\(\int \frac {x^2 (a+b \log (c x^n))}{d+e x} \, dx\) [32]

Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=-\frac {a d x}{e^2}+\frac {b d n x}{e^2}-\frac {b n x^2}{4 e}-\frac {b d x \log \left (c x^n\right )}{e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}+\frac {b d^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3} \] Output:

-a*d*x/e^2+b*d*n*x/e^2-1/4*b*n*x^2/e-b*d*x*ln(c*x^n)/e^2+1/2*x^2*(a+b*ln(c 
*x^n))/e+d^2*(a+b*ln(c*x^n))*ln(1+e*x/d)/e^3+b*d^2*n*polylog(2,-e*x/d)/e^3
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\frac {-4 a d e x+4 b d e n x+2 a e^2 x^2-b e^2 n x^2+4 a d^2 \log \left (1+\frac {e x}{d}\right )+2 b \log \left (c x^n\right ) \left (e x (-2 d+e x)+2 d^2 \log \left (1+\frac {e x}{d}\right )\right )+4 b d^2 n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 e^3} \] Input:

Integrate[(x^2*(a + b*Log[c*x^n]))/(d + e*x),x]
 

Output:

(-4*a*d*e*x + 4*b*d*e*n*x + 2*a*e^2*x^2 - b*e^2*n*x^2 + 4*a*d^2*Log[1 + (e 
*x)/d] + 2*b*Log[c*x^n]*(e*x*(-2*d + e*x) + 2*d^2*Log[1 + (e*x)/d]) + 4*b* 
d^2*n*PolyLog[2, -((e*x)/d)])/(4*e^3)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 107, 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.

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

\(\Big \downarrow \) 2793

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

\(\Big \downarrow \) 2009

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

Input:

Int[(x^2*(a + b*Log[c*x^n]))/(d + e*x),x]
 

Output:

-((a*d*x)/e^2) + (b*d*n*x)/e^2 - (b*n*x^2)/(4*e) - (b*d*x*Log[c*x^n])/e^2 
+ (x^2*(a + b*Log[c*x^n]))/(2*e) + (d^2*(a + b*Log[c*x^n])*Log[1 + (e*x)/d 
])/e^3 + (b*d^2*n*PolyLog[2, -((e*x)/d)])/e^3
 

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]))
 

Maple [C] (warning: unable to verify)

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

Time = 0.46 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.18

method	result	size
risch	\(\frac {b \ln \left (x^{n}\right ) x^{2}}{2 e}-\frac {b \ln \left (x^{n}\right ) d x}{e^{2}}+\frac {b \ln \left (x^{n}\right ) d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {b n \,d^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{3}}-\frac {b n \,d^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{3}}-\frac {b n \,x^{2}}{4 e}+\frac {b d n x}{e^{2}}+\frac {5 b n \,d^{2}}{4 e^{3}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{2} e \,x^{2}-d x}{e^{2}}+\frac {d^{2} \ln \left (e x +d \right )}{e^{3}}\right )\)	\(233\)

Input:

int(x^2*(a+b*ln(c*x^n))/(e*x+d),x,method=_RETURNVERBOSE)
 

Output:

1/2*b*ln(x^n)/e*x^2-b*ln(x^n)/e^2*d*x+b*ln(x^n)*d^2/e^3*ln(e*x+d)-b*n*d^2/ 
e^3*ln(e*x+d)*ln(-e*x/d)-b*n*d^2/e^3*dilog(-e*x/d)-1/4*b*n*x^2/e+b*d*n*x/e 
^2+5/4*b*n*d^2/e^3+(1/2*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*Pi*b*csgn 
(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/2*I*Pi*b*csgn(I*c*x^n)^3+1/2*I*Pi*b*csgn 
(I*c*x^n)^2*csgn(I*c)+b*ln(c)+a)*(1/e^2*(1/2*e*x^2-d*x)+d^2/e^3*ln(e*x+d))
 

Fricas [F]

\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{e x + d} \,d x } \] Input:

integrate(x^2*(a+b*log(c*x^n))/(e*x+d),x, algorithm="fricas")
 

Output:

integral((b*x^2*log(c*x^n) + a*x^2)/(e*x + d), x)
 

Sympy [A] (verification not implemented)

Time = 14.95 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.04 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\frac {a d^{2} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}} - \frac {a d x}{e^{2}} + \frac {a x^{2}}{2 e} - \frac {b d^{2} n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{2}} + \frac {b d n x}{e^{2}} - \frac {b d x \log {\left (c x^{n} \right )}}{e^{2}} - \frac {b n x^{2}}{4 e} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2 e} \] Input:

integrate(x**2*(a+b*ln(c*x**n))/(e*x+d),x)
 

Output:

a*d**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**2 - a*d*x/e** 
2 + a*x**2/(2*e) - b*d**2*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((-polylo 
g(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) - po 
lylog(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) - po 
lylog(2, e*x*exp_polar(I*pi)/d), True))/e, True))/e**2 + b*d**2*Piecewise( 
(x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e**2 + b*d*n*x/e**2 - 
 b*d*x*log(c*x**n)/e**2 - b*n*x**2/(4*e) + b*x**2*log(c*x**n)/(2*e)
 

Maxima [F]

\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{e x + d} \,d x } \] Input:

integrate(x^2*(a+b*log(c*x^n))/(e*x+d),x, algorithm="maxima")
 

Output:

1/2*a*(2*d^2*log(e*x + d)/e^3 + (e*x^2 - 2*d*x)/e^2) + b*integrate((x^2*lo 
g(c) + x^2*log(x^n))/(e*x + d), x)
 

Giac [F]

\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{e x + d} \,d x } \] Input:

integrate(x^2*(a+b*log(c*x^n))/(e*x+d),x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)*x^2/(e*x + d), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\int \frac {x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{d+e\,x} \,d x \] Input:

int((x^2*(a + b*log(c*x^n)))/(d + e*x),x)
 

Output:

int((x^2*(a + b*log(c*x^n)))/(d + e*x), x)
 

Reduce [F]

\[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\frac {-4 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e \,x^{2}+d x}d x \right ) b \,d^{3} n +4 \,\mathrm {log}\left (e x +d \right ) a \,d^{2} n +2 \mathrm {log}\left (x^{n} c \right )^{2} b \,d^{2}-4 \,\mathrm {log}\left (x^{n} c \right ) b d e n x +2 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{2} n \,x^{2}-4 a d e n x +2 a \,e^{2} n \,x^{2}+4 b d e \,n^{2} x -b \,e^{2} n^{2} x^{2}}{4 e^{3} n} \] Input:

int(x^2*(a+b*log(c*x^n))/(e*x+d),x)
 

Output:

( - 4*int(log(x**n*c)/(d*x + e*x**2),x)*b*d**3*n + 4*log(d + e*x)*a*d**2*n 
 + 2*log(x**n*c)**2*b*d**2 - 4*log(x**n*c)*b*d*e*n*x + 2*log(x**n*c)*b*e** 
2*n*x**2 - 4*a*d*e*n*x + 2*a*e**2*n*x**2 + 4*b*d*e*n**2*x - b*e**2*n**2*x* 
*2)/(4*e**3*n)