Integrand size = 23, antiderivative size = 185 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=-\frac {a e^3 x}{d^4}+\frac {b e^3 n x}{d^4}-\frac {b e^2 n x^2}{4 d^3}+\frac {b e n x^3}{9 d^2}-\frac {b n x^4}{16 d}-\frac {b e^3 x \log \left (c x^n\right )}{d^4}+\frac {e^2 x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^3}-\frac {e x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d}+\frac {e^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{d^5}+\frac {b e^4 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^5} \]
-a*e^3*x/d^4+b*e^3*n*x/d^4-1/4*b*e^2*n*x^2/d^3+1/9*b*e*n*x^3/d^2-1/16*b*n* x^4/d-b*e^3*x*ln(c*x^n)/d^4+1/2*e^2*x^2*(a+b*ln(c*x^n))/d^3-1/3*e*x^3*(a+b *ln(c*x^n))/d^2+1/4*x^4*(a+b*ln(c*x^n))/d+e^4*(a+b*ln(c*x^n))*ln(1+d*x/e)/ d^5+b*e^4*n*polylog(2,-d*x/e)/d^5
Time = 0.07 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {-144 a d e^3 x+144 b d e^3 n x-36 b d^2 e^2 n x^2+16 b d^3 e n x^3-9 b d^4 n x^4-144 b d e^3 x \log \left (c x^n\right )+72 d^2 e^2 x^2 \left (a+b \log \left (c x^n\right )\right )-48 d^3 e x^3 \left (a+b \log \left (c x^n\right )\right )+36 d^4 x^4 \left (a+b \log \left (c x^n\right )\right )+144 e^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )+144 b e^4 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{144 d^5} \]
(-144*a*d*e^3*x + 144*b*d*e^3*n*x - 36*b*d^2*e^2*n*x^2 + 16*b*d^3*e*n*x^3 - 9*b*d^4*n*x^4 - 144*b*d*e^3*x*Log[c*x^n] + 72*d^2*e^2*x^2*(a + b*Log[c*x ^n]) - 48*d^3*e*x^3*(a + b*Log[c*x^n]) + 36*d^4*x^4*(a + b*Log[c*x^n]) + 1 44*e^4*(a + b*Log[c*x^n])*Log[1 + (d*x)/e] + 144*b*e^4*n*PolyLog[2, -((d*x )/e)])/(144*d^5)
Time = 0.42 (sec) , antiderivative size = 185, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx\) |
| \(\Big \downarrow \) 2005 |
| \(\displaystyle \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{d x+e}dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (\frac {e^4 \left (a+b \log \left (c x^n\right )\right )}{d^4 (d x+e)}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )}{d^4}+\frac {e^2 x \left (a+b \log \left (c x^n\right )\right )}{d^3}-\frac {e x^2 \left (a+b \log \left (c x^n\right )\right )}{d^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^4 \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^5}+\frac {e^2 x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^3}-\frac {e x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d}-\frac {a e^3 x}{d^4}-\frac {b e^3 x \log \left (c x^n\right )}{d^4}+\frac {b e^4 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^5}+\frac {b e^3 n x}{d^4}-\frac {b e^2 n x^2}{4 d^3}+\frac {b e n x^3}{9 d^2}-\frac {b n x^4}{16 d}\) |
-((a*e^3*x)/d^4) + (b*e^3*n*x)/d^4 - (b*e^2*n*x^2)/(4*d^3) + (b*e*n*x^3)/( 9*d^2) - (b*n*x^4)/(16*d) - (b*e^3*x*Log[c*x^n])/d^4 + (e^2*x^2*(a + b*Log [c*x^n]))/(2*d^3) - (e*x^3*(a + b*Log[c*x^n]))/(3*d^2) + (x^4*(a + b*Log[c *x^n]))/(4*d) + (e^4*(a + b*Log[c*x^n])*Log[1 + (d*x)/e])/d^5 + (b*e^4*n*P olyLog[2, -((d*x)/e)])/d^5
3.4.29.3.1 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 = 0.34 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.69
