\(\int x^3 (d+e x^2)^2 (a+b \log (c x^n)) \, dx\) [184]

Optimal result

Integrand size = 23, antiderivative size = 74 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} b d^2 n x^4-\frac {1}{18} b d e n x^6-\frac {1}{64} b e^2 n x^8+\frac {1}{24} \left (6 d^2 x^4+8 d e x^6+3 e^2 x^8\right ) \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

-1/16*b*d^2*n*x^4-1/18*b*d*e*n*x^6-1/64*b*e^2*n*x^8+1/24*(3*e^2*x^8+8*d*e*x^6+6*d^2*x^4)*(a+b*ln(c*x^n))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {272, 45, 2371, 12, 14} \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{24} \left (6 d^2 x^4+8 d e x^6+3 e^2 x^8\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{16} b d^2 n x^4-\frac {1}{18} b d e n x^6-\frac {1}{64} b e^2 n x^8 \]

[In]

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

[Out]

-1/16*(b*d^2*n*x^4) - (b*d*e*n*x^6)/18 - (b*e^2*n*x^8)/64 + ((6*d^2*x^4 + 8*d*e*x^6 + 3*e^2*x^8)*(a + b*Log[c*
x^n]))/24

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

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 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2371

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{24} \left (6 d^2 x^4+8 d e x^6+3 e^2 x^8\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {1}{24} x^3 \left (6 d^2+8 d e x^2+3 e^2 x^4\right ) \, dx \\ & = \frac {1}{24} \left (6 d^2 x^4+8 d e x^6+3 e^2 x^8\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{24} (b n) \int x^3 \left (6 d^2+8 d e x^2+3 e^2 x^4\right ) \, dx \\ & = \frac {1}{24} \left (6 d^2 x^4+8 d e x^6+3 e^2 x^8\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{24} (b n) \int \left (6 d^2 x^3+8 d e x^5+3 e^2 x^7\right ) \, dx \\ & = -\frac {1}{16} b d^2 n x^4-\frac {1}{18} b d e n x^6-\frac {1}{64} b e^2 n x^8+\frac {1}{24} \left (6 d^2 x^4+8 d e x^6+3 e^2 x^8\right ) \left (a+b \log \left (c x^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{576} x^4 \left (24 a \left (6 d^2+8 d e x^2+3 e^2 x^4\right )-b n \left (36 d^2+32 d e x^2+9 e^2 x^4\right )+24 b \left (6 d^2+8 d e x^2+3 e^2 x^4\right ) \log \left (c x^n\right )\right ) \]

[In]

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

[Out]

(x^4*(24*a*(6*d^2 + 8*d*e*x^2 + 3*e^2*x^4) - b*n*(36*d^2 + 32*d*e*x^2 + 9*e^2*x^4) + 24*b*(6*d^2 + 8*d*e*x^2 +
3*e^2*x^4)*Log[c*x^n]))/576

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.36

method	result	size
parallelrisch	\(\frac {x^{8} \ln \left (c \,x^{n}\right ) b \,e^{2}}{8}-\frac {b \,e^{2} n \,x^{8}}{64}+\frac {a \,e^{2} x^{8}}{8}+\frac {x^{6} \ln \left (c \,x^{n}\right ) b d e}{3}-\frac {b d e n \,x^{6}}{18}+\frac {a d e \,x^{6}}{3}+\frac {x^{4} \ln \left (c \,x^{n}\right ) b \,d^{2}}{4}-\frac {b \,d^{2} n \,x^{4}}{16}+\frac {a \,d^{2} x^{4}}{4}\)	\(101\)
risch	\(\frac {b \,x^{4} \left (3 e^{2} x^{4}+8 d e \,x^{2}+6 d^{2}\right ) \ln \left (x^{n}\right )}{24}+\frac {i \pi b \,d^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b d e \,x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{6}+\frac {i \pi b \,e^{2} x^{8} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {i \pi b d e \,x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{6}+\frac {\ln \left (c \right ) b \,e^{2} x^{8}}{8}-\frac {b \,e^{2} n \,x^{8}}{64}+\frac {a \,e^{2} x^{8}}{8}-\frac {i \pi b \,e^{2} x^{8} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{16}+\frac {i \pi b \,d^{2} x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b \,d^{2} x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{8}+\frac {i \pi b \,e^{2} x^{8} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {\ln \left (c \right ) b d e \,x^{6}}{3}-\frac {b d e n \,x^{6}}{18}+\frac {a d e \,x^{6}}{3}+\frac {i \pi b d e \,x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{6}-\frac {i \pi b d e \,x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{6}-\frac {i \pi b \,e^{2} x^{8} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{16}-\frac {i \pi b \,d^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{8}+\frac {\ln \left (c \right ) b \,d^{2} x^{4}}{4}-\frac {b \,d^{2} n \,x^{4}}{16}+\frac {a \,d^{2} x^{4}}{4}\)	\(434\)

[In]

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

[Out]

