Integrand size = 27, antiderivative size = 397 \[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=-\frac {b d f n x}{2 e g^2}+\frac {b d^3 n x}{4 e^3 g}+\frac {b f n x^2}{4 g^2}-\frac {b d^2 n x^2}{8 e^2 g}+\frac {b d n x^3}{12 e g}-\frac {b n x^4}{16 g}+\frac {b d^2 f n \log (d+e x)}{2 e^2 g^2}-\frac {b d^4 n \log (d+e x)}{4 e^4 g}-\frac {f x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 g}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g^3}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^3}+\frac {b f^2 n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^3}+\frac {b f^2 n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g^3} \]
-1/2*b*d*f*n*x/e/g^2+1/4*b*d^3*n*x/e^3/g+1/4*b*f*n*x^2/g^2-1/8*b*d^2*n*x^2 /e^2/g+1/12*b*d*n*x^3/e/g-1/16*b*n*x^4/g+1/2*b*d^2*f*n*ln(e*x+d)/e^2/g^2-1 /4*b*d^4*n*ln(e*x+d)/e^4/g-1/2*f*x^2*(a+b*ln(c*(e*x+d)^n))/g^2+1/4*x^4*(a+ b*ln(c*(e*x+d)^n))/g+1/2*f^2*(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^3+1/2*f^2*(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^3+1/2*b*f^2*n*polylog(2,-(e *x+d)*g^(1/2)/(e*(-f)^(1/2)-d*g^(1/2)))/g^3+1/2*b*f^2*n*polylog(2,(e*x+d)* g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2)))/g^3
Time = 0.19 (sec) , antiderivative size = 331, normalized size of antiderivative = 0.83 \[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\frac {\frac {12 b f g n \left (e x (-2 d+e x)+2 d^2 \log (d+e x)\right )}{e^2}-\frac {b g^2 n \left (e x \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )+12 d^4 \log (d+e x)\right )}{e^4}-24 f g x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )+12 g^2 x^4 \left (a+b \log \left (c (d+e x)^n\right )\right )+24 f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )+24 f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )+24 b f^2 n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+24 b f^2 n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{48 g^3} \]
((12*b*f*g*n*(e*x*(-2*d + e*x) + 2*d^2*Log[d + e*x]))/e^2 - (b*g^2*n*(e*x* (-12*d^3 + 6*d^2*e*x - 4*d*e^2*x^2 + 3*e^3*x^3) + 12*d^4*Log[d + e*x]))/e^ 4 - 24*f*g*x^2*(a + b*Log[c*(d + e*x)^n]) + 12*g^2*x^4*(a + b*Log[c*(d + e *x)^n]) + 24*f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x)) /(e*Sqrt[-f] + d*Sqrt[g])] + 24*f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqr t[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])] + 24*b*f^2*n*PolyLog[2, -((S qrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))] + 24*b*f^2*n*PolyLog[2, (Sqrt [g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(48*g^3)
Time = 0.73 (sec) , antiderivative size = 397, 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.
\(\displaystyle \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \int \left (\frac {f^2 x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 \left (f+g x^2\right )}-\frac {f x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}+\frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {f^2 \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3}+\frac {f^2 \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3}-\frac {f x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 g}-\frac {b d^4 n \log (d+e x)}{4 e^4 g}+\frac {b d^3 n x}{4 e^3 g}+\frac {b d^2 f n \log (d+e x)}{2 e^2 g^2}-\frac {b d^2 n x^2}{8 e^2 g}+\frac {b f^2 n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^3}+\frac {b f^2 n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 g^3}-\frac {b d f n x}{2 e g^2}+\frac {b d n x^3}{12 e g}+\frac {b f n x^2}{4 g^2}-\frac {b n x^4}{16 g}\) |
-1/2*(b*d*f*n*x)/(e*g^2) + (b*d^3*n*x)/(4*e^3*g) + (b*f*n*x^2)/(4*g^2) - ( b*d^2*n*x^2)/(8*e^2*g) + (b*d*n*x^3)/(12*e*g) - (b*n*x^4)/(16*g) + (b*d^2* f*n*Log[d + e*x])/(2*e^2*g^2) - (b*d^4*n*Log[d + e*x])/(4*e^4*g) - (f*x^2* (a + b*Log[c*(d + e*x)^n]))/(2*g^2) + (x^4*(a + b*Log[c*(d + e*x)^n]))/(4* g) + (f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqr t[-f] + d*Sqrt[g])])/(2*g^3) + (f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqr t[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/(2*g^3) + (b*f^2*n*PolyLog[ 2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(2*g^3) + (b*f^2*n*Po lyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*g^3)
3.3.56.3.1 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 = 1.70 (sec) , antiderivative size = 549, normalized size of antiderivative = 1.38
method | result | size |
risch | \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x^{4}}{4 g}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f \,x^{2}}{2 g^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f^{2} \ln \left (g \,x^{2}+f \right )}{2 g^{3}}-\frac {b n \,f^{2} \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{2 g^{3}}+\frac {b n \,f^{2} \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 g^{3}}+\frac {b n \,f^{2} \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 g^{3}}+\frac {b n \,f^{2} \operatorname {dilog}\left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 g^{3}}+\frac {b n \,f^{2} \operatorname {dilog}\left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 g^{3}}-\frac {b n \,x^{4}}{16 g}+\frac {b d n \,x^{3}}{12 e g}-\frac {b \,d^{2} n \,x^{2}}{8 e^{2} g}+\frac {b f n \,x^{2}}{4 g^{2}}+\frac {b \,d^{3} n x}{4 e^{3} g}-\frac {b d f n x}{2 e \,g^{2}}-\frac {b \,d^{4} n \ln \left (e x +d \right )}{4 e^{4} g}+\frac {b \,d^{2} f n \ln \left (e x +d \right )}{2 e^{2} g^{2}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{2} g \,x^{4}-f \,x^{2}}{2 g^{2}}+\frac {f^{2} \ln \left (g \,x^{2}+f \right )}{2 g^{3}}\right )\) | \(549\) |
1/4*b*ln((e*x+d)^n)/g*x^4-1/2*b*ln((e*x+d)^n)/g^2*f*x^2+1/2*b*ln((e*x+d)^n )*f^2/g^3*ln(g*x^2+f)-1/2*b*n*f^2/g^3*ln(e*x+d)*ln(g*x^2+f)+1/2*b*n*f^2/g^ 3*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^3*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^3*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^3*dilog((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e*(-f*g)^(1/2)-d* g))-1/16*b*n*x^4/g+1/12*b*d*n*x^3/e/g-1/8*b*d^2*n*x^2/e^2/g+1/4*b*f*n*x^2/ g^2+1/4*b*d^3*n*x/e^3/g-1/2*b*d*f*n*x/e/g^2-1/4*b*d^4*n*ln(e*x+d)/e^4/g+1/ 2*b*d^2*f*n*ln(e*x+d)/e^2/g^2+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csg n(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*csg n(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/g^2*(1/2*g*x^4-f*x^2)+1/2*f^2/g^3*ln(g*x^2+f))
\[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{5}}{g x^{2} + f} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\text {Timed out} \]
\[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{5}}{g x^{2} + f} \,d x } \]
1/4*a*(2*f^2*log(g*x^2 + f)/g^3 + (g*x^4 - 2*f*x^2)/g^2) + b*integrate((x^ 5*log((e*x + d)^n) + x^5*log(c))/(g*x^2 + f), x)
\[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{5}}{g x^{2} + f} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{g\,x^2+f} \,d x \]