\(\int x^2 (d+e x^2)^3 (a+b \log (c x^n)) \, dx\) [203]

Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int x^2 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{9} b d^3 n x^3-\frac {3}{25} b d^2 e n x^5-\frac {3}{49} b d e^2 n x^7-\frac {1}{81} b e^3 n x^9+\frac {1}{315} \left (105 d^3 x^3+189 d^2 e x^5+135 d e^2 x^7+35 e^3 x^9\right ) \left (a+b \log \left (c x^n\right )\right ) \] Output:

-1/9*b*d^3*n*x^3-3/25*b*d^2*e*n*x^5-3/49*b*d*e^2*n*x^7-1/81*b*e^3*n*x^9+1/ 
315*(35*e^3*x^9+135*d*e^2*x^7+189*d^2*e*x^5+105*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 \[ \int x^2 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{9} b d^3 n x^3-\frac {3}{25} b d^2 e n x^5-\frac {3}{49} b d e^2 n x^7-\frac {1}{81} b e^3 n x^9+\frac {1}{3} d^3 x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {3}{5} d^2 e x^5 \left (a+b \log \left (c x^n\right )\right )+\frac {3}{7} d e^2 x^7 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{9} e^3 x^9 \left (a+b \log \left (c x^n\right )\right ) \] Input:

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

Output:

-1/9*(b*d^3*n*x^3) - (3*b*d^2*e*n*x^5)/25 - (3*b*d*e^2*n*x^7)/49 - (b*e^3* 
n*x^9)/81 + (d^3*x^3*(a + b*Log[c*x^n]))/3 + (3*d^2*e*x^5*(a + b*Log[c*x^n 
]))/5 + (3*d*e^2*x^7*(a + b*Log[c*x^n]))/7 + (e^3*x^9*(a + b*Log[c*x^n]))/ 
9
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2771, 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.

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

\(\Big \downarrow \) 2771

\(\displaystyle \frac {1}{315} \left (105 d^3 x^3+189 d^2 e x^5+135 d e^2 x^7+35 e^3 x^9\right ) \left (a+b \log \left (c x^n\right )\right )-b n \int \left (\frac {e^3 x^8}{9}+\frac {3}{7} d e^2 x^6+\frac {3}{5} d^2 e x^4+\frac {d^3 x^2}{3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{315} \left (105 d^3 x^3+189 d^2 e x^5+135 d e^2 x^7+35 e^3 x^9\right ) \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {d^3 x^3}{9}+\frac {3}{25} d^2 e x^5+\frac {3}{49} d e^2 x^7+\frac {e^3 x^9}{81}\right )\)

Input:

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

Output:

-(b*n*((d^3*x^3)/9 + (3*d^2*e*x^5)/25 + (3*d*e^2*x^7)/49 + (e^3*x^9)/81)) 
+ ((105*d^3*x^3 + 189*d^2*e*x^5 + 135*d*e^2*x^7 + 35*e^3*x^9)*(a + b*Log[c 
*x^n]))/315
 

Defintions of rubi rules used

rule 2009

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

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 = 2.07 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.44