1/8*x^8*ln(c*x^n)*b*e^2-1/64*b*e^2*n*x^8+1/8*a*e^2*x^8+1/3*x^6*ln(c*x^n)*b*d*e-1/18*b*d*e*n*x^6+1/3*a*d*e*x^6+
1/4*x^4*ln(c*x^n)*b*d^2-1/16*b*d^2*n*x^4+1/4*a*d^2*x^4

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.59 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{64} \, {\left (b e^{2} n - 8 \, a e^{2}\right )} x^{8} - \frac {1}{18} \, {\left (b d e n - 6 \, a d e\right )} x^{6} - \frac {1}{16} \, {\left (b d^{2} n - 4 \, a d^{2}\right )} x^{4} + \frac {1}{24} \, {\left (3 \, b e^{2} x^{8} + 8 \, b d e x^{6} + 6 \, b d^{2} x^{4}\right )} \log \left (c\right ) + \frac {1}{24} \, {\left (3 \, b e^{2} n x^{8} + 8 \, b d e n x^{6} + 6 \, b d^{2} n x^{4}\right )} \log \left (x\right ) \]

[In]

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

[Out]

-1/64*(b*e^2*n - 8*a*e^2)*x^8 - 1/18*(b*d*e*n - 6*a*d*e)*x^6 - 1/16*(b*d^2*n - 4*a*d^2)*x^4 + 1/24*(3*b*e^2*x^
8 + 8*b*d*e*x^6 + 6*b*d^2*x^4)*log(c) + 1/24*(3*b*e^2*n*x^8 + 8*b*d*e*n*x^6 + 6*b*d^2*n*x^4)*log(x)

Sympy [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d^{2} x^{4}}{4} + \frac {a d e x^{6}}{3} + \frac {a e^{2} x^{8}}{8} - \frac {b d^{2} n x^{4}}{16} + \frac {b d^{2} x^{4} \log {\left (c x^{n} \right )}}{4} - \frac {b d e n x^{6}}{18} + \frac {b d e x^{6} \log {\left (c x^{n} \right )}}{3} - \frac {b e^{2} n x^{8}}{64} + \frac {b e^{2} x^{8} \log {\left (c x^{n} \right )}}{8} \]

[In]

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

[Out]

a*d**2*x**4/4 + a*d*e*x**6/3 + a*e**2*x**8/8 - b*d**2*n*x**4/16 + b*d**2*x**4*log(c*x**n)/4 - b*d*e*n*x**6/18
+ b*d*e*x**6*log(c*x**n)/3 - b*e**2*n*x**8/64 + b*e**2*x**8*log(c*x**n)/8

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{64} \, b e^{2} n x^{8} + \frac {1}{8} \, b e^{2} x^{8} \log \left (c x^{n}\right ) + \frac {1}{8} \, a e^{2} x^{8} - \frac {1}{18} \, b d e n x^{6} + \frac {1}{3} \, b d e x^{6} \log \left (c x^{n}\right ) + \frac {1}{3} \, a d e x^{6} - \frac {1}{16} \, b d^{2} n x^{4} + \frac {1}{4} \, b d^{2} x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a d^{2} x^{4} \]

[In]

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

[Out]

-1/64*b*e^2*n*x^8 + 1/8*b*e^2*x^8*log(c*x^n) + 1/8*a*e^2*x^8 - 1/18*b*d*e*n*x^6 + 1/3*b*d*e*x^6*log(c*x^n) + 1
/3*a*d*e*x^6 - 1/16*b*d^2*n*x^4 + 1/4*b*d^2*x^4*log(c*x^n) + 1/4*a*d^2*x^4

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.66 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{8} \, b e^{2} n x^{8} \log \left (x\right ) - \frac {1}{64} \, b e^{2} n x^{8} + \frac {1}{8} \, b e^{2} x^{8} \log \left (c\right ) + \frac {1}{8} \, a e^{2} x^{8} + \frac {1}{3} \, b d e n x^{6} \log \left (x\right ) - \frac {1}{18} \, b d e n x^{6} + \frac {1}{3} \, b d e x^{6} \log \left (c\right ) + \frac {1}{3} \, a d e x^{6} + \frac {1}{4} \, b d^{2} n x^{4} \log \left (x\right ) - \frac {1}{16} \, b d^{2} n x^{4} + \frac {1}{4} \, b d^{2} x^{4} \log \left (c\right ) + \frac {1}{4} \, a d^{2} x^{4} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11 \[ \int x^3 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2\,x^4}{4}+\frac {b\,d\,e\,x^6}{3}+\frac {b\,e^2\,x^8}{8}\right )+\frac {d^2\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {e^2\,x^8\,\left (8\,a-b\,n\right )}{64}+\frac {d\,e\,x^6\,\left (6\,a-b\,n\right )}{18} \]

[In]

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

[Out]

log(c*x^n)*((b*d^2*x^4)/4 + (b*e^2*x^8)/8 + (b*d*e*x^6)/3) + (d^2*x^4*(4*a - b*n))/16 + (e^2*x^8*(8*a - b*n))/
64 + (d*e*x^6*(6*a - b*n))/18