Integrand size = 21, antiderivative size = 154 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=-\frac {b n}{4 d^2 x^2}+\frac {2 b e n}{d^3 x}-\frac {a+b \log \left (c x^n\right )}{2 d^2 x^2}+\frac {2 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}-\frac {3 e^2 \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}+\frac {b e^2 n \log (d+e x)}{d^4}+\frac {3 b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^4} \]
-1/4*b*n/d^2/x^2+2*b*e*n/d^3/x+1/2*(-a-b*ln(c*x^n))/d^2/x^2+2*e*(a+b*ln(c* x^n))/d^3/x-e^3*x*(a+b*ln(c*x^n))/d^4/(e*x+d)-3*e^2*ln(1+d/e/x)*(a+b*ln(c* x^n))/d^4+b*e^2*n*ln(e*x+d)/d^4+3*b*e^2*n*polylog(2,-d/e/x)/d^4
Time = 0.14 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.07 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=-\frac {\frac {b d^2 n}{x^2}-\frac {8 b d e n}{x}+\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac {8 d e \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {4 d e^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}-\frac {6 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+4 b e^2 n (\log (x)-\log (d+e x))+12 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+12 b e^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 d^4} \]
-1/4*((b*d^2*n)/x^2 - (8*b*d*e*n)/x + (2*d^2*(a + b*Log[c*x^n]))/x^2 - (8* d*e*(a + b*Log[c*x^n]))/x - (4*d*e^2*(a + b*Log[c*x^n]))/(d + e*x) - (6*e^ 2*(a + b*Log[c*x^n])^2)/(b*n) + 4*b*e^2*n*(Log[x] - Log[d + e*x]) + 12*e^2 *(a + b*Log[c*x^n])*Log[1 + (e*x)/d] + 12*b*e^2*n*PolyLog[2, -((e*x)/d)])/ d^4
Time = 0.42 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {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.
| \(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)^2}+\frac {3 e^2 \left (a+b \log \left (c x^n\right )\right )}{d^3 x (d+e x)}-\frac {2 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x^2}+\frac {a+b \log \left (c x^n\right )}{d^2 x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}-\frac {3 e^2 \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}+\frac {2 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {a+b \log \left (c x^n\right )}{2 d^2 x^2}+\frac {3 b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^4}+\frac {b e^2 n \log (d+e x)}{d^4}+\frac {2 b e n}{d^3 x}-\frac {b n}{4 d^2 x^2}\) |
-1/4*(b*n)/(d^2*x^2) + (2*b*e*n)/(d^3*x) - (a + b*Log[c*x^n])/(2*d^2*x^2) + (2*e*(a + b*Log[c*x^n]))/(d^3*x) - (e^3*x*(a + b*Log[c*x^n]))/(d^4*(d + e*x)) - (3*e^2*Log[1 + d/(e*x)]*(a + b*Log[c*x^n]))/d^4 + (b*e^2*n*Log[d + e*x])/d^4 + (3*b*e^2*n*PolyLog[2, -(d/(e*x))])/d^4
3.1.45.3.1 Defintions of rubi rules used
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 = 0.50 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.15
| method | result | size |
| risch | \(-\frac {3 b \ln \left (x^{n}\right ) e^{2} \ln \left (e x +d \right )}{d^{4}}+\frac {b \ln \left (x^{n}\right ) e^{2}}{d^{3} \left (e x +d \right )}-\frac {b \ln \left (x^{n}\right )}{2 d^{2} x^{2}}+\frac {3 b \ln \left (x^{n}\right ) e^{2} \ln \left (x \right )}{d^{4}}+\frac {2 b \ln \left (x^{n}\right ) e}{d^{3} x}+\frac {b \,e^{2} n \ln \left (e x +d \right )}{d^{4}}-\frac {b n}{4 d^{2} x^{2}}+\frac {2 b e n}{d^{3} x}-\frac {b n \,e^{2} \ln \left (x \right )}{d^{4}}-\frac {3 b n \,e^{2} \ln \left (x \right )^{2}}{2 d^{4}}+\frac {3 b n \,e^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{4}}+\frac {3 b n \,e^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{4}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {3 e^{2} \ln \left (e x +d \right )}{d^{4}}+\frac {e^{2}}{d^{3} \left (e x +d \right )}-\frac {1}{2 d^{2} x^{2}}+\frac {3 e^{2} \ln \left (x \right )}{d^{4}}+\frac {2 e}{d^{3} x}\right )\) | \(331\) |
