Integrand size = 21, antiderivative size = 149 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx=-\frac {3 b n x}{e^3}+\frac {(6 a+5 b n) x}{2 e^3}+\frac {3 b x \log \left (c x^n\right )}{e^3}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{2 e (d+e x)^2}-\frac {x^2 \left (3 a+b n+3 b \log \left (c x^n\right )\right )}{2 e^2 (d+e x)}-\frac {d \left (6 a+5 b n+6 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{2 e^4}-\frac {3 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^4} \]
-3*b*n*x/e^3+1/2*(5*b*n+6*a)*x/e^3+3*b*x*ln(c*x^n)/e^3-1/2*x^3*(a+b*ln(c*x ^n))/e/(e*x+d)^2-1/2*x^2*(3*a+b*n+3*b*ln(c*x^n))/e^2/(e*x+d)-1/2*d*(6*a+5* b*n+6*b*ln(c*x^n))*ln(1+e*x/d)/e^4-3*b*d*n*polylog(2,-e*x/d)/e^4
Time = 0.09 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.01 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx=\frac {2 a e x-2 b e n x+2 b e x \log \left (c x^n\right )+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}-\frac {6 d^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}+6 b d n (\log (x)-\log (d+e x))-b d n \left (\frac {d}{d+e x}+\log (x)-\log (d+e x)\right )-6 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-6 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{2 e^4} \]
(2*a*e*x - 2*b*e*n*x + 2*b*e*x*Log[c*x^n] + (d^3*(a + b*Log[c*x^n]))/(d + e*x)^2 - (6*d^2*(a + b*Log[c*x^n]))/(d + e*x) + 6*b*d*n*(Log[x] - Log[d + e*x]) - b*d*n*(d/(d + e*x) + Log[x] - Log[d + e*x]) - 6*d*(a + b*Log[c*x^n ])*Log[1 + (e*x)/d] - 6*b*d*n*PolyLog[2, -((e*x)/d)])/(2*e^4)
Time = 0.49 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2784, 2784, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {\int \frac {x^2 \left (3 a+b n+3 b \log \left (c x^n\right )\right )}{(d+e x)^2}dx}{2 e}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{2 e (d+e x)^2}\) |
| \(\Big \downarrow \) 2784 |
| \(\displaystyle \frac {\frac {\int \frac {x \left (6 a+5 b n+6 b \log \left (c x^n\right )\right )}{d+e x}dx}{e}-\frac {x^2 \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{e (d+e x)}}{2 e}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{2 e (d+e x)^2}\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \frac {\frac {\int \left (\frac {6 a+5 b n+6 b \log \left (c x^n\right )}{e}-\frac {d \left (6 a+5 b n+6 b \log \left (c x^n\right )\right )}{e (d+e x)}\right )dx}{e}-\frac {x^2 \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{e (d+e x)}}{2 e}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{2 e (d+e x)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {-\frac {d \log \left (\frac {e x}{d}+1\right ) \left (6 a+6 b \log \left (c x^n\right )+5 b n\right )}{e^2}+\frac {x (6 a+5 b n)}{e}+\frac {6 b x \log \left (c x^n\right )}{e}-\frac {6 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^2}-\frac {6 b n x}{e}}{e}-\frac {x^2 \left (3 a+3 b \log \left (c x^n\right )+b n\right )}{e (d+e x)}}{2 e}-\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{2 e (d+e x)^2}\) |
-1/2*(x^3*(a + b*Log[c*x^n]))/(e*(d + e*x)^2) + (-((x^2*(3*a + b*n + 3*b*L og[c*x^n]))/(e*(d + e*x))) + ((-6*b*n*x)/e + ((6*a + 5*b*n)*x)/e + (6*b*x* Log[c*x^n])/e - (d*(6*a + 5*b*n + 6*b*Log[c*x^n])*Log[1 + (e*x)/d])/e^2 - (6*b*d*n*PolyLog[2, -((e*x)/d)])/e^2)/e)/(2*e)
3.1.46.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_))^(q_.), x_Symbol] :> Simp[(f*x)^m*(d + e*x)^(q + 1)*((a + b*Log[c*x^n] )/(e*(q + 1))), x] - Simp[f/(e*(q + 1)) Int[(f*x)^(m - 1)*(d + e*x)^(q + 1)*(a*m + b*n + b*m*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && ILtQ[q, -1] && GtQ[m, 0]
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 = 0.44 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.03