method	result	size
parallelrisch	\(\frac {x^{9} b \ln \left (c \,x^{n}\right ) e^{3}}{9}-\frac {b \,e^{3} n \,x^{9}}{81}+\frac {x^{9} a \,e^{3}}{9}+\frac {3 x^{7} b \ln \left (c \,x^{n}\right ) e^{2} d}{7}-\frac {3 b d \,e^{2} n \,x^{7}}{49}+\frac {3 x^{7} a \,e^{2} d}{7}+\frac {3 x^{5} b \ln \left (c \,x^{n}\right ) d^{2} e}{5}-\frac {3 b \,d^{2} e n \,x^{5}}{25}+\frac {3 a \,d^{2} e \,x^{5}}{5}+\frac {x^{3} b \ln \left (c \,x^{n}\right ) d^{3}}{3}-\frac {b \,d^{3} n \,x^{3}}{9}+\frac {x^{3} a \,d^{3}}{3}\)	\(144\)
risch	\(\frac {b \,x^{3} \left (35 e^{3} x^{6}+135 e^{2} d \,x^{4}+189 d^{2} e \,x^{2}+105 d^{3}\right ) \ln \left (x^{n}\right )}{315}+\frac {3 a \,d^{2} e \,x^{5}}{5}+\frac {x^{3} a \,d^{3}}{3}+\frac {\ln \left (c \right ) b \,e^{3} x^{9}}{9}+\frac {x^{9} a \,e^{3}}{9}+\frac {3 x^{7} a \,e^{2} d}{7}+\frac {3 \ln \left (c \right ) b d \,e^{2} x^{7}}{7}+\frac {i \pi b \,e^{3} x^{9} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{18}-\frac {3 i \pi b d \,e^{2} x^{7} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{14}+\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^{9} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{18}-\frac {3 i \pi b \,d^{2} e \,x^{5} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{10}-\frac {i \pi b \,e^{3} x^{9} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{18}-\frac {i \pi b \,e^{3} x^{9} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{18}+\frac {3 i \pi b \,d^{2} e \,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^{7} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{14}+\frac {3 i \pi b \,d^{2} e \,x^{5} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{10}+\frac {3 i \pi b d \,e^{2} x^{7} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{14}-\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 {i \pi b \,d^{3} x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{6}-\frac {3 i \pi b \,d^{2} e \,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 {3 i \pi b d \,e^{2} x^{7} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{14}-\frac {b \,e^{3} n \,x^{9}}{81}-\frac {3 b \,d^{2} e n \,x^{5}}{25}+\frac {\ln \left (c \right ) b \,d^{3} x^{3}}{3}-\frac {b \,d^{3} n \,x^{3}}{9}+\frac {3 \ln \left (c \right ) b \,d^{2} e \,x^{5}}{5}-\frac {3 b d \,e^{2} n \,x^{7}}{49}\)	\(602\)

Input:

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

Output:

1/9*x^9*b*ln(c*x^n)*e^3-1/81*b*e^3*n*x^9+1/9*x^9*a*e^3+3/7*x^7*b*ln(c*x^n) 
*e^2*d-3/49*b*d*e^2*n*x^7+3/7*x^7*a*e^2*d+3/5*x^5*b*ln(c*x^n)*d^2*e-3/25*b 
*d^2*e*n*x^5+3/5*a*d^2*e*x^5+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 \[ \int x^2 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{81} \, {\left (b e^{3} n - 9 \, a e^{3}\right )} x^{9} - \frac {3}{49} \, {\left (b d e^{2} n - 7 \, a d e^{2}\right )} x^{7} - \frac {3}{25} \, {\left (b d^{2} e n - 5 \, a d^{2} e\right )} x^{5} - \frac {1}{9} \, {\left (b d^{3} n - 3 \, a d^{3}\right )} x^{3} + \frac {1}{315} \, {\left (35 \, b e^{3} x^{9} + 135 \, b d e^{2} x^{7} + 189 \, b d^{2} e x^{5} + 105 \, b d^{3} x^{3}\right )} \log \left (c\right ) + \frac {1}{315} \, {\left (35 \, b e^{3} n x^{9} + 135 \, b d e^{2} n x^{7} + 189 \, b d^{2} e n x^{5} + 105 \, b d^{3} n x^{3}\right )} \log \left (x\right ) \] Input:

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

Output:

-1/81*(b*e^3*n - 9*a*e^3)*x^9 - 3/49*(b*d*e^2*n - 7*a*d*e^2)*x^7 - 3/25*(b 
*d^2*e*n - 5*a*d^2*e)*x^5 - 1/9*(b*d^3*n - 3*a*d^3)*x^3 + 1/315*(35*b*e^3* 
x^9 + 135*b*d*e^2*x^7 + 189*b*d^2*e*x^5 + 105*b*d^3*x^3)*log(c) + 1/315*(3 
5*b*e^3*n*x^9 + 135*b*d*e^2*n*x^7 + 189*b*d^2*e*n*x^5 + 105*b*d^3*n*x^3)*l 
og(x)
 

