3.1.52 \(\int \frac {a+b \log (c x^n)}{x^3 (d+e x)^3} \, dx\) [52]

3.1.52.1 Optimal result
3.1.52.2 Mathematica [A] (verified)
3.1.52.3 Rubi [A] (verified)
3.1.52.4 Maple [C] (warning: unable to verify)
3.1.52.5 Fricas [F]
3.1.52.6 Sympy [A] (verification not implemented)
3.1.52.7 Maxima [F]
3.1.52.8 Giac [F]
3.1.52.9 Mupad [F(-1)]

3.1.52.1 Optimal result

Integrand size = 21, antiderivative size = 217 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^3} \, dx=-\frac {b n}{4 d^3 x^2}+\frac {3 b e n}{d^4 x}-\frac {b e^2 n}{2 d^4 (d+e x)}-\frac {b e^2 n \log (x)}{2 d^5}-\frac {a+b \log \left (c x^n\right )}{2 d^3 x^2}+\frac {3 e \left (a+b \log \left (c x^n\right )\right )}{d^4 x}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^3 (d+e x)^2}-\frac {3 e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^5 (d+e x)}-\frac {6 e^2 \log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^5}+\frac {7 b e^2 n \log (d+e x)}{2 d^5}+\frac {6 b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^5} \]

output
-1/4*b*n/d^3/x^2+3*b*e*n/d^4/x-1/2*b*e^2*n/d^4/(e*x+d)-1/2*b*e^2*n*ln(x)/d 
^5+1/2*(-a-b*ln(c*x^n))/d^3/x^2+3*e*(a+b*ln(c*x^n))/d^4/x+1/2*e^2*(a+b*ln( 
c*x^n))/d^3/(e*x+d)^2-3*e^3*x*(a+b*ln(c*x^n))/d^5/(e*x+d)-6*e^2*ln(1+d/e/x 
)*(a+b*ln(c*x^n))/d^5+7/2*b*e^2*n*ln(e*x+d)/d^5+6*b*e^2*n*polylog(2,-d/e/x 
)/d^5
 
3.1.52.2 Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^3} \, dx=-\frac {\frac {b d^2 n}{x^2}-\frac {12 b d e n}{x}+\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac {12 d e \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {2 d^2 e^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}-\frac {12 d e^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}-\frac {12 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+12 b e^2 n (\log (x)-\log (d+e x))+\frac {2 b e^2 n (d+(d+e x) \log (x)-(d+e x) \log (d+e x))}{d+e x}+24 e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )+24 b e^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 d^5} \]

input
Integrate[(a + b*Log[c*x^n])/(x^3*(d + e*x)^3),x]
 
output
-1/4*((b*d^2*n)/x^2 - (12*b*d*e*n)/x + (2*d^2*(a + b*Log[c*x^n]))/x^2 - (1 
2*d*e*(a + b*Log[c*x^n]))/x - (2*d^2*e^2*(a + b*Log[c*x^n]))/(d + e*x)^2 - 
 (12*d*e^2*(a + b*Log[c*x^n]))/(d + e*x) - (12*e^2*(a + b*Log[c*x^n])^2)/( 
b*n) + 12*b*e^2*n*(Log[x] - Log[d + e*x]) + (2*b*e^2*n*(d + (d + e*x)*Log[ 
x] - (d + e*x)*Log[d + e*x]))/(d + e*x) + 24*e^2*(a + b*Log[c*x^n])*Log[1 
+ (e*x)/d] + 24*b*e^2*n*PolyLog[2, -((e*x)/d)])/d^5
 
3.1.52.3 Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 217, 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.

\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^3} \, dx\)

\(\Big \downarrow \) 2793

\(\displaystyle \int \left (-\frac {3 e^3 \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)^2}+\frac {6 e^2 \left (a+b \log \left (c x^n\right )\right )}{d^4 x (d+e x)}-\frac {3 e \left (a+b \log \left (c x^n\right )\right )}{d^4 x^2}-\frac {e^3 \left (a+b \log \left (c x^n\right )\right )}{d^3 (d+e x)^3}+\frac {a+b \log \left (c x^n\right )}{d^3 x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 e^3 x \left (a+b \log \left (c x^n\right )\right )}{d^5 (d+e x)}-\frac {6 e^2 \log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^5}+\frac {3 e \left (a+b \log \left (c x^n\right )\right )}{d^4 x}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{2 d^3 (d+e x)^2}-\frac {a+b \log \left (c x^n\right )}{2 d^3 x^2}+\frac {6 b e^2 n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^5}-\frac {b e^2 n \log (x)}{2 d^5}+\frac {7 b e^2 n \log (d+e x)}{2 d^5}-\frac {b e^2 n}{2 d^4 (d+e x)}+\frac {3 b e n}{d^4 x}-\frac {b n}{4 d^3 x^2}\)

