Integrand size = 24, antiderivative size = 503 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+g x^2\right )^2} \, dx=\frac {b e n \log (d+e x)}{4 f \left (e \sqrt {-f}+d \sqrt {g}\right ) \sqrt {g}}+\frac {b e n \log (d+e x)}{4 \left (e (-f)^{3/2}+d f \sqrt {g}\right ) \sqrt {g}}-\frac {a+b \log \left (c (d+e x)^n\right )}{4 f \sqrt {g} \left (\sqrt {-f}-\sqrt {g} x\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{4 f \sqrt {g} \left (\sqrt {-f}+\sqrt {g} x\right )}-\frac {b e n \log \left (\sqrt {-f}-\sqrt {g} x\right )}{4 f \left (e \sqrt {-f}+d \sqrt {g}\right ) \sqrt {g}}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{4 (-f)^{3/2} \sqrt {g}}-\frac {b e n \log \left (\sqrt {-f}+\sqrt {g} x\right )}{4 \left (e (-f)^{3/2}+d f \sqrt {g}\right ) \sqrt {g}}+\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{4 (-f)^{3/2} \sqrt {g}}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{4 (-f)^{3/2} \sqrt {g}}-\frac {b n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{4 (-f)^{3/2} \sqrt {g}} \]
-1/4*(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)))/(-f)^(3/2)/g^(1/2)+1/4*(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)))/(-f)^(3/2)/g^(1/2)+1/4*b*n*polylog(2,-(e*x +d)*g^(1/2)/(e*(-f)^(1/2)-d*g^(1/2)))/(-f)^(3/2)/g^(1/2)-1/4*b*n*polylog(2 ,(e*x+d)*g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2)))/(-f)^(3/2)/g^(1/2)+1/4*b*e*n*ln (e*x+d)/f/g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2))-1/4*b*e*n*ln((-f)^(1/2)-x*g^(1/ 2))/f/g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2))+1/4*b*e*n*ln(e*x+d)/g^(1/2)/(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)/(e*(-f)^(3/ 2)+d*f*g^(1/2))+1/4*(-a-b*ln(c*(e*x+d)^n))/f/g^(1/2)/((-f)^(1/2)-x*g^(1/2) )+1/4*(a+b*ln(c*(e*x+d)^n))/f/g^(1/2)/((-f)^(1/2)+x*g^(1/2))
Time = 0.55 (sec) , antiderivative size = 407, normalized size of antiderivative = 0.81 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+g x^2\right )^2} \, dx=\frac {1}{4} \left (\frac {a+b \log \left (c (d+e x)^n\right )}{f \left (\sqrt {-f} \sqrt {g}+g x\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{(-f)^{3/2} \sqrt {g}+f g x}+\frac {b e n \left (\log (d+e x)-\log \left (\sqrt {-f}-\sqrt {g} x\right )\right )}{e \sqrt {-f} f \sqrt {g}+d f g}+\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2} \sqrt {g}}+\frac {b e n \left (\log (d+e x)-\log \left (\sqrt {-f}+\sqrt {g} x\right )\right )}{e (-f)^{3/2} \sqrt {g}+d f g}+\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{3/2} \sqrt {g}}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{3/2} \sqrt {g}}+\frac {b f n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2} \sqrt {g}}\right ) \]
((a + b*Log[c*(d + e*x)^n])/(f*(Sqrt[-f]*Sqrt[g] + g*x)) + (a + b*Log[c*(d + e*x)^n])/((-f)^(3/2)*Sqrt[g] + f*g*x) + (b*e*n*(Log[d + e*x] - Log[Sqrt [-f] - Sqrt[g]*x]))/(e*Sqrt[-f]*f*Sqrt[g] + d*f*g) + (f*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/((-f)^( 5/2)*Sqrt[g]) + (b*e*n*(Log[d + e*x] - Log[Sqrt[-f] + Sqrt[g]*x]))/(e*(-f) ^(3/2)*Sqrt[g] + d*f*g) + ((a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + S qrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/((-f)^(3/2)*Sqrt[g]) + (b*n*PolyLog[ 2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/((-f)^(3/2)*Sqrt[g]) + (b*f*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/((-f)^( 5/2)*Sqrt[g]))/4
Time = 0.71 (sec) , antiderivative size = 503, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2856, 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.
