3.178 \(\int x^2 (d+e x^2) (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=48 \[ \frac {1}{15} \left (5 d x^3+3 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b d n x^3-\frac {1}{25} b e n x^5 \]

[Out]

-1/9*b*d*n*x^3-1/25*b*e*n*x^5+1/15*(3*e*x^5+5*d*x^3)*(a+b*ln(c*x^n))

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Rubi [A]  time = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {14, 2334} \[ \frac {1}{15} \left (5 d x^3+3 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b d n x^3-\frac {1}{25} b e n x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d*n*x^3)/9 - (b*e*n*x^5)/25 + ((5*d*x^3 + 3*e*x^5)*(a + b*Log[c*x^n]))/15

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 2334

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] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {align*} \int x^2 \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {1}{15} \left (5 d x^3+3 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac {d x^2}{3}+\frac {e x^4}{5}\right ) \, dx\\ &=-\frac {1}{9} b d n x^3-\frac {1}{25} b e n x^5+\frac {1}{15} \left (5 d x^3+3 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 69, normalized size = 1.44 \[ \frac {1}{3} a d x^3+\frac {1}{5} a e x^5+\frac {1}{3} b d x^3 \log \left (c x^n\right )+\frac {1}{5} b e x^5 \log \left (c x^n\right )-\frac {1}{9} b d n x^3-\frac {1}{25} b e n x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 69, normalized size = 1.44 \[ -\frac {1}{25} \, {\left (b e n - 5 \, a e\right )} x^{5} - \frac {1}{9} \, {\left (b d n - 3 \, a d\right )} x^{3} + \frac {1}{15} \, {\left (3 \, b e x^{5} + 5 \, b d x^{3}\right )} \log \relax (c) + \frac {1}{15} \, {\left (3 \, b e n x^{5} + 5 \, b d n x^{3}\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/25*(b*e*n - 5*a*e)*x^5 - 1/9*(b*d*n - 3*a*d)*x^3 + 1/15*(3*b*e*x^5 + 5*b*d*x^3)*log(c) + 1/15*(3*b*e*n*x^5
+ 5*b*d*n*x^3)*log(x)

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giac [A]  time = 0.36, size = 73, normalized size = 1.52 \[ \frac {1}{5} \, b n x^{5} e \log \relax (x) - \frac {1}{25} \, b n x^{5} e + \frac {1}{5} \, b x^{5} e \log \relax (c) + \frac {1}{5} \, a x^{5} e + \frac {1}{3} \, b d n x^{3} \log \relax (x) - \frac {1}{9} \, b d n x^{3} + \frac {1}{3} \, b d x^{3} \log \relax (c) + \frac {1}{3} \, a d x^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.21, size = 266, normalized size = 5.54 \[ -\frac {i \pi b e \,x^{5} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{10}+\frac {i \pi b e \,x^{5} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{10}+\frac {i \pi b e \,x^{5} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{10}-\frac {i \pi b e \,x^{5} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{10}-\frac {i \pi b d \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{6}+\frac {i \pi b d \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6}+\frac {i \pi b d \,x^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6}-\frac {i \pi b d \,x^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{6}-\frac {b e n \,x^{5}}{25}+\frac {b e \,x^{5} \ln \relax (c )}{5}+\frac {a e \,x^{5}}{5}-\frac {b d n \,x^{3}}{9}+\frac {b d \,x^{3} \ln \relax (c )}{3}+\frac {a d \,x^{3}}{3}+\frac {\left (3 e \,x^{2}+5 d \right ) b \,x^{3} \ln \left (x^{n}\right )}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*b*x^3*(3*e*x^2+5*d)*ln(x^n)+1/10*I*Pi*b*e*x^5*csgn(I*x^n)*csgn(I*c*x^n)^2-1/10*I*Pi*b*e*x^5*csgn(I*x^n)*c
sgn(I*c*x^n)*csgn(I*c)-1/10*I*Pi*b*e*x^5*csgn(I*c*x^n)^3+1/10*I*Pi*b*e*x^5*csgn(I*c*x^n)^2*csgn(I*c)+1/5*b*e*x
^5*ln(c)-1/25*b*e*n*x^5+1/5*a*e*x^5+1/6*I*Pi*b*d*x^3*csgn(I*x^n)*csgn(I*c*x^n)^2-1/6*I*Pi*b*d*x^3*csgn(I*x^n)*
csgn(I*c*x^n)*csgn(I*c)-1/6*I*Pi*b*d*x^3*csgn(I*c*x^n)^3+1/6*I*Pi*b*d*x^3*csgn(I*c*x^n)^2*csgn(I*c)+1/3*b*d*x^
3*ln(c)-1/9*b*d*n*x^3+1/3*a*d*x^3

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maxima [A]  time = 0.46, size = 57, normalized size = 1.19 \[ -\frac {1}{25} \, b e n x^{5} + \frac {1}{5} \, b e x^{5} \log \left (c x^{n}\right ) + \frac {1}{5} \, a e x^{5} - \frac {1}{9} \, b d n x^{3} + \frac {1}{3} \, b d x^{3} \log \left (c x^{n}\right ) + \frac {1}{3} \, a d x^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.31, size = 51, normalized size = 1.06 \[ \ln \left (c\,x^n\right )\,\left (\frac {b\,e\,x^5}{5}+\frac {b\,d\,x^3}{3}\right )+\frac {d\,x^3\,\left (3\,a-b\,n\right )}{9}+\frac {e\,x^5\,\left (5\,a-b\,n\right )}{25} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 2.31, size = 87, normalized size = 1.81 \[ \frac {a d x^{3}}{3} + \frac {a e x^{5}}{5} + \frac {b d n x^{3} \log {\relax (x )}}{3} - \frac {b d n x^{3}}{9} + \frac {b d x^{3} \log {\relax (c )}}{3} + \frac {b e n x^{5} \log {\relax (x )}