Integrand size = 21, antiderivative size = 150 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 (d+e x)} \, dx=-\frac {b n}{9 d x^3}+\frac {b e n}{4 d^2 x^2}-\frac {b e^2 n}{d^3 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3 x}+\frac {e^3 \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac {b e^3 n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^4} \]
-1/9*b*n/d/x^3+1/4*b*e*n/d^2/x^2-b*e^2*n/d^3/x+1/3*(-a-b*ln(c*x^n))/d/x^3+ 1/2*e*(a+b*ln(c*x^n))/d^2/x^2-e^2*(a+b*ln(c*x^n))/d^3/x+e^3*ln(1+d/e/x)*(a +b*ln(c*x^n))/d^4-b*e^3*n*polylog(2,-d/e/x)/d^4
Time = 0.11 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.06 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 (d+e x)} \, dx=\frac {-\frac {4 b d^3 n}{x^3}+\frac {9 b d^2 e n}{x^2}-\frac {36 b d e^2 n}{x}-\frac {12 d^3 \left (a+b \log \left (c x^n\right )\right )}{x^3}+\frac {18 d^2 e \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac {36 d e^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {18 e^3 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+36 e^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+36 b e^3 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{36 d^4} \]
((-4*b*d^3*n)/x^3 + (9*b*d^2*e*n)/x^2 - (36*b*d*e^2*n)/x - (12*d^3*(a + b* Log[c*x^n]))/x^3 + (18*d^2*e*(a + b*Log[c*x^n]))/x^2 - (36*d*e^2*(a + b*Lo g[c*x^n]))/x - (18*e^3*(a + b*Log[c*x^n])^2)/(b*n) + 36*e^3*(a + b*Log[c*x ^n])*Log[1 + (e*x)/d] + 36*b*e^3*n*PolyLog[2, -((e*x)/d)])/(36*d^4)
Time = 0.77 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {2780, 2741, 2780, 2741, 2780, 2741, 2779, 2838}
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.
| \(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^4 (d+e x)} \, dx\) |
| \(\Big \downarrow \) 2780 |
| \(\displaystyle \frac {\int \frac {a+b \log \left (c x^n\right )}{x^4}dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)}dx}{d}\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)}dx}{d}\) |
| \(\Big \downarrow \) 2780 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {\int \frac {a+b \log \left (c x^n\right )}{x^3}dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)}dx}{d}\right )}{d}\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{2 x^2}-\frac {b n}{4 x^2}}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x^2 (d+e x)}dx}{d}\right )}{d}\) |
| \(\Big \downarrow \) 2780 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{2 x^2}-\frac {b n}{4 x^2}}{d}-\frac {e \left (\frac {\int \frac {a+b \log \left (c x^n\right )}{x^2}dx}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)}dx}{d}\right )}{d}\right )}{d}\) |
| \(\Big \downarrow \) 2741 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{2 x^2}-\frac {b n}{4 x^2}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{x}-\frac {b n}{x}}{d}-\frac {e \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)}dx}{d}\right )}{d}\right )}{d}\) |
| \(\Big \downarrow \) 2779 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{2 x^2}-\frac {b n}{4 x^2}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{x}-\frac {b n}{x}}{d}-\frac {e \left (\frac {b n \int \frac {\log \left (\frac {d}{e x}+1\right )}{x}dx}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d}\right )}{d}\right )}{d}\right )}{d}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {-\frac {a+b \log \left (c x^n\right )}{3 x^3}-\frac {b n}{9 x^3}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{2 x^2}-\frac {b n}{4 x^2}}{d}-\frac {e \left (\frac {-\frac {a+b \log \left (c x^n\right )}{x}-\frac {b n}{x}}{d}-\frac {e \left (\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d}\right )}{d}\right )}{d}\right )}{d}\) |
(-1/9*(b*n)/x^3 - (a + b*Log[c*x^n])/(3*x^3))/d - (e*((-1/4*(b*n)/x^2 - (a + b*Log[c*x^n])/(2*x^2))/d - (e*((-((b*n)/x) - (a + b*Log[c*x^n])/x)/d - (e*(-((Log[1 + d/(e*x)]*(a + b*Log[c*x^n]))/d) + (b*n*PolyLog[2, -(d/(e*x) )])/d))/d))/d))/d
