\(\int \frac {x^2 (a+b \log (c x^n))}{(d+e x)^2} \, dx\) [40]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=-\frac {b n x}{e^2}+\frac {2 x \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}-\frac {d \left (2 a+b n+2 b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^3}-\frac {2 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\frac {a e x-b e n x+b e x \log \left (c x^n\right )-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{d+e x}+b d n (\log (x)-\log (d+e x))-2 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-2 b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {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.

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

\(\Big \downarrow \) 2784

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

\(\Big \downarrow \) 2793

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2784

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]
 

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 = 0.47 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.55

method	result	size
risch	\(\frac {b \ln \left (x^{n}\right ) x}{e^{2}}-\frac {b \ln \left (x^{n}\right ) d^{2}}{e^{3} \left (e x +d \right )}-\frac {2 b \ln \left (x^{n}\right ) d \ln \left (e x +d \right )}{e^{3}}-\frac {b n d \ln \left (e x +d \right )}{e^{3}}+\frac {b n d \ln \left (e x \right )}{e^{3}}-\frac {b n x}{e^{2}}-\frac {b n d}{e^{3}}+\frac {2 b n d \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{3}}+\frac {2 b n d \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{3}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {x}{e^{2}}-\frac {d^{2}}{e^{3} \left (e x +d \right )}-\frac {2 d \ln \left (e x +d \right )}{e^{3}}\right )\)	\(250\)

Input:

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

Output:

b*ln(x^n)/e^2*x-b*ln(x^n)/e^3*d^2/(e*x+d)-2*b*ln(x^n)/e^3*d*ln(e*x+d)-b*n/ 
e^3*d*ln(e*x+d)+b*n/e^3*d*ln(e*x)-b*n*x/e^2-b*n/e^3*d+2*b*n/e^3*d*ln(e*x+d 
)*ln(-e*x/d)+2*b*n/e^3*d*dilog(-e*x/d)+(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)*(x/e^2-1/e^3*d^2/(e*x 
+d)-2/e^3*d*ln(e*x+d))
 

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 13.21 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.74 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2} \, dx=\frac {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^{2}} - \frac {2 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^{2}} + \frac {a x}{e^{2}} - \frac {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^{2}} + \frac {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^{2}} + \frac {2 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^{2}} - \frac {2 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^{2}} - \frac {b n x}{e^{2}} + \frac {b x \log {\left (c x^{n} \right )}}{e^{2}} \] Input:

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

Output:

a*d**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/e**2 - 2*a 
*d*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**2 + a*x/e**2 - b* 
d**2*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x)/(d*e), 
True))/e**2 + b*d**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), Tru 
e))*log(c*x**n)/e**2 + 2*b*d*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((-pol 
ylog(2, e*x*exp_polar(I*pi)/d), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d)*lo 
g(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**2 - 2*b*d*Piecewis 
e((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e**2 - b*n*x/e**2 + 
 b*x*log(c*x**n)/e**2
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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