Integrand size = 25, antiderivative size = 335 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=-\frac {b e n}{2 d f^2 x}-\frac {b e^2 n \log (x)}{2 d^2 f^2}-\frac {2 b e g n \log (x)}{d f^3}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}+\frac {2 b e g n \log (d+e x)}{d f^3}-\frac {b e g^2 n \log (d+e x)}{f^3 (e f-d g)}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac {b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f^4}-\frac {3 b g^2 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{f^4} \]
-1/2*b*e*n/d/f^2/x-1/2*b*e^2*n*ln(x)/d^2/f^2-2*b*e*g*n*ln(x)/d/f^3+1/2*b*e ^2*n*ln(e*x+d)/d^2/f^2+2*b*e*g*n*ln(e*x+d)/d/f^3-b*e*g^2*n*ln(e*x+d)/f^3/( -d*g+e*f)+1/2*(-a-b*ln(c*(e*x+d)^n))/f^2/x^2+2*g*(a+b*ln(c*(e*x+d)^n))/f^3 /x+g^2*(a+b*ln(c*(e*x+d)^n))/f^3/(g*x+f)+3*g^2*ln(-e*x/d)*(a+b*ln(c*(e*x+d )^n))/f^4+b*e*g^2*n*ln(g*x+f)/f^3/(-d*g+e*f)-3*g^2*(a+b*ln(c*(e*x+d)^n))*l n(e*(g*x+f)/(-d*g+e*f))/f^4-3*b*g^2*n*polylog(2,-g*(e*x+d)/(-d*g+e*f))/f^4 +3*b*g^2*n*polylog(2,1+e*x/d)/f^4
Time = 0.25 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.80 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=-\frac {\frac {4 b e f g n (\log (x)-\log (d+e x))}{d}+\frac {b e f^2 n (d+e x \log (x)-e x \log (d+e x))}{d^2 x}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^2}-\frac {4 f g \left (a+b \log \left (c (d+e x)^n\right )\right )}{x}-\frac {2 f g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}-6 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {2 b e f g^2 n (\log (d+e x)-\log (f+g x))}{e f-d g}+6 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )+6 b g^2 n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )-6 b g^2 n \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{2 f^4} \]
-1/2*((4*b*e*f*g*n*(Log[x] - Log[d + e*x]))/d + (b*e*f^2*n*(d + e*x*Log[x] - e*x*Log[d + e*x]))/(d^2*x) + (f^2*(a + b*Log[c*(d + e*x)^n]))/x^2 - (4* f*g*(a + b*Log[c*(d + e*x)^n]))/x - (2*f*g^2*(a + b*Log[c*(d + e*x)^n]))/( f + g*x) - 6*g^2*Log[-((e*x)/d)]*(a + b*Log[c*(d + e*x)^n]) + (2*b*e*f*g^2 *n*(Log[d + e*x] - Log[f + g*x]))/(e*f - d*g) + 6*g^2*(a + b*Log[c*(d + e* x)^n])*Log[(e*(f + g*x))/(e*f - d*g)] + 6*b*g^2*n*PolyLog[2, (g*(d + e*x)) /(-(e*f) + d*g)] - 6*b*g^2*n*PolyLog[2, 1 + (e*x)/d])/f^4
Time = 0.59 (sec) , antiderivative size = 335, 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.
\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \int \left (-\frac {3 g^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4 (f+g x)}+\frac {3 g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4 x}-\frac {g^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)^2}-\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x^2}+\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 g^2 \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}-\frac {3 g^2 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^4}+\frac {g^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 (f+g x)}+\frac {2 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^3 x}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 f^2 x^2}-\frac {b e^2 n \log (x)}{2 d^2 f^2}+\frac {b e^2 n \log (d+e x)}{2 d^2 f^2}-\frac {3 b g^2 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^4}+\frac {3 b g^2 n \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^4}-\frac {b e g^2 n \log (d+e x)}{f^3 (e f-d g)}+\frac {b e g^2 n \log (f+g x)}{f^3 (e f-d g)}-\frac {2 b e g n \log (x)}{d f^3}+\frac {2 b e g n \log (d+e x)}{d f^3}-\frac {b e n}{2 d f^2 x}\) |
