Integrand size = 22, antiderivative size = 141 \[ \int (e x)^{-1+2 n} \left (a+b \csc \left (c+d x^n\right )\right ) \, dx=\frac {a (e x)^{2 n}}{2 e n}-\frac {2 b x^{-n} (e x)^{2 n} \text {arctanh}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {i b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {i b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,e^{i \left (c+d x^n\right )}\right )}{d^2 e n} \]
1/2*a*(e*x)^(2*n)/e/n-2*b*(e*x)^(2*n)*arctanh(exp(I*(c+d*x^n)))/d/e/n/(x^n )+I*b*(e*x)^(2*n)*polylog(2,-exp(I*(c+d*x^n)))/d^2/e/n/(x^(2*n))-I*b*(e*x) ^(2*n)*polylog(2,exp(I*(c+d*x^n)))/d^2/e/n/(x^(2*n))
Time = 0.49 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.31 \[ \int (e x)^{-1+2 n} \left (a+b \csc \left (c+d x^n\right )\right ) \, dx=\frac {x^{-2 n} (e x)^{2 n} \left (a d^2 x^{2 n}+2 b c \log \left (1-e^{i \left (c+d x^n\right )}\right )+2 b d x^n \log \left (1-e^{i \left (c+d x^n\right )}\right )-2 b c \log \left (1+e^{i \left (c+d x^n\right )}\right )-2 b d x^n \log \left (1+e^{i \left (c+d x^n\right )}\right )-2 b c \log \left (\tan \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )+2 i b \operatorname {PolyLog}\left (2,-e^{i \left (c+d x^n\right )}\right )-2 i b \operatorname {PolyLog}\left (2,e^{i \left (c+d x^n\right )}\right )\right )}{2 d^2 e n} \]
((e*x)^(2*n)*(a*d^2*x^(2*n) + 2*b*c*Log[1 - E^(I*(c + d*x^n))] + 2*b*d*x^n *Log[1 - E^(I*(c + d*x^n))] - 2*b*c*Log[1 + E^(I*(c + d*x^n))] - 2*b*d*x^n *Log[1 + E^(I*(c + d*x^n))] - 2*b*c*Log[Tan[(c + d*x^n)/2]] + (2*I)*b*Poly Log[2, -E^(I*(c + d*x^n))] - (2*I)*b*PolyLog[2, E^(I*(c + d*x^n))]))/(2*d^ 2*e*n*x^(2*n))
Time = 0.31 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2010, 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.
\(\displaystyle \int (e x)^{2 n-1} \left (a+b \csc \left (c+d x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (a (e x)^{2 n-1}+b (e x)^{2 n-1} \csc \left (c+d x^n\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a (e x)^{2 n}}{2 e n}-\frac {2 b x^{-n} (e x)^{2 n} \text {arctanh}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {i b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-e^{i \left (d x^n+c\right )}\right )}{d^2 e n}-\frac {i b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,e^{i \left (d x^n+c\right )}\right )}{d^2 e n}\) |
(a*(e*x)^(2*n))/(2*e*n) - (2*b*(e*x)^(2*n)*ArcTanh[E^(I*(c + d*x^n))])/(d* e*n*x^n) + (I*b*(e*x)^(2*n)*PolyLog[2, -E^(I*(c + d*x^n))])/(d^2*e*n*x^(2* n)) - (I*b*(e*x)^(2*n)*PolyLog[2, E^(I*(c + d*x^n))])/(d^2*e*n*x^(2*n))