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) x^{4}}{4 d}-\frac {b \ln \left (x^{n}\right ) e \,x^{3}}{3 d^{2}}+\frac {b \ln \left (x^{n}\right ) x^{2} e^{2}}{2 d^{3}}-\frac {b \ln \left (x^{n}\right ) x \,e^{3}}{d^{4}}+\frac {b \ln \left (x^{n}\right ) e^{4} \ln \left (d x +e \right )}{d^{5}}-\frac {b n \,e^{4} \ln \left (d x +e \right ) \ln \left (-\frac {d x}{e}\right )}{d^{5}}-\frac {b n \,e^{4} \operatorname {dilog}\left (-\frac {d x}{e}\right )}{d^{5}}-\frac {b n \,x^{4}}{16 d}+\frac {b e n \,x^{3}}{9 d^{2}}-\frac {b \,e^{2} n \,x^{2}}{4 d^{3}}+\frac {b \,e^{3} n x}{d^{4}}+\frac {205 b n \,e^{4}}{144 d^{5}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{4} d^{3} x^{4}-\frac {1}{3} e \,d^{2} x^{3}+\frac {1}{2} d \,e^{2} x^{2}-x \,e^{3}}{d^{4}}+\frac {e^{4} \ln \left (d x +e \right )}{d^{5}}\right )\) | \(313\) |
1/4*b*ln(x^n)/d*x^4-1/3*b*ln(x^n)/d^2*e*x^3+1/2*b*ln(x^n)/d^3*x^2*e^2-b*ln (x^n)/d^4*x*e^3+b*ln(x^n)*e^4/d^5*ln(d*x+e)-b*n*e^4/d^5*ln(d*x+e)*ln(-d*x/ e)-b*n*e^4/d^5*dilog(-d*x/e)-1/16*b*n*x^4/d+1/9*b*e*n*x^3/d^2-1/4*b*e^2*n* x^2/d^3+b*e^3*n*x/d^4+205/144*b*n*e^4/d^5+(-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*ln(c)+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))
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{d + \frac {e}{x}} \,d x } \]
Time = 75.86 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.71 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {a x^{4}}{4 d} - \frac {a e x^{3}}{3 d^{2}} + \frac {a e^{2} x^{2}}{2 d^{3}} + \frac {a e^{4} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{4}} - \frac {a e^{3} x}{d^{4}} - \frac {b n x^{4}}{16 d} + \frac {b x^{4} \log {\left (c x^{n} \right )}}{4 d} + \frac {b e n x^{3}}{9 d^{2}} - \frac {b e x^{3} \log {\left (c x^{n} \right )}}{3 d^{2}} - \frac {b e^{2} n x^{2}}{4 d^{3}} + \frac {b e^{2} x^{2} \log {\left (c x^{n} \right )}}{2 d^{3}} - \frac {b e^{4} n \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{4}} + \frac {b e^{4} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{4}} + \frac {b e^{3} n x}{d^{4}} - \frac {b e^{3} x \log {\left (c x^{n} \right )}}{d^{4}} \]
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*n*x**4/ (16*d) + b*x**4*log(c*x**n)/(4*d) + b*e*n*x**3/(9*d**2) - b*e*x**3*log(c*x **n)/(3*d**2) - b*e**2*n*x**2/(4*d**3) + b*e**2*x**2*log(c*x**n)/(2*d**3) - b*e**4*n*Piecewise((x/e, Eq(d, 0)), (Piecewise((-polylog(2, d*x*exp_pola r(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)*lo g(e) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(e) - polylog(2, d*x*exp_ polar(I*pi)/e), True))/d, True))/d**4 + b*e**4*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))*log(c*x**n)/d**4 + b*e**3*n*x/d**4 - b*e**3*x*log( c*x**n)/d**4
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{d + \frac {e}{x}} \,d x } \]
1/12*a*(12*e^4*log(d*x + e)/d^5 + (3*d^3*x^4 - 4*d^2*e*x^3 + 6*d*e^2*x^2 - 12*e^3*x)/d^4) + b*integrate((x^4*log(c) + x^4*log(x^n))/(d*x + e), x)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{d + \frac {e}{x}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{d+\frac {e}{x}} \,d x \]