Integrand size = 27, antiderivative size = 417 \[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\frac {b d n x}{2 e g^2}-\frac {b n x^2}{4 g^2}+\frac {b d e f^{3/2} n \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 g^{5/2} \left (e^2 f+d^2 g\right )}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}+\frac {b e^2 f^2 n \log (d+e x)}{2 g^3 \left (e^2 f+d^2 g\right )}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{g^3}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{g^3}-\frac {b e^2 f^2 n \log \left (f+g x^2\right )}{4 g^3 \left (e^2 f+d^2 g\right )}-\frac {b f n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{g^3}-\frac {b f n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{g^3} \]
1/2*b*d*n*x/e/g^2-1/4*b*n*x^2/g^2+1/2*b*d*e*f^(3/2)*n*arctan(x*g^(1/2)/f^( 1/2))/g^(5/2)/(d^2*g+e^2*f)-1/2*b*d^2*n*ln(e*x+d)/e^2/g^2+1/2*b*e^2*f^2*n* ln(e*x+d)/g^3/(d^2*g+e^2*f)+1/2*x^2*(a+b*ln(c*(e*x+d)^n))/g^2-1/2*f^2*(a+b *ln(c*(e*x+d)^n))/g^3/(g*x^2+f)-1/4*b*e^2*f^2*n*ln(g*x^2+f)/g^3/(d^2*g+e^2 *f)-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^3-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^3-b*f*n*polylog(2,-(e*x+d)*g^(1/2)/(e*(-f)^(1/2)-d*g^(1/ 2)))/g^3-b*f*n*polylog(2,(e*x+d)*g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2)))/g^3
Result contains complex when optimal does not.
Time = 0.93 (sec) , antiderivative size = 530, normalized size of antiderivative = 1.27 \[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\frac {2 g x^2 \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )-\frac {2 f^2 \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2}-4 f \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right ) \log \left (f+g x^2\right )+b n \left (\frac {g \left (e x (2 d-e x)-2 \left (d^2-e^2 x^2\right ) \log (d+e x)\right )}{e^2}+\frac {f^{3/2} \left (i \sqrt {g} (d+e x) \log (d+e x)-e \left (\sqrt {f}+i \sqrt {g} x\right ) \log \left (i \sqrt {f}-\sqrt {g} x\right )\right )}{\left (e \sqrt {f}-i d \sqrt {g}\right ) \left (\sqrt {f}+i \sqrt {g} x\right )}+\frac {i f^{3/2} \left (-\sqrt {g} (d+e x) \log (d+e x)+e \left (i \sqrt {f}+\sqrt {g} x\right ) \log \left (i \sqrt {f}+\sqrt {g} x\right )\right )}{\left (e \sqrt {f}+i d \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}-4 f \left (\log (d+e x) \log \left (\frac {e \left (\sqrt {f}+i \sqrt {g} x\right )}{e \sqrt {f}-i d \sqrt {g}}\right )+\operatorname {PolyLog}\left (2,-\frac {i \sqrt {g} (d+e x)}{e \sqrt {f}-i d \sqrt {g}}\right )\right )-4 f \left (\log (d+e x) \log \left (\frac {e \left (\sqrt {f}-i \sqrt {g} x\right )}{e \sqrt {f}+i d \sqrt {g}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {g} (d+e x)}{e \sqrt {f}+i d \sqrt {g}}\right )\right )\right )}{4 g^3} \]
(2*g*x^2*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n]) - (2*f^2*(a - b*n*L og[d + e*x] + b*Log[c*(d + e*x)^n]))/(f + g*x^2) - 4*f*(a - b*n*Log[d + e* x] + b*Log[c*(d + e*x)^n])*Log[f + g*x^2] + b*n*((g*(e*x*(2*d - e*x) - 2*( d^2 - e^2*x^2)*Log[d + e*x]))/e^2 + (f^(3/2)*(I*Sqrt[g]*(d + e*x)*Log[d + e*x] - e*(Sqrt[f] + I*Sqrt[g]*x)*Log[I*Sqrt[f] - Sqrt[g]*x]))/((e*Sqrt[f] - I*d*Sqrt[g])*(Sqrt[f] + I*Sqrt[g]*x)) + (I*f^(3/2)*(-(Sqrt[g]*(d + e*x)* Log[d + e*x]) + e*(I*Sqrt[f] + Sqrt[g]*x)*Log[I*Sqrt[f] + Sqrt[g]*x]))/((e *Sqrt[f] + I*d*Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x)) - 4*f*(Log[d + e*x]*Log[( e*(Sqrt[f] + I*Sqrt[g]*x))/(e*Sqrt[f] - I*d*Sqrt[g])] + PolyLog[2, ((-I)*S qrt[g]*(d + e*x))/(e*Sqrt[f] - I*d*Sqrt[g])]) - 4*f*(Log[d + e*x]*Log[(e*( Sqrt[f] - I*Sqrt[g]*x))/(e*Sqrt[f] + I*d*Sqrt[g])] + PolyLog[2, (I*Sqrt[g] *(d + e*x))/(e*Sqrt[f] + I*d*Sqrt[g])])))/(4*g^3)
Time = 0.78 (sec) , antiderivative size = 417, 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.
