Integrand size = 31, antiderivative size = 162 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\frac {5 x}{32 a c^4}-\frac {i}{16 a f (c-i c \tan (e+f x))^4}-\frac {i}{12 a c f (c-i c \tan (e+f x))^3}-\frac {3 i}{32 a f \left (c^2-i c^2 \tan (e+f x)\right )^2}-\frac {i}{8 a f \left (c^4-i c^4 \tan (e+f x)\right )}+\frac {i}{32 a f \left (c^4+i c^4 \tan (e+f x)\right )} \]
5/32*x/a/c^4-1/16*I/a/f/(c-I*c*tan(f*x+e))^4-1/12*I/a/c/f/(c-I*c*tan(f*x+e ))^3-3/32*I/a/f/(c^2-I*c^2*tan(f*x+e))^2-1/8*I/a/f/(c^4-I*c^4*tan(f*x+e))+ 1/32*I/a/f/(c^4+I*c^4*tan(f*x+e))
Time = 0.91 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\frac {i c^2 \left (-\frac {5 i \arctan (\tan (e+f x))}{32 c^6}-\frac {1}{16 c^2 (c-i c \tan (e+f x))^4}-\frac {1}{12 c^3 (c-i c \tan (e+f x))^3}-\frac {3}{32 c^4 (c-i c \tan (e+f x))^2}-\frac {1}{8 c^5 (c-i c \tan (e+f x))}+\frac {1}{32 c^5 (c+i c \tan (e+f x))}\right )}{a f} \]
(I*c^2*((((-5*I)/32)*ArcTan[Tan[e + f*x]])/c^6 - 1/(16*c^2*(c - I*c*Tan[e + f*x])^4) - 1/(12*c^3*(c - I*c*Tan[e + f*x])^3) - 3/(32*c^4*(c - I*c*Tan[ e + f*x])^2) - 1/(8*c^5*(c - I*c*Tan[e + f*x])) + 1/(32*c^5*(c + I*c*Tan[e + f*x]))))/(a*f)
Time = 0.41 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.86, 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, 54, 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.
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle \frac {\int \frac {\cos ^2(e+f x)}{(c-i c \tan (e+f x))^3}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{\sec (e+f x)^2 (c-i c \tan (e+f x))^3}dx}{a c}\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i c^2 \int \frac {1}{(c-i c \tan (e+f x))^5 (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{a f}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {i c^2 \int \left (\frac {1}{8 c^5 (c-i c \tan (e+f x))^2}+\frac {1}{32 c^5 (i \tan (e+f x) c+c)^2}+\frac {3}{16 c^4 (c-i c \tan (e+f x))^3}+\frac {1}{4 c^3 (c-i c \tan (e+f x))^4}+\frac {1}{4 c^2 (c-i c \tan (e+f x))^5}+\frac {5}{32 c^5 \left (\tan ^2(e+f x) c^2+c^2\right )}\right )d(-i c \tan (e+f x))}{a f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i c^2 \left (-\frac {5 i \arctan (\tan (e+f x))}{32 c^6}-\frac {1}{8 c^5 (c-i c \tan (e+f x))}+\frac {1}{32 c^5 (c+i c \tan (e+f x))}-\frac {3}{32 c^4 (c-i c \tan (e+f x))^2}-\frac {1}{12 c^3 (c-i c \tan (e+f x))^3}-\frac {1}{16 c^2 (c-i c \tan (e+f x))^4}\right )}{a f}\) |
(I*c^2*((((-5*I)/32)*ArcTan[Tan[e + f*x]])/c^6 - 1/(16*c^2*(c - I*c*Tan[e + f*x])^4) - 1/(12*c^3*(c - I*c*Tan[e + f*x])^3) - 3/(32*c^4*(c - I*c*Tan[ e + f*x])^2) - 1/(8*c^5*(c - I*c*Tan[e + f*x])) + 1/(32*c^5*(c + I*c*Tan[e + f*x]))))/(a*f)
3.10.53.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[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 = 0.31 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.71
method | result | size |
risch | \(\frac {5 x}{32 a \,c^{4}}-\frac {i {\mathrm e}^{8 i \left (f x +e \right )}}{256 a \,c^{4} f}-\frac {5 i {\mathrm e}^{6 i \left (f x +e \right )}}{192 a \,c^{4} f}-\frac {5 i {\mathrm e}^{4 i \left (f x +e \right )}}{64 a \,c^{4} f}-\frac {9 i \cos \left (2 f x +2 e \right )}{64 a \,c^{4} f}+\frac {11 \sin \left (2 f x +2 e \right )}{64 a \,c^{4} f}\) | \(115\) |
derivativedivides | \(\frac {5 \arctan \left (\tan \left (f x +e \right )\right )}{32 f a \,c^{4}}+\frac {1}{32 f a \,c^{4} \left (\tan \left (f x +e \right )-i\right )}+\frac {3 i}{32 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {i}{16 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{4}}-\frac {1}{12 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {1}{8 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )}\) | \(132\) |
default | \(\frac {5 \arctan \left (\tan \left (f x +e \right )\right )}{32 f a \,c^{4}}+\frac {1}{32 f a \,c^{4} \left (\tan \left (f x +e \right )-i\right )}+\frac {3 i}{32 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {i}{16 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{4}}-\frac {1}{12 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{3}}+\frac {1}{8 f a \,c^{4} \left (\tan \left (f x +e \right )+i\right )}\) | \(132\) |
