Integrand size = 27, antiderivative size = 560 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\frac {b e n \log (x)}{d f^2}-\frac {b e n \log (d+e x)}{d f^2}-\frac {b e \sqrt {g} n \log (d+e x)}{4 f^2 \left (e \sqrt {-f}+d \sqrt {g}\right )}-\frac {b e \sqrt {g} n \log (d+e x)}{4 f \left (e (-f)^{3/2}+d f \sqrt {g}\right )}-\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x}+\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f^2 \left (\sqrt {-f}-\sqrt {g} x\right )}-\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f^2 \left (\sqrt {-f}+\sqrt {g} x\right )}+\frac {b e \sqrt {g} n \log \left (\sqrt {-f}-\sqrt {g} x\right )}{4 f^2 \left (e \sqrt {-f}+d \sqrt {g}\right )}-\frac {3 \sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{4 (-f)^{5/2}}+\frac {b e \sqrt {g} n \log \left (\sqrt {-f}+\sqrt {g} x\right )}{4 f \left (e (-f)^{3/2}+d f \sqrt {g}\right )}+\frac {3 \sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{4 (-f)^{5/2}}+\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{4 (-f)^{5/2}}-\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{4 (-f)^{5/2}} \]
b*e*n*ln(x)/d/f^2-b*e*n*ln(e*x+d)/d/f^2+(-a-b*ln(c*(e*x+d)^n))/f^2/x-3/4*( 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^(1/2)/(-f)^(5/2)+3/4*(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^(1/2)/(-f)^(5/2)+3/4*b*n*polylog(2,-(e*x+d)*g^ (1/2)/(e*(-f)^(1/2)-d*g^(1/2)))*g^(1/2)/(-f)^(5/2)-3/4*b*n*polylog(2,(e*x+ d)*g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2)))*g^(1/2)/(-f)^(5/2)-1/4*b*e*n*ln(e*x+d )*g^(1/2)/f^2/(e*(-f)^(1/2)+d*g^(1/2))+1/4*b*e*n*ln((-f)^(1/2)-x*g^(1/2))* g^(1/2)/f^2/(e*(-f)^(1/2)+d*g^(1/2))-1/4*b*e*n*ln(e*x+d)*g^(1/2)/f/(e*(-f) ^(3/2)+d*f*g^(1/2))+1/4*b*e*n*ln((-f)^(1/2)+x*g^(1/2))*g^(1/2)/f/(e*(-f)^( 3/2)+d*f*g^(1/2))+1/4*(a+b*ln(c*(e*x+d)^n))*g^(1/2)/f^2/((-f)^(1/2)-x*g^(1 /2))-1/4*(a+b*ln(c*(e*x+d)^n))*g^(1/2)/f^2/((-f)^(1/2)+x*g^(1/2))
Time = 0.46 (sec) , antiderivative size = 475, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\frac {1}{4} \left (\frac {4 b e n (\log (x)-\log (d+e x))}{d f^2}-\frac {4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 \left (\sqrt {-f}-\sqrt {g} x\right )}-\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 \left (\sqrt {-f}+\sqrt {g} x\right )}+\frac {b e \sqrt {g} n \left (-\log (d+e x)+\log \left (\sqrt {-f}-\sqrt {g} x\right )\right )}{f^2 \left (e \sqrt {-f}+d \sqrt {g}\right )}-\frac {3 \sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2}}+\frac {b e \sqrt {g} n \left (\log (d+e x)-\log \left (\sqrt {-f}+\sqrt {g} x\right )\right )}{f^2 \left (e \sqrt {-f}-d \sqrt {g}\right )}+\frac {3 \sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{5/2}}+\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{5/2}}-\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2}}\right ) \]
((4*b*e*n*(Log[x] - Log[d + e*x]))/(d*f^2) - (4*(a + b*Log[c*(d + e*x)^n]) )/(f^2*x) + (Sqrt[g]*(a + b*Log[c*(d + e*x)^n]))/(f^2*(Sqrt[-f] - Sqrt[g]* x)) - (Sqrt[g]*(a + b*Log[c*(d + e*x)^n]))/(f^2*(Sqrt[-f] + Sqrt[g]*x)) + (b*e*Sqrt[g]*n*(-Log[d + e*x] + Log[Sqrt[-f] - Sqrt[g]*x]))/(f^2*(e*Sqrt[- f] + d*Sqrt[g])) - (3*Sqrt[g]*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(-f)^(5/2) + (b*e*Sqrt[g]*n*(Log[ d + e*x] - Log[Sqrt[-f] + Sqrt[g]*x]))/(f^2*(e*Sqrt[-f] - d*Sqrt[g])) + (3 *Sqrt[g]*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt [-f] - d*Sqrt[g])])/(-f)^(5/2) + (3*b*Sqrt[g]*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(-f)^(5/2) - (3*b*Sqrt[g]*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(-f)^(5/2))/4
Time = 1.11 (sec) , antiderivative size = 560, 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.