| \(\displaystyle \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \int \left (\frac {f^2 x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 \left (f+g x^2\right )^2}-\frac {2 f x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 \left (f+g x^2\right )}+\frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\frac {f \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}-\frac {f \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {b d e f^{3/2} n \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 g^{5/2} \left (d^2 g+e^2 f\right )}-\frac {b e^2 f^2 n \log \left (f+g x^2\right )}{4 g^3 \left (d^2 g+e^2 f\right )}+\frac {b e^2 f^2 n \log (d+e x)}{2 g^3 \left (d^2 g+e^2 f\right )}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}-\frac {b f n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{g^3}-\frac {b f n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{g^3}+\frac {b d n x}{2 e g^2}-\frac {b n x^2}{4 g^2}\) |
(b*d*n*x)/(2*e*g^2) - (b*n*x^2)/(4*g^2) + (b*d*e*f^(3/2)*n*ArcTan[(Sqrt[g] *x)/Sqrt[f]])/(2*g^(5/2)*(e^2*f + d^2*g)) - (b*d^2*n*Log[d + e*x])/(2*e^2* g^2) + (b*e^2*f^2*n*Log[d + e*x])/(2*g^3*(e^2*f + d^2*g)) + (x^2*(a + b*Lo g[c*(d + e*x)^n]))/(2*g^2) - (f^2*(a + b*Log[c*(d + e*x)^n]))/(2*g^3*(f + g*x^2)) - (f*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e* Sqrt[-f] + d*Sqrt[g])])/g^3 - (f*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[- f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/g^3 - (b*e^2*f^2*n*Log[f + g*x ^2])/(4*g^3*(e^2*f + d^2*g)) - (b*f*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e* Sqrt[-f] - d*Sqrt[g]))])/g^3 - (b*f*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sq rt[-f] + d*Sqrt[g])])/g^3
3.3.66.3.1 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 = 1.57 (sec) , antiderivative size = 624, normalized size of antiderivative = 1.50
| method | result | size |
| risch | \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x^{2}}{2 g^{2}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f^{2}}{2 g^{3} \left (g \,x^{2}+f \right )}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f \ln \left (g \,x^{2}+f \right )}{g^{3}}-\frac {b n \,x^{2}}{4 g^{2}}+\frac {b d n x}{2 e \,g^{2}}-\frac {b n \ln \left (e x +d \right ) d^{4}}{2 e^{2} g \left (d^{2} g +f \,e^{2}\right )}-\frac {b n \ln \left (e x +d \right ) d^{2} f}{2 g^{2} \left (d^{2} g +f \,e^{2}\right )}+\frac {b \,e^{2} f^{2} n \ln \left (e x +d \right )}{2 g^{3} \left (d^{2} g +f \,e^{2}\right )}-\frac {b \,e^{2} f^{2} n \ln \left (g \,x^{2}+f \right )}{4 g^{3} \left (d^{2} g +f \,e^{2}\right )}+\frac {b e n \,f^{2} d \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{2 g^{2} \left (d^{2} g +f \,e^{2}\right ) \sqrt {f g}}+\frac {b n f \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{g^{3}}-\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 )}{g^{3}}-\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 )}{g^{3}}-\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 )}{g^{3}}-\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 )}{g^{3}}+\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^{2}}-\frac {f \left (\frac {f}{g \left (g \,x^{2}+f \right )}+\frac {2 \ln \left (g \,x^{2}+f \right )}{g}\right )}{2 g^{2}}\right )\) | \(624\) |
1/2*b*ln((e*x+d)^n)/g^2*x^2-1/2*b*ln((e*x+d)^n)*f^2/g^3/(g*x^2+f)-b*ln((e* x+d)^n)*f/g^3*ln(g*x^2+f)-1/4*b*n*x^2/g^2+1/2*b*d*n*x/e/g^2-1/2*b/e^2*n/g/ (d^2*g+e^2*f)*ln(e*x+d)*d^4-1/2*b*n/g^2/(d^2*g+e^2*f)*ln(e*x+d)*d^2*f+1/2* b*e^2*f^2*n*ln(e*x+d)/g^3/(d^2*g+e^2*f)-1/4*b*e^2*f^2*n*ln(g*x^2+f)/g^3/(d ^2*g+e^2*f)+1/2*b*e*n/g^2*f^2/(d^2*g+e^2*f)*d/(f*g)^(1/2)*arctan(g*x/(f*g) ^(1/2))+b*n*f/g^3*ln(e*x+d)*ln(g*x^2+f)-b*n*f/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))-b*n*f/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))-b*n*f/g^3*dilog((e*(-f*g)^(1/2 )-g*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+d*g))-b*n*f/g^3*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^2-1/2*f/g^2*(f/g/(g*x^2+f)+2/g*ln(g*x^2+f)))
\[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{5}}{{\left (g x^{2} + f\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{5}}{{\left (g x^{2} + f\right )}^{2}} \,d x } \]
-1/2*a*(f^2/(g^4*x^2 + f*g^3) - x^2/g^2 + 2*f*log(g*x^2 + f)/g^3) + b*inte grate((x^5*log((e*x + d)^n) + x^5*log(c))/(g^2*x^4 + 2*f*g*x^2 + f^2), x)
\[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{5}}{{\left (g x^{2} + f\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{{\left (g\,x^2+f\right )}^2} \,d x \]