Integrand size = 27, antiderivative size = 278 \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\frac {b d n x}{2 e g}-\frac {b n x^2}{4 g}-\frac {b d^2 n \log (d+e x)}{2 e^2 g}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g^2}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^2}-\frac {b f n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^2}-\frac {b f n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g^2} \]
1/2*b*d*n*x/e/g-1/4*b*n*x^2/g-1/2*b*d^2*n*ln(e*x+d)/e^2/g+1/2*x^2*(a+b*ln( c*(e*x+d)^n))/g-1/2*f*(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)))/g^2-1/2*f*(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)))/g^2-1/2*b*f*n*polylog(2,-(e*x+d)*g^(1/ 2)/(e*(-f)^(1/2)-d*g^(1/2)))/g^2-1/2*b*f*n*polylog(2,(e*x+d)*g^(1/2)/(e*(- f)^(1/2)+d*g^(1/2)))/g^2
Time = 0.11 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.87 \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=-\frac {\frac {b g n \left (e x (-2 d+e x)+2 d^2 \log (d+e x)\right )}{e^2}-2 g x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )+2 f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )+2 f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )+2 b f n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+2 b f n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{4 g^2} \]
-1/4*((b*g*n*(e*x*(-2*d + e*x) + 2*d^2*Log[d + e*x]))/e^2 - 2*g*x^2*(a + b *Log[c*(d + e*x)^n]) + 2*f*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - S qrt[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*b*f*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))] + 2*b*f*n*PolyLog[2, (Sqr t[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/g^2
Time = 0.56 (sec) , antiderivative size = 278, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \int \left (\frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac {f x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \left (f+g x^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {f \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {f \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {b d^2 n \log (d+e x)}{2 e^2 g}-\frac {b f n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^2}-\frac {b f n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 g^2}+\frac {b d n x}{2 e g}-\frac {b n x^2}{4 g}\) |
(b*d*n*x)/(2*e*g) - (b*n*x^2)/(4*g) - (b*d^2*n*Log[d + e*x])/(2*e^2*g) + ( x^2*(a + b*Log[c*(d + e*x)^n]))/(2*g) - (f*(a + b*Log[c*(d + e*x)^n])*Log[ (e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*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*g^2) - (b*f*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt [g]))])/(2*g^2) - (b*f*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sq rt[g])])/(2*g^2)
3.3.57.3.1 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 = 0.94 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.55
| method | result | size |
| risch | \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x^{2}}{2 g}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f \ln \left (g \,x^{2}+f \right )}{2 g^{2}}-\frac {b n \,x^{2}}{4 g}+\frac {b d n x}{2 e g}-\frac {b \,d^{2} n \ln \left (e x +d \right )}{2 e^{2} g}+\frac {b n f \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{2 g^{2}}-\frac {b n f \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 g^{2}}-\frac {b n f \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 g^{2}}-\frac {b n f \operatorname {dilog}\left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 g^{2}}-\frac {b n f \operatorname {dilog}\left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 g^{2}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b}{2}+b \ln \left (c \right )+a \right ) \left (\frac {x^{2}}{2 g}-\frac {f \ln \left (g \,x^{2}+f \right )}{2 g^{2}}\right )\) | \(431\) |
1/2*b*ln((e*x+d)^n)/g*x^2-1/2*b*ln((e*x+d)^n)*f/g^2*ln(g*x^2+f)-1/4*b*n*x^ 2/g+1/2*b*d*n*x/e/g-1/2*b*d^2*n*ln(e*x+d)/e^2/g+1/2*b*n*f/g^2*ln(e*x+d)*ln (g*x^2+f)-1/2*b*n*f/g^2*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/g^2*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/g^2*dilog((e*(-f*g)^(1/2)-g*(e*x+d)+d*g)/( e*(-f*g)^(1/2)+d*g))-1/2*b*n*f/g^2*dilog((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e *(-f*g)^(1/2)-d*g))+(-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*x^2/g-1/2*f/g^2*ln(g*x^2+f))
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3}}{g x^{2} + f} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3}}{g x^{2} + f} \,d x } \]
1/2*a*(x^2/g - f*log(g*x^2 + f)/g^2) + b*integrate((x^3*log((e*x + d)^n) + x^3*log(c))/(g*x^2 + f), x)
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3}}{g x^{2} + f} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{g\,x^2+f} \,d x \]