| \(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \int \left (-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 \left (f+g x^2\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x^2}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f \left (f+g x^2\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f^2 \left (\sqrt {-f}-\sqrt {g} x\right )}-\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f^2 \left (\sqrt {-f}+\sqrt {g} x\right )}-\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x}-\frac {3 \sqrt {g} \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 (-f)^{5/2}}+\frac {3 \sqrt {g} \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 (-f)^{5/2}}-\frac {b e \sqrt {g} n \log (d+e x)}{4 f^2 \left (d \sqrt {g}+e \sqrt {-f}\right )}+\frac {b e \sqrt {g} n \log \left (\sqrt {-f}-\sqrt {g} x\right )}{4 f^2 \left (d \sqrt {g}+e \sqrt {-f}\right )}+\frac {b e n \log (x)}{d f^2}-\frac {b e n \log (d+e x)}{d f^2}+\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{4 (-f)^{5/2}}-\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{4 (-f)^{5/2}}-\frac {b e \sqrt {g} n \log (d+e x)}{4 f \left (d f \sqrt {g}+e (-f)^{3/2}\right )}+\frac {b e \sqrt {g} n \log \left (\sqrt {-f}+\sqrt {g} x\right )}{4 f \left (d f \sqrt {g}+e (-f)^{3/2}\right )}\) |
(b*e*n*Log[x])/(d*f^2) - (b*e*n*Log[d + e*x])/(d*f^2) - (b*e*Sqrt[g]*n*Log [d + e*x])/(4*f^2*(e*Sqrt[-f] + d*Sqrt[g])) - (b*e*Sqrt[g]*n*Log[d + e*x]) /(4*f*(e*(-f)^(3/2) + d*f*Sqrt[g])) - (a + b*Log[c*(d + e*x)^n])/(f^2*x) + (Sqrt[g]*(a + b*Log[c*(d + e*x)^n]))/(4*f^2*(Sqrt[-f] - Sqrt[g]*x)) - (Sq rt[g]*(a + b*Log[c*(d + e*x)^n]))/(4*f^2*(Sqrt[-f] + Sqrt[g]*x)) + (b*e*Sq rt[g]*n*Log[Sqrt[-f] - Sqrt[g]*x])/(4*f^2*(e*Sqrt[-f] + d*Sqrt[g])) - (3*S qrt[g]*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[- f] + d*Sqrt[g])])/(4*(-f)^(5/2)) + (b*e*Sqrt[g]*n*Log[Sqrt[-f] + Sqrt[g]*x ])/(4*f*(e*(-f)^(3/2) + d*f*Sqrt[g])) + (3*Sqrt[g]*(a + b*Log[c*(d + e*x)^ n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/(4*(-f)^(5/2 )) + (3*b*Sqrt[g]*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[ g]))])/(4*(-f)^(5/2)) - (3*b*Sqrt[g]*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*S qrt[-f] + d*Sqrt[g])])/(4*(-f)^(5/2))
3.3.74.3.1 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 = 2.25 (sec) , antiderivative size = 1619, normalized size of antiderivative = 2.89
3/2*b/f^2*g/(f*g)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/(f*g)^(1/2))*n*ln (e*x+d)-1/2*b*e^2/f^2*g*x/(e^2*g*x^2+e^2*f)*ln((e*x+d)^n)-b*ln((e*x+d)^n)/ f^2/x-1/2*b*e^3*n/f^2*g^2*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)*x^2*d- 1/2*b*e^2*n/f^2*g^2*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)*x*d^2-1/2*b* n/f^2*g*ln(e*x+d)/(-f*g)^(1/2)*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*ln(e*x+d)/(-f*g)^(1/2)*ln((e*(-f*g)^(1/2)+g*(e* x+d)-d*g)/(e*(-f*g)^(1/2)-d*g))+1/4*b*e*n/f^2*g/(d^2*g+e^2*f)*d*ln(g*(e*x+ d)^2-2*(e*x+d)*d*g+d^2*g+f*e^2)+1/2*b*e^2*n/f*g/(d^2*g+e^2*f)/(f*g)^(1/2)* arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/(f*g)^(1/2))+1/2*b*e^2/f^2*g*x/(e^2*g*x^2 +e^2*f)*n*ln(e*x+d)-1/2*b*e^3*n/f*g*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2 *f)*d-1/2*b*e^4*n/f*g*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)*x-1/4*b*e^ 4*n*g*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/(-f*g)^(1/2)*ln((e*(-f*g)^ (1/2)-g*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+d*g))+1/4*b*e^4*n*g*ln(e*x+d)/(d^2*g+ e^2*f)/(e^2*g*x^2+e^2*f)/(-f*g)^(1/2)*ln((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e *(-f*g)^(1/2)-d*g))-1/4*b*e^4*n/f*g^2*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e ^2*f)/(-f*g)^(1/2)*ln((e*(-f*g)^(1/2)-g*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+d*g)) *x^2+1/4*b*e^4*n/f*g^2*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/(-f*g)^(1 /2)*ln((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e*(-f*g)^(1/2)-d*g))*x^2-1/4*b*e^2* n/f*g^2*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/(-f*g)^(1/2)*ln((e*(-f*g )^(1/2)-g*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+d*g))*d^2+1/4*b*e^2*n/f*g^2*ln(e...
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x^{2}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x^{2}} \,d x } \]
-1/2*a*((3*g*x^2 + 2*f)/(f^2*g*x^3 + f^3*x) + 3*g*arctan(g*x/sqrt(f*g))/(s qrt(f*g)*f^2)) + b*integrate((log((e*x + d)^n) + log(c))/(g^2*x^6 + 2*f*g* x^4 + f^2*x^2), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x^{2}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^2\,{\left (g\,x^2+f\right )}^2} \,d x \]