-3*b*ln(x^n)/d^4*e^2*ln(e*x+d)+b*ln(x^n)/d^3*e^2/(e*x+d)-1/2*b*ln(x^n)/d^2 /x^2+3*b*ln(x^n)/d^4*e^2*ln(x)+2*b*ln(x^n)/d^3*e/x+b*e^2*n*ln(e*x+d)/d^4-1 /4*b*n/d^2/x^2+2*b*e*n/d^3/x-b*n/d^4*e^2*ln(x)-3/2*b*n/d^4*e^2*ln(x)^2+3*b *n/d^4*e^2*ln(e*x+d)*ln(-e*x/d)+3*b*n/d^4*e^2*dilog(-e*x/d)+(-1/2*I*b*Pi*c sgn(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) *(-3/d^4*e^2*ln(e*x+d)+1/d^3*e^2/(e*x+d)-1/2/d^2/x^2+3/d^4*e^2*ln(x)+2/d^3 *e/x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{2} x^{3}} \,d x } \]
Time = 43.49 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.44 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=- \frac {a}{2 d^{2} x^{2}} - \frac {a e^{3} \left (\begin {cases} \frac {x}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x} & \text {otherwise} \end {cases}\right )}{d^{3}} + \frac {2 a e}{d^{3} x} - \frac {3 a e^{3} \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 {3 a e^{2} \log {\left (x \right )}}{d^{4}} - \frac {b n}{4 d^{2} x^{2}} - \frac {b \log {\left (c x^{n} \right )}}{2 d^{2} x^{2}} + \frac {b e^{3} 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 )}{d^{3}} - \frac {b e^{3} \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 )}}{d^{3}} + \frac {2 b e n}{d^{3} x} + \frac {2 b e \log {\left (c x^{n} \right )}}{d^{3} x} + \frac {3 b e^{3} 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 {3 b e^{3} \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 {3 b e^{2} n \log {\left (x \right )}^{2}}{2 d^{4}} + \frac {3 b e^{2} \log {\left (x \right )} \log {\left (c x^{n} \right )}}{d^{4}} \]
-a/(2*d**2*x**2) - a*e**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x) , True))/d**3 + 2*a*e/(d**3*x) - 3*a*e**3*Piecewise((x/d, Eq(e, 0)), (log( d + e*x)/e, True))/d**4 + 3*a*e**2*log(x)/d**4 - b*n/(4*d**2*x**2) - b*log (c*x**n)/(2*d**2*x**2) + b*e**3*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/( d*e) + log(d/e + x)/(d*e), True))/d**3 - b*e**3*Piecewise((x/d**2, Eq(e, 0 )), (-1/(d*e + e**2*x), True))*log(c*x**n)/d**3 + 2*b*e*n/(d**3*x) + 2*b*e *log(c*x**n)/(d**3*x) + 3*b*e**3*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_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))/d**4 - 3*b*e**3*P iecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/d**4 - 3*b*e **2*n*log(x)**2/(2*d**4) + 3*b*e**2*log(x)*log(c*x**n)/d**4
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{2} x^{3}} \,d x } \]
1/2*a*((6*e^2*x^2 + 3*d*e*x - d^2)/(d^3*e*x^3 + d^4*x^2) - 6*e^2*log(e*x + d)/d^4 + 6*e^2*log(x)/d^4) + b*integrate((log(c) + log(x^n))/(e^2*x^5 + 2 *d*e*x^4 + d^2*x^3), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{2} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^2} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^3\,{\left (d+e\,x\right )}^2} \,d x \]