Integrand size = 31, antiderivative size = 113 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=-\frac {a^4 x}{c^3}+\frac {i a^4 \log (\cos (e+f x))}{c^3 f}-\frac {8 i a^4}{3 c^3 f (1-i \tan (e+f x))^3}+\frac {6 i a^4}{c^3 f (1-i \tan (e+f x))^2}-\frac {6 i a^4}{c^3 f (1-i \tan (e+f x))} \] Output:
-a^4*x/c^3+I*a^4*ln(cos(f*x+e))/c^3/f-8/3*I*a^4/c^3/f/(1-I*tan(f*x+e))^3+6 *I*a^4/c^3/f/(1-I*tan(f*x+e))^2-6*I*a^4/c^3/f/(1-I*tan(f*x+e))
Time = 1.82 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.59 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=\frac {i a^4 \left (-\log (i+\tan (e+f x))+\frac {2 \left (4 i+9 \tan (e+f x)-9 i \tan ^2(e+f x)\right )}{3 (i+\tan (e+f x))^3}\right )}{c^3 f} \] Input:
Integrate[(a + I*a*Tan[e + f*x])^4/(c - I*c*Tan[e + f*x])^3,x]
Output:
(I*a^4*(-Log[I + Tan[e + f*x]] + (2*(4*I + 9*Tan[e + f*x] - (9*I)*Tan[e + f*x]^2))/(3*(I + Tan[e + f*x])^3)))/(c^3*f)
Time = 0.37 (sec) , antiderivative size = 90, 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.
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle a^4 c^4 \int \frac {\sec ^8(e+f x)}{(c-i c \tan (e+f x))^7}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^4 c^4 \int \frac {\sec (e+f x)^8}{(c-i c \tan (e+f x))^7}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i a^4 \int \frac {(i \tan (e+f x) c+c)^3}{(c-i c \tan (e+f x))^4}d(-i c \tan (e+f x))}{c^3 f}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {i a^4 \int \left (\frac {8 c^3}{(c-i c \tan (e+f x))^4}-\frac {12 c^2}{(c-i c \tan (e+f x))^3}+\frac {6 c}{(c-i c \tan (e+f x))^2}+\frac {1}{i c \tan (e+f x)-c}\right )d(-i c \tan (e+f x))}{c^3 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i a^4 \left (-\frac {8 c^3}{3 (c-i c \tan (e+f x))^3}+\frac {6 c^2}{(c-i c \tan (e+f x))^2}-\frac {6 c}{c-i c \tan (e+f x)}-\log (c-i c \tan (e+f x))\right )}{c^3 f}\) |
Input:
Int[(a + I*a*Tan[e + f*x])^4/(c - I*c*Tan[e + f*x])^3,x]
Output:
(I*a^4*(-Log[c - I*c*Tan[e + f*x]] - (8*c^3)/(3*(c - I*c*Tan[e + f*x])^3) + (6*c^2)/(c - I*c*Tan[e + f*x])^2 - (6*c)/(c - I*c*Tan[e + f*x])))/(c^3*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 = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89
method | result | size |
risch | \(-\frac {i a^{4} {\mathrm e}^{6 i \left (f x +e \right )}}{3 c^{3} f}+\frac {i a^{4} {\mathrm e}^{4 i \left (f x +e \right )}}{2 c^{3} f}-\frac {i a^{4} {\mathrm e}^{2 i \left (f x +e \right )}}{c^{3} f}+\frac {2 a^{4} e}{c^{3} f}+\frac {i a^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{c^{3} f}\) | \(101\) |
derivativedivides | \(-\frac {6 i a^{4}}{f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {6 a^{4}}{f \,c^{3} \left (i+\tan \left (f x +e \right )\right )}-\frac {8 a^{4}}{3 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {i a^{4} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{3}}-\frac {a^{4} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{3}}\) | \(110\) |
default | \(-\frac {6 i a^{4}}{f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {6 a^{4}}{f \,c^{3} \left (i+\tan \left (f x +e \right )\right )}-\frac {8 a^{4}}{3 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {i a^{4} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{3}}-\frac {a^{4} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{3}}\) | \(110\) |
norman | \(\frac {-\frac {4 i a^{4} \tan \left (f x +e \right )^{2}}{c f}-\frac {a^{4} x}{c}-\frac {8 i a^{4}}{3 c f}-\frac {3 a^{4} x \tan \left (f x +e \right )^{2}}{c}-\frac {3 a^{4} x \tan \left (f x +e \right )^{4}}{c}-\frac {a^{4} x \tan \left (f x +e \right )^{6}}{c}+\frac {2 a^{4} \tan \left (f x +e \right )}{c f}-\frac {8 a^{4} \tan \left (f x +e \right )^{3}}{3 c f}+\frac {6 a^{4} \tan \left (f x +e \right )^{5}}{c f}-\frac {12 i a^{4} \tan \left (f x +e \right )^{4}}{c f}}{c^{2} \left (1+\tan \left (f x +e \right )^{2}\right )^{3}}-\frac {i a^{4} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,c^{3}}\) | \(209\) |
Input:
int((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
Output:
-1/3*I/c^3/f*a^4*exp(6*I*(f*x+e))+1/2*I/c^3/f*a^4*exp(4*I*(f*x+e))-I/c^3/f *a^4*exp(2*I*(f*x+e))+2*a^4/c^3/f*e+I*a^4/c^3/f*ln(exp(2*I*(f*x+e))+1)
