Integrand size = 23, antiderivative size = 297 \[ \int \frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=-\frac {b d p x}{2 a e^2}-\frac {b^2 p x}{3 a^2 e}+\frac {b p x^2}{6 a e}+\frac {d^2 x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^3}-\frac {d x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e^2}+\frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 e}+\frac {b d^2 p \log (b+a x)}{a e^3}+\frac {b^2 d p \log (b+a x)}{2 a^2 e^2}+\frac {b^3 p \log (b+a x)}{3 a^3 e}-\frac {d^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e^4}-\frac {d^3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^4}+\frac {d^3 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e^4}+\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e^4}-\frac {d^3 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e^4} \]
-1/2*b*d*p*x/a/e^2-1/3*b^2*p*x/a^2/e+1/6*b*p*x^2/a/e+d^2*x*ln(c*(a+b/x)^p) /e^3-1/2*d*x^2*ln(c*(a+b/x)^p)/e^2+1/3*x^3*ln(c*(a+b/x)^p)/e+b*d^2*p*ln(a* x+b)/a/e^3+1/2*b^2*d*p*ln(a*x+b)/a^2/e^2+1/3*b^3*p*ln(a*x+b)/a^3/e-d^3*ln( c*(a+b/x)^p)*ln(e*x+d)/e^4-d^3*p*ln(-e*x/d)*ln(e*x+d)/e^4+d^3*p*ln(-e*(a*x +b)/(a*d-b*e))*ln(e*x+d)/e^4+d^3*p*polylog(2,a*(e*x+d)/(a*d-b*e))/e^4-d^3* p*polylog(2,1+e*x/d)/e^4
Time = 0.10 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.04 \[ \int \frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {b d^2 p \log \left (a+\frac {b}{x}\right )}{a e^3}+\frac {d^2 x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^3}-\frac {d x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e^2}+\frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 e}+\frac {b d^2 p \log (x)}{a e^3}-\frac {b p \left (\frac {2 b x}{a^2}-\frac {x^2}{a}-\frac {2 b^2 \log \left (a+\frac {b}{x}\right )}{a^3}-\frac {2 b^2 \log (x)}{a^3}\right )}{6 e}-\frac {b d p \left (\frac {x}{a}-\frac {b \log (b+a x)}{a^2}\right )}{2 e^2}-\frac {d^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e^4}-\frac {d^3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^4}+\frac {d^3 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e^4}-\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{e^4}+\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e^4} \]
(b*d^2*p*Log[a + b/x])/(a*e^3) + (d^2*x*Log[c*(a + b/x)^p])/e^3 - (d*x^2*L og[c*(a + b/x)^p])/(2*e^2) + (x^3*Log[c*(a + b/x)^p])/(3*e) + (b*d^2*p*Log [x])/(a*e^3) - (b*p*((2*b*x)/a^2 - x^2/a - (2*b^2*Log[a + b/x])/a^3 - (2*b ^2*Log[x])/a^3))/(6*e) - (b*d*p*(x/a - (b*Log[b + a*x])/a^2))/(2*e^2) - (d ^3*Log[c*(a + b/x)^p]*Log[d + e*x])/e^4 - (d^3*p*Log[-((e*x)/d)]*Log[d + e *x])/e^4 + (d^3*p*Log[-((e*(b + a*x))/(a*d - b*e))]*Log[d + e*x])/e^4 - (d ^3*p*PolyLog[2, (d + e*x)/d])/e^4 + (d^3*p*PolyLog[2, (a*(d + e*x))/(a*d - b*e)])/e^4
Time = 0.57 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2916, 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.
