Integrand size = 27, antiderivative size = 245 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )} \, dx=\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f}-\frac {b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f}-\frac {b n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f}+\frac {b n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f} \] Output:
ln(-e*x/d)*(a+b*ln(c*(e*x+d)^n))/f-1/2*(a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1 /2)-g^(1/2)*x)/(e*(-f)^(1/2)+d*g^(1/2)))/f-1/2*(a+b*ln(c*(e*x+d)^n))*ln(e* ((-f)^(1/2)+g^(1/2)*x)/(e*(-f)^(1/2)-d*g^(1/2)))/f-1/2*b*n*polylog(2,-g^(1 /2)*(e*x+d)/(e*(-f)^(1/2)-d*g^(1/2)))/f-1/2*b*n*polylog(2,g^(1/2)*(e*x+d)/ (e*(-f)^(1/2)+d*g^(1/2)))/f+b*n*polylog(2,1+e*x/d)/f
Time = 0.10 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )} \, dx=-\frac {-2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )+\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )+b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+b n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )-2 b n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{2 f} \] Input:
Integrate[(a + b*Log[c*(d + e*x)^n])/(x*(f + g*x^2)),x]
Output:
-1/2*(-2*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]) + (a + b*Log[c*(d + e* x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])] + (a + b*L og[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g]) ] + b*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))] + b*n* PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])] - 2*b*n*PolyLog[2 , 1 + (e*x)/d])/f
Time = 0.96 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2863, 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.
\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )} \, dx\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \int \left (\frac {a+b \log \left (c (d+e x)^n\right )}{f x}-\frac {g x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f \left (f+g x^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f}-\frac {\log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f}-\frac {b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f}-\frac {b n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 f}+\frac {b n \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f}\) |
Input:
Int[(a + b*Log[c*(d + e*x)^n])/(x*(f + g*x^2)),x]
Output:
(Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f - ((a + b*Log[c*(d + e*x)^n ])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*f) - ((a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt [g])])/(2*f) - (b*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[ g]))])/(2*f) - (b*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g] )])/(2*f) + (b*n*PolyLog[2, 1 + (e*x)/d])/f
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) ^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c , d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.32 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.69
method | result | size |
risch | \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (x \right )}{f}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g \,x^{2}+f \right )}{2 f}-\frac {b n \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f}-\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f}+\frac {b n \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{2 f}-\frac {b n \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-g f}-g \left (e x +d \right )+d g}{e \sqrt {-g f}+d g}\right )}{2 f}-\frac {b n \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-g f}+g \left (e x +d \right )-d g}{e \sqrt {-g f}-d g}\right )}{2 f}-\frac {b n \operatorname {dilog}\left (\frac {e \sqrt {-g f}-g \left (e x +d \right )+d g}{e \sqrt {-g f}+d g}\right )}{2 f}-\frac {b n \operatorname {dilog}\left (\frac {e \sqrt {-g f}+g \left (e x +d \right )-d g}{e \sqrt {-g f}-d g}\right )}{2 f}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\ln \left (x \right )}{f}-\frac {\ln \left (g \,x^{2}+f \right )}{2 f}\right )\) | \(415\) |
Input:
int((a+b*ln(c*(e*x+d)^n))/x/(g*x^2+f),x,method=_RETURNVERBOSE)
Output:
b*ln((e*x+d)^n)/f*ln(x)-1/2*b*ln((e*x+d)^n)/f*ln(g*x^2+f)-b*n/f*dilog((e*x +d)/d)-b*n/f*ln(x)*ln((e*x+d)/d)+1/2*b*n/f*ln(e*x+d)*ln(g*x^2+f)-1/2*b*n/f *ln(e*x+d)*ln((e*(-g*f)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1/2)+d*g))-1/2*b*n /f*ln(e*x+d)*ln((e*(-g*f)^(1/2)+g*(e*x+d)-d*g)/(e*(-g*f)^(1/2)-d*g))-1/2*b *n/f*dilog((e*(-g*f)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1/2)+d*g))-1/2*b*n/f* dilog((e*(-g*f)^(1/2)+g*(e*x+d)-d*g)/(e*(-g*f)^(1/2)-d*g))+(1/2*I*b*Pi*csg n(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c *(e*x+d)^n)*csgn(I*c)-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3+1/2*I*b*Pi*csgn(I*c *(e*x+d)^n)^2*csgn(I*c)+b*ln(c)+a)*(1/f*ln(x)-1/2/f*ln(g*x^2+f))
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x} \,d x } \] Input:
integrate((a+b*log(c*(e*x+d)^n))/x/(g*x^2+f),x, algorithm="fricas")
Output:
integral((b*log((e*x + d)^n*c) + a)/(g*x^3 + f*x), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )} \, dx=\int \frac {a + b \log {\left (c \left (d + e x\right )^{n} \right )}}{x \left (f + g x^{2}\right )}\, dx \] Input:
integrate((a+b*ln(c*(e*x+d)**n))/x/(g*x**2+f),x)
Output:
Integral((a + b*log(c*(d + e*x)**n))/(x*(f + g*x**2)), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x} \,d x } \] Input:
integrate((a+b*log(c*(e*x+d)^n))/x/(g*x^2+f),x, algorithm="maxima")
Output:
-1/2*a*(log(g*x^2 + f)/f - 2*log(x)/f) + b*integrate((log((e*x + d)^n) + l og(c))/(g*x^3 + f*x), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x} \,d x } \] Input:
integrate((a+b*log(c*(e*x+d)^n))/x/(g*x^2+f),x, algorithm="giac")
Output:
integrate((b*log((e*x + d)^n*c) + a)/((g*x^2 + f)*x), x)
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x\,\left (g\,x^2+f\right )} \,d x \] Input:
int((a + b*log(c*(d + e*x)^n))/(x*(f + g*x^2)),x)
Output:
int((a + b*log(c*(d + e*x)^n))/(x*(f + g*x^2)), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )} \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (\left (e x +d \right )^{n} c \right )}{g \,x^{3}+f x}d x \right ) b f -\mathrm {log}\left (g \,x^{2}+f \right ) a +2 \,\mathrm {log}\left (x \right ) a}{2 f} \] Input:
int((a+b*log(c*(e*x+d)^n))/x/(g*x^2+f),x)
Output:
(2*int(log((d + e*x)**n*c)/(f*x + g*x**3),x)*b*f - log(f + g*x**2)*a + 2*l og(x)*a)/(2*f)