input
Int[(a + b*Log[c*x^n])/(x^3*(d + e*x)^3),x]
 
output
-1/4*(b*n)/(d^3*x^2) + (3*b*e*n)/(d^4*x) - (b*e^2*n)/(2*d^4*(d + e*x)) - ( 
b*e^2*n*Log[x])/(2*d^5) - (a + b*Log[c*x^n])/(2*d^3*x^2) + (3*e*(a + b*Log 
[c*x^n]))/(d^4*x) + (e^2*(a + b*Log[c*x^n]))/(2*d^3*(d + e*x)^2) - (3*e^3* 
x*(a + b*Log[c*x^n]))/(d^5*(d + e*x)) - (6*e^2*Log[1 + d/(e*x)]*(a + b*Log 
[c*x^n]))/d^5 + (7*b*e^2*n*Log[d + e*x])/(2*d^5) + (6*b*e^2*n*PolyLog[2, - 
(d/(e*x))])/d^5
 

3.1.52.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2793
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]))
 
3.1.52.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.68 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.78

method result size
risch \(-\frac {6 b \ln \left (x^{n}\right ) e^{2} \ln \left (e x +d \right )}{d^{5}}+\frac {3 b \ln \left (x^{n}\right ) e^{2}}{d^{4} \left (e x +d \right )}+\frac {b \ln \left (x^{n}\right ) e^{2}}{2 d^{3} \left (e x +d \right )^{2}}-\frac {b \ln \left (x^{n}\right )}{2 d^{3} x^{2}}+\frac {6 b \ln \left (x^{n}\right ) e^{2} \ln \left (x \right )}{d^{5}}+\frac {3 b \ln \left (x^{n}\right ) e}{d^{4} x}-\frac {b \,e^{2} n}{2 d^{4} \left (e x +d \right )}+\frac {7 b \,e^{2} n \ln \left (e x +d \right )}{2 d^{5}}-\frac {b n}{4 d^{3} x^{2}}+\frac {3 b e n}{d^{4} x}-\frac {7 b \,e^{2} n \ln \left (x \right )}{2 d^{5}}-\frac {3 b n \,e^{2} \ln \left (x \right )^{2}}{d^{5}}+\frac {6 b n \,e^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{5}}+\frac {6 b n \,e^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{5}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {6 e^{2} \ln \left (e x +d \right )}{d^{5}}+\frac {3 e^{2}}{d^{4} \left (e x +d \right )}+\frac {e^{2}}{2 d^{3} \left (e x +d \right )^{2}}-\frac {1}{2 d^{3} x^{2}}+\frac {6 e^{2} \ln \left (x \right )}{d^{5}}+\frac {3 e}{d^{4} x}\right )\) \(386\)

input
int((a+b*ln(c*x^n))/x^3/(e*x+d)^3,x,method=_RETURNVERBOSE)
 
output
-6*b*ln(x^n)/d^5*e^2*ln(e*x+d)+3*b*ln(x^n)/d^4*e^2/(e*x+d)+1/2*b*ln(x^n)/d 
^3*e^2/(e*x+d)^2-1/2*b*ln(x^n)/d^3/x^2+6*b*ln(x^n)/d^5*e^2*ln(x)+3*b*ln(x^ 
n)/d^4*e/x-1/2*b*e^2*n/d^4/(e*x+d)+7/2*b*e^2*n*ln(e*x+d)/d^5-1/4*b*n/d^3/x 
^2+3*b*e*n/d^4/x-7/2*b*e^2*n*ln(x)/d^5-3*b*n/d^5*e^2*ln(x)^2+6*b*n/d^5*e^2 
*ln(e*x+d)*ln(-e*x/d)+6*b*n/d^5*e^2*dilog(-e*x/d)+(-1/2*I*b*Pi*csgn(I*c)*c 
sgn(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*c 
sgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*b*Pi*csgn(I*c*x^n)^3+b*ln(c)+a)*(-6/d^5*e 
^2*ln(e*x+d)+3/d^4*e^2/(e*x+d)+1/2/d^3*e^2/(e*x+d)^2-1/2/d^3/x^2+6/d^5*e^2 
*ln(x)+3/d^4*e/x)
 
