Integrand size = 25, antiderivative size = 265 \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=-\frac {2 a f x}{g^3}+\frac {2 b f n x}{g^3}+\frac {b d n x}{2 e g^2}-\frac {b n x^2}{4 g^2}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}-\frac {b e f^3 n \log (d+e x)}{g^4 (e f-d g)}-\frac {2 b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac {b e f^3 n \log (f+g x)}{g^4 (e f-d g)}+\frac {3 f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^4}+\frac {3 b f^2 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^4} \]
-2*a*f*x/g^3+2*b*f*n*x/g^3+1/2*b*d*n*x/e/g^2-1/4*b*n*x^2/g^2-1/2*b*d^2*n*l n(e*x+d)/e^2/g^2-b*e*f^3*n*ln(e*x+d)/g^4/(-d*g+e*f)-2*b*f*(e*x+d)*ln(c*(e* x+d)^n)/e/g^3+1/2*x^2*(a+b*ln(c*(e*x+d)^n))/g^2+f^3*(a+b*ln(c*(e*x+d)^n))/ g^4/(g*x+f)+b*e*f^3*n*ln(g*x+f)/g^4/(-d*g+e*f)+3*f^2*(a+b*ln(c*(e*x+d)^n)) *ln(e*(g*x+f)/(-d*g+e*f))/g^4+3*b*f^2*n*polylog(2,-g*(e*x+d)/(-d*g+e*f))/g ^4
Time = 0.18 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.83 \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\frac {-8 a f g x+8 b f g n x-\frac {b g^2 n \left (e x (-2 d+e x)+2 d^2 \log (d+e x)\right )}{e^2}-\frac {8 b f g (d+e x) \log \left (c (d+e x)^n\right )}{e}+2 g^2 x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {4 f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}-\frac {4 b e f^3 n (\log (d+e x)-\log (f+g x))}{e f-d g}+12 f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )+12 b f^2 n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )}{4 g^4} \]
(-8*a*f*g*x + 8*b*f*g*n*x - (b*g^2*n*(e*x*(-2*d + e*x) + 2*d^2*Log[d + e*x ]))/e^2 - (8*b*f*g*(d + e*x)*Log[c*(d + e*x)^n])/e + 2*g^2*x^2*(a + b*Log[ c*(d + e*x)^n]) + (4*f^3*(a + b*Log[c*(d + e*x)^n]))/(f + g*x) - (4*b*e*f^ 3*n*(Log[d + e*x] - Log[f + g*x]))/(e*f - d*g) + 12*f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)] + 12*b*f^2*n*PolyLog[2, (g*(d + e* x))/(-(e*f) + d*g)])/(4*g^4)
Time = 0.50 (sec) , antiderivative size = 265, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx\) |
| \(\Big \downarrow \) 2863 |
| \(\displaystyle \int \left (-\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)^2}+\frac {3 f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3 (f+g x)}-\frac {2 f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4 (f+g x)}+\frac {3 f^2 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {2 a f x}{g^3}-\frac {2 b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}-\frac {b e f^3 n \log (d+e x)}{g^4 (e f-d g)}+\frac {b e f^3 n \log (f+g x)}{g^4 (e f-d g)}+\frac {3 b f^2 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^4}+\frac {b d n x}{2 e g^2}+\frac {2 b f n x}{g^3}-\frac {b n x^2}{4 g^2}\) |
(-2*a*f*x)/g^3 + (2*b*f*n*x)/g^3 + (b*d*n*x)/(2*e*g^2) - (b*n*x^2)/(4*g^2) - (b*d^2*n*Log[d + e*x])/(2*e^2*g^2) - (b*e*f^3*n*Log[d + e*x])/(g^4*(e*f - d*g)) - (2*b*f*(d + e*x)*Log[c*(d + e*x)^n])/(e*g^3) + (x^2*(a + b*Log[ c*(d + e*x)^n]))/(2*g^2) + (f^3*(a + b*Log[c*(d + e*x)^n]))/(g^4*(f + g*x) ) + (b*e*f^3*n*Log[f + g*x])/(g^4*(e*f - d*g)) + (3*f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/g^4 + (3*b*f^2*n*PolyLog[2, -((g* (d + e*x))/(e*f - d*g))])/g^4
