Integrand size = 21, antiderivative size = 136 \[ \int \frac {x^2 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\frac {a e^2 x}{d^3}-\frac {b e^2 x}{d^3}+\frac {b e x^2}{4 d^2}-\frac {b x^3}{9 d}+\frac {b e^2 x \log (c x)}{d^3}-\frac {e x^2 (a+b \log (c x))}{2 d^2}+\frac {x^3 (a+b \log (c x))}{3 d}-\frac {e^3 (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{d^4}-\frac {b e^3 \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^4} \]
a*e^2*x/d^3-b*e^2*x/d^3+1/4*b*e*x^2/d^2-1/9*b*x^3/d+b*e^2*x*ln(c*x)/d^3-1/ 2*e*x^2*(a+b*ln(c*x))/d^2+1/3*x^3*(a+b*ln(c*x))/d-e^3*(a+b*ln(c*x))*ln(1+d *x/e)/d^4-b*e^3*polylog(2,-d*x/e)/d^4
Time = 0.04 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.92 \[ \int \frac {x^2 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\frac {36 a d e^2 x-36 b d e^2 x+9 b d^2 e x^2-4 b d^3 x^3+36 b d e^2 x \log (c x)-18 d^2 e x^2 (a+b \log (c x))+12 d^3 x^3 (a+b \log (c x))-36 e^3 (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )-36 b e^3 \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{36 d^4} \]
(36*a*d*e^2*x - 36*b*d*e^2*x + 9*b*d^2*e*x^2 - 4*b*d^3*x^3 + 36*b*d*e^2*x* Log[c*x] - 18*d^2*e*x^2*(a + b*Log[c*x]) + 12*d^3*x^3*(a + b*Log[c*x]) - 3 6*e^3*(a + b*Log[c*x])*Log[1 + (d*x)/e] - 36*b*e^3*PolyLog[2, -((d*x)/e)]) /(36*d^4)
Time = 0.36 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2005, 2793, 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.
| \(\displaystyle \int \frac {x^2 (a+b \log (c x))}{d+\frac {e}{x}} \, dx\) |
| \(\Big \downarrow \) 2005 |
| \(\displaystyle \int \frac {x^3 (a+b \log (c x))}{d x+e}dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (-\frac {e^3 (a+b \log (c x))}{d^3 (d x+e)}+\frac {e^2 (a+b \log (c x))}{d^3}-\frac {e x (a+b \log (c x))}{d^2}+\frac {x^2 (a+b \log (c x))}{d}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {e^3 \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))}{d^4}-\frac {e x^2 (a+b \log (c x))}{2 d^2}+\frac {x^3 (a+b \log (c x))}{3 d}+\frac {a e^2 x}{d^3}+\frac {b e^2 x \log (c x)}{d^3}-\frac {b e^3 \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^4}-\frac {b e^2 x}{d^3}+\frac {b e x^2}{4 d^2}-\frac {b x^3}{9 d}\) |
(a*e^2*x)/d^3 - (b*e^2*x)/d^3 + (b*e*x^2)/(4*d^2) - (b*x^3)/(9*d) + (b*e^2 *x*Log[c*x])/d^3 - (e*x^2*(a + b*Log[c*x]))/(2*d^2) + (x^3*(a + b*Log[c*x] ))/(3*d) - (e^3*(a + b*Log[c*x])*Log[1 + (d*x)/e])/d^4 - (b*e^3*PolyLog[2, -((d*x)/e)])/d^4
3.4.38.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg Q[n]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && Integer Q[r]))
Time = 0.05 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.24
| method | result | size |
| risch | \(\frac {a \,x^{3}}{3 d}-\frac {a e \,x^{2}}{2 d^{2}}+\frac {a \,e^{2} x}{d^{3}}-\frac {a \,e^{3} \ln \left (d x +e \right )}{d^{4}}+\frac {b \,x^{3} \ln \left (x c \right )}{3 d}-\frac {b \,x^{3}}{9 d}-\frac {b e \,x^{2} \ln \left (x c \right )}{2 d^{2}}+\frac {b e \,x^{2}}{4 d^{2}}+\frac {b \,e^{2} x \ln \left (x c \right )}{d^{3}}-\frac {b \,e^{2} x}{d^{3}}-\frac {b \,e^{3} \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d^{4}}-\frac {b \,e^{3} \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d^{4}}\) | \(168\) |
| parts | \(\frac {a \,x^{3}}{3 d}-\frac {a e \,x^{2}}{2 d^{2}}+\frac {a \,e^{2} x}{d^{3}}-\frac {a \,e^{3} \ln \left (d x +e \right )}{d^{4}}+\frac {b \,x^{3} \ln \left (x c \right )}{3 d}-\frac {b \,x^{3}}{9 d}-\frac {b e \,x^{2} \ln \left (x c \right )}{2 d^{2}}+\frac {b e \,x^{2}}{4 d^{2}}+\frac {b \,e^{2} x \ln \left (x c \right )}{d^{3}}-\frac {b \,e^{2} x}{d^{3}}-\frac {b \,e^{3} \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d^{4}}-\frac {b \,e^{3} \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d^{4}}\) | \(168\) |
| derivativedivides | \(\frac {\frac {a \,c^{3} e^{2} x}{d^{3}}-\frac {a \,c^{3} e \,x^{2}}{2 d^{2}}+\frac {a \,x^{3} c^{3}}{3 d}-\frac {a \,c^{3} e^{3} \ln \left (c d x +c e \right )}{d^{4}}+b \left (\frac {\frac {x^{3} c^{3} \ln \left (x c \right )}{3}-\frac {x^{3} c^{3}}{9}}{d}-\frac {c e \left (\frac {x^{2} c^{2} \ln \left (x c \right )}{2}-\frac {x^{2} c^{2}}{4}\right )}{d^{2}}+\frac {c^{2} e^{2} \left (x c \ln \left (x c \right )-x c \right )}{d^{3}}-\frac {c^{3} e^{3} \left (\frac {\operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {\ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\right )}{d^{3}}\right )}{c^{3}}\) | \(198\) |