3.1.74.3.1 Defintions of rubi rules used
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.17 (sec) , antiderivative size = 699, normalized size of antiderivative = 4.96
1/2*a/n*x*exp(1/2*(-1+2*n)*(-I*csgn(I*e)*csgn(I*x)*csgn(I*e*x)*Pi+I*csgn(I *e)*csgn(I*e*x)^2*Pi+I*csgn(I*x)*csgn(I*e*x)^2*Pi-I*csgn(I*e*x)^3*Pi+2*ln( e)+2*ln(x)))+1/d/n/e*(e^n)^2*b*ln(1-exp(I*(c+d*x^n)))*x^n*(-1)^(1/2*csgn(I *e)*csgn(I*x)*csgn(I*e*x))*exp(1/2*I*Pi*csgn(I*e*x)*(-2*csgn(I*e*x)^2*n+2* csgn(I*e)*csgn(I*e*x)*n+2*csgn(I*x)*csgn(I*e*x)*n-2*csgn(I*e)*csgn(I*x)*n+ csgn(I*e*x)^2-csgn(I*e)*csgn(I*e*x)-csgn(I*x)*csgn(I*e*x)))-1/d/n/e*(e^n)^ 2*b*ln(exp(I*(c+d*x^n))+1)*x^n*(-1)^(1/2*csgn(I*e)*csgn(I*x)*csgn(I*e*x))* exp(1/2*I*Pi*csgn(I*e*x)*(-2*csgn(I*e*x)^2*n+2*csgn(I*e)*csgn(I*e*x)*n+2*c sgn(I*x)*csgn(I*e*x)*n-2*csgn(I*e)*csgn(I*x)*n+csgn(I*e*x)^2-csgn(I*e)*csg n(I*e*x)-csgn(I*x)*csgn(I*e*x)))-I/d^2/n/e*(e^n)^2*b*dilog(1-exp(I*(c+d*x^ n)))*(-1)^(1/2*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*exp(1/2*I*Pi*csgn(I*e*x)*( -2*csgn(I*e*x)^2*n+2*csgn(I*e)*csgn(I*e*x)*n+2*csgn(I*x)*csgn(I*e*x)*n-2*c sgn(I*e)*csgn(I*x)*n+csgn(I*e*x)^2-csgn(I*e)*csgn(I*e*x)-csgn(I*x)*csgn(I* e*x)))+I/d^2/n/e*(e^n)^2*b*dilog(exp(I*(c+d*x^n))+1)*(-1)^(1/2*csgn(I*e)*c sgn(I*x)*csgn(I*e*x))*exp(1/2*I*Pi*csgn(I*e*x)*(-2*csgn(I*e*x)^2*n+2*csgn( I*e)*csgn(I*e*x)*n+2*csgn(I*x)*csgn(I*e*x)*n-2*csgn(I*e)*csgn(I*x)*n+csgn( I*e*x)^2-csgn(I*e)*csgn(I*e*x)-csgn(I*x)*csgn(I*e*x)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (131) = 262\).
Time = 0.27 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.72 \[ \int (e x)^{-1+2 n} \left (a+b \csc \left (c+d x^n\right )\right ) \, dx=\frac {a d^{2} e^{2 \, n - 1} x^{2 \, n} - b d e^{2 \, n - 1} x^{n} \log \left (\cos \left (d x^{n} + c\right ) + i \, \sin \left (d x^{n} + c\right ) + 1\right ) - b d e^{2 \, n - 1} x^{n} \log \left (\cos \left (d x^{n} + c\right ) - i \, \sin \left (d x^{n} + c\right ) + 1\right ) - b c e^{2 \, n - 1} \log \left (-\frac {1}{2} \, \cos \left (d x^{n} + c\right ) + \frac {1}{2} i \, \sin \left (d x^{n} + c\right ) + \frac {1}{2}\right ) - b c e^{2 \, n - 1} \log \left (-\frac {1}{2} \, \cos \left (d x^{n} + c\right ) - \frac {1}{2} i \, \sin \left (d x^{n} + c\right ) + \frac {1}{2}\right ) - i \, b e^{2 \, n - 1} {\rm Li}_2\left (\cos \left (d x^{n} + c\right ) + i \, \sin \left (d x^{n} + c\right )\right ) + i \, b e^{2 \, n - 1} {\rm Li}_2\left (\cos \left (d x^{n} + c\right ) - i \, \sin \left (d x^{n} + c\right )\right ) - i \, b e^{2 \, n - 1} {\rm Li}_2\left (-\cos \left (d x^{n} + c\right ) + i \, \sin \left (d x^{n} + c\right )\right ) + i \, b e^{2 \, n - 1} {\rm Li}_2\left (-\cos \left (d x^{n} + c\right ) - i \, \sin \left (d x^{n} + c\right )\right ) + {\left (b d e^{2 \, n - 1} x^{n} + b c e^{2 \, n - 1}\right )} \log \left (-\cos \left (d x^{n} + c\right ) + i \, \sin \left (d x^{n} + c\right ) + 1\right ) + {\left (b d e^{2 \, n - 1} x^{n} + b c e^{2 \, n - 1}\right )} \log \left (-\cos \left (d x^{n} + c\right ) - i \, \sin \left (d x^{n} + c\right ) + 1\right )}{2 \, d^{2} n} \]
1/2*(a*d^2*e^(2*n - 1)*x^(2*n) - b*d*e^(2*n - 1)*x^n*log(cos(d*x^n + c) + I*sin(d*x^n + c) + 1) - b*d*e^(2*n - 1)*x^n*log(cos(d*x^n + c) - I*sin(d*x ^n + c) + 1) - b*c*e^(2*n - 1)*log(-1/2*cos(d*x^n + c) + 1/2*I*sin(d*x^n + c) + 1/2) - b*c*e^(2*n - 1)*log(-1/2*cos(d*x^n + c) - 1/2*I*sin(d*x^n + c ) + 1/2) - I*b*e^(2*n - 1)*dilog(cos(d*x^n + c) + I*sin(d*x^n + c)) + I*b* e^(2*n - 1)*dilog(cos(d*x^n + c) - I*sin(d*x^n + c)) - I*b*e^(2*n - 1)*dil og(-cos(d*x^n + c) + I*sin(d*x^n + c)) + I*b*e^(2*n - 1)*dilog(-cos(d*x^n + c) - I*sin(d*x^n + c)) + (b*d*e^(2*n - 1)*x^n + b*c*e^(2*n - 1))*log(-co s(d*x^n + c) + I*sin(d*x^n + c) + 1) + (b*d*e^(2*n - 1)*x^n + b*c*e^(2*n - 1))*log(-cos(d*x^n + c) - I*sin(d*x^n + c) + 1))/(d^2*n)
\[ \int (e x)^{-1+2 n} \left (a+b \csc \left (c+d x^n\right )\right ) \, dx=\int \left (e x\right )^{2 n - 1} \left (a + b \csc {\left (c + d x^{n} \right )}\right )\, dx \]
\[ \int (e x)^{-1+2 n} \left (a+b \csc \left (c+d x^n\right )\right ) \, dx=\int { {\left (b \csc \left (d x^{n} + c\right ) + a\right )} \left (e x\right )^{2 \, n - 1} \,d x } \]
(e^(2*n + 1)*integrate(x^(2*n)*sin(d*x^n + c)/(e^2*x*cos(d*x^n + c)^2 + e^ 2*x*sin(d*x^n + c)^2 + 2*e^2*x*cos(d*x^n + c) + e^2*x), x) + e^(2*n + 1)*i ntegrate(x^(2*n)*sin(d*x^n + c)/(e^2*x*cos(d*x^n + c)^2 + e^2*x*sin(d*x^n + c)^2 - 2*e^2*x*cos(d*x^n + c) + e^2*x), x))*b + 1/2*(e*x)^(2*n)*a/(e*n)
\[ \int (e x)^{-1+2 n} \left (a+b \csc \left (c+d x^n\right )\right ) \, dx=\int { {\left (b \csc \left (d x^{n} + c\right ) + a\right )} \left (e x\right )^{2 \, n - 1} \,d x } \]
Timed out. \[ \int (e x)^{-1+2 n} \left (a+b \csc \left (c+d x^n\right )\right ) \, dx=\int \left (a+\frac {b}{\sin \left (c+d\,x^n\right )}\right )\,{\left (e\,x\right )}^{2\,n-1} \,d x \]