Integrand size = 21, antiderivative size = 170 \[ \int \frac {x^3 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=-\frac {a e^3 x}{d^4}+\frac {b e^3 x}{d^4}-\frac {b e^2 x^2}{4 d^3}+\frac {b e x^3}{9 d^2}-\frac {b x^4}{16 d}-\frac {b e^3 x \log (c x)}{d^4}+\frac {e^2 x^2 (a+b \log (c x))}{2 d^3}-\frac {e x^3 (a+b \log (c x))}{3 d^2}+\frac {x^4 (a+b \log (c x))}{4 d}+\frac {e^4 (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{d^5}+\frac {b e^4 \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^5} \]
-a*e^3*x/d^4+b*e^3*x/d^4-1/4*b*e^2*x^2/d^3+1/9*b*e*x^3/d^2-1/16*b*x^4/d-b* e^3*x*ln(c*x)/d^4+1/2*e^2*x^2*(a+b*ln(c*x))/d^3-1/3*e*x^3*(a+b*ln(c*x))/d^ 2+1/4*x^4*(a+b*ln(c*x))/d+e^4*(a+b*ln(c*x))*ln(1+d*x/e)/d^5+b*e^4*polylog( 2,-d*x/e)/d^5
Time = 0.06 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\frac {-144 a d e^3 x+144 b d e^3 x-36 b d^2 e^2 x^2+16 b d^3 e x^3-9 b d^4 x^4-144 b d e^3 x \log (c x)+72 d^2 e^2 x^2 (a+b \log (c x))-48 d^3 e x^3 (a+b \log (c x))+36 d^4 x^4 (a+b \log (c x))+144 e^4 (a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )+144 b e^4 \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{144 d^5} \]
(-144*a*d*e^3*x + 144*b*d*e^3*x - 36*b*d^2*e^2*x^2 + 16*b*d^3*e*x^3 - 9*b* d^4*x^4 - 144*b*d*e^3*x*Log[c*x] + 72*d^2*e^2*x^2*(a + b*Log[c*x]) - 48*d^ 3*e*x^3*(a + b*Log[c*x]) + 36*d^4*x^4*(a + b*Log[c*x]) + 144*e^4*(a + b*Lo g[c*x])*Log[1 + (d*x)/e] + 144*b*e^4*PolyLog[2, -((d*x)/e)])/(144*d^5)
Time = 0.39 (sec) , antiderivative size = 170, 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.
| \(\displaystyle \int \frac {x^3 (a+b \log (c x))}{d+\frac {e}{x}} \, dx\) |
| \(\Big \downarrow \) 2005 |
| \(\displaystyle \int \frac {x^4 (a+b \log (c x))}{d x+e}dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (\frac {e^4 (a+b \log (c x))}{d^4 (d x+e)}-\frac {e^3 (a+b \log (c x))}{d^4}+\frac {e^2 x (a+b \log (c x))}{d^3}-\frac {e x^2 (a+b \log (c x))}{d^2}+\frac {x^3 (a+b \log (c x))}{d}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^4 \log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))}{d^5}+\frac {e^2 x^2 (a+b \log (c x))}{2 d^3}-\frac {e x^3 (a+b \log (c x))}{3 d^2}+\frac {x^4 (a+b \log (c x))}{4 d}-\frac {a e^3 x}{d^4}-\frac {b e^3 x \log (c x)}{d^4}+\frac {b e^4 \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^5}+\frac {b e^3 x}{d^4}-\frac {b e^2 x^2}{4 d^3}+\frac {b e x^3}{9 d^2}-\frac {b x^4}{16 d}\) |
-((a*e^3*x)/d^4) + (b*e^3*x)/d^4 - (b*e^2*x^2)/(4*d^3) + (b*e*x^3)/(9*d^2) - (b*x^4)/(16*d) - (b*e^3*x*Log[c*x])/d^4 + (e^2*x^2*(a + b*Log[c*x]))/(2 *d^3) - (e*x^3*(a + b*Log[c*x]))/(3*d^2) + (x^4*(a + b*Log[c*x]))/(4*d) + (e^4*(a + b*Log[c*x])*Log[1 + (d*x)/e])/d^5 + (b*e^4*PolyLog[2, -((d*x)/e) ])/d^5
3.4.37.3.1 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 = 0.13 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.20
| method | result | size |