-1/2*(b*e*n)/(d*f^2*x) - (b*e^2*n*Log[x])/(2*d^2*f^2) - (2*b*e*g*n*Log[x]) /(d*f^3) + (b*e^2*n*Log[d + e*x])/(2*d^2*f^2) + (2*b*e*g*n*Log[d + e*x])/( d*f^3) - (b*e*g^2*n*Log[d + e*x])/(f^3*(e*f - d*g)) - (a + b*Log[c*(d + e* x)^n])/(2*f^2*x^2) + (2*g*(a + b*Log[c*(d + e*x)^n]))/(f^3*x) + (g^2*(a + b*Log[c*(d + e*x)^n]))/(f^3*(f + g*x)) + (3*g^2*Log[-((e*x)/d)]*(a + b*Log [c*(d + e*x)^n]))/f^4 + (b*e*g^2*n*Log[f + g*x])/(f^3*(e*f - d*g)) - (3*g^ 2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/f^4 - (3*b*g^ 2*n*PolyLog[2, -((g*(d + e*x))/(e*f - d*g))])/f^4 + (3*b*g^2*n*PolyLog[2, 1 + (e*x)/d])/f^4
3.3.55.3.1 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 = 1.66 (sec) , antiderivative size = 549, normalized size of antiderivative = 1.64
method | result | size |
risch | \(-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{2 f^{2} x^{2}}+\frac {3 b \ln \left (\left (e x +d \right )^{n}\right ) g^{2} \ln \left (x \right )}{f^{4}}+\frac {2 b \ln \left (\left (e x +d \right )^{n}\right ) g}{f^{3} x}-\frac {3 b \ln \left (\left (e x +d \right )^{n}\right ) g^{2} \ln \left (g x +f \right )}{f^{4}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g^{2}}{f^{3} \left (g x +f \right )}+\frac {3 b e n \ln \left (e x +d \right ) g^{2}}{f^{3} \left (d g -e f \right )}-\frac {3 b \,e^{2} n \ln \left (e x +d \right ) g}{2 f^{2} \left (d g -e f \right ) d}-\frac {b \,e^{3} n \ln \left (e x +d \right )}{2 f \left (d g -e f \right ) d^{2}}-\frac {2 b e g n \ln \left (x \right )}{d \,f^{3}}-\frac {b \,e^{2} n \ln \left (x \right )}{2 d^{2} f^{2}}-\frac {b e n}{2 d \,f^{2} x}-\frac {b e n \,g^{2} \ln \left (g x +f \right )}{f^{3} \left (d g -e f \right )}-\frac {3 b n \,g^{2} \operatorname {dilog}\left (\frac {e x +d}{d}\right )}{f^{4}}-\frac {3 b n \,g^{2} \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{4}}+\frac {3 b n \,g^{2} \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f^{4}}+\frac {3 b n \,g^{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 )}{f^{4}}+\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 {3 g^{2} \ln \left (x \right )}{f^{4}}+\frac {2 g}{f^{3} x}-\frac {3 g^{2} \ln \left (g x +f \right )}{f^{4}}+\frac {g^{2}}{f^{3} \left (g x +f \right )}\right )\) | \(549\) |
-1/2*b*ln((e*x+d)^n)/f^2/x^2+3*b*ln((e*x+d)^n)/f^4*g^2*ln(x)+2*b*ln((e*x+d )^n)/f^3*g/x-3*b*ln((e*x+d)^n)/f^4*g^2*ln(g*x+f)+b*ln((e*x+d)^n)/f^3*g^2/( g*x+f)+3*b*e*n/f^3/(d*g-e*f)*ln(e*x+d)*g^2-3/2*b*e^2*n/f^2/(d*g-e*f)/d*ln( e*x+d)*g-1/2*b*e^3*n/f/(d*g-e*f)/d^2*ln(e*x+d)-2*b*e*g*n*ln(x)/d/f^3-1/2*b *e^2*n*ln(x)/d^2/f^2-1/2*b*e*n/d/f^2/x-b*e*n/f^3*g^2/(d*g-e*f)*ln(g*x+f)-3 *b*n/f^4*g^2*dilog((e*x+d)/d)-3*b*n/f^4*g^2*ln(x)*ln((e*x+d)/d)+3*b*n/f^4* g^2*dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+3*b*n/f^4*g^2*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 )*(-1/2/f^2/x^2+3/f^4*g^2*ln(x)+2/f^3*g/x-3/f^4*g^2*ln(g*x+f)+1/f^3*g^2/(g *x+f))
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x^{3}} \,d x } \]
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int \frac {a + b \log {\left (c \left (d + e x\right )^{n} \right )}}{x^{3} \left (f + g x\right )^{2}}\, dx \]
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x^{3}} \,d x } \]
1/2*a*((6*g^2*x^2 + 3*f*g*x - f^2)/(f^3*g*x^3 + f^4*x^2) - 6*g^2*log(g*x + f)/f^4 + 6*g^2*log(x)/f^4) + b*integrate((log((e*x + d)^n) + log(c))/(g^2 *x^5 + 2*f*g*x^4 + f^2*x^3), x)
\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^3 (f+g x)^2} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^3\,{\left (f+g\,x\right )}^2} \,d x \]