Integrand size = 27, antiderivative size = 460 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )^2} \, dx=-\frac {b e n}{2 d f^2 x}+\frac {b d e g^{3/2} n \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 f^{5/2} \left (e^2 f+d^2 g\right )}-\frac {b e^2 n \log (x)}{2 d^2 f^2}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}+\frac {b e^2 g n \log (d+e x)}{2 f^2 \left (e^2 f+d^2 g\right )}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2 \left (f+g x^2\right )}-\frac {2 g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{f^3}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{f^3}-\frac {b e^2 g n \log \left (f+g x^2\right )}{4 f^2 \left (e^2 f+d^2 g\right )}+\frac {b g n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{f^3}+\frac {b g n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{f^3}-\frac {2 b g n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^3} \]
-1/2*b*e*n/d/f^2/x+1/2*b*d*e*g^(3/2)*n*arctan(x*g^(1/2)/f^(1/2))/f^(5/2)/( d^2*g+e^2*f)-1/2*b*e^2*n*ln(x)/d^2/f^2+1/2*b*e^2*n*ln(e*x+d)/d^2/f^2+1/2*b *e^2*g*n*ln(e*x+d)/f^2/(d^2*g+e^2*f)+1/2*(-a-b*ln(c*(e*x+d)^n))/f^2/x^2-1/ 2*g*(a+b*ln(c*(e*x+d)^n))/f^2/(g*x^2+f)-2*g*ln(-e*x/d)*(a+b*ln(c*(e*x+d)^n ))/f^3-1/4*b*e^2*g*n*ln(g*x^2+f)/f^2/(d^2*g+e^2*f)+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^3+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^3-2*b*g*n *polylog(2,1+e*x/d)/f^3+b*g*n*polylog(2,-(e*x+d)*g^(1/2)/(e*(-f)^(1/2)-d*g ^(1/2)))/f^3+b*g*n*polylog(2,(e*x+d)*g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2)))/f^3
Result contains complex when optimal does not.
Time = 1.01 (sec) , antiderivative size = 596, normalized size of antiderivative = 1.30 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\frac {-\frac {2 f \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )}{x^2}-\frac {2 f g \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2}-8 g \log (x) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )+4 g \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right ) \log \left (f+g x^2\right )+b n \left (-\frac {2 f \left (d e x+e^2 x^2 \log (x)+\left (d^2-e^2 x^2\right ) \log (d+e x)\right )}{d^2 x^2}+\frac {i \sqrt {f} g \left (\sqrt {g} (d+e x) \log (d+e x)+i e \left (\sqrt {f}+i \sqrt {g} x\right ) \log \left (i \sqrt {f}-\sqrt {g} x\right )\right )}{\left (e \sqrt {f}-i d \sqrt {g}\right ) \left (\sqrt {f}+i \sqrt {g} x\right )}+\frac {i \sqrt {f} g \left (-\sqrt {g} (d+e x) \log (d+e x)+e \left (i \sqrt {f}+\sqrt {g} x\right ) \log \left (i \sqrt {f}+\sqrt {g} x\right )\right )}{\left (e \sqrt {f}+i d \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}+4 g \left (\log (d+e x) \log \left (\frac {e \left (\sqrt {f}+i \sqrt {g} x\right )}{e \sqrt {f}-i d \sqrt {g}}\right )+\operatorname {PolyLog}\left (2,-\frac {i \sqrt {g} (d+e x)}{e \sqrt {f}-i d \sqrt {g}}\right )\right )+4 g \left (\log (d+e x) \log \left (\frac {e \left (\sqrt {f}-i \sqrt {g} x\right )}{e \sqrt {f}+i d \sqrt {g}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {g} (d+e x)}{e \sqrt {f}+i d \sqrt {g}}\right )\right )-8 g \left (\log \left (-\frac {e x}{d}\right ) \log (d+e x)+\operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )\right )\right )}{4 f^3} \]
((-2*f*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n]))/x^2 - (2*f*g*(a - b* n*Log[d + e*x] + b*Log[c*(d + e*x)^n]))/(f + g*x^2) - 8*g*Log[x]*(a - b*n* Log[d + e*x] + b*Log[c*(d + e*x)^n]) + 4*g*(a - b*n*Log[d + e*x] + b*Log[c *(d + e*x)^n])*Log[f + g*x^2] + b*n*((-2*f*(d*e*x + e^2*x^2*Log[x] + (d^2 - e^2*x^2)*Log[d + e*x]))/(d^2*x^2) + (I*Sqrt[f]*g*(Sqrt[g]*(d + e*x)*Log[ d + e*x] + I*e*(Sqrt[f] + I*Sqrt[g]*x)*Log[I*Sqrt[f] - Sqrt[g]*x]))/((e*Sq rt[f] - I*d*Sqrt[g])*(Sqrt[f] + I*Sqrt[g]*x)) + (I*Sqrt[f]*g*(-(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*g*(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*(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])]) - 8*g*(Log[-((e*x)/d)]*Log [d + e*x] + PolyLog[2, 1 + (e*x)/d])))/(4*f^3)
Time = 0.84 (sec) , antiderivative size = 460, 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.