3.3.49.3.1 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 = 0.94 (sec) , antiderivative size = 550, normalized size of antiderivative = 2.08
| method | result | size |
| risch | \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x^{2}}{2 g^{2}}-\frac {2 b \ln \left (\left (e x +d \right )^{n}\right ) x f}{g^{3}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f^{3}}{g^{4} \left (g x +f \right )}+\frac {3 b \ln \left (\left (e x +d \right )^{n}\right ) f^{2} \ln \left (g x +f \right )}{g^{4}}-\frac {3 b n \,f^{2} \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{4}}-\frac {3 b n \,f^{2} \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^{4}}-\frac {b n \,x^{2}}{4 g^{2}}+\frac {2 b f n x}{g^{3}}+\frac {9 b n \,f^{2}}{4 g^{4}}+\frac {b d n x}{2 e \,g^{2}}+\frac {b n d f}{2 e \,g^{3}}-\frac {b e n \,f^{3} \ln \left (g x +f \right )}{g^{4} \left (d g -e f \right )}-\frac {b n \ln \left (\left (g x +f \right ) e +d g -e f \right ) d^{3}}{2 e^{2} g \left (d g -e f \right )}-\frac {3 b n \ln \left (\left (g x +f \right ) e +d g -e f \right ) d^{2} f}{2 e \,g^{2} \left (d g -e f \right )}+\frac {2 b n \ln \left (\left (g x +f \right ) e +d g -e f \right ) d \,f^{2}}{g^{3} \left (d g -e f \right )}+\frac {b e n \ln \left (\left (g x +f \right ) e +d g -e f \right ) f^{3}}{g^{4} \left (d g -e f \right )}+\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 {\frac {1}{2} g \,x^{2}-2 f x}{g^{3}}+\frac {f^{3}}{g^{4} \left (g x +f \right )}+\frac {3 f^{2} \ln \left (g x +f \right )}{g^{4}}\right )\) | \(550\) |
1/2*b*ln((e*x+d)^n)/g^2*x^2-2*b*ln((e*x+d)^n)/g^3*x*f+b*ln((e*x+d)^n)*f^3/ g^4/(g*x+f)+3*b*ln((e*x+d)^n)/g^4*f^2*ln(g*x+f)-3*b*n/g^4*f^2*dilog(((g*x+ f)*e+d*g-e*f)/(d*g-e*f))-3*b*n/g^4*f^2*ln(g*x+f)*ln(((g*x+f)*e+d*g-e*f)/(d *g-e*f))-1/4*b*n*x^2/g^2+2*b*f*n*x/g^3+9/4*b*n/g^4*f^2+1/2*b*d*n*x/e/g^2+1 /2*b/e*n/g^3*d*f-b*e*n/g^4*f^3/(d*g-e*f)*ln(g*x+f)-1/2*b/e^2*n/g/(d*g-e*f) *ln((g*x+f)*e+d*g-e*f)*d^3-3/2*b/e*n/g^2/(d*g-e*f)*ln((g*x+f)*e+d*g-e*f)*d ^2*f+2*b*n/g^3/(d*g-e*f)*ln((g*x+f)*e+d*g-e*f)*d*f^2+b*e*n/g^4/(d*g-e*f)*l n((g*x+f)*e+d*g-e*f)*f^3+(-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/g^3*(1/2*g*x^2-2*f*x)+f^3/g^4/(g*x+f)+3/g^4*f^2*ln(g*x+f))
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3}}{{\left (g x + f\right )}^{2}} \,d x } \]
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\int \frac {x^{3} \left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )}{\left (f + g x\right )^{2}}\, dx \]
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3}}{{\left (g x + f\right )}^{2}} \,d x } \]
1/2*(2*f^3/(g^5*x + f*g^4) + 6*f^2*log(g*x + f)/g^4 + (g*x^2 - 4*f*x)/g^3) *a + b*integrate((x^3*log((e*x + d)^n) + x^3*log(c))/(g^2*x^2 + 2*f*g*x + f^2), x)
\[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{3}}{{\left (g x + f\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{(f+g x)^2} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{{\left (f+g\,x\right )}^2} \,d x \]