norman | \(\frac {\frac {5 x}{32 a c}+\frac {27 \tan \left (f x +e \right )}{32 a c f}+\frac {73 \left (\tan ^{3}\left (f x +e \right )\right )}{96 a c f}+\frac {55 \left (\tan ^{5}\left (f x +e \right )\right )}{96 a c f}+\frac {5 \left (\tan ^{7}\left (f x +e \right )\right )}{32 a c f}+\frac {5 x \left (\tan ^{2}\left (f x +e \right )\right )}{8 a c}+\frac {15 x \left (\tan ^{4}\left (f x +e \right )\right )}{16 a c}+\frac {5 x \left (\tan ^{6}\left (f x +e \right )\right )}{8 a c}+\frac {5 x \left (\tan ^{8}\left (f x +e \right )\right )}{32 a c}-\frac {i}{3 a c f}+\frac {i \left (\tan ^{2}\left (f x +e \right )\right )}{6 a c f}}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{4} c^{3}}\) | \(201\) |
5/32*x/a/c^4-1/256*I/a/c^4/f*exp(8*I*(f*x+e))-5/192*I/a/c^4/f*exp(6*I*(f*x +e))-5/64*I/a/c^4/f*exp(4*I*(f*x+e))-9/64*I/a/c^4/f*cos(2*f*x+2*e)+11/64/a /c^4/f*sin(2*f*x+2*e)
Time = 0.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\frac {{\left (120 \, f x e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 20 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 60 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 120 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{768 \, a c^{4} f} \]
1/768*(120*f*x*e^(2*I*f*x + 2*I*e) - 3*I*e^(10*I*f*x + 10*I*e) - 20*I*e^(8 *I*f*x + 8*I*e) - 60*I*e^(6*I*f*x + 6*I*e) - 120*I*e^(4*I*f*x + 4*I*e) + 1 2*I)*e^(-2*I*f*x - 2*I*e)/(a*c^4*f)
Time = 0.26 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.52 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\begin {cases} \frac {\left (- 25165824 i a^{4} c^{16} f^{4} e^{10 i e} e^{8 i f x} - 167772160 i a^{4} c^{16} f^{4} e^{8 i e} e^{6 i f x} - 503316480 i a^{4} c^{16} f^{4} e^{6 i e} e^{4 i f x} - 1006632960 i a^{4} c^{16} f^{4} e^{4 i e} e^{2 i f x} + 100663296 i a^{4} c^{16} f^{4} e^{- 2 i f x}\right ) e^{- 2 i e}}{6442450944 a^{5} c^{20} f^{5}} & \text {for}\: a^{5} c^{20} f^{5} e^{2 i e} \neq 0 \\x \left (\frac {\left (e^{10 i e} + 5 e^{8 i e} + 10 e^{6 i e} + 10 e^{4 i e} + 5 e^{2 i e} + 1\right ) e^{- 2 i e}}{32 a c^{4}} - \frac {5}{32 a c^{4}}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{32 a c^{4}} \]
Piecewise(((-25165824*I*a**4*c**16*f**4*exp(10*I*e)*exp(8*I*f*x) - 1677721 60*I*a**4*c**16*f**4*exp(8*I*e)*exp(6*I*f*x) - 503316480*I*a**4*c**16*f**4 *exp(6*I*e)*exp(4*I*f*x) - 1006632960*I*a**4*c**16*f**4*exp(4*I*e)*exp(2*I *f*x) + 100663296*I*a**4*c**16*f**4*exp(-2*I*f*x))*exp(-2*I*e)/(6442450944 *a**5*c**20*f**5), Ne(a**5*c**20*f**5*exp(2*I*e), 0)), (x*((exp(10*I*e) + 5*exp(8*I*e) + 10*exp(6*I*e) + 10*exp(4*I*e) + 5*exp(2*I*e) + 1)*exp(-2*I* e)/(32*a*c**4) - 5/(32*a*c**4)), True)) + 5*x/(32*a*c**4)
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.54 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=-\frac {-\frac {60 i \, \log \left (\tan \left (f x + e\right ) + i\right )}{a c^{4}} + \frac {60 i \, \log \left (\tan \left (f x + e\right ) - i\right )}{a c^{4}} - \frac {12 \, {\left (5 \, \tan \left (f x + e\right ) - 7 i\right )}}{a c^{4} {\left (-i \, \tan \left (f x + e\right ) - 1\right )}} + \frac {125 i \, \tan \left (f x + e\right )^{4} - 596 \, \tan \left (f x + e\right )^{3} - 1110 i \, \tan \left (f x + e\right )^{2} + 996 \, \tan \left (f x + e\right ) + 405 i}{a c^{4} {\left (\tan \left (f x + e\right ) + i\right )}^{4}}}{768 \, f} \]
-1/768*(-60*I*log(tan(f*x + e) + I)/(a*c^4) + 60*I*log(tan(f*x + e) - I)/( a*c^4) - 12*(5*tan(f*x + e) - 7*I)/(a*c^4*(-I*tan(f*x + e) - 1)) + (125*I* tan(f*x + e)^4 - 596*tan(f*x + e)^3 - 1110*I*tan(f*x + e)^2 + 996*tan(f*x + e) + 405*I)/(a*c^4*(tan(f*x + e) + I)^4))/f
Time = 6.86 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^4} \, dx=\frac {5\,x}{32\,a\,c^4}-\frac {-\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,5{}\mathrm {i}}{32}+\frac {15\,{\mathrm {tan}\left (e+f\,x\right )}^3}{32}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,35{}\mathrm {i}}{96}+\frac {5\,\mathrm {tan}\left (e+f\,x\right )}{32}+\frac {1}{3}{}\mathrm {i}}{a\,c^4\,f\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^4} \]