\(\displaystyle \int \frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx\) |
\(\Big \downarrow \) 2916 |
\(\displaystyle \int \left (-\frac {d^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^3 (d+e x)}+\frac {d^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^3}-\frac {d x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^2}+\frac {x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^3 p \log (a x+b)}{3 a^3 e}+\frac {b^2 d p \log (a x+b)}{2 a^2 e^2}-\frac {b^2 p x}{3 a^2 e}-\frac {d^3 \log (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^4}+\frac {d^2 x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^3}-\frac {d x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e^2}+\frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 e}+\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e^4}+\frac {d^3 p \log (d+e x) \log \left (-\frac {e (a x+b)}{a d-b e}\right )}{e^4}+\frac {b d^2 p \log (a x+b)}{a e^3}-\frac {b d p x}{2 a e^2}+\frac {b p x^2}{6 a e}-\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{e^4}-\frac {d^3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^4}\) |
-1/2*(b*d*p*x)/(a*e^2) - (b^2*p*x)/(3*a^2*e) + (b*p*x^2)/(6*a*e) + (d^2*x* Log[c*(a + b/x)^p])/e^3 - (d*x^2*Log[c*(a + b/x)^p])/(2*e^2) + (x^3*Log[c* (a + b/x)^p])/(3*e) + (b*d^2*p*Log[b + a*x])/(a*e^3) + (b^2*d*p*Log[b + a* x])/(2*a^2*e^2) + (b^3*p*Log[b + a*x])/(3*a^3*e) - (d^3*Log[c*(a + b/x)^p] *Log[d + e*x])/e^4 - (d^3*p*Log[-((e*x)/d)]*Log[d + e*x])/e^4 + (d^3*p*Log [-((e*(b + a*x))/(a*d - b*e))]*Log[d + e*x])/e^4 + (d^3*p*PolyLog[2, (a*(d + e*x))/(a*d - b*e)])/e^4 - (d^3*p*PolyLog[2, 1 + (e*x)/d])/e^4
3.3.40.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log [c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e, f, g , n, p, q}, x] && IntegerQ[m] && IntegerQ[r]
Time = 1.84 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.01
method | result | size |
parts | \(\frac {x^{3} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{3 e}-\frac {d \,x^{2} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{2 e^{2}}+\frac {d^{2} x \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{e^{3}}-\frac {d^{3} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) \ln \left (e x +d \right )}{e^{4}}+p b e \left (-\frac {\frac {5 a d \left (e x +d \right )-a \left (e x +d \right )^{2}+2 \left (e x +d \right ) b e}{a^{2}}+\frac {\left (-6 a^{2} d^{2}-3 a d e b -2 e^{2} b^{2}\right ) \ln \left (a d -a \left (e x +d \right )-b e \right )}{a^{3}}}{6 e^{4}}+\frac {d^{3} \operatorname {dilog}\left (\frac {-a d +a \left (e x +d \right )+b e}{-a d +b e}\right )}{e^{5} b}+\frac {d^{3} \ln \left (e x +d \right ) \ln \left (\frac {-a d +a \left (e x +d \right )+b e}{-a d +b e}\right )}{e^{5} b}-\frac {d^{3} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{5} b}-\frac {d^{3} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{5} b}\right )\) | \(301\) |
1/3*x^3*ln(c*(a+b/x)^p)/e-1/2*d*x^2*ln(c*(a+b/x)^p)/e^2+d^2*x*ln(c*(a+b/x) ^p)/e^3-d^3*ln(c*(a+b/x)^p)*ln(e*x+d)/e^4+p*b*e*(-1/6/e^4*(1/a^2*(5*a*d*(e *x+d)-a*(e*x+d)^2+2*(e*x+d)*b*e)+(-6*a^2*d^2-3*a*b*d*e-2*b^2*e^2)/a^3*ln(a *d-a*(e*x+d)-b*e))+1/e^5*d^3/b*dilog((-a*d+a*(e*x+d)+b*e)/(-a*d+b*e))+1/e^ 5*d^3/b*ln(e*x+d)*ln((-a*d+a*(e*x+d)+b*e)/(-a*d+b*e))-1/e^5*d^3/b*ln(e*x+d )*ln(-e*x/d)-1/e^5*d^3/b*dilog(-e*x/d))
\[ \int \frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\int { \frac {x^{3} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{e x + d} \,d x } \]
Timed out. \[ \int \frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\int { \frac {x^{3} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{e x + d} \,d x } \]
\[ \int \frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\int { \frac {x^{3} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{e x + d} \,d x } \]
Timed out. \[ \int \frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\int \frac {x^3\,\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{d+e\,x} \,d x \]