3.2.96 \(\int x^5 (d+e x^2)^3 (a+b \log (c x^n)) \, dx\) [196]

3.2.96.1 Optimal result
3.2.96.2 Mathematica [A] (verified)
3.2.96.3 Rubi [A] (verified)
3.2.96.4 Maple [A] (verified)
3.2.96.5 Fricas [A] (verification not implemented)
3.2.96.6 Sympy [A] (verification not implemented)
3.2.96.7 Maxima [A] (verification not implemented)
3.2.96.8 Giac [A] (verification not implemented)
3.2.96.9 Mupad [B] (verification not implemented)

3.2.96.1 Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int x^5 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{36} b d^3 n x^6-\frac {3}{64} b d^2 e n x^8-\frac {3}{100} b d e^2 n x^{10}-\frac {1}{144} b e^3 n x^{12}+\frac {1}{120} \left (20 d^3 x^6+45 d^2 e x^8+36 d e^2 x^{10}+10 e^3 x^{12}\right ) \left (a+b \log \left (c x^n\right )\right ) \]

output
-1/36*b*d^3*n*x^6-3/64*b*d^2*e*n*x^8-3/100*b*d*e^2*n*x^10-1/144*b*e^3*n*x^ 
12+1/120*(10*e^3*x^12+36*d*e^2*x^10+45*d^2*e*x^8+20*d^3*x^6)*(a+b*ln(c*x^n 
))
 
3.2.96.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.20 \[ \int x^5 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^6 \left (120 a \left (20 d^3+45 d^2 e x^2+36 d e^2 x^4+10 e^3 x^6\right )-b n \left (400 d^3+675 d^2 e x^2+432 d e^2 x^4+100 e^3 x^6\right )+120 b \left (20 d^3+45 d^2 e x^2+36 d e^2 x^4+10 e^3 x^6\right ) \log \left (c x^n\right )\right )}{14400} \]

input
Integrate[x^5*(d + e*x^2)^3*(a + b*Log[c*x^n]),x]
 
output
(x^6*(120*a*(20*d^3 + 45*d^2*e*x^2 + 36*d*e^2*x^4 + 10*e^3*x^6) - b*n*(400 
*d^3 + 675*d^2*e*x^2 + 432*d*e^2*x^4 + 100*e^3*x^6) + 120*b*(20*d^3 + 45*d 
^2*e*x^2 + 36*d*e^2*x^4 + 10*e^3*x^6)*Log[c*x^n]))/14400
 
3.2.96.3 Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, 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.

\(\displaystyle \int x^5 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx\)

\(\Big \downarrow \) 2771

\(\displaystyle \frac {1}{120} \left (20 d^3 x^6+45 d^2 e x^8+36 d e^2 x^{10}+10 e^3 x^{12}\right ) \left (a+b \log \left (c x^n\right )\right )-b n \int \frac {1}{120} x^5 \left (10 e^3 x^6+36 d e^2 x^4+45 d^2 e x^2+20 d^3\right )dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{120} \left (20 d^3 x^6+45 d^2 e x^8+36 d e^2 x^{10}+10 e^3 x^{12}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{120} b n \int x^5 \left (10 e^3 x^6+36 d e^2 x^4+45 d^2 e x^2+20 d^3\right )dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \frac {1}{120} \left (20 d^3 x^6+45 d^2 e x^8+36 d e^2 x^{10}+10 e^3 x^{12}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{120} b n \int \left (10 e^3 x^{11}+36 d e^2 x^9+45 d^2 e x^7+20 d^3 x^5\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{120} \left (20 d^3 x^6+45 d^2 e x^8+36 d e^2 x^{10}+10 e^3 x^{12}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{120} b n \left (\frac {10 d^3 x^6}{3}+\frac {45}{8} d^2 e x^8+\frac {18}{5} d e^2 x^{10}+\frac {5 e^3 x^{12}}{6}\right )\)

input
Int[x^5*(d + e*x^2)^3*(a + b*Log[c*x^n]),x]
 
output
-1/120*(b*n*((10*d^3*x^6)/3 + (45*d^2*e*x^8)/8 + (18*d*e^2*x^10)/5 + (5*e^ 
3*x^12)/6)) + ((20*d^3*x^6 + 45*d^2*e*x^8 + 36*d*e^2*x^10 + 10*e^3*x^12)*( 
a + b*Log[c*x^n]))/120
 

