3.2.97 \(\int x^3 (d+e x^2)^3 (a+b \log (c x^n)) \, dx\) [197]

3.2.97.1 Optimal result

Integrand size = 23, antiderivative size = 130 \[ \int x^3 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b d^4 n x^2}{20 e}+\frac {3}{80} b d^3 n x^4+\frac {1}{60} b d^2 e n x^6+\frac {1}{320} b d e^2 n x^8-\frac {b n \left (d+e x^2\right )^5}{100 e^2}+\frac {b d^5 n \log (x)}{40 e^2}-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right ) \]

output

1/20*b*d^4*n*x^2/e+3/80*b*d^3*n*x^4+1/60*b*d^2*e*n*x^6+1/320*b*d*e^2*n*x^8 
-1/100*b*n*(e*x^2+d)^5/e^2+1/40*b*d^5*n*ln(x)/e^2-1/40*(5*d*(e*x^2+d)^4/e^ 
2-4*(e*x^2+d)^5/e^2)*(a+b*ln(c*x^n))
 

3.2.97.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.92 \[ \int x^3 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^4 \left (120 a \left (10 d^3+20 d^2 e x^2+15 d e^2 x^4+4 e^3 x^6\right )-b n \left (300 d^3+400 d^2 e x^2+225 d e^2 x^4+48 e^3 x^6\right )+120 b \left (10 d^3+20 d^2 e x^2+15 d e^2 x^4+4 e^3 x^6\right ) \log \left (c x^n\right )\right )}{4800} \]

input

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

output

(x^4*(120*a*(10*d^3 + 20*d^2*e*x^2 + 15*d*e^2*x^4 + 4*e^3*x^6) - b*n*(300* 
d^3 + 400*d^2*e*x^2 + 225*d*e^2*x^4 + 48*e^3*x^6) + 120*b*(10*d^3 + 20*d^2 
*e*x^2 + 15*d*e^2*x^4 + 4*e^3*x^6)*Log[c*x^n]))/4800
 

3.2.97.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2771, 27, 354, 90, 49, 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.

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

\(\Big \downarrow \) 2771

\(\displaystyle -b n \int -\frac {\left (d-4 e x^2\right ) \left (e x^2+d\right )^4}{40 e^2 x}dx-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b n \int \frac {\left (d-4 e x^2\right ) \left (e x^2+d\right )^4}{x}dx}{40 e^2}-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {b n \int \frac {\left (d-4 e x^2\right ) \left (e x^2+d\right )^4}{x^2}dx^2}{80 e^2}-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {b n \left (d \int \frac {\left (e x^2+d\right )^4}{x^2}dx^2-\frac {4}{5} \left (d+e x^2\right )^5\right )}{80 e^2}-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {b n \left (d \int \left (e^4 x^6+4 d e^3 x^4+6 d^2 e^2 x^2+4 d^3 e+\frac {d^4}{x^2}\right )dx^2-\frac {4}{5} \left (d+e x^2\right )^5\right )}{80 e^2}-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b n \left (d \left (d^4 \log \left (x^2\right )+4 d^3 e x^2+3 d^2 e^2 x^4+\frac {4}{3} d e^3 x^6+\frac {e^4 x^8}{4}\right )-\frac {4}{5} \left (d+e x^2\right )^5\right )}{80 e^2}-\frac {1}{40} \left (\frac {5 d \left (d+e x^2\right )^4}{e^2}-\frac {4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )\)

input

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

output

(b*n*((-4*(d + e*x^2)^5)/5 + d*(4*d^3*e*x^2 + 3*d^2*e^2*x^4 + (4*d*e^3*x^6 
)/3 + (e^4*x^8)/4 + d^4*Log[x^2])))/(80*e^2) - (((5*d*(d + e*x^2)^4)/e^2 - 
 (4*(d + e*x^2)^5)/e^2)*(a + b*Log[c*x^n]))/40
 

3.2.97.3.1 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 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

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]
 

