Integrand size = 27, antiderivative size = 331 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )} \, dx=-\frac {b e n}{2 d f x}-\frac {b e^2 n \log (x)}{2 d^2 f}+\frac {b e^2 n \log (d+e x)}{2 d^2 f}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f x^2}-\frac {g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f^2}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}+\frac {b g n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}+\frac {b g n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f^2}-\frac {b g n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^2} \]
-1/2*b*e*n/d/f/x-1/2*b*e^2*n*ln(x)/d^2/f+1/2*b*e^2*n*ln(e*x+d)/d^2/f+1/2*( -a-b*ln(c*(e*x+d)^n))/f/x^2-g*ln(-e*x/d)*(a+b*ln(c*(e*x+d)^n))/f^2+1/2*g*( a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1/2)-x*g^(1/2))/(e*(-f)^(1/2)+d*g^(1/2))) /f^2+1/2*g*(a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1/2)+x*g^(1/2))/(e*(-f)^(1/2) -d*g^(1/2)))/f^2-b*g*n*polylog(2,1+e*x/d)/f^2+1/2*b*g*n*polylog(2,-(e*x+d) *g^(1/2)/(e*(-f)^(1/2)-d*g^(1/2)))/f^2+1/2*b*g*n*polylog(2,(e*x+d)*g^(1/2) /(e*(-f)^(1/2)+d*g^(1/2)))/f^2
Time = 0.11 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.84 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )} \, dx=\frac {-\frac {b e f n (d+e x \log (x)-e x \log (d+e x))}{d^2 x}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^2}-2 g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )+g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )+b g n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+b g n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )-2 b g n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{2 f^2} \]
(-((b*e*f*n*(d + e*x*Log[x] - e*x*Log[d + e*x]))/(d^2*x)) - (f*(a + b*Log[ c*(d + e*x)^n]))/x^2 - 2*g*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]) + g* (a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d* Sqrt[g])] + g*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e *Sqrt[-f] - d*Sqrt[g])] + b*g*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[- f] - d*Sqrt[g]))] + b*g*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*S qrt[g])] - 2*b*g*n*PolyLog[2, 1 + (e*x)/d])/(2*f^2)
Time = 0.62 (sec) , antiderivative size = 331, 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.
| \(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \int \left (\frac {g^2 x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 \left (f+g x^2\right )}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {a+b \log \left (c (d+e x)^n\right )}{f x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {g \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}+\frac {g \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f x^2}-\frac {b e^2 n \log (x)}{2 d^2 f}+\frac {b e^2 n \log (d+e x)}{2 d^2 f}+\frac {b g n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}+\frac {b g n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 f^2}-\frac {b g n \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^2}-\frac {b e n}{2 d f x}\) |
-1/2*(b*e*n)/(d*f*x) - (b*e^2*n*Log[x])/(2*d^2*f) + (b*e^2*n*Log[d + e*x]) /(2*d^2*f) - (a + b*Log[c*(d + e*x)^n])/(2*f*x^2) - (g*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f^2 + (g*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[ -f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*f^2) + (g*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/(2*f^ 2) + (b*g*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/( 2*f^2) + (b*g*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/ (2*f^2) - (b*g*n*PolyLog[2, 1 + (e*x)/d])/f^2
3.3.60.3.1 Defintions of rubi rules used
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 = 1.20 (sec) , antiderivative size = 499, normalized size of antiderivative = 1.51
| method | result | size |
| risch | \(-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{2 f \,x^{2}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g \ln \left (x \right )}{f^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g \ln \left (g \,x^{2}+f \right )}{2 f^{2}}-\frac {b n g \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{2 f^{2}}+\frac {b n g \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 f^{2}}+\frac {b n g \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 f^{2}}+\frac {b n g \operatorname {dilog}\left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 f^{2}}+\frac {b n g \operatorname {dilog}\left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 f^{2}}+\frac {b \,e^{2} n \ln \left (e x +d \right )}{2 d^{2} f}-\frac {b e n}{2 d f x}-\frac {b \,e^{2} n \ln \left (x \right )}{2 d^{2} f}+\frac {b n g \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f^{2}}+\frac {b n g \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{2}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {1}{2 f \,x^{2}}-\frac {g \ln \left (x \right )}{f^{2}}+\frac {g \ln \left (g \,x^{2}+f \right )}{2 f^{2}}\right )\) | \(499\) |
-1/2*b*ln((e*x+d)^n)/f/x^2-b*ln((e*x+d)^n)/f^2*g*ln(x)+1/2*b*ln((e*x+d)^n) *g/f^2*ln(g*x^2+f)-1/2*b*n/f^2*g*ln(e*x+d)*ln(g*x^2+f)+1/2*b*n/f^2*g*ln(e* x+d)*ln((e*(-f*g)^(1/2)-g*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+d*g))+1/2*b*n/f^2*g *ln(e*x+d)*ln((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e*(-f*g)^(1/2)-d*g))+1/2*b*n /f^2*g*dilog((e*(-f*g)^(1/2)-g*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+d*g))+1/2*b*n/ f^2*g*dilog((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e*(-f*g)^(1/2)-d*g))+1/2*b*e^2 *n*ln(e*x+d)/d^2/f-1/2*b*e*n/d/f/x-1/2*b*e^2*n*ln(x)/d^2/f+b*n/f^2*g*dilog ((e*x+d)/d)+b*n/f^2*g*ln(x)*ln((e*x+d)/d)+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*(e *x+d)^n)*csgn(I*c*(e*x+d)^n)+1/2*I*b*Pi*csgn(I*c)*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)^2-1/2*I*b*Pi*csgn(I*c*(e*x+ d)^n)^3+b*ln(c)+a)*(-1/2/f/x^2-1/f^2*g*ln(x)+1/2*g/f^2*ln(g*x^2+f))
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{3}} \,d x } \]
1/2*a*(g*log(g*x^2 + f)/f^2 - 2*g*log(x)/f^2 - 1/(f*x^2)) + b*integrate((l og((e*x + d)^n) + log(c))/(g*x^5 + f*x^3), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^3\,\left (g\,x^2+f\right )} \,d x \]