Integrand size = 25, antiderivative size = 186 \[ \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\frac {a x}{g^2}-\frac {b n x}{g^2}+\frac {b e f^2 n \log (d+e x)}{g^3 (e f-d g)}+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}-\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)}-\frac {b e f^2 n \log (f+g x)}{g^3 (e f-d g)}-\frac {2 f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^3}-\frac {2 b f n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^3} \]
a*x/g^2-b*n*x/g^2+b*e*f^2*n*ln(e*x+d)/g^3/(-d*g+e*f)+b*(e*x+d)*ln(c*(e*x+d )^n)/e/g^2-f^2*(a+b*ln(c*(e*x+d)^n))/g^3/(g*x+f)-b*e*f^2*n*ln(g*x+f)/g^3/( -d*g+e*f)-2*f*(a+b*ln(c*(e*x+d)^n))*ln(e*(g*x+f)/(-d*g+e*f))/g^3-2*b*f*n*p olylog(2,-g*(e*x+d)/(-d*g+e*f))/g^3
Time = 0.09 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.82 \[ \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\frac {a g x-b g n x+\frac {b g (d+e x) \log \left (c (d+e x)^n\right )}{e}-\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}+\frac {b e f^2 n (\log (d+e x)-\log (f+g x))}{e f-d g}-2 f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )-2 b f n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )}{g^3} \]
(a*g*x - b*g*n*x + (b*g*(d + e*x)*Log[c*(d + e*x)^n])/e - (f^2*(a + b*Log[ c*(d + e*x)^n]))/(f + g*x) + (b*e*f^2*n*(Log[d + e*x] - Log[f + g*x]))/(e* f - d*g) - 2*f*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)] - 2*b*f*n*PolyLog[2, (g*(d + e*x))/(-(e*f) + d*g)])/g^3
Time = 0.42 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.
| \(\displaystyle \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \int \left (\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 (f+g x)^2}-\frac {2 f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 (f+g x)}+\frac {a+b \log \left (c (d+e x)^n\right )}{g^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)}-\frac {2 f \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac {a x}{g^2}+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}+\frac {b e f^2 n \log (d+e x)}{g^3 (e f-d g)}-\frac {b e f^2 n \log (f+g x)}{g^3 (e f-d g)}-\frac {2 b f n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^3}-\frac {b n x}{g^2}\) |
(a*x)/g^2 - (b*n*x)/g^2 + (b*e*f^2*n*Log[d + e*x])/(g^3*(e*f - d*g)) + (b* (d + e*x)*Log[c*(d + e*x)^n])/(e*g^2) - (f^2*(a + b*Log[c*(d + e*x)^n]))/( g^3*(f + g*x)) - (b*e*f^2*n*Log[f + g*x])/(g^3*(e*f - d*g)) - (2*f*(a + b* Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/g^3 - (2*b*f*n*PolyLog [2, -((g*(d + e*x))/(e*f - d*g))])/g^3
3.3.50.3.1 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 = 0.84 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.34
| method | result | size |
| risch | \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x}{g^{2}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f^{2}}{g^{3} \left (g x +f \right )}-\frac {2 b \ln \left (\left (e x +d \right )^{n}\right ) f \ln \left (g x +f \right )}{g^{3}}-\frac {b n x}{g^{2}}-\frac {b f n}{g^{3}}+\frac {b e n \,f^{2} \ln \left (g x +f \right )}{g^{3} \left (d g -e f \right )}+\frac {b n \ln \left (\left (g x +f \right ) e +d g -e f \right ) d^{2}}{e g \left (d g -e f \right )}-\frac {b n \ln \left (\left (g x +f \right ) e +d g -e f \right ) d f}{g^{2} \left (d g -e f \right )}-\frac {b e n \ln \left (\left (g x +f \right ) e +d g -e f \right ) f^{2}}{g^{3} \left (d g -e f \right )}+\frac {2 b n f \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{3}}+\frac {2 b n f \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b}{2}+b \ln \left (c \right )+a \right ) \left (\frac {x}{g^{2}}-\frac {f^{2}}{g^{3} \left (g x +f \right )}-\frac {2 f \ln \left (g x +f \right )}{g^{3}}\right )\) | \(435\) |
b*ln((e*x+d)^n)/g^2*x-b*ln((e*x+d)^n)/g^3*f^2/(g*x+f)-2*b*ln((e*x+d)^n)/g^ 3*f*ln(g*x+f)-b*n*x/g^2-b*f*n/g^3+b*e*n/g^3*f^2/(d*g-e*f)*ln(g*x+f)+b/e*n/ g/(d*g-e*f)*ln((g*x+f)*e+d*g-e*f)*d^2-b*n/g^2/(d*g-e*f)*ln((g*x+f)*e+d*g-e *f)*d*f-b*e*n/g^3/(d*g-e*f)*ln((g*x+f)*e+d*g-e*f)*f^2+2*b*n/g^3*f*dilog((( g*x+f)*e+d*g-e*f)/(d*g-e*f))+2*b*n/g^3*f*ln(g*x+f)*ln(((g*x+f)*e+d*g-e*f)/ (d*g-e*f))+(-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)*(x/g^2-1/g^3 *f^2/(g*x+f)-2/g^3*f*ln(g*x+f))
\[ \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{2}}{{\left (g x + f\right )}^{2}} \,d x } \]
\[ \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\int \frac {x^{2} \left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )}{\left (f + g x\right )^{2}}\, dx \]
\[ \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{2}}{{\left (g x + f\right )}^{2}} \,d x } \]
-a*(f^2/(g^4*x + f*g^3) - x/g^2 + 2*f*log(g*x + f)/g^3) + b*integrate((x^2 *log((e*x + d)^n) + x^2*log(c))/(g^2*x^2 + 2*f*g*x + f^2), x)
\[ \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{2}}{{\left (g x + f\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\int \frac {x^2\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{{\left (f+g\,x\right )}^2} \,d x \]