| parts | \(a \left (\frac {\frac {1}{4} d^{3} x^{4}-\frac {1}{3} e \,d^{2} x^{3}+\frac {1}{2} d \,e^{2} x^{2}-x \,e^{3}}{d^{4}}+\frac {e^{4} \ln \left (d x +e \right )}{d^{5}}\right )+\frac {b \,x^{4} \ln \left (x c \right )}{4 d}-\frac {b \,x^{4}}{16 d}-\frac {b e \,x^{3} \ln \left (x c \right )}{3 d^{2}}+\frac {b e \,x^{3}}{9 d^{2}}+\frac {b \,e^{2} x^{2} \ln \left (x c \right )}{2 d^{3}}-\frac {b \,e^{2} x^{2}}{4 d^{3}}-\frac {b \,e^{3} x \ln \left (x c \right )}{d^{4}}+\frac {b \,e^{3} x}{d^{4}}+\frac {b \,e^{4} \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d^{5}}+\frac {b \,e^{4} \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d^{5}}\) | \(204\) |
| risch | \(\frac {a \,x^{4}}{4 d}-\frac {a e \,x^{3}}{3 d^{2}}+\frac {a \,e^{2} x^{2}}{2 d^{3}}-\frac {a \,e^{3} x}{d^{4}}+\frac {a \,e^{4} \ln \left (d x +e \right )}{d^{5}}+\frac {b \,x^{4} \ln \left (x c \right )}{4 d}-\frac {b \,x^{4}}{16 d}-\frac {b e \,x^{3} \ln \left (x c \right )}{3 d^{2}}+\frac {b e \,x^{3}}{9 d^{2}}+\frac {b \,e^{2} x^{2} \ln \left (x c \right )}{2 d^{3}}-\frac {b \,e^{2} x^{2}}{4 d^{3}}-\frac {b \,e^{3} x \ln \left (x c \right )}{d^{4}}+\frac {b \,e^{3} x}{d^{4}}+\frac {b \,e^{4} \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d^{5}}+\frac {b \,e^{4} \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d^{5}}\) | \(206\) |
| derivativedivides | \(\frac {a \left (-\frac {c^{4} e^{3} x -\frac {1}{2} d \,c^{4} e^{2} x^{2}+\frac {1}{3} e \,c^{4} x^{3} d^{2}-\frac {1}{4} x^{4} c^{4} d^{3}}{d^{4}}+\frac {c^{4} e^{4} \ln \left (c d x +c e \right )}{d^{5}}\right )+b \left (\frac {\frac {x^{4} c^{4} \ln \left (x c \right )}{4}-\frac {x^{4} c^{4}}{16}}{d}-\frac {c e \left (\frac {x^{3} c^{3} \ln \left (x c \right )}{3}-\frac {x^{3} c^{3}}{9}\right )}{d^{2}}+\frac {c^{2} e^{2} \left (\frac {x^{2} c^{2} \ln \left (x c \right )}{2}-\frac {x^{2} c^{2}}{4}\right )}{d^{3}}-\frac {c^{3} e^{3} \left (x c \ln \left (x c \right )-x c \right )}{d^{4}}+\frac {c^{4} e^{4} \left (\frac {\operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {\ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\right )}{d^{4}}\right )}{c^{4}}\) | \(242\) |
| default | \(\frac {a \left (-\frac {c^{4} e^{3} x -\frac {1}{2} d \,c^{4} e^{2} x^{2}+\frac {1}{3} e \,c^{4} x^{3} d^{2}-\frac {1}{4} x^{4} c^{4} d^{3}}{d^{4}}+\frac {c^{4} e^{4} \ln \left (c d x +c e \right )}{d^{5}}\right )+b \left (\frac {\frac {x^{4} c^{4} \ln \left (x c \right )}{4}-\frac {x^{4} c^{4}}{16}}{d}-\frac {c e \left (\frac {x^{3} c^{3} \ln \left (x c \right )}{3}-\frac {x^{3} c^{3}}{9}\right )}{d^{2}}+\frac {c^{2} e^{2} \left (\frac {x^{2} c^{2} \ln \left (x c \right )}{2}-\frac {x^{2} c^{2}}{4}\right )}{d^{3}}-\frac {c^{3} e^{3} \left (x c \ln \left (x c \right )-x c \right )}{d^{4}}+\frac {c^{4} e^{4} \left (\frac {\operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {\ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\right )}{d^{4}}\right )}{c^{4}}\) | \(242\) |
a*(1/d^4*(1/4*d^3*x^4-1/3*e*d^2*x^3+1/2*d*e^2*x^2-x*e^3)+e^4/d^5*ln(d*x+e) )+1/4*b/d*x^4*ln(x*c)-1/16*b*x^4/d-1/3*b/d^2*e*x^3*ln(x*c)+1/9*b*e*x^3/d^2 +1/2*b/d^3*e^2*x^2*ln(x*c)-1/4*b*e^2*x^2/d^3-b*e^3*x*ln(x*c)/d^4+b*e^3*x/d ^4+b*e^4/d^5*dilog((c*d*x+c*e)/e/c)+b*e^4/d^5*ln(x*c)*ln((c*d*x+c*e)/e/c)