| \(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+g x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 2856 |
| \(\displaystyle \int \left (-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f \left (-f g-g^2 x^2\right )}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f \left (\sqrt {-f} \sqrt {g}-g x\right )^2}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f \left (\sqrt {-f} \sqrt {g}+g x\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a+b \log \left (c (d+e x)^n\right )}{4 f \sqrt {g} \left (\sqrt {-f}-\sqrt {g} x\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{4 f \sqrt {g} \left (\sqrt {-f}+\sqrt {g} x\right )}-\frac {\log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 (-f)^{3/2} \sqrt {g}}+\frac {\log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 (-f)^{3/2} \sqrt {g}}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{4 (-f)^{3/2} \sqrt {g}}-\frac {b n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{4 (-f)^{3/2} \sqrt {g}}+\frac {b e n \log (d+e x)}{4 f \sqrt {g} \left (d \sqrt {g}+e \sqrt {-f}\right )}+\frac {b e n \log (d+e x)}{4 \sqrt {g} \left (d f \sqrt {g}+e (-f)^{3/2}\right )}-\frac {b e n \log \left (\sqrt {-f}-\sqrt {g} x\right )}{4 f \sqrt {g} \left (d \sqrt {g}+e \sqrt {-f}\right )}-\frac {b e n \log \left (\sqrt {-f}+\sqrt {g} x\right )}{4 \sqrt {g} \left (d f \sqrt {g}+e (-f)^{3/2}\right )}\) |
(b*e*n*Log[d + e*x])/(4*f*(e*Sqrt[-f] + d*Sqrt[g])*Sqrt[g]) + (b*e*n*Log[d + e*x])/(4*(e*(-f)^(3/2) + d*f*Sqrt[g])*Sqrt[g]) - (a + b*Log[c*(d + e*x) ^n])/(4*f*Sqrt[g]*(Sqrt[-f] - Sqrt[g]*x)) + (a + b*Log[c*(d + e*x)^n])/(4* f*Sqrt[g]*(Sqrt[-f] + Sqrt[g]*x)) - (b*e*n*Log[Sqrt[-f] - Sqrt[g]*x])/(4*f *(e*Sqrt[-f] + d*Sqrt[g])*Sqrt[g]) - ((a + b*Log[c*(d + e*x)^n])*Log[(e*(S qrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(4*(-f)^(3/2)*Sqrt[g]) - (b*e*n*Log[Sqrt[-f] + Sqrt[g]*x])/(4*(e*(-f)^(3/2) + d*f*Sqrt[g])*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)^(3/2)*Sqrt[g]) + (b*n*PolyLog[2, -((Sqrt[g]*(d + e* x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(4*(-f)^(3/2)*Sqrt[g]) - (b*n*PolyLog[2, ( Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(4*(-f)^(3/2)*Sqrt[g])
3.3.73.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. )*(x_)^(r_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x) ^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x] && I GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.31 (sec) , antiderivative size = 1406, normalized size of antiderivative = 2.80
-1/2*b*e^2/f/(e^2*g*x^2+e^2*f)*x*n*ln(e*x+d)+1/2*b*e^2/f/(e^2*g*x^2+e^2*f) *x*ln((e*x+d)^n)-1/2*b/f/(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/f/(f*g)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/ (f*g)^(1/2))*ln((e*x+d)^n)-1/4*b*e*n/f/(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/(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))+1/4*b*e^2*n*ln(e*x+d)/f/(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*d^2*g^2+1/4*b*e^4*n*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 *g-1/4*b*e^2*n*ln(e*x+d)/f/(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*d^2*g^2-1/4*b*e^ 4*n*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*g+1/2*b*e^3*n*ln(e*x+d)/f/(d^ 2*g+e^2*f)/(e^2*g*x^2+e^2*f)*x^2*d*g+1/4*b*e^2*n*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*g+1/4*b*e^4*n*ln(e*x+d)*f/(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^2*n*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*g-1/4*b*e^4*n*ln(e*x+d)*f/ (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+...
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+g x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2}} \,d x } \]
1/2*a*(x/(f*g*x^2 + f^2) + arctan(g*x/sqrt(f*g))/(sqrt(f*g)*f)) + b*integr ate((log((e*x + d)^n) + log(c))/(g^2*x^4 + 2*f*g*x^2 + f^2), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{\left (f+g x^2\right )^2} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{{\left (g\,x^2+f\right )}^2} \,d x \]