| default | \(\frac {\frac {a \,c^{3} e^{2} x}{d^{3}}-\frac {a \,c^{3} e \,x^{2}}{2 d^{2}}+\frac {a \,x^{3} c^{3}}{3 d}-\frac {a \,c^{3} e^{3} \ln \left (c d x +c e \right )}{d^{4}}+b \left (\frac {\frac {x^{3} c^{3} \ln \left (x c \right )}{3}-\frac {x^{3} c^{3}}{9}}{d}-\frac {c e \left (\frac {x^{2} c^{2} \ln \left (x c \right )}{2}-\frac {x^{2} c^{2}}{4}\right )}{d^{2}}+\frac {c^{2} e^{2} \left (x c \ln \left (x c \right )-x c \right )}{d^{3}}-\frac {c^{3} e^{3} \left (\frac {\operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {\ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\right )}{d^{3}}\right )}{c^{3}}\) | \(198\) |
1/3*a/d*x^3-1/2*a*e/d^2*x^2+a*e^2*x/d^3-a*e^3/d^4*ln(d*x+e)+1/3*b/d*x^3*ln (x*c)-1/9*b*x^3/d-1/2*b/d^2*e*x^2*ln(x*c)+1/4*b*e*x^2/d^2+b*e^2*x*ln(x*c)/ d^3-b*e^2*x/d^3-b*e^3/d^4*dilog((c*d*x+c*e)/e/c)-b*e^3/d^4*ln(x*c)*ln((c*d *x+c*e)/e/c)
\[ \int \frac {x^2 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x\right ) + a\right )} x^{2}}{d + \frac {e}{x}} \,d x } \]
Time = 79.52 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.86 \[ \int \frac {x^2 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\frac {a x^{3}}{3 d} - \frac {a e x^{2}}{2 d^{2}} - \frac {a e^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {a e^{2} x}{d^{3}} + \frac {b x^{3} \log {\left (c x \right )}}{3 d} - \frac {b x^{3}}{9 d} - \frac {b e x^{2} \log {\left (c x \right )}}{2 d^{2}} + \frac {b e x^{2}}{4 d^{2}} + \frac {b e^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}} - \frac {b e^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x \right )}}{d^{3}} + \frac {b e^{2} x \log {\left (c x \right )}}{d^{3}} - \frac {b e^{2} x}{d^{3}} \]
a*x**3/(3*d) - a*e*x**2/(2*d**2) - a*e**3*Piecewise((x/e, Eq(d, 0)), (log( d*x + e)/d, True))/d**3 + a*e**2*x/d**3 + b*x**3*log(c*x)/(3*d) - b*x**3/( 9*d) - b*e*x**2*log(c*x)/(2*d**2) + b*e*x**2/(4*d**2) + b*e**3*Piecewise(( x/e, Eq(d, 0)), (Piecewise((-polylog(2, d*x*exp_polar(I*pi)/e), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(e)*log(x) - polylog(2, d*x*exp_polar(I*pi)/e), Abs(x) < 1), (-log(e)*log(1/x) - polylog(2, d*x*exp_polar(I*pi)/e), 1/Abs( x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(e) + meijerg(((1, 1) , ()), ((), (0, 0)), x)*log(e) - polylog(2, d*x*exp_polar(I*pi)/e), True)) /d, True))/d**3 - b*e**3*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True) )*log(c*x)/d**3 + b*e**2*x*log(c*x)/d**3 - b*e**2*x/d**3
Time = 0.23 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.21 \[ \int \frac {x^2 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=-\frac {{\left (\log \left (\frac {d x}{e} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {d x}{e}\right )\right )} b e^{3}}{d^{4}} + \frac {4 \, {\left (3 \, a d^{2} + {\left (3 \, d^{2} \log \left (c\right ) - d^{2}\right )} b\right )} x^{3} - 9 \, {\left (2 \, a d e + {\left (2 \, d e \log \left (c\right ) - d e\right )} b\right )} x^{2} + 36 \, {\left (a e^{2} + {\left (e^{2} \log \left (c\right ) - e^{2}\right )} b\right )} x + 6 \, {\left (2 \, b d^{2} x^{3} - 3 \, b d e x^{2} + 6 \, b e^{2} x\right )} \log \left (x\right )}{36 \, d^{3}} - \frac {{\left (b e^{3} \log \left (c\right ) + a e^{3}\right )} \log \left (d x + e\right )}{d^{4}} \]
-(log(d*x/e + 1)*log(x) + dilog(-d*x/e))*b*e^3/d^4 + 1/36*(4*(3*a*d^2 + (3 *d^2*log(c) - d^2)*b)*x^3 - 9*(2*a*d*e + (2*d*e*log(c) - d*e)*b)*x^2 + 36* (a*e^2 + (e^2*log(c) - e^2)*b)*x + 6*(2*b*d^2*x^3 - 3*b*d*e*x^2 + 6*b*e^2* x)*log(x))/d^3 - (b*e^3*log(c) + a*e^3)*log(d*x + e)/d^4
\[ \int \frac {x^2 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x\right ) + a\right )} x^{2}}{d + \frac {e}{x}} \,d x } \]
Timed out. \[ \int \frac {x^2 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\int \frac {x^2\,\left (a+b\,\ln \left (c\,x\right )\right )}{d+\frac {e}{x}} \,d x \]