\(\int x^2 (d+e x)^3 (a+b \log (c x^n)) \, dx\) [20]

Optimal result

Integrand size = 21, antiderivative size = 100 \[ \int x^2 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{9} b d^3 n x^3-\frac {3}{16} b d^2 e n x^4-\frac {3}{25} b d e^2 n x^5-\frac {1}{36} b e^3 n x^6+\frac {1}{60} \left (20 d^3 x^3+45 d^2 e x^4+36 d e^2 x^5+10 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right ) \] Output:

-1/9*b*d^3*n*x^3-3/16*b*d^2*e*n*x^4-3/25*b*d*e^2*n*x^5-1/36*b*e^3*n*x^6+1/ 
60*(10*e^3*x^6+36*d*e^2*x^5+45*d^2*e*x^4+20*d^3*x^3)*(a+b*ln(c*x^n))
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.33 \[ \int x^2 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{9} b d^3 n x^3-\frac {3}{16} b d^2 e n x^4-\frac {3}{25} b d e^2 n x^5-\frac {1}{36} b e^3 n x^6+\frac {1}{3} d^3 x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {3}{4} d^2 e x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {3}{5} d e^2 x^5 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{6} e^3 x^6 \left (a+b \log \left (c x^n\right )\right ) \] Input:

Integrate[x^2*(d + e*x)^3*(a + b*Log[c*x^n]),x]
 

Output:

-1/9*(b*d^3*n*x^3) - (3*b*d^2*e*n*x^4)/16 - (3*b*d*e^2*n*x^5)/25 - (b*e^3* 
n*x^6)/36 + (d^3*x^3*(a + b*Log[c*x^n]))/3 + (3*d^2*e*x^4*(a + b*Log[c*x^n 
]))/4 + (3*d*e^2*x^5*(a + b*Log[c*x^n]))/5 + (e^3*x^6*(a + b*Log[c*x^n]))/ 
6
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2771, 27, 2010, 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.

\(\displaystyle \int x^2 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx\)

\(\Big \downarrow \) 2771

\(\displaystyle \frac {1}{60} \left (20 d^3 x^3+45 d^2 e x^4+36 d e^2 x^5+10 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-b n \int \frac {1}{60} x^2 \left (20 d^3+45 e x d^2+36 e^2 x^2 d+10 e^3 x^3\right )dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{60} \left (20 d^3 x^3+45 d^2 e x^4+36 d e^2 x^5+10 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{60} b n \int x^2 \left (20 d^3+45 e x d^2+36 e^2 x^2 d+10 e^3 x^3\right )dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \frac {1}{60} \left (20 d^3 x^3+45 d^2 e x^4+36 d e^2 x^5+10 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{60} b n \int \left (10 e^3 x^5+36 d e^2 x^4+45 d^2 e x^3+20 d^3 x^2\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{60} \left (20 d^3 x^3+45 d^2 e x^4+36 d e^2 x^5+10 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{60} b n \left (\frac {20 d^3 x^3}{3}+\frac {45}{4} d^2 e x^4+\frac {36}{5} d e^2 x^5+\frac {5 e^3 x^6}{3}\right )\)

Input:

Int[x^2*(d + e*x)^3*(a + b*Log[c*x^n]),x]
 

Output:

-1/60*(b*n*((20*d^3*x^3)/3 + (45*d^2*e*x^4)/4 + (36*d*e^2*x^5)/5 + (5*e^3* 
x^6)/3)) + ((20*d^3*x^3 + 45*d^2*e*x^4 + 36*d*e^2*x^5 + 10*e^3*x^6)*(a + b 
*Log[c*x^n]))/60
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

rule 2771

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_ 
.))^(q_.), x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a 
 + b*Log[c*x^n]), x] - Simp[b*n   Int[SimplifyIntegrand[u/x, x], x], x]] /; 
 FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IGtQ[m, 0]
 

Maple [A] (verified)

