Integrand size = 23, antiderivative size = 49 \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )} \, dx=-\frac {\log \left (1+\frac {d}{e x^2}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{4 d} \]
Leaf count is larger than twice the leaf count of optimal. \(126\) vs. \(2(49)=98\).
Time = 0.06 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.57 \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )} \, dx=-\frac {-\left (\left (a+b \log \left (c x^n\right )\right ) \left (a+b \log \left (c x^n\right )-b n \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )-b n \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )\right )\right )+b^2 n^2 \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )+b^2 n^2 \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{2 b d n} \]
-1/2*(-((a + b*Log[c*x^n])*(a + b*Log[c*x^n] - b*n*Log[1 + (Sqrt[e]*x)/Sqr t[-d]] - b*n*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)])) + b^2*n^2*PolyLog[2, (Sqr t[e]*x)/Sqrt[-d]] + b^2*n^2*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/(b*d*n)
Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {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.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )} \, dx\) |
\(\Big \downarrow \) 2779 |
\(\displaystyle \frac {b n \int \frac {\log \left (\frac {d}{e x^2}+1\right )}{x}dx}{2 d}-\frac {\log \left (\frac {d}{e x^2}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x^2}\right )}{4 d}-\frac {\log \left (\frac {d}{e x^2}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d}\) |
3.3.13.3.1 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[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 = 0.42 (sec) , antiderivative size = 274, normalized size of antiderivative = 5.59
method | result | size |
risch | \(-\frac {b \ln \left (x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{2 d}+\frac {b \ln \left (x^{n}\right ) \ln \left (x \right )}{d}-\frac {b n \ln \left (x \right )^{2}}{2 d}+\frac {b n \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 d}-\frac {b n \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d}-\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d}-\frac {b n \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d}-\frac {b n \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 d}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {\ln \left (e \,x^{2}+d \right )}{2 d}+\frac {\ln \left (x \right )}{d}\right )\) | \(274\) |
-1/2*b*ln(x^n)/d*ln(e*x^2+d)+b*ln(x^n)/d*ln(x)-1/2*b*n/d*ln(x)^2+1/2*b*n/d *ln(x)*ln(e*x^2+d)-1/2*b*n/d*ln(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/ 2*b*n/d*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n/d*dilog((-e*x+(- d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n/d*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)*csg n(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*csgn(I*c*x^ n)^3+b*ln(c)+a)*(-1/2/d*ln(e*x^2+d)+1/d*ln(x))
\[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x} \,d x } \]
Time = 6.00 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.94 \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )} \, dx=\frac {a \log {\left (x \right )}}{d} - \frac {a \log {\left (d + e x^{2} \right )}}{2 d} + \frac {b n \left (\begin {cases} \frac {\operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x^{2}}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} + \frac {\operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x^{2}}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} + \frac {\operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x^{2}}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + \frac {\operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x^{2}}\right )}{2} & \text {otherwise} \end {cases}\right )}{2 d} - \frac {b \log {\left (c x^{n} \right )} \log {\left (\frac {d}{x^{2}} + e \right )}}{2 d} \]
a*log(x)/d - a*log(d + e*x**2)/(2*d) + b*n*Piecewise((polylog(2, d*exp_pol ar(I*pi)/(e*x**2))/2, (Abs(x) < 1) & (1/Abs(x) < 1)), (log(e)*log(x) + pol ylog(2, d*exp_polar(I*pi)/(e*x**2))/2, Abs(x) < 1), (-log(e)*log(1/x) + po lylog(2, d*exp_polar(I*pi)/(e*x**2))/2, 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(e) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log( e) + polylog(2, d*exp_polar(I*pi)/(e*x**2))/2, True))/(2*d) - b*log(c*x**n )*log(d/x**2 + e)/(2*d)
\[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x} \,d x } \]
\[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x\,\left (e\,x^2+d\right )} \,d x \]