Integrand size = 23, antiderivative size = 82 \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^2} \, dx=\frac {a+b \log \left (c x^n\right )}{2 d \left (d+e x^2\right )}-\frac {\log \left (1+\frac {d}{e x^2}\right ) \left (2 a-b n+2 b \log \left (c x^n\right )\right )}{4 d^2}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{4 d^2} \]
1/2*(a+b*ln(c*x^n))/d/(e*x^2+d)-1/4*ln(1+d/e/x^2)*(2*a-b*n+2*b*ln(c*x^n))/ d^2+1/4*b*n*polylog(2,-d/e/x^2)/d^2
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 279, normalized size of antiderivative = 3.40 \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^2} \, dx=\frac {a-b n \log (x)+b \log \left (c x^n\right )}{2 d^2+2 d e x^2}+\frac {\log (x) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{d^2}-\frac {\left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+e x^2\right )}{2 d^2}+\frac {b n \left (\frac {\sqrt {e} x \log (x)}{i \sqrt {d}-\sqrt {e} x}-\frac {\sqrt {e} x \log (x)}{i \sqrt {d}+\sqrt {e} x}+2 \log ^2(x)+\log \left (i \sqrt {d}-\sqrt {e} x\right )+\log \left (i \sqrt {d}+\sqrt {e} x\right )-2 \left (\log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )-2 \left (\log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )\right )}{4 d^2} \]
(a - b*n*Log[x] + b*Log[c*x^n])/(2*d^2 + 2*d*e*x^2) + (Log[x]*(a - b*n*Log [x] + b*Log[c*x^n]))/d^2 - ((a - b*n*Log[x] + b*Log[c*x^n])*Log[d + e*x^2] )/(2*d^2) + (b*n*((Sqrt[e]*x*Log[x])/(I*Sqrt[d] - Sqrt[e]*x) - (Sqrt[e]*x* Log[x])/(I*Sqrt[d] + Sqrt[e]*x) + 2*Log[x]^2 + Log[I*Sqrt[d] - Sqrt[e]*x] + Log[I*Sqrt[d] + Sqrt[e]*x] - 2*(Log[x]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]] + PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]]) - 2*(Log[x]*Log[1 - (I*Sqrt[e]*x)/Sq rt[d]] + PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]])))/(4*d^2)
Time = 0.36 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2785, 25, 2779, 2838}
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.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 2785 |
\(\displaystyle \frac {a+b \log \left (c x^n\right )}{2 d \left (d+e x^2\right )}-\frac {\int -\frac {2 a-b n+2 b \log \left (c x^n\right )}{x \left (e x^2+d\right )}dx}{2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {2 a-b n+2 b \log \left (c x^n\right )}{x \left (e x^2+d\right )}dx}{2 d}+\frac {a+b \log \left (c x^n\right )}{2 d \left (d+e x^2\right )}\) |
\(\Big \downarrow \) 2779 |
\(\displaystyle \frac {\frac {b n \int \frac {\log \left (\frac {d}{e x^2}+1\right )}{x}dx}{d}-\frac {\log \left (\frac {d}{e x^2}+1\right ) \left (2 a+2 b \log \left (c x^n\right )-b n\right )}{2 d}}{2 d}+\frac {a+b \log \left (c x^n\right )}{2 d \left (d+e x^2\right )}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{2 d}-\frac {\log \left (\frac {d}{e x^2}+1\right ) \left (2 a+2 b \log \left (c x^n\right )-b n\right )}{2 d}}{2 d}+\frac {a+b \log \left (c x^n\right )}{2 d \left (d+e x^2\right )}\) |
(a + b*Log[c*x^n])/(2*d*(d + e*x^2)) + (-1/2*(Log[1 + d/(e*x^2)]*(2*a - b* n + 2*b*Log[c*x^n]))/d + (b*n*PolyLog[2, -(d/(e*x^2))])/(2*d))/(2*d)
3.3.24.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r _.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) , x] + Simp[b*n*(p/(d*r)) Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^2)^(q_.), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(d + e*x^2)^(q + 1)*((a + b*Log[c*x^n])/(2*d*f*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(f*x)^m*(d + e*x^2)^(q + 1)*(a*(m + 2*q + 3) + b*n + b*(m + 2*q + 3)*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && ILtQ[q, -1] && ILtQ[m, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.59 (sec) , antiderivative size = 338, normalized size of antiderivative = 4.12
method | result | size |
risch | \(\frac {b \ln \left (x^{n}\right )}{2 d \left (e \,x^{2}+d \right )}-\frac {b \ln \left (x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{2 d^{2}}+\frac {b \ln \left (x^{n}\right ) \ln \left (x \right )}{d^{2}}+\frac {b n \ln \left (e \,x^{2}+d \right )}{4 d^{2}}-\frac {b n \ln \left (x \right )}{2 d^{2}}-\frac {b n \ln \left (x \right )^{2}}{2 d^{2}}+\frac {b n \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 d^{2}}-\frac {b n \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2}}-\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2}}-\frac {b n \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2}}-\frac {b n \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d^{2}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {e \left (-\frac {d}{e \left (e \,x^{2}+d \right )}+\frac {\ln \left (e \,x^{2}+d \right )}{e}\right )}{2 d^{2}}+\frac {\ln \left (x \right )}{d^{2}}\right )\) | \(338\) |
1/2*b*ln(x^n)/d/(e*x^2+d)-1/2*b*ln(x^n)/d^2*ln(e*x^2+d)+b*ln(x^n)/d^2*ln(x )+1/4*b*n/d^2*ln(e*x^2+d)-1/2*b*n/d^2*ln(x)-1/2*b*n/d^2*ln(x)^2+1/2*b*n/d^ 2*ln(x)*ln(e*x^2+d)-1/2*b*n/d^2*ln(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2)) -1/2*b*n/d^2*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n/d^2*dilog(( -e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n/d^2*dilog((e*x+(-d*e)^(1/2))/(-d* e)^(1/2))+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2*I*b*Pi*csgn (I*c)*csgn(I*c*x^n)^2+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*b*Pi*cs gn(I*c*x^n)^3+b*ln(c)+a)*(-1/2*e/d^2*(-d/e/(e*x^2+d)+1/e*ln(e*x^2+d))+1/d^ 2*ln(x))
\[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{2} x} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{2} x} \,d x } \]
1/2*a*(1/(d*e*x^2 + d^2) - log(e*x^2 + d)/d^2 + 2*log(x)/d^2) + b*integrat e((log(c) + log(x^n))/(e^2*x^5 + 2*d*e*x^3 + d^2*x), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{2} x} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^2} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x\,{\left (e\,x^2+d\right )}^2} \,d x \]