Integrand size = 23, antiderivative size = 148 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {a e^2 x}{d^3}-\frac {b e^2 n x}{d^3}+\frac {b e n x^2}{4 d^2}-\frac {b n x^3}{9 d}+\frac {b e^2 x \log \left (c x^n\right )}{d^3}-\frac {e x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{d^4}-\frac {b e^3 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^4} \]
a*e^2*x/d^3-b*e^2*n*x/d^3+1/4*b*e*n*x^2/d^2-1/9*b*n*x^3/d+b*e^2*x*ln(c*x^n )/d^3-1/2*e*x^2*(a+b*ln(c*x^n))/d^2+1/3*x^3*(a+b*ln(c*x^n))/d-e^3*(a+b*ln( c*x^n))*ln(1+d*x/e)/d^4-b*e^3*n*polylog(2,-d*x/e)/d^4
Time = 0.05 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.96 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {36 a d e^2 x-36 b d e^2 n x-18 a d^2 e x^2+9 b d^2 e n x^2+12 a d^3 x^3-4 b d^3 n x^3-36 a e^3 \log \left (1+\frac {d x}{e}\right )+6 b \log \left (c x^n\right ) \left (d x \left (6 e^2-3 d e x+2 d^2 x^2\right )-6 e^3 \log \left (1+\frac {d x}{e}\right )\right )-36 b e^3 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{36 d^4} \]
(36*a*d*e^2*x - 36*b*d*e^2*n*x - 18*a*d^2*e*x^2 + 9*b*d^2*e*n*x^2 + 12*a*d ^3*x^3 - 4*b*d^3*n*x^3 - 36*a*e^3*Log[1 + (d*x)/e] + 6*b*Log[c*x^n]*(d*x*( 6*e^2 - 3*d*e*x + 2*d^2*x^2) - 6*e^3*Log[1 + (d*x)/e]) - 36*b*e^3*n*PolyLo g[2, -((d*x)/e)])/(36*d^4)
Time = 0.38 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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.
| \(\displaystyle \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx\) |
| \(\Big \downarrow \) 2005 |
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d x+e}dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )}{d^3 (d x+e)}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {e^3 \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac {e x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d}+\frac {a e^2 x}{d^3}+\frac {b e^2 x \log \left (c x^n\right )}{d^3}-\frac {b e^3 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^4}-\frac {b e^2 n x}{d^3}+\frac {b e n x^2}{4 d^2}-\frac {b n x^3}{9 d}\) |
(a*e^2*x)/d^3 - (b*e^2*n*x)/d^3 + (b*e*n*x^2)/(4*d^2) - (b*n*x^3)/(9*d) + (b*e^2*x*Log[c*x^n])/d^3 - (e*x^2*(a + b*Log[c*x^n]))/(2*d^2) + (x^3*(a + b*Log[c*x^n]))/(3*d) - (e^3*(a + b*Log[c*x^n])*Log[1 + (d*x)/e])/d^4 - (b* e^3*n*PolyLog[2, -((d*x)/e)])/d^4
3.4.30.3.1 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]))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.18 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.84
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) x^{3}}{3 d}-\frac {b \ln \left (x^{n}\right ) e \,x^{2}}{2 d^{2}}+\frac {b \ln \left (x^{n}\right ) x \,e^{2}}{d^{3}}-\frac {b \ln \left (x^{n}\right ) e^{3} \ln \left (d x +e \right )}{d^{4}}-\frac {b n \,x^{3}}{9 d}+\frac {b e n \,x^{2}}{4 d^{2}}-\frac {b \,e^{2} n x}{d^{3}}-\frac {49 b n \,e^{3}}{36 d^{4}}+\frac {b n \,e^{3} \ln \left (d x +e \right ) \ln \left (-\frac {d x}{e}\right )}{d^{4}}+\frac {b n \,e^{3} \operatorname {dilog}\left (-\frac {d x}{e}\right )}{d^{4}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{3} d^{2} x^{3}-\frac {1}{2} d e \,x^{2}+x \,e^{2}}{d^{3}}-\frac {e^{3} \ln \left (d x +e \right )}{d^{4}}\right )\) | \(272\) |
1/3*b*ln(x^n)/d*x^3-1/2*b*ln(x^n)/d^2*e*x^2+b*ln(x^n)/d^3*x*e^2-b*ln(x^n)* e^3/d^4*ln(d*x+e)-1/9*b*n*x^3/d+1/4*b*e*n*x^2/d^2-b*e^2*n*x/d^3-49/36*b*n* e^3/d^4+b*n*e^3/d^4*ln(d*x+e)*ln(-d*x/e)+b*n*e^3/d^4*dilog(-d*x/e)+(-1/2*I *b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*x^ n)^2+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*b*Pi*csgn(I*c*x^n)^3+b*l n(c)+a)*(1/d^3*(1/3*d^2*x^3-1/2*d*e*x^2+x*e^2)-e^3/d^4*ln(d*x+e))
\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{d + \frac {e}{x}} \,d x } \]
Time = 80.12 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.80 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {a x^{3}}{3 d} - \frac {a e x^{2}}{2 d^{2}} - \frac {a e^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {a e^{2} x}{d^{3}} - \frac {b n x^{3}}{9 d} + \frac {b x^{3} \log {\left (c x^{n} \right )}}{3 d} + \frac {b e n x^{2}}{4 d^{2}} - \frac {b e x^{2} \log {\left (c x^{n} \right )}}{2 d^{2}} + \frac {b e^{3} n \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{3}} - \frac {b e^{3} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{3}} - \frac {b e^{2} n x}{d^{3}} + \frac {b e^{2} x \log {\left (c x^{n} \right )}}{d^{3}} \]
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*n*x**3/(9*d) + b*x**3*log(c*x* *n)/(3*d) + b*e*n*x**2/(4*d**2) - b*e*x**2*log(c*x**n)/(2*d**2) + b*e**3*n *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_pola r(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) + mei jerg(((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**n)/d**3 - b*e**2*n*x/d**3 + b*e**2*x*log(c*x**n)/d* *3
\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{d + \frac {e}{x}} \,d x } \]
-1/6*a*(6*e^3*log(d*x + e)/d^4 - (2*d^2*x^3 - 3*d*e*x^2 + 6*e^2*x)/d^3) + b*integrate((x^3*log(c) + x^3*log(x^n))/(d*x + e), x)
\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{d + \frac {e}{x}} \,d x } \]
Timed out. \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int \frac {x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{d+\frac {e}{x}} \,d x \]