Time = 143.70 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.44

method	result	size
parallelrisch	\(\frac {x^{6} b \ln \left (c \,x^{n}\right ) e^{3}}{6}-\frac {b \,e^{3} n \,x^{6}}{36}+\frac {x^{6} a \,e^{3}}{6}+\frac {3 x^{5} b \ln \left (c \,x^{n}\right ) d \,e^{2}}{5}-\frac {3 b d \,e^{2} n \,x^{5}}{25}+\frac {3 x^{5} a \,e^{2} d}{5}+\frac {3 x^{4} b \ln \left (c \,x^{n}\right ) d^{2} e}{4}-\frac {3 b \,d^{2} e n \,x^{4}}{16}+\frac {3 x^{4} a \,d^{2} e}{4}+\frac {x^{3} b \ln \left (c \,x^{n}\right ) d^{3}}{3}-\frac {b \,d^{3} n \,x^{3}}{9}+\frac {x^{3} a \,d^{3}}{3}\)	\(144\)
risch	\(\frac {3 x^{5} a \,e^{2} d}{5}+\frac {3 x^{4} a \,d^{2} e}{4}+\frac {x^{3} a \,d^{3}}{3}+\frac {x^{6} a \,e^{3}}{6}+\frac {i \pi b \,d^{3} x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{6}+\frac {i \pi b \,d^{3} x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{6}+\frac {i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{12}+\frac {i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{12}-\frac {3 i \pi b \,d^{2} e \,x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{8}-\frac {3 i \pi b d \,e^{2} x^{5} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{10}+\frac {3 \ln \left (c \right ) b \,d^{2} e \,x^{4}}{4}+\frac {3 \ln \left (c \right ) b d \,e^{2} x^{5}}{5}-\frac {3 b \,d^{2} e n \,x^{4}}{16}-\frac {3 b d \,e^{2} n \,x^{5}}{25}+\frac {\ln \left (c \right ) b \,d^{3} x^{3}}{3}+\frac {\ln \left (c \right ) b \,e^{3} x^{6}}{6}+\frac {b \,x^{3} \left (10 e^{3} x^{3}+36 d \,e^{2} x^{2}+45 d^{2} e x +20 d^{3}\right ) \ln \left (x^{n}\right )}{60}-\frac {b \,d^{3} n \,x^{3}}{9}-\frac {b \,e^{3} n \,x^{6}}{36}-\frac {3 i \pi b \,d^{2} e \,x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{8}-\frac {i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{12}-\frac {i \pi b \,d^{3} x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{6}-\frac {3 i \pi b d \,e^{2} x^{5} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{10}-\frac {i \pi b \,d^{3} x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{6}+\frac {3 i \pi b \,d^{2} e \,x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {3 i \pi b \,d^{2} e \,x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{8}-\frac {i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{12}+\frac {3 i \pi b d \,e^{2} x^{5} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{10}+\frac {3 i \pi b d \,e^{2} x^{5} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{10}\)	\(600\)

Input:

int(x^2*(e*x+d)^3*(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)
 

Output:

1/6*x^6*b*ln(c*x^n)*e^3-1/36*b*e^3*n*x^6+1/6*x^6*a*e^3+3/5*x^5*b*ln(c*x^n) 
*d*e^2-3/25*b*d*e^2*n*x^5+3/5*x^5*a*e^2*d+3/4*x^4*b*ln(c*x^n)*d^2*e-3/16*b 
*d^2*e*n*x^4+3/4*x^4*a*d^2*e+1/3*x^3*b*ln(c*x^n)*d^3-1/9*b*d^3*n*x^3+1/3*x 
^3*a*d^3
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.67 \[ \int x^2 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{36} \, {\left (b e^{3} n - 6 \, a e^{3}\right )} x^{6} - \frac {3}{25} \, {\left (b d e^{2} n - 5 \, a d e^{2}\right )} x^{5} - \frac {3}{16} \, {\left (b d^{2} e n - 4 \, a d^{2} e\right )} x^{4} - \frac {1}{9} \, {\left (b d^{3} n - 3 \, a d^{3}\right )} x^{3} + \frac {1}{60} \, {\left (10 \, b e^{3} x^{6} + 36 \, b d e^{2} x^{5} + 45 \, b d^{2} e x^{4} + 20 \, b d^{3} x^{3}\right )} \log \left (c\right ) + \frac {1}{60} \, {\left (10 \, b e^{3} n x^{6} + 36 \, b d e^{2} n x^{5} + 45 \, b d^{2} e n x^{4} + 20 \, b d^{3} n x^{3}\right )} \log \left (x\right ) \] Input:

integrate(x^2*(e*x+d)^3*(a+b*log(c*x^n)),x, algorithm="fricas")
 

