Integrand size = 27, antiderivative size = 383 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )^2} \, dx=-\frac {b d e \sqrt {g} n \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 f^{3/2} \left (e^2 f+d^2 g\right )}-\frac {b e^2 n \log (d+e x)}{2 f \left (e^2 f+d^2 g\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{2 f \left (f+g x^2\right )}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f^2}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}+\frac {b e^2 n \log \left (f+g x^2\right )}{4 f \left (e^2 f+d^2 g\right )}-\frac {b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}-\frac {b n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f^2}+\frac {b n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^2} \]
-1/2*b*e^2*n*ln(e*x+d)/f/(d^2*g+e^2*f)+1/2*(a+b*ln(c*(e*x+d)^n))/f/(g*x^2+ f)+ln(-e*x/d)*(a+b*ln(c*(e*x+d)^n))/f^2+1/4*b*e^2*n*ln(g*x^2+f)/f/(d^2*g+e ^2*f)-1/2*(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)))/f^2-1/2*(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)))/f^2+b*n*polylog(2,1+e*x/d)/f^2-1/2*b*n*polylog(2,-(e *x+d)*g^(1/2)/(e*(-f)^(1/2)-d*g^(1/2)))/f^2-1/2*b*n*polylog(2,(e*x+d)*g^(1 /2)/(e*(-f)^(1/2)+d*g^(1/2)))/f^2-1/2*b*d*e*n*arctan(x*g^(1/2)/f^(1/2))*g^ (1/2)/f^(3/2)/(d^2*g+e^2*f)
Result contains complex when optimal does not.
Time = 0.82 (sec) , antiderivative size = 521, normalized size of antiderivative = 1.36 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )^2} \, dx=\frac {a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )}{2 f^2+2 f g x^2}+\frac {\log (x) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )}{f^2}-\frac {\left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right ) \log \left (f+g x^2\right )}{2 f^2}+\frac {b n \left (\frac {\sqrt {f} \left (-i \sqrt {g} (d+e x) \log (d+e x)+e \left (\sqrt {f}+i \sqrt {g} x\right ) \log \left (i \sqrt {f}-\sqrt {g} x\right )\right )}{\left (e \sqrt {f}-i d \sqrt {g}\right ) \left (\sqrt {f}+i \sqrt {g} x\right )}+\frac {\sqrt {f} \left (i \sqrt {g} (d+e x) \log (d+e x)+e \left (\sqrt {f}-i \sqrt {g} x\right ) \log \left (i \sqrt {f}+\sqrt {g} x\right )\right )}{\left (e \sqrt {f}+i d \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}-2 \left (\log (d+e x) \log \left (\frac {e \left (\sqrt {f}+i \sqrt {g} x\right )}{e \sqrt {f}-i d \sqrt {g}}\right )+\operatorname {PolyLog}\left (2,-\frac {i \sqrt {g} (d+e x)}{e \sqrt {f}-i d \sqrt {g}}\right )\right )-2 \left (\log (d+e x) \log \left (\frac {e \left (\sqrt {f}-i \sqrt {g} x\right )}{e \sqrt {f}+i d \sqrt {g}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {g} (d+e x)}{e \sqrt {f}+i d \sqrt {g}}\right )\right )+4 \left (\log \left (-\frac {e x}{d}\right ) \log (d+e x)+\operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )\right )\right )}{4 f^2} \]
(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])/(2*f^2 + 2*f*g*x^2) + (Log[x ]*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n]))/f^2 - ((a - b*n*Log[d + e *x] + b*Log[c*(d + e*x)^n])*Log[f + g*x^2])/(2*f^2) + (b*n*((Sqrt[f]*((-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)) + (Sqrt [f]*(I*Sqrt[g]*(d + e*x)*Log[d + e*x] + e*(Sqrt[f] - I*Sqrt[g]*x)*Log[I*Sq rt[f] + Sqrt[g]*x]))/((e*Sqrt[f] + I*d*Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x)) - 2*(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])]) - 2*(L og[d + e*x]*Log[(e*(Sqrt[f] - I*Sqrt[g]*x))/(e*Sqrt[f] + I*d*Sqrt[g])] + P olyLog[2, (I*Sqrt[g]*(d + e*x))/(e*Sqrt[f] + I*d*Sqrt[g])]) + 4*(Log[-((e* x)/d)]*Log[d + e*x] + PolyLog[2, 1 + (e*x)/d])))/(4*f^2)
Time = 0.75 (sec) , antiderivative size = 383, 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.
