- \frac {i a^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{2} f} + \begin {cases} \frac {- i a^{3} c^{2} f e^{4 i e} e^{4 i f x} + 2 i a^{3} c^{2} f e^{2 i e} e^{2 i f x}}{2 c^{4} f^{2}} & \text {for}\: c^{4} f^{2} \neq 0 \\\frac {x \left (2 a^{3} e^{4 i e} - 2 a^{3} e^{2 i e}\right )}{c^{2}} & \text {otherwise} \end {cases} \] Input:
integrate((a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**2,x)
Output:
-I*a**3*log(exp(2*I*f*x) + exp(-2*I*e))/(c**2*f) + Piecewise(((-I*a**3*c** 2*f*exp(4*I*e)*exp(4*I*f*x) + 2*I*a**3*c**2*f*exp(2*I*e)*exp(2*I*f*x))/(2* c**4*f**2), Ne(c**4*f**2, 0)), (x*(2*a**3*exp(4*I*e) - 2*a**3*exp(2*I*e))/ c**2, True))
Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.50 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.67 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=\frac {i \, a^{3} \log \left (\tan \left (f x + e\right ) + i\right )}{c^{2} f} - \frac {2 \, {\left (2 \, a^{3} \tan \left (f x + e\right ) + i \, a^{3}\right )}}{c^{2} f {\left (\tan \left (f x + e\right ) + i\right )}^{2}} \] Input:
integrate((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")
Output:
I*a^3*log(tan(f*x + e) + I)/(c^2*f) - 2*(2*a^3*tan(f*x + e) + I*a^3)/(c^2* f*(tan(f*x + e) + I)^2)
Time = 1.85 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=-\frac {\frac {4\,a^3\,\mathrm {tan}\left (e+f\,x\right )}{c^2}+\frac {a^3\,2{}\mathrm {i}}{c^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}-1\right )}+\frac {a^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^2\,f} \] Input:
int((a + a*tan(e + f*x)*1i)^3/(c - c*tan(e + f*x)*1i)^2,x)
Output:
(a^3*log(tan(e + f*x) + 1i)*1i)/(c^2*f) - ((a^3*2i)/c^2 + (4*a^3*tan(e + f *x))/c^2)/(f*(tan(e + f*x)*2i + tan(e + f*x)^2 - 1))
\[ \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^2} \, dx=\frac {a^{3} \left (\left (\int \frac {\tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i -1}d x \right ) i +3 \left (\int \frac {\tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i -1}d x \right )-3 \left (\int \frac {\tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i -1}d x \right ) i -\left (\int \frac {1}{\tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i -1}d x \right )\right )}{c^{2}} \] Input:
int((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^2,x)
Output:
(a**3*(int(tan(e + f*x)**3/(tan(e + f*x)**2 + 2*tan(e + f*x)*i - 1),x)*i + 3*int(tan(e + f*x)**2/(tan(e + f*x)**2 + 2*tan(e + f*x)*i - 1),x) - 3*int (tan(e + f*x)/(tan(e + f*x)**2 + 2*tan(e + f*x)*i - 1),x)*i - int(1/(tan(e + f*x)**2 + 2*tan(e + f*x)*i - 1),x)))/c**2