3.2.96.3.1 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]
 
3.2.96.4 Maple [A] (verified)

Time = 9.55 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.44

method result size
parallelrisch \(\frac {e^{3} b \ln \left (c \,x^{n}\right ) x^{12}}{12}-\frac {b \,e^{3} n \,x^{12}}{144}+\frac {a \,e^{3} x^{12}}{12}+\frac {3 e^{2} d b \ln \left (c \,x^{n}\right ) x^{10}}{10}-\frac {3 b d \,e^{2} n \,x^{10}}{100}+\frac {3 a d \,e^{2} x^{10}}{10}+\frac {3 e \,d^{2} b \ln \left (c \,x^{n}\right ) x^{8}}{8}-\frac {3 b \,d^{2} e n \,x^{8}}{64}+\frac {3 a \,d^{2} e \,x^{8}}{8}+\frac {b \,d^{3} \ln \left (c \,x^{n}\right ) x^{6}}{6}-\frac {b \,d^{3} n \,x^{6}}{36}+\frac {a \,d^{3} x^{6}}{6}\) \(144\)
risch \(\frac {b \,x^{6} \left (10 e^{3} x^{6}+36 e^{2} d \,x^{4}+45 d^{2} e \,x^{2}+20 d^{3}\right ) \ln \left (x^{n}\right )}{120}+\frac {a \,e^{3} x^{12}}{12}+\frac {a \,d^{3} x^{6}}{6}-\frac {3 i \pi b d \,e^{2} x^{10} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{20}+\frac {i \pi b \,e^{3} x^{12} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{24}+\frac {i \pi b \,e^{3} x^{12} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{24}+\frac {i \pi b \,d^{3} x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{12}+\frac {i \pi b \,d^{3} x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{12}-\frac {3 i \pi b \,d^{2} e \,x^{8} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{16}-\frac {i \pi b \,e^{3} x^{12} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{24}-\frac {i \pi b \,d^{3} x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{12}+\frac {3 i \pi b d \,e^{2} x^{10} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{20}+\frac {3 i \pi b d \,e^{2} x^{10} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{20}+\frac {3 i \pi b \,d^{2} e \,x^{8} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {3 i \pi b \,d^{2} e \,x^{8} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {3 \ln \left (c \right ) b \,d^{2} e \,x^{8}}{8}+\frac {3 \ln \left (c \right ) b d \,e^{2} x^{10}}{10}-\frac {i \pi b \,d^{3} x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{12}-\frac {i \pi b \,e^{3} x^{12} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{24}-\frac {3 b \,d^{2} e n \,x^{8}}{64}-\frac {3 b d \,e^{2} n \,x^{10}}{100}+\frac {3 a d \,e^{2} x^{10}}{10}+\frac {3 a \,d^{2} e \,x^{8}}{8}+\frac {\ln \left (c \right ) b \,e^{3} x^{12}}{12}+\frac {\ln \left (c \right ) b \,d^{3} x^{6}}{6}-\frac {b \,d^{3} n \,x^{6}}{36}-\frac {b \,e^{3} n \,x^{12}}{144}-\frac {3 i \pi b \,d^{2} e \,x^{8} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{16}-\frac {3 i \pi b d \,e^{2} x^{10} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{20}\) \(602\)

input
int(x^5*(e*x^2+d)^3*(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)
 
output
1/12*e^3*b*ln(c*x^n)*x^12-1/144*b*e^3*n*x^12+1/12*a*e^3*x^12+3/10*e^2*d*b* 
ln(c*x^n)*x^10-3/100*b*d*e^2*n*x^10+3/10*a*d*e^2*x^10+3/8*e*d^2*b*ln(c*x^n 
)*x^8-3/64*b*d^2*e*n*x^8+3/8*a*d^2*e*x^8+1/6*b*d^3*ln(c*x^n)*x^6-1/36*b*d^ 
3*n*x^6+1/6*a*d^3*x^6
 