3.2.97.4 Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.11

method	result	size
parallelrisch	\(\frac {x^{10} \ln \left (c \,x^{n}\right ) b \,e^{3}}{10}-\frac {b \,e^{3} n \,x^{10}}{100}+\frac {a \,e^{3} x^{10}}{10}+\frac {3 x^{8} \ln \left (c \,x^{n}\right ) b d \,e^{2}}{8}-\frac {3 b d \,e^{2} n \,x^{8}}{64}+\frac {3 a d \,e^{2} x^{8}}{8}+\frac {x^{6} \ln \left (c \,x^{n}\right ) b \,d^{2} e}{2}-\frac {b \,d^{2} e n \,x^{6}}{12}+\frac {a \,d^{2} e \,x^{6}}{2}+\frac {x^{4} \ln \left (c \,x^{n}\right ) b \,d^{3}}{4}-\frac {b \,d^{3} n \,x^{4}}{16}+\frac {a \,d^{3} x^{4}}{4}\)	\(144\)
risch	\(\frac {a \,e^{3} x^{10}}{10}+\frac {a \,d^{3} x^{4}}{4}+\frac {i \pi b \,e^{3} 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 \,e^{2} x^{8} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{16}+\frac {i \pi b \,d^{2} e \,x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {i \pi b \,d^{2} e \,x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4}-\frac {i \pi b \,e^{3} x^{10} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{20}+\frac {\ln \left (c \right ) b \,d^{2} e \,x^{6}}{2}+\frac {3 \ln \left (c \right ) b d \,e^{2} x^{8}}{8}+\frac {\ln \left (c \right ) b \,d^{3} x^{4}}{4}-\frac {i \pi b \,e^{3} x^{10} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{20}-\frac {3 i \pi b d \,e^{2} 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 {i \pi b \,d^{2} e \,x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{4}+\frac {i \pi b \,e^{3} x^{10} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{20}-\frac {i \pi b \,d^{3} x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{8}-\frac {i \pi b \,d^{2} e \,x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{4}+\frac {i \pi b \,d^{3} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {i \pi b \,d^{3} 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 \,e^{2} x^{8} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {3 i \pi b d \,e^{2} x^{8} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {a \,d^{2} e \,x^{6}}{2}-\frac {b \,d^{2} e n \,x^{6}}{12}-\frac {3 b d \,e^{2} n \,x^{8}}{64}-\frac {b \,d^{3} n \,x^{4}}{16}-\frac {i \pi b \,d^{3} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{8}-\frac {b \,e^{3} n \,x^{10}}{100}+\frac {3 a d \,e^{2} x^{8}}{8}+\frac {\ln \left (c \right ) b \,e^{3} x^{10}}{10}+\frac {b \,x^{4} \left (4 e^{3} x^{6}+15 e^{2} d \,x^{4}+20 d^{2} e \,x^{2}+10 d^{3}\right ) \ln \left (x^{n}\right )}{40}\)	\(602\)

input

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

output

1/10*x^10*ln(c*x^n)*b*e^3-1/100*b*e^3*n*x^10+1/10*a*e^3*x^10+3/8*x^8*ln(c* 
x^n)*b*d*e^2-3/64*b*d*e^2*n*x^8+3/8*a*d*e^2*x^8+1/2*x^6*ln(c*x^n)*b*d^2*e- 
1/12*b*d^2*e*n*x^6+1/2*a*d^2*e*x^6+1/4*x^4*ln(c*x^n)*b*d^3-1/16*b*d^3*n*x^ 
4+1/4*a*d^3*x^4
 

3.2.97.5 Fricas [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.28 \[ \int x^3 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{100} \, {\left (b e^{3} n - 10 \, a e^{3}\right )} x^{10} - \frac {3}{64} \, {\left (b d e^{2} n - 8 \, a d e^{2}\right )} x^{8} - \frac {1}{12} \, {\left (b d^{2} e n - 6 \, a d^{2} e\right )} x^{6} - \frac {1}{16} \, {\left (b d^{3} n - 4 \, a d^{3}\right )} x^{4} + \frac {1}{40} \, {\left (4 \, b e^{3} x^{10} + 15 \, b d e^{2} x^{8} + 20 \, b d^{2} e x^{6} + 10 \, b d^{3} x^{4}\right )} \log \left (c\right ) + \frac {1}{40} \, {\left (4 \, b e^{3} n x^{10} + 15 \, b d e^{2} n x^{8} + 20 \, b d^{2} e n x^{6} + 10 \, b d^{3} n x^{4}\right )} \log \left (x\right ) \]

input

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

output

-1/100*(b*e^3*n - 10*a*e^3)*x^10 - 3/64*(b*d*e^2*n - 8*a*d*e^2)*x^8 - 1/12 
*(b*d^2*e*n - 6*a*d^2*e)*x^6 - 1/16*(b*d^3*n - 4*a*d^3)*x^4 + 1/40*(4*b*e^ 
3*x^10 + 15*b*d*e^2*x^8 + 20*b*d^2*e*x^6 + 10*b*d^3*x^4)*log(c) + 1/40*(4* 
b*e^3*n*x^10 + 15*b*d*e^2*n*x^8 + 20*b*d^2*e*n*x^6 + 10*b*d^3*n*x^4)*log(x 
)
 

3.2.97.6 Sympy [A] (verification not implemented)

Time = 1.73 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.31 \[ \int x^3 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d^{3} x^{4}}{4} + \frac {a d^{2} e x^{6}}{2} + \frac {3 a d e^{2} x^{8}}{8} + \frac {a e^{3} x^{10}}{10} - \frac {b d^{3} n x^{4}}{16} + \frac {b d^{3} x^{4} \log {\left (c x^{n} \right )}}{4} - \frac {b d^{2} e n x^{6}}{12} + \frac {b d^{2} e x^{6} \log {\left (c x^{n} \right )}}{2} - \frac {3 b d e^{2} n x^{8}}{64} + \frac {3 b d e^{2} x^{8} \log {\left (c x^{n} \right )}}{8} - \frac {b e^{3} n x^{10}}{100} + \frac {b e^{3} x^{10} \log {\left (c x^{n} \right )}}{10} \]

input

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

output

a*d**3*x**4/4 + a*d**2*e*x**6/2 + 3*a*d*e**2*x**8/8 + a*e**3*x**10/10 - b* 
d**3*n*x**4/16 + b*d**3*x**4*log(c*x**n)/4 - b*d**2*e*n*x**6/12 + b*d**2*e 
*x**6*log(c*x**n)/2 - 3*b*d*e**2*n*x**8/64 + 3*b*d*e**2*x**8*log(c*x**n)/8 
 - b*e**3*n*x**10/100 + b*e**3*x**10*log(c*x**n)/10
 