\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \int \left (\frac {2 g^2 x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 \left (f+g x^2\right )}-\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac {g^2 x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 \left (f+g x^2\right )^2}+\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 g \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}+\frac {g \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}+\frac {g \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2 \left (f+g x^2\right )}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac {b d e g^{3/2} n \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 f^{5/2} \left (d^2 g+e^2 f\right )}-\frac {b e^2 g n \log \left (f+g x^2\right )}{4 f^2 \left (d^2 g+e^2 f\right )}+\frac {b e^2 g n \log (d+e x)}{2 f^2 \left (d^2 g+e^2 f\right )}-\frac {b e^2 n \log (x)}{2 d^2 f^2}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}+\frac {b g n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{f^3}+\frac {b g n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{f^3}-\frac {2 b g n \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^3}-\frac {b e n}{2 d f^2 x}\) |
-1/2*(b*e*n)/(d*f^2*x) + (b*d*e*g^(3/2)*n*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/(2* f^(5/2)*(e^2*f + d^2*g)) - (b*e^2*n*Log[x])/(2*d^2*f^2) + (b*e^2*n*Log[d + e*x])/(2*d^2*f^2) + (b*e^2*g*n*Log[d + e*x])/(2*f^2*(e^2*f + d^2*g)) - (a + b*Log[c*(d + e*x)^n])/(2*f^2*x^2) - (g*(a + b*Log[c*(d + e*x)^n]))/(2*f ^2*(f + g*x^2)) - (2*g*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]))/f^3 + ( g*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/f^3 + (g*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g ]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/f^3 - (b*e^2*g*n*Log[f + g*x^2])/(4*f^2*( e^2*f + d^2*g)) + (b*g*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d* Sqrt[g]))])/f^3 + (b*g*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sq rt[g])])/f^3 - (2*b*g*n*PolyLog[2, 1 + (e*x)/d])/f^3
3.3.70.3.1 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 = 2.91 (sec) , antiderivative size = 656, normalized size of antiderivative = 1.43
method | result | size |
risch | \(-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{2 f^{2} x^{2}}-\frac {2 b \ln \left (\left (e x +d \right )^{n}\right ) g \ln \left (x \right )}{f^{3}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g}{2 f^{2} \left (g \,x^{2}+f \right )}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g \ln \left (g \,x^{2}+f \right )}{f^{3}}+\frac {2 b n g \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f^{3}}+\frac {2 b n g \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{3}}-\frac {b n g \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{f^{3}}+\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 )}{f^{3}}+\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 )}{f^{3}}+\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 )}{f^{3}}+\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 )}{f^{3}}+\frac {b \,e^{2} g n \ln \left (e x +d \right )}{f^{2} \left (d^{2} g +f \,e^{2}\right )}+\frac {b \,e^{4} n \ln \left (e x +d \right )}{2 f \left (d^{2} g +f \,e^{2}\right ) d^{2}}-\frac {b e n}{2 d \,f^{2} x}-\frac {b \,e^{2} n \ln \left (x \right )}{2 d^{2} f^{2}}-\frac {b \,e^{2} g n \ln \left (g \,x^{2}+f \right )}{4 f^{2} \left (d^{2} g +f \,e^{2}\right )}+\frac {b e n \,g^{2} d \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{2 f^{2} \left (d^{2} g +f \,e^{2}\right ) \sqrt {f g}}+\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^{2} x^{2}}-\frac {2 g \ln \left (x \right )}{f^{3}}+\frac {g^{2} \left (-\frac {f}{g \left (g \,x^{2}+f \right )}+\frac {2 \ln \left (g \,x^{2}+f \right )}{g}\right )}{2 f^{3}}\right )\) | \(656\) |
-1/2*b*ln((e*x+d)^n)/f^2/x^2-2*b*ln((e*x+d)^n)/f^3*g*ln(x)-1/2*b*ln((e*x+d )^n)*g/f^2/(g*x^2+f)+b*ln((e*x+d)^n)*g/f^3*ln(g*x^2+f)+2*b*n/f^3*g*dilog(( e*x+d)/d)+2*b*n/f^3*g*ln(x)*ln((e*x+d)/d)-b*n/f^3*g*ln(e*x+d)*ln(g*x^2+f)+ b*n/f^3*g*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^3*g*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^3*g*dilog((e*(-f*g)^(1/2)-g*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+d*g))+b *n/f^3*g*dilog((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e*(-f*g)^(1/2)-d*g))+b*e^2* g*n*ln(e*x+d)/f^2/(d^2*g+e^2*f)+1/2*b*e^4*n/f/(d^2*g+e^2*f)/d^2*ln(e*x+d)- 1/2*b*e*n/d/f^2/x-1/2*b*e^2*n*ln(x)/d^2/f^2-1/4*b*e^2*g*n*ln(g*x^2+f)/f^2/ (d^2*g+e^2*f)+1/2*b*e*n/f^2/(d^2*g+e^2*f)*g^2*d/(f*g)^(1/2)*arctan(g*x/(f* g)^(1/2))+(-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^2/x^2 -2/f^3*g*ln(x)+1/2*g^2/f^3*(-f/g/(g*x^2+f)+2/g*ln(g*x^2+f)))
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x^{3}} \,d x } \]
-1/2*a*((2*g*x^2 + f)/(f^2*g*x^4 + f^3*x^2) - 2*g*log(g*x^2 + f)/f^3 + 4*g *log(x)/f^3) + b*integrate((log((e*x + d)^n) + log(c))/(g^2*x^7 + 2*f*g*x^ 5 + f^2*x^3), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^3\,{\left (g\,x^2+f\right )}^2} \,d x \]