3.2.96.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.67 \[ \int x^5 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{144} \, {\left (b e^{3} n - 12 \, a e^{3}\right )} x^{12} - \frac {3}{100} \, {\left (b d e^{2} n - 10 \, a d e^{2}\right )} x^{10} - \frac {3}{64} \, {\left (b d^{2} e n - 8 \, a d^{2} e\right )} x^{8} - \frac {1}{36} \, {\left (b d^{3} n - 6 \, a d^{3}\right )} x^{6} + \frac {1}{120} \, {\left (10 \, b e^{3} x^{12} + 36 \, b d e^{2} x^{10} + 45 \, b d^{2} e x^{8} + 20 \, b d^{3} x^{6}\right )} \log \left (c\right ) + \frac {1}{120} \, {\left (10 \, b e^{3} n x^{12} + 36 \, b d e^{2} n x^{10} + 45 \, b d^{2} e n x^{8} + 20 \, b d^{3} n x^{6}\right )} \log \left (x\right ) \]

input
integrate(x^5*(e*x^2+d)^3*(a+b*log(c*x^n)),x, algorithm="fricas")
 
output
-1/144*(b*e^3*n - 12*a*e^3)*x^12 - 3/100*(b*d*e^2*n - 10*a*d*e^2)*x^10 - 3 
/64*(b*d^2*e*n - 8*a*d^2*e)*x^8 - 1/36*(b*d^3*n - 6*a*d^3)*x^6 + 1/120*(10 
*b*e^3*x^12 + 36*b*d*e^2*x^10 + 45*b*d^2*e*x^8 + 20*b*d^3*x^6)*log(c) + 1/ 
120*(10*b*e^3*n*x^12 + 36*b*d*e^2*n*x^10 + 45*b*d^2*e*n*x^8 + 20*b*d^3*n*x 
^6)*log(x)
 
3.2.96.6 Sympy [A] (verification not implemented)

Time = 2.96 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.75 \[ \int x^5 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d^{3} x^{6}}{6} + \frac {3 a d^{2} e x^{8}}{8} + \frac {3 a d e^{2} x^{10}}{10} + \frac {a e^{3} x^{12}}{12} - \frac {b d^{3} n x^{6}}{36} + \frac {b d^{3} x^{6} \log {\left (c x^{n} \right )}}{6} - \frac {3 b d^{2} e n x^{8}}{64} + \frac {3 b d^{2} e x^{8} \log {\left (c x^{n} \right )}}{8} - \frac {3 b d e^{2} n x^{10}}{100} + \frac {3 b d e^{2} x^{10} \log {\left (c x^{n} \right )}}{10} - \frac {b e^{3} n x^{12}}{144} + \frac {b e^{3} x^{12} \log {\left (c x^{n} \right )}}{12} \]

input
integrate(x**5*(e*x**2+d)**3*(a+b*ln(c*x**n)),x)
 
output
a*d**3*x**6/6 + 3*a*d**2*e*x**8/8 + 3*a*d*e**2*x**10/10 + a*e**3*x**12/12 
- b*d**3*n*x**6/36 + b*d**3*x**6*log(c*x**n)/6 - 3*b*d**2*e*n*x**8/64 + 3* 
b*d**2*e*x**8*log(c*x**n)/8 - 3*b*d*e**2*n*x**10/100 + 3*b*d*e**2*x**10*lo 
g(c*x**n)/10 - b*e**3*n*x**12/144 + b*e**3*x**12*log(c*x**n)/12
 
3.2.96.7 Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.43 \[ \int x^5 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{144} \, b e^{3} n x^{12} + \frac {1}{12} \, b e^{3} x^{12} \log \left (c x^{n}\right ) + \frac {1}{12} \, a e^{3} x^{12} - \frac {3}{100} \, b d e^{2} n x^{10} + \frac {3}{10} \, b d e^{2} x^{10} \log \left (c x^{n}\right ) + \frac {3}{10} \, a d e^{2} x^{10} - \frac {3}{64} \, b d^{2} e n x^{8} + \frac {3}{8} \, b d^{2} e x^{8} \log \left (c x^{n}\right ) + \frac {3}{8} \, a d^{2} e x^{8} - \frac {1}{36} \, b d^{3} n x^{6} + \frac {1}{6} \, b d^{3} x^{6} \log \left (c x^{n}\right ) + \frac {1}{6} \, a d^{3} x^{6} \]