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.60 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=\frac {-2 i \, a^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 i \, a^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - 6 i \, a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, a^{4} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{6 \, c^{3} f} \] Input:
integrate((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^3,x, algorithm="fricas")
Output:
1/6*(-2*I*a^4*e^(6*I*f*x + 6*I*e) + 3*I*a^4*e^(4*I*f*x + 4*I*e) - 6*I*a^4* e^(2*I*f*x + 2*I*e) + 6*I*a^4*log(e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)
Time = 0.33 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.49 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=\frac {i a^{4} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{3} f} + \begin {cases} \frac {- 2 i a^{4} c^{6} f^{2} e^{6 i e} e^{6 i f x} + 3 i a^{4} c^{6} f^{2} e^{4 i e} e^{4 i f x} - 6 i a^{4} c^{6} f^{2} e^{2 i e} e^{2 i f x}}{6 c^{9} f^{3}} & \text {for}\: c^{9} f^{3} \neq 0 \\\frac {x \left (2 a^{4} e^{6 i e} - 2 a^{4} e^{4 i e} + 2 a^{4} e^{2 i e}\right )}{c^{3}} & \text {otherwise} \end {cases} \] Input:
integrate((a+I*a*tan(f*x+e))**4/(c-I*c*tan(f*x+e))**3,x)
Output:
I*a**4*log(exp(2*I*f*x) + exp(-2*I*e))/(c**3*f) + Piecewise(((-2*I*a**4*c* *6*f**2*exp(6*I*e)*exp(6*I*f*x) + 3*I*a**4*c**6*f**2*exp(4*I*e)*exp(4*I*f* x) - 6*I*a**4*c**6*f**2*exp(2*I*e)*exp(2*I*f*x))/(6*c**9*f**3), Ne(c**9*f* *3, 0)), (x*(2*a**4*exp(6*I*e) - 2*a**4*exp(4*I*e) + 2*a**4*exp(2*I*e))/c* *3, True))
Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((a+I*a*tan(f*x+e))^4/(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.
Time = 0.57 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.61 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=-\frac {i \, a^{4} \log \left (\tan \left (f x + e\right ) + i\right )}{c^{3} f} + \frac {2 \, {\left (9 \, a^{4} \tan \left (f x + e\right )^{2} + 9 i \, a^{4} \tan \left (f x + e\right ) - 4 \, a^{4}\right )}}{3 \, c^{3} f {\left (\tan \left (f x + e\right ) + i\right )}^{3}} \] Input:
integrate((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^3,x, algorithm="giac")
Output:
-I*a^4*log(tan(f*x + e) + I)/(c^3*f) + 2/3*(9*a^4*tan(f*x + e)^2 + 9*I*a^4 *tan(f*x + e) - 4*a^4)/(c^3*f*(tan(f*x + e) + I)^3)
Time = 1.95 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.90 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=-\frac {\frac {6\,a^4\,{\mathrm {tan}\left (e+f\,x\right )}^2}{c^3}-\frac {8\,a^4}{3\,c^3}+\frac {a^4\,\mathrm {tan}\left (e+f\,x\right )\,6{}\mathrm {i}}{c^3}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {a^4\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^3\,f} \] Input:
int((a + a*tan(e + f*x)*1i)^4/(c - c*tan(e + f*x)*1i)^3,x)
Output:
- ((a^4*tan(e + f*x)*6i)/c^3 - (8*a^4)/(3*c^3) + (6*a^4*tan(e + f*x)^2)/c^ 3)/(f*(3*tan(e + f*x) - tan(e + f*x)^2*3i - tan(e + f*x)^3 + 1i)) - (a^4*l og(tan(e + f*x) + 1i)*1i)/(c^3*f)
\[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=\frac {a^{4} \left (\int \frac {\tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{3} i -3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i +1}d x -4 \left (\int \frac {\tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{3} i -3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i +1}d x \right ) i -6 \left (\int \frac {\tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3} i -3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i +1}d x \right )+4 \left (\int \frac {\tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3} i -3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i +1}d x \right ) i +\int \frac {1}{\tan \left (f x +e \right )^{3} i -3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i +1}d x \right )}{c^{3}} \] Input:
int((a+I*a*tan(f*x+e))^4/(c-I*c*tan(f*x+e))^3,x)
Output:
(a**4*(int(tan(e + f*x)**4/(tan(e + f*x)**3*i - 3*tan(e + f*x)**2 - 3*tan( e + f*x)*i + 1),x) - 4*int(tan(e + f*x)**3/(tan(e + f*x)**3*i - 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i + 1),x)*i - 6*int(tan(e + f*x)**2/(tan(e + f*x) **3*i - 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i + 1),x) + 4*int(tan(e + f*x)/ (tan(e + f*x)**3*i - 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i + 1),x)*i + int( 1/(tan(e + f*x)**3*i - 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i + 1),x)))/c**3