\[ \int \frac {x^3 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x\right ) + a\right )} x^{3}}{d + \frac {e}{x}} \,d x } \]
Time = 82.83 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.76 \[ \int \frac {x^3 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\frac {a x^{4}}{4 d} - \frac {a e x^{3}}{3 d^{2}} + \frac {a e^{2} x^{2}}{2 d^{3}} + \frac {a e^{4} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{4}} - \frac {a e^{3} x}{d^{4}} + \frac {b x^{4} \log {\left (c x \right )}}{4 d} - \frac {b x^{4}}{16 d} - \frac {b e x^{3} \log {\left (c x \right )}}{3 d^{2}} + \frac {b e x^{3}}{9 d^{2}} + \frac {b e^{2} x^{2} \log {\left (c x \right )}}{2 d^{3}} - \frac {b e^{2} x^{2}}{4 d^{3}} - \frac {b e^{4} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{4}} + \frac {b e^{4} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x \right )}}{d^{4}} - \frac {b e^{3} x \log {\left (c x \right )}}{d^{4}} + \frac {b e^{3} x}{d^{4}} \]
a*x**4/(4*d) - a*e*x**3/(3*d**2) + a*e**2*x**2/(2*d**3) + a*e**4*Piecewise ((x/e, Eq(d, 0)), (log(d*x + e)/d, True))/d**4 - a*e**3*x/d**4 + b*x**4*lo g(c*x)/(4*d) - b*x**4/(16*d) - b*e*x**3*log(c*x)/(3*d**2) + b*e*x**3/(9*d* *2) + b*e**2*x**2*log(c*x)/(2*d**3) - b*e**2*x**2/(4*d**3) - b*e**4*Piecew ise((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), T rue))/d, True))/d**4 + b*e**4*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))*log(c*x)/d**4 - b*e**3*x*log(c*x)/d**4 + b*e**3*x/d**4
Time = 0.24 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.24 \[ \int \frac {x^3 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\frac {{\left (\log \left (\frac {d x}{e} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {d x}{e}\right )\right )} b e^{4}}{d^{5}} + \frac {9 \, {\left (4 \, a d^{3} + {\left (4 \, d^{3} \log \left (c\right ) - d^{3}\right )} b\right )} x^{4} - 16 \, {\left (3 \, a d^{2} e + {\left (3 \, d^{2} e \log \left (c\right ) - d^{2} e\right )} b\right )} x^{3} + 36 \, {\left (2 \, a d e^{2} + {\left (2 \, d e^{2} \log \left (c\right ) - d e^{2}\right )} b\right )} x^{2} - 144 \, {\left (a e^{3} + {\left (e^{3} \log \left (c\right ) - e^{3}\right )} b\right )} x + 12 \, {\left (3 \, b d^{3} x^{4} - 4 \, b d^{2} e x^{3} + 6 \, b d e^{2} x^{2} - 12 \, b e^{3} x\right )} \log \left (x\right )}{144 \, d^{4}} + \frac {{\left (b e^{4} \log \left (c\right ) + a e^{4}\right )} \log \left (d x + e\right )}{d^{5}} \]
(log(d*x/e + 1)*log(x) + dilog(-d*x/e))*b*e^4/d^5 + 1/144*(9*(4*a*d^3 + (4 *d^3*log(c) - d^3)*b)*x^4 - 16*(3*a*d^2*e + (3*d^2*e*log(c) - d^2*e)*b)*x^ 3 + 36*(2*a*d*e^2 + (2*d*e^2*log(c) - d*e^2)*b)*x^2 - 144*(a*e^3 + (e^3*lo g(c) - e^3)*b)*x + 12*(3*b*d^3*x^4 - 4*b*d^2*e*x^3 + 6*b*d*e^2*x^2 - 12*b* e^3*x)*log(x))/d^4 + (b*e^4*log(c) + a*e^4)*log(d*x + e)/d^5
\[ \int \frac {x^3 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x\right ) + a\right )} x^{3}}{d + \frac {e}{x}} \,d x } \]
Timed out. \[ \int \frac {x^3 (a+b \log (c x))}{d+\frac {e}{x}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x\right )\right )}{d+\frac {e}{x}} \,d x \]