input
integrate(x^5*(e*x^2+d)^3*(a+b*log(c*x^n)),x, algorithm="maxima")
 
output
-1/144*b*e^3*n*x^12 + 1/12*b*e^3*x^12*log(c*x^n) + 1/12*a*e^3*x^12 - 3/100 
*b*d*e^2*n*x^10 + 3/10*b*d*e^2*x^10*log(c*x^n) + 3/10*a*d*e^2*x^10 - 3/64* 
b*d^2*e*n*x^8 + 3/8*b*d^2*e*x^8*log(c*x^n) + 3/8*a*d^2*e*x^8 - 1/36*b*d^3* 
n*x^6 + 1/6*b*d^3*x^6*log(c*x^n) + 1/6*a*d^3*x^6
 
3.2.96.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.77 \[ \int x^5 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{12} \, b e^{3} n x^{12} \log \left (x\right ) - \frac {1}{144} \, b e^{3} n x^{12} + \frac {1}{12} \, b e^{3} x^{12} \log \left (c\right ) + \frac {1}{12} \, a e^{3} x^{12} + \frac {3}{10} \, b d e^{2} n x^{10} \log \left (x\right ) - \frac {3}{100} \, b d e^{2} n x^{10} + \frac {3}{10} \, b d e^{2} x^{10} \log \left (c\right ) + \frac {3}{10} \, a d e^{2} x^{10} + \frac {3}{8} \, b d^{2} e n x^{8} \log \left (x\right ) - \frac {3}{64} \, b d^{2} e n x^{8} + \frac {3}{8} \, b d^{2} e x^{8} \log \left (c\right ) + \frac {3}{8} \, a d^{2} e x^{8} + \frac {1}{6} \, b d^{3} n x^{6} \log \left (x\right ) - \frac {1}{36} \, b d^{3} n x^{6} + \frac {1}{6} \, b d^{3} x^{6} \log \left (c\right ) + \frac {1}{6} \, a d^{3} x^{6} \]

input
integrate(x^5*(e*x^2+d)^3*(a+b*log(c*x^n)),x, algorithm="giac")
 
output
1/12*b*e^3*n*x^12*log(x) - 1/144*b*e^3*n*x^12 + 1/12*b*e^3*x^12*log(c) + 1 
/12*a*e^3*x^12 + 3/10*b*d*e^2*n*x^10*log(x) - 3/100*b*d*e^2*n*x^10 + 3/10* 
b*d*e^2*x^10*log(c) + 3/10*a*d*e^2*x^10 + 3/8*b*d^2*e*n*x^8*log(x) - 3/64* 
b*d^2*e*n*x^8 + 3/8*b*d^2*e*x^8*log(c) + 3/8*a*d^2*e*x^8 + 1/6*b*d^3*n*x^6 
*log(x) - 1/36*b*d^3*n*x^6 + 1/6*b*d^3*x^6*log(c) + 1/6*a*d^3*x^6
 
3.2.96.9 Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.13 \[ \int x^5 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,d^3\,x^6}{6}+\frac {3\,b\,d^2\,e\,x^8}{8}+\frac {3\,b\,d\,e^2\,x^{10}}{10}+\frac {b\,e^3\,x^{12}}{12}\right )+\frac {d^3\,x^6\,\left (6\,a-b\,n\right )}{36}+\frac {e^3\,x^{12}\,\left (12\,a-b\,n\right )}{144}+\frac {3\,d^2\,e\,x^8\,\left (8\,a-b\,n\right )}{64}+\frac {3\,d\,e^2\,x^{10}\,\left (10\,a-b\,n\right )}{100} \]

input
int(x^5*(d + e*x^2)^3*(a + b*log(c*x^n)),x)
 
output
log(c*x^n)*((b*d^3*x^6)/6 + (b*e^3*x^12)/12 + (3*b*d^2*e*x^8)/8 + (3*b*d*e 
^2*x^10)/10) + (d^3*x^6*(6*a - b*n))/36 + (e^3*x^12*(12*a - b*n))/144 + (3 
*d^2*e*x^8*(8*a - b*n))/64 + (3*d*e^2*x^10*(10*a - b*n))/100