Integrand size = 31, antiderivative size = 114 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=-\frac {a^4 x}{c^3}+\frac {i a^4 \log (\cos (e+f x))}{c^3 f}-\frac {8 i a^4}{3 f (c-i c \tan (e+f x))^3}+\frac {6 i a^4}{c f (c-i c \tan (e+f x))^2}-\frac {6 i a^4}{f \left (c^3-i c^3 \tan (e+f x)\right )} \]
-a^4*x/c^3+I*a^4*ln(cos(f*x+e))/c^3/f-8/3*I*a^4/f/(c-I*c*tan(f*x+e))^3+6*I *a^4/c/f/(c-I*c*tan(f*x+e))^2-6*I*a^4/f/(c^3-I*c^3*tan(f*x+e))
Time = 2.95 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.59 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=\frac {i a^4 \left (-\log (i+\tan (e+f x))+\frac {2 \left (4 i+9 \tan (e+f x)-9 i \tan ^2(e+f x)\right )}{3 (i+\tan (e+f x))^3}\right )}{c^3 f} \]
(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 = 0.35 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.79, 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 |
\(\displaystyle \frac {i a^4 \left (-\frac {8 c^3}{3 (c-i c \tan (e+f x))^3}+\frac {6 c^2}{(c-i c \tan (e+f x))^2}-\frac {6 c}{c-i c \tan (e+f x)}-\log (c-i c \tan (e+f x))\right )}{c^3 f}\) |
(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 )
3.10.40.3.1 Defintions of rubi rules used
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 = 0.30 (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 | \(-\frac {i a^{4} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,c^{3}}-\frac {a^{4} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{3}}-\frac {8 a^{4}}{3 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{3}}-\frac {6 i a^{4}}{f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{2}}+\frac {6 a^{4}}{f \,c^{3} \left (\tan \left (f x +e \right )+i\right )}\) | \(110\) |
default | \(-\frac {i a^{4} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,c^{3}}-\frac {a^{4} \arctan \left (\tan \left (f x +e \right )\right )}{f \,c^{3}}-\frac {8 a^{4}}{3 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{3}}-\frac {6 i a^{4}}{f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{2}}+\frac {6 a^{4}}{f \,c^{3} \left (\tan \left (f x +e \right )+i\right )}\) | \(110\) |
norman | \(\frac {-\frac {4 i a^{4} \left (\tan ^{2}\left (f x +e \right )\right )}{c f}-\frac {a^{4} x}{c}-\frac {8 i a^{4}}{3 c f}-\frac {3 a^{4} x \left (\tan ^{2}\left (f x +e \right )\right )}{c}-\frac {3 a^{4} x \left (\tan ^{4}\left (f x +e \right )\right )}{c}-\frac {a^{4} x \left (\tan ^{6}\left (f x +e \right )\right )}{c}+\frac {2 a^{4} \tan \left (f x +e \right )}{c f}-\frac {8 a^{4} \left (\tan ^{3}\left (f x +e \right )\right )}{3 c f}+\frac {6 a^{4} \left (\tan ^{5}\left (f x +e \right )\right )}{c f}-\frac {12 i a^{4} \left (\tan ^{4}\left (f x +e \right )\right )}{c f}}{c^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {i a^{4} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,c^{3}}\) | \(209\) |
-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 = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.60 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=\frac {-2 i \, a^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 i \, a^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - 6 i \, a^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, a^{4} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{6 \, c^{3} f} \]
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 = 0.30 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.47 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=\frac {i a^{4} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c^{3} f} + \begin {cases} \frac {- 2 i a^{4} c^{6} f^{2} e^{6 i e} e^{6 i f x} + 3 i a^{4} c^{6} f^{2} e^{4 i e} e^{4 i f x} - 6 i a^{4} c^{6} f^{2} e^{2 i e} e^{2 i f x}}{6 c^{9} f^{3}} & \text {for}\: c^{9} f^{3} \neq 0 \\\frac {x \left (2 a^{4} e^{6 i e} - 2 a^{4} e^{4 i e} + 2 a^{4} e^{2 i e}\right )}{c^{3}} & \text {otherwise} \end {cases} \]
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. \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.79 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.61 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=-\frac {-\frac {30 i \, a^{4} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{3}} + \frac {60 i \, a^{4} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c^{3}} - \frac {30 i \, a^{4} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c^{3}} + \frac {-147 i \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 1002 \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 2445 i \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 3820 \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2445 i \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1002 \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 147 i \, a^{4}}{c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{6}}}{30 \, f} \]
-1/30*(-30*I*a^4*log(tan(1/2*f*x + 1/2*e) + 1)/c^3 + 60*I*a^4*log(tan(1/2* f*x + 1/2*e) + I)/c^3 - 30*I*a^4*log(tan(1/2*f*x + 1/2*e) - 1)/c^3 + (-147 *I*a^4*tan(1/2*f*x + 1/2*e)^6 + 1002*a^4*tan(1/2*f*x + 1/2*e)^5 + 2445*I*a ^4*tan(1/2*f*x + 1/2*e)^4 - 3820*a^4*tan(1/2*f*x + 1/2*e)^3 - 2445*I*a^4*t an(1/2*f*x + 1/2*e)^2 + 1002*a^4*tan(1/2*f*x + 1/2*e) + 147*I*a^4)/(c^3*(t an(1/2*f*x + 1/2*e) + I)^6))/f
Time = 5.93 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^3} \, dx=-\frac {\frac {6\,a^4\,{\mathrm {tan}\left (e+f\,x\right )}^2}{c^3}-\frac {8\,a^4}{3\,c^3}+\frac {a^4\,\mathrm {tan}\left (e+f\,x\right )\,6{}\mathrm {i}}{c^3}}{f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {a^4\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^3\,f} \]