Output:

-1/36*(b*e^3*n - 6*a*e^3)*x^6 - 3/25*(b*d*e^2*n - 5*a*d*e^2)*x^5 - 3/16*(b 
*d^2*e*n - 4*a*d^2*e)*x^4 - 1/9*(b*d^3*n - 3*a*d^3)*x^3 + 1/60*(10*b*e^3*x 
^6 + 36*b*d*e^2*x^5 + 45*b*d^2*e*x^4 + 20*b*d^3*x^3)*log(c) + 1/60*(10*b*e 
^3*n*x^6 + 36*b*d*e^2*n*x^5 + 45*b*d^2*e*n*x^4 + 20*b*d^3*n*x^3)*log(x)
 

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.75 \[ \int x^2 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d^{3} x^{3}}{3} + \frac {3 a d^{2} e x^{4}}{4} + \frac {3 a d e^{2} x^{5}}{5} + \frac {a e^{3} x^{6}}{6} - \frac {b d^{3} n x^{3}}{9} + \frac {b d^{3} x^{3} \log {\left (c x^{n} \right )}}{3} - \frac {3 b d^{2} e n x^{4}}{16} + \frac {3 b d^{2} e x^{4} \log {\left (c x^{n} \right )}}{4} - \frac {3 b d e^{2} n x^{5}}{25} + \frac {3 b d e^{2} x^{5} \log {\left (c x^{n} \right )}}{5} - \frac {b e^{3} n x^{6}}{36} + \frac {b e^{3} x^{6} \log {\left (c x^{n} \right )}}{6} \] Input:

integrate(x**2*(e*x+d)**3*(a+b*ln(c*x**n)),x)
 

Output:

a*d**3*x**3/3 + 3*a*d**2*e*x**4/4 + 3*a*d*e**2*x**5/5 + a*e**3*x**6/6 - b* 
d**3*n*x**3/9 + b*d**3*x**3*log(c*x**n)/3 - 3*b*d**2*e*n*x**4/16 + 3*b*d** 
2*e*x**4*log(c*x**n)/4 - 3*b*d*e**2*n*x**5/25 + 3*b*d*e**2*x**5*log(c*x**n 
)/5 - b*e**3*n*x**6/36 + b*e**3*x**6*log(c*x**n)/6
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.43 \[ \int x^2 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{36} \, b e^{3} n x^{6} + \frac {1}{6} \, b e^{3} x^{6} \log \left (c x^{n}\right ) - \frac {3}{25} \, b d e^{2} n x^{5} + \frac {1}{6} \, a e^{3} x^{6} + \frac {3}{5} \, b d e^{2} x^{5} \log \left (c x^{n}\right ) - \frac {3}{16} \, b d^{2} e n x^{4} + \frac {3}{5} \, a d e^{2} x^{5} + \frac {3}{4} \, b d^{2} e x^{4} \log \left (c x^{n}\right ) - \frac {1}{9} \, b d^{3} n x^{3} + \frac {3}{4} \, a d^{2} e x^{4} + \frac {1}{3} \, b d^{3} x^{3} \log \left (c x^{n}\right ) + \frac {1}{3} \, a d^{3} x^{3} \] Input:

integrate(x^2*(e*x+d)^3*(a+b*log(c*x^n)),x, algorithm="maxima")
 

Output:

-1/36*b*e^3*n*x^6 + 1/6*b*e^3*x^6*log(c*x^n) - 3/25*b*d*e^2*n*x^5 + 1/6*a* 
e^3*x^6 + 3/5*b*d*e^2*x^5*log(c*x^n) - 3/16*b*d^2*e*n*x^4 + 3/5*a*d*e^2*x^ 
5 + 3/4*b*d^2*e*x^4*log(c*x^n) - 1/9*b*d^3*n*x^3 + 3/4*a*d^2*e*x^4 + 1/3*b 
*d^3*x^3*log(c*x^n) + 1/3*a*d^3*x^3
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.77 \[ \int x^2 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{6} \, b e^{3} n x^{6} \log \left (x\right ) - \frac {1}{36} \, b e^{3} n x^{6} + \frac {1}{6} \, b e^{3} x^{6} \log \left (c\right ) + \frac {3}{5} \, b d e^{2} n x^{5} \log \left (x\right ) - \frac {3}{25} \, b d e^{2} n x^{5} + \frac {1}{6} \, a e^{3} x^{6} + \frac {3}{5} \, b d e^{2} x^{5} \log \left (c\right ) + \frac {3}{4} \, b d^{2} e n x^{4} \log \left (x\right ) - \frac {3}{16} \, b d^{2} e n x^{4} + \frac {3}{5} \, a d e^{2} x^{5} + \frac {3}{4} \, b d^{2} e x^{4} \log \left (c\right ) + \frac {1}{3} \, b d^{3} n x^{3} \log \left (x\right ) - \frac {1}{9} \, b d^{3} n x^{3} + \frac {3}{4} \, a d^{2} e x^{4} + \frac {1}{3} \, b d^{3} x^{3} \log \left (c\right ) + \frac {1}{3} \, a d^{3} x^{3} \] Input:

integrate(x^2*(e*x+d)^3*(a+b*log(c*x^n)),x, algorithm="giac")
 

Output:

1/6*b*e^3*n*x^6*log(x) - 1/36*b*e^3*n*x^6 + 1/6*b*e^3*x^6*log(c) + 3/5*b*d 
*e^2*n*x^5*log(x) - 3/25*b*d*e^2*n*x^5 + 1/6*a*e^3*x^6 + 3/5*b*d*e^2*x^5*l 
og(c) + 3/4*b*d^2*e*n*x^4*log(x) - 3/16*b*d^2*e*n*x^4 + 3/5*a*d*e^2*x^5 + 
3/4*b*d^2*e*x^4*log(c) + 1/3*b*d^3*n*x^3*log(x) - 1/9*b*d^3*n*x^3 + 3/4*a* 
d^2*e*x^4 + 1/3*b*d^3*x^3*log(c) + 1/3*a*d^3*x^3
 

Mupad [B] (verification not implemented)

Time = 28.68 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.13 \[ \int x^2 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,d^3\,x^3}{3}+\frac {3\,b\,d^2\,e\,x^4}{4}+\frac {3\,b\,d\,e^2\,x^5}{5}+\frac {b\,e^3\,x^6}{6}\right )+\frac {d^3\,x^3\,\left (3\,a-b\,n\right )}{9}+\frac {e^3\,x^6\,\left (6\,a-b\,n\right )}{36}+\frac {3\,d^2\,e\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {3\,d\,e^2\,x^5\,\left (5\,a-b\,n\right )}{25} \] Input:

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

Output:

log(c*x^n)*((b*d^3*x^3)/3 + (b*e^3*x^6)/6 + (3*b*d^2*e*x^4)/4 + (3*b*d*e^2 
*x^5)/5) + (d^3*x^3*(3*a - b*n))/9 + (e^3*x^6*(6*a - b*n))/36 + (3*d^2*e*x 
^4*(4*a - b*n))/16 + (3*d*e^2*x^5*(5*a - b*n))/25
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.33 \[ \int x^2 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{3} \left (1200 \,\mathrm {log}\left (x^{n} c \right ) b \,d^{3}+2700 \,\mathrm {log}\left (x^{n} c \right ) b \,d^{2} e x +2160 \,\mathrm {log}\left (x^{n} c \right ) b d \,e^{2} x^{2}+600 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{3} x^{3}+1200 a \,d^{3}+2700 a \,d^{2} e x +2160 a d \,e^{2} x^{2}+600 a \,e^{3} x^{3}-400 b \,d^{3} n -675 b \,d^{2} e n x -432 b d \,e^{2} n \,x^{2}-100 b \,e^{3} n \,x^{3}\right )}{3600} \] Input:

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

Output:

(x**3*(1200*log(x**n*c)*b*d**3 + 2700*log(x**n*c)*b*d**2*e*x + 2160*log(x* 
*n*c)*b*d*e**2*x**2 + 600*log(x**n*c)*b*e**3*x**3 + 1200*a*d**3 + 2700*a*d 
**2*e*x + 2160*a*d*e**2*x**2 + 600*a*e**3*x**3 - 400*b*d**3*n - 675*b*d**2 
*e*n*x - 432*b*d*e**2*n*x**2 - 100*b*e**3*n*x**3))/3600