| \(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \int \left (-\frac {g x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 \left (f+g x^2\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x}-\frac {g x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f \left (f+g x^2\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}-\frac {\log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {a+b \log \left (c (d+e x)^n\right )}{2 f \left (f+g x^2\right )}-\frac {b d e \sqrt {g} n \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 f^{3/2} \left (d^2 g+e^2 f\right )}+\frac {b e^2 n \log \left (f+g x^2\right )}{4 f \left (d^2 g+e^2 f\right )}-\frac {b e^2 n \log (d+e x)}{2 f \left (d^2 g+e^2 f\right )}-\frac {b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}-\frac {b n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 f^2}+\frac {b n \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^2}\) |
-1/2*(b*d*e*Sqrt[g]*n*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/(f^(3/2)*(e^2*f + d^2*g )) - (b*e^2*n*Log[d + e*x])/(2*f*(e^2*f + d^2*g)) + (a + b*Log[c*(d + e*x) ^n])/(2*f*(f + g*x^2)) + (Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f^2 - ((a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*f^2) - ((a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqr t[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/(2*f^2) + (b*e^2*n*Log[f + g*x^2])/(4* f*(e^2*f + d^2*g)) - (b*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d *Sqrt[g]))])/(2*f^2) - (b*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d *Sqrt[g])])/(2*f^2) + (b*n*PolyLog[2, 1 + (e*x)/d])/f^2
3.3.69.3.1 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 = 0.91 (sec) , antiderivative size = 556, normalized size of antiderivative = 1.45
| method | result | size |
| risch | \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (x \right )}{f^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{2 f \left (g \,x^{2}+f \right )}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g \,x^{2}+f \right )}{2 f^{2}}-\frac {b \,e^{2} n \ln \left (e x +d \right )}{2 f \left (d^{2} g +f \,e^{2}\right )}+\frac {b \,e^{2} n \ln \left (g \,x^{2}+f \right )}{4 f \left (d^{2} g +f \,e^{2}\right )}-\frac {b e n g d \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{2 f \left (d^{2} g +f \,e^{2}\right ) \sqrt {f g}}-\frac {b n \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f^{2}}-\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{2}}+\frac {b n \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{2 f^{2}}-\frac {b n \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 \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 \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 \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}}+\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 {\ln \left (x \right )}{f^{2}}-\frac {g \left (-\frac {f}{g \left (g \,x^{2}+f \right )}+\frac {\ln \left (g \,x^{2}+f \right )}{g}\right )}{2 f^{2}}\right )\) | \(556\) |
b*ln((e*x+d)^n)/f^2*ln(x)+1/2*b*ln((e*x+d)^n)/f/(g*x^2+f)-1/2*b*ln((e*x+d) ^n)/f^2*ln(g*x^2+f)-1/2*b*e^2*n*ln(e*x+d)/f/(d^2*g+e^2*f)+1/4*b*e^2*n*ln(g *x^2+f)/f/(d^2*g+e^2*f)-1/2*b*e*n/f/(d^2*g+e^2*f)*g*d/(f*g)^(1/2)*arctan(g *x/(f*g)^(1/2))-b*n/f^2*dilog((e*x+d)/d)-b*n/f^2*ln(x)*ln((e*x+d)/d)+1/2*b *n/f^2*ln(e*x+d)*ln(g*x^2+f)-1/2*b*n/f^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^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^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^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)*cs gn(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*cs gn(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*l n(c)+a)*(1/f^2*ln(x)-1/2*g/f^2*(-f/g/(g*x^2+f)+1/g*ln(g*x^2+f)))
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x} \,d x } \]
1/2*a*(1/(f*g*x^2 + f^2) - log(g*x^2 + f)/f^2 + 2*log(x)/f^2) + b*integrat e((log((e*x + d)^n) + log(c))/(g^2*x^5 + 2*f*g*x^3 + f^2*x), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x \left (f+g x^2\right )^2} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x\,{\left (g\,x^2+f\right )}^2} \,d x \]