Sympy [A] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.75 \[ \int x^2 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d^{3} x^{3}}{3} + \frac {3 a d^{2} e x^{5}}{5} + \frac {3 a d e^{2} x^{7}}{7} + \frac {a e^{3} x^{9}}{9} - \frac {b d^{3} n x^{3}}{9} + \frac {b d^{3} x^{3} \log {\left (c x^{n} \right )}}{3} - \frac {3 b d^{2} e n x^{5}}{25} + \frac {3 b d^{2} e x^{5} \log {\left (c x^{n} \right )}}{5} - \frac {3 b d e^{2} n x^{7}}{49} + \frac {3 b d e^{2} x^{7} \log {\left (c x^{n} \right )}}{7} - \frac {b e^{3} n x^{9}}{81} + \frac {b e^{3} x^{9} \log {\left (c x^{n} \right )}}{9} \] Input:

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

Output:

a*d**3*x**3/3 + 3*a*d**2*e*x**5/5 + 3*a*d*e**2*x**7/7 + a*e**3*x**9/9 - b* 
d**3*n*x**3/9 + b*d**3*x**3*log(c*x**n)/3 - 3*b*d**2*e*n*x**5/25 + 3*b*d** 
2*e*x**5*log(c*x**n)/5 - 3*b*d*e**2*n*x**7/49 + 3*b*d*e**2*x**7*log(c*x**n 
)/7 - b*e**3*n*x**9/81 + b*e**3*x**9*log(c*x**n)/9
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.43 \[ \int x^2 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{81} \, b e^{3} n x^{9} + \frac {1}{9} \, b e^{3} x^{9} \log \left (c x^{n}\right ) + \frac {1}{9} \, a e^{3} x^{9} - \frac {3}{49} \, b d e^{2} n x^{7} + \frac {3}{7} \, b d e^{2} x^{7} \log \left (c x^{n}\right ) + \frac {3}{7} \, a d e^{2} x^{7} - \frac {3}{25} \, b d^{2} e n x^{5} + \frac {3}{5} \, b d^{2} e x^{5} \log \left (c x^{n}\right ) + \frac {3}{5} \, a d^{2} e x^{5} - \frac {1}{9} \, b d^{3} n x^{3} + \frac {1}{3} \, b d^{3} x^{3} \log \left (c x^{n}\right ) + \frac {1}{3} \, a d^{3} x^{3} \] Input:

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

Output:

-1/81*b*e^3*n*x^9 + 1/9*b*e^3*x^9*log(c*x^n) + 1/9*a*e^3*x^9 - 3/49*b*d*e^ 
2*n*x^7 + 3/7*b*d*e^2*x^7*log(c*x^n) + 3/7*a*d*e^2*x^7 - 3/25*b*d^2*e*n*x^ 
5 + 3/5*b*d^2*e*x^5*log(c*x^n) + 3/5*a*d^2*e*x^5 - 1/9*b*d^3*n*x^3 + 1/3*b 
*d^3*x^3*log(c*x^n) + 1/3*a*d^3*x^3
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.77 \[ \int x^2 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{9} \, b e^{3} n x^{9} \log \left (x\right ) - \frac {1}{81} \, b e^{3} n x^{9} + \frac {1}{9} \, b e^{3} x^{9} \log \left (c\right ) + \frac {1}{9} \, a e^{3} x^{9} + \frac {3}{7} \, b d e^{2} n x^{7} \log \left (x\right ) - \frac {3}{49} \, b d e^{2} n x^{7} + \frac {3}{7} \, b d e^{2} x^{7} \log \left (c\right ) + \frac {3}{7} \, a d e^{2} x^{7} + \frac {3}{5} \, b d^{2} e n x^{5} \log \left (x\right ) - \frac {3}{25} \, b d^{2} e n x^{5} + \frac {3}{5} \, b d^{2} e x^{5} \log \left (c\right ) + \frac {3}{5} \, a d^{2} e x^{5} + \frac {1}{3} \, b d^{3} n x^{3} \log \left (x\right ) - \frac {1}{9} \, b d^{3} n x^{3} + \frac {1}{3} \, b d^{3} x^{3} \log \left (c\right ) + \frac {1}{3} \, a d^{3} x^{3} \] Input:

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

Output:

1/9*b*e^3*n*x^9*log(x) - 1/81*b*e^3*n*x^9 + 1/9*b*e^3*x^9*log(c) + 1/9*a*e 
^3*x^9 + 3/7*b*d*e^2*n*x^7*log(x) - 3/49*b*d*e^2*n*x^7 + 3/7*b*d*e^2*x^7*l 
og(c) + 3/7*a*d*e^2*x^7 + 3/5*b*d^2*e*n*x^5*log(x) - 3/25*b*d^2*e*n*x^5 + 
3/5*b*d^2*e*x^5*log(c) + 3/5*a*d^2*e*x^5 + 1/3*b*d^3*n*x^3*log(x) - 1/9*b* 
d^3*n*x^3 + 1/3*b*d^3*x^3*log(c) + 1/3*a*d^3*x^3
 

Mupad [B] (verification not implemented)

Time = 25.62 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.13 \[ \int x^2 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,d^3\,x^3}{3}+\frac {3\,b\,d^2\,e\,x^5}{5}+\frac {3\,b\,d\,e^2\,x^7}{7}+\frac {b\,e^3\,x^9}{9}\right )+\frac {d^3\,x^3\,\left (3\,a-b\,n\right )}{9}+\frac {e^3\,x^9\,\left (9\,a-b\,n\right )}{81}+\frac {3\,d^2\,e\,x^5\,\left (5\,a-b\,n\right )}{25}+\frac {3\,d\,e^2\,x^7\,\left (7\,a-b\,n\right )}{49} \] Input:

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

Output:

log(c*x^n)*((b*d^3*x^3)/3 + (b*e^3*x^9)/9 + (3*b*d^2*e*x^5)/5 + (3*b*d*e^2 
*x^7)/7) + (d^3*x^3*(3*a - b*n))/9 + (e^3*x^9*(9*a - b*n))/81 + (3*d^2*e*x 
^5*(5*a - b*n))/25 + (3*d*e^2*x^7*(7*a - b*n))/49
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.39 \[ \int x^2 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{3} \left (33075 \,\mathrm {log}\left (x^{n} c \right ) b \,d^{3}+59535 \,\mathrm {log}\left (x^{n} c \right ) b \,d^{2} e \,x^{2}+42525 \,\mathrm {log}\left (x^{n} c \right ) b d \,e^{2} x^{4}+11025 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{3} x^{6}+33075 a \,d^{3}+59535 a \,d^{2} e \,x^{2}+42525 a d \,e^{2} x^{4}+11025 a \,e^{3} x^{6}-11025 b \,d^{3} n -11907 b \,d^{2} e n \,x^{2}-6075 b d \,e^{2} n \,x^{4}-1225 b \,e^{3} n \,x^{6}\right )}{99225} \] Input:

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

Output:

(x**3*(33075*log(x**n*c)*b*d**3 + 59535*log(x**n*c)*b*d**2*e*x**2 + 42525* 
log(x**n*c)*b*d*e**2*x**4 + 11025*log(x**n*c)*b*e**3*x**6 + 33075*a*d**3 + 
 59535*a*d**2*e*x**2 + 42525*a*d*e**2*x**4 + 11025*a*e**3*x**6 - 11025*b*d 
**3*n - 11907*b*d**2*e*n*x**2 - 6075*b*d*e**2*n*x**4 - 1225*b*e**3*n*x**6) 
)/99225