3.2.97.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.10 \[ \int x^3 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{100} \, b e^{3} n x^{10} + \frac {1}{10} \, b e^{3} x^{10} \log \left (c x^{n}\right ) + \frac {1}{10} \, a e^{3} x^{10} - \frac {3}{64} \, b d e^{2} n x^{8} + \frac {3}{8} \, b d e^{2} x^{8} \log \left (c x^{n}\right ) + \frac {3}{8} \, a d e^{2} x^{8} - \frac {1}{12} \, b d^{2} e n x^{6} + \frac {1}{2} \, b d^{2} e x^{6} \log \left (c x^{n}\right ) + \frac {1}{2} \, a d^{2} e x^{6} - \frac {1}{16} \, b d^{3} n x^{4} + \frac {1}{4} \, b d^{3} x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a d^{3} x^{4} \]

input

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

output

-1/100*b*e^3*n*x^10 + 1/10*b*e^3*x^10*log(c*x^n) + 1/10*a*e^3*x^10 - 3/64* 
b*d*e^2*n*x^8 + 3/8*b*d*e^2*x^8*log(c*x^n) + 3/8*a*d*e^2*x^8 - 1/12*b*d^2* 
e*n*x^6 + 1/2*b*d^2*e*x^6*log(c*x^n) + 1/2*a*d^2*e*x^6 - 1/16*b*d^3*n*x^4 
+ 1/4*b*d^3*x^4*log(c*x^n) + 1/4*a*d^3*x^4
 

3.2.97.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.36 \[ \int x^3 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{10} \, b e^{3} n x^{10} \log \left (x\right ) - \frac {1}{100} \, b e^{3} n x^{10} + \frac {1}{10} \, b e^{3} x^{10} \log \left (c\right ) + \frac {1}{10} \, a e^{3} x^{10} + \frac {3}{8} \, b d e^{2} n x^{8} \log \left (x\right ) - \frac {3}{64} \, b d e^{2} n x^{8} + \frac {3}{8} \, b d e^{2} x^{8} \log \left (c\right ) + \frac {3}{8} \, a d e^{2} x^{8} + \frac {1}{2} \, b d^{2} e n x^{6} \log \left (x\right ) - \frac {1}{12} \, b d^{2} e n x^{6} + \frac {1}{2} \, b d^{2} e x^{6} \log \left (c\right ) + \frac {1}{2} \, a d^{2} e x^{6} + \frac {1}{4} \, b d^{3} n x^{4} \log \left (x\right ) - \frac {1}{16} \, b d^{3} n x^{4} + \frac {1}{4} \, b d^{3} x^{4} \log \left (c\right ) + \frac {1}{4} \, a d^{3} x^{4} \]

input

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

output

1/10*b*e^3*n*x^10*log(x) - 1/100*b*e^3*n*x^10 + 1/10*b*e^3*x^10*log(c) + 1 
/10*a*e^3*x^10 + 3/8*b*d*e^2*n*x^8*log(x) - 3/64*b*d*e^2*n*x^8 + 3/8*b*d*e 
^2*x^8*log(c) + 3/8*a*d*e^2*x^8 + 1/2*b*d^2*e*n*x^6*log(x) - 1/12*b*d^2*e* 
n*x^6 + 1/2*b*d^2*e*x^6*log(c) + 1/2*a*d^2*e*x^6 + 1/4*b*d^3*n*x^4*log(x) 
- 1/16*b*d^3*n*x^4 + 1/4*b*d^3*x^4*log(c) + 1/4*a*d^3*x^4
 

3.2.97.9 Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.87 \[ \int x^3 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,d^3\,x^4}{4}+\frac {b\,d^2\,e\,x^6}{2}+\frac {3\,b\,d\,e^2\,x^8}{8}+\frac {b\,e^3\,x^{10}}{10}\right )+\frac {d^3\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {e^3\,x^{10}\,\left (10\,a-b\,n\right )}{100}+\frac {d^2\,e\,x^6\,\left (6\,a-b\,n\right )}{12}+\frac {3\,d\,e^2\,x^8\,\left (8\,a-b\,n\right )}{64} \]

input

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

output

log(c*x^n)*((b*d^3*x^4)/4 + (b*e^3*x^10)/10 + (b*d^2*e*x^6)/2 + (3*b*d*e^2 
*x^8)/8) + (d^3*x^4*(4*a - b*n))/16 + (e^3*x^10*(10*a - b*n))/100 + (d^2*e 
*x^6*(6*a - b*n))/12 + (3*d*e^2*x^8*(8*a - b*n))/64