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) x}{e^{3}}-\frac {3 b \ln \left (x^{n}\right ) d \ln \left (e x +d \right )}{e^{4}}-\frac {3 b \ln \left (x^{n}\right ) d^{2}}{e^{4} \left (e x +d \right )}+\frac {b \ln \left (x^{n}\right ) d^{3}}{2 e^{4} \left (e x +d \right )^{2}}-\frac {b n x}{e^{3}}-\frac {b n d}{e^{4}}-\frac {5 b n d \ln \left (e x +d \right )}{2 e^{4}}-\frac {b n \,d^{2}}{2 e^{4} \left (e x +d \right )}+\frac {5 b n d \ln \left (e x \right )}{2 e^{4}}+\frac {3 b n d \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{4}}+\frac {3 b n d \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{4}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {x}{e^{3}}-\frac {3 d \ln \left (e x +d \right )}{e^{4}}-\frac {3 d^{2}}{e^{4} \left (e x +d \right )}+\frac {d^{3}}{2 e^{4} \left (e x +d \right )^{2}}\right )\) | \(303\) |
b*ln(x^n)*x/e^3-3*b*ln(x^n)/e^4*d*ln(e*x+d)-3*b*ln(x^n)/e^4*d^2/(e*x+d)+1/ 2*b*ln(x^n)*d^3/e^4/(e*x+d)^2-b*n*x/e^3-b*n/e^4*d-5/2*b*n/e^4*d*ln(e*x+d)- 1/2*b*n/e^4*d^2/(e*x+d)+5/2*b*n/e^4*d*ln(e*x)+3*b*n/e^4*d*ln(e*x+d)*ln(-e* x/d)+3*b*n/e^4*d*dilog(-e*x/d)+(-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)*(x/e^3-3/e^4*d*ln(e*x+d)-3/e ^4*d^2/(e*x+d)+1/2*d^3/e^4/(e*x+d)^2)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{3}} \,d x } \]
Time = 27.60 (sec) , antiderivative size = 391, normalized size of antiderivative = 2.62 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx=- \frac {a d^{3} \left (\begin {cases} \frac {x}{d^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 e \left (d + e x\right )^{2}} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {3 a d^{2} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {3 a d \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {a x}{e^{3}} + \frac {b d^{3} n \left (\begin {cases} \frac {x}{d^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 d^{2} e + 2 d e^{2} x} - \frac {\log {\left (x \right )}}{2 d^{2} e} + \frac {\log {\left (\frac {d}{e} + x \right )}}{2 d^{2} e} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {b d^{3} \left (\begin {cases} \frac {x}{d^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 e \left (d + e x\right )^{2}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{3}} - \frac {3 b d^{2} n \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {\log {\left (x \right )}}{d e} + \frac {\log {\left (\frac {d}{e} + x \right )}}{d e} & \text {otherwise} \end {cases}\right )}{e^{3}} + \frac {3 b d^{2} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{3}} + \frac {3 b d n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\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 (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{e^{3}} - \frac {3 b d \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{3}} - \frac {b n x}{e^{3}} + \frac {b x \log {\left (c x^{n} \right )}}{e^{3}} \]
-a*d**3*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))/e**3 + 3*a*d**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/e**3 - 3*a*d*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**3 + a*x/e**3 + b*d**3*n*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*d**2*e + 2*d*e**2*x) - log (x)/(2*d**2*e) + log(d/e + x)/(2*d**2*e), True))/e**3 - b*d**3*Piecewise(( x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))*log(c*x**n)/e**3 - 3*b*d **2*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x)/(d*e), T rue))/e**3 + 3*b*d**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), Tr ue))*log(c*x**n)/e**3 + 3*b*d*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((-po lylog(2, e*x*exp_polar(I*pi)/d), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d)*l og(x) - polylog(2, e*x*exp_polar(I*pi)/d), Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x*exp_polar(I*pi)/d), 1/Abs(x) < 1), (-meijerg(((), (1, 1)) , ((0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x*exp_polar(I*pi)/d), True))/e, True))/e**3 - 3*b*d*Piecewi se((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e**3 - b*n*x/e**3 + b*x*log(c*x**n)/e**3
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{3}} \,d x } \]
-1/2*a*((6*d^2*e*x + 5*d^3)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) - 2*x/e^3 + 6* d*log(e*x + d)/e^4) + b*integrate((x^3*log(c) + x^3*log(x^n))/(e^3*x^3 + 3 *d*e^2*x^2 + 3*d^2*e*x + d^3), x)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^3} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^3} \,d x \]