3.1.52.5 Fricas [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{3} x^{3}} \,d x } \]

input
integrate((a+b*log(c*x^n))/x^3/(e*x+d)^3,x, algorithm="fricas")
 
output
integral((b*log(c*x^n) + a)/(e^3*x^6 + 3*d*e^2*x^5 + 3*d^2*e*x^4 + d^3*x^3 
), x)
 
3.1.52.6 Sympy [A] (verification not implemented)

Time = 41.53 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.29 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^3} \, dx=- \frac {a e^{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 )}{d^{3}} - \frac {a}{2 d^{3} x^{2}} - \frac {3 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^{4}} + \frac {3 a e}{d^{4} x} - \frac {6 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^{5}} + \frac {6 a e^{2} \log {\left (x \right )}}{d^{5}} + \frac {b e^{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 )}{d^{3}} - \frac {b e^{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 )}}{d^{3}} - \frac {b n}{4 d^{3} x^{2}} - \frac {b \log {\left (c x^{n} \right )}}{2 d^{3} x^{2}} + \frac {3 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^{4}} - \frac {3 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^{4}} + \frac {3 b e n}{d^{4} x} + \frac {3 b e \log {\left (c x^{n} \right )}}{d^{4} x} + \frac {6 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^{5}} - \frac {6 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^{5}} - \frac {3 b e^{2} n \log {\left (x \right )}^{2}}{d^{5}} + \frac {6 b e^{2} \log {\left (x \right )} \log {\left (c x^{n} \right )}}{d^{5}} \]

input
integrate((a+b*ln(c*x**n))/x**3/(e*x+d)**3,x)
 
output
-a*e**3*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))/d**3 
- a/(2*d**3*x**2) - 3*a*e**3*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2 
*x), True))/d**4 + 3*a*e/(d**4*x) - 6*a*e**3*Piecewise((x/d, Eq(e, 0)), (l 
og(d + e*x)/e, True))/d**5 + 6*a*e**2*log(x)/d**5 + b*e**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))/d**3 - b*e**3*Piecewise((x/d**3, Eq(e, 0)), (-1/( 
2*e*(d + e*x)**2), True))*log(c*x**n)/d**3 - b*n/(4*d**3*x**2) - b*log(c*x 
**n)/(2*d**3*x**2) + 3*b*e**3*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(d* 
e) + log(d/e + x)/(d*e), True))/d**4 - 3*b*e**3*Piecewise((x/d**2, Eq(e, 0 
)), (-1/(d*e + e**2*x), True))*log(c*x**n)/d**4 + 3*b*e*n/(d**4*x) + 3*b*e 
*log(c*x**n)/(d**4*x) + 6*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**5 - 6*b*e**3*P 
iecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/d**5 - 3*b*e 
**2*n*log(x)**2/d**5 + 6*b*e**2*log(x)*log(c*x**n)/d**5
 
3.1.52.7 Maxima [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{3} x^{3}} \,d x } \]

input
integrate((a+b*log(c*x^n))/x^3/(e*x+d)^3,x, algorithm="maxima")
 
output
1/2*a*((12*e^3*x^3 + 18*d*e^2*x^2 + 4*d^2*e*x - d^3)/(d^4*e^2*x^4 + 2*d^5* 
e*x^3 + d^6*x^2) - 12*e^2*log(e*x + d)/d^5 + 12*e^2*log(x)/d^5) + b*integr 
ate((log(c) + log(x^n))/(e^3*x^6 + 3*d*e^2*x^5 + 3*d^2*e*x^4 + d^3*x^3), x 
)
 
3.1.52.8 Giac [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^3} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{3} x^{3}} \,d x } \]

input
integrate((a+b*log(c*x^n))/x^3/(e*x+d)^3,x, algorithm="giac")
 
output
integrate((b*log(c*x^n) + a)/((e*x + d)^3*x^3), x)
 
3.1.52.9 Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 (d+e x)^3} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^3\,{\left (d+e\,x\right )}^3} \,d x \]

input
int((a + b*log(c*x^n))/(x^3*(d + e*x)^3),x)
 
output
int((a + b*log(c*x^n))/(x^3*(d + e*x)^3), x)