3.1.38.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^( m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r _.))), x_Symbol] :> Simp[(-Log[1 + d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)) , x] + Simp[b*n*(p/(d*r)) Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.))/((d_) + (e_.)* (x_)^(r_.)), x_Symbol] :> Simp[1/d Int[x^m*(a + b*Log[c*x^n])^p, x], x] - Simp[e/d Int[(x^(m + r)*(a + b*Log[c*x^n])^p)/(d + e*x^r), x], x] /; Fre eQ[{a, b, c, d, e, m, n, r}, x] && IGtQ[p, 0] && IGtQ[r, 0] && ILtQ[m, -1]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.47 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.06
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) e^{3} \ln \left (e x +d \right )}{d^{4}}-\frac {b \ln \left (x^{n}\right )}{3 d \,x^{3}}-\frac {b \ln \left (x^{n}\right ) e^{2}}{d^{3} x}+\frac {b \ln \left (x^{n}\right ) e}{2 d^{2} x^{2}}-\frac {b \ln \left (x^{n}\right ) e^{3} \ln \left (x \right )}{d^{4}}-\frac {b \,e^{2} n}{d^{3} x}+\frac {b e n}{4 d^{2} x^{2}}-\frac {b n}{9 d \,x^{3}}+\frac {b n \,e^{3} \ln \left (x \right )^{2}}{2 d^{4}}-\frac {b n \,e^{3} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{4}}-\frac {b n \,e^{3} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{4}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {e^{3} \ln \left (e x +d \right )}{d^{4}}-\frac {1}{3 d \,x^{3}}-\frac {e^{2}}{d^{3} x}+\frac {e}{2 d^{2} x^{2}}-\frac {e^{3} \ln \left (x \right )}{d^{4}}\right )\) | \(309\) |
b*ln(x^n)*e^3/d^4*ln(e*x+d)-1/3*b*ln(x^n)/d/x^3-b*ln(x^n)*e^2/d^3/x+1/2*b* ln(x^n)*e/d^2/x^2-b*ln(x^n)*e^3/d^4*ln(x)-b*e^2*n/d^3/x+1/4*b*e*n/d^2/x^2- 1/9*b*n/d/x^3+1/2*b*n*e^3/d^4*ln(x)^2-b*n*e^3/d^4*ln(e*x+d)*ln(-e*x/d)-b*n *e^3/d^4*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)*(e^3/d^4*ln(e*x+d)-1/3/d/x^3-e^2/d^3 /x+1/2*e/d^2/x^2-e^3/d^4*ln(x))
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 (d+e x)} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )} x^{4}} \,d x } \]
Time = 52.48 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.09 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 (d+e x)} \, dx=- \frac {a}{3 d x^{3}} + \frac {a e}{2 d^{2} x^{2}} - \frac {a e^{2}}{d^{3} x} + \frac {a e^{4} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{d^{4}} - \frac {a e^{3} \log {\left (x \right )}}{d^{4}} - \frac {b n}{9 d x^{3}} - \frac {b \log {\left (c x^{n} \right )}}{3 d x^{3}} + \frac {b e n}{4 d^{2} x^{2}} + \frac {b e \log {\left (c x^{n} \right )}}{2 d^{2} x^{2}} - \frac {b e^{2} n}{d^{3} x} - \frac {b e^{2} \log {\left (c x^{n} \right )}}{d^{3} x} - \frac {b e^{4} n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{d^{4}} + \frac {b e^{4} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{4}} + \frac {b e^{3} n \log {\left (x \right )}^{2}}{2 d^{4}} - \frac {b e^{3} \log {\left (x \right )} \log {\left (c x^{n} \right )}}{d^{4}} \]
-a/(3*d*x**3) + a*e/(2*d**2*x**2) - a*e**2/(d**3*x) + a*e**4*Piecewise((x/ d, Eq(e, 0)), (log(d + e*x)/e, True))/d**4 - a*e**3*log(x)/d**4 - b*n/(9*d *x**3) - b*log(c*x**n)/(3*d*x**3) + b*e*n/(4*d**2*x**2) + b*e*log(c*x**n)/ (2*d**2*x**2) - b*e**2*n/(d**3*x) - b*e**2*log(c*x**n)/(d**3*x) - b*e**4*n *Piecewise((x/d, Eq(e, 0)), (Piecewise((-polylog(2, e*x*exp_polar(I*pi)/d) , (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d)*log(x) - polylog(2, e*x*exp_pola r(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) + mei jerg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x*exp_polar(I*pi )/d), True))/e, True))/d**4 + b*e**4*Piecewise((x/d, Eq(e, 0)), (log(d + e *x)/e, True))*log(c*x**n)/d**4 + b*e**3*n*log(x)**2/(2*d**4) - b*e**3*log( x)*log(c*x**n)/d**4
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 (d+e x)} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )} x^{4}} \,d x } \]
1/6*a*(6*e^3*log(e*x + d)/d^4 - 6*e^3*log(x)/d^4 - (6*e^2*x^2 - 3*d*e*x + 2*d^2)/(d^3*x^3)) + b*integrate((log(c) + log(x^n))/(e*x^5 + d*x^4), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 (d+e x)} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )} x^{4}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 (d+e x)} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^4\,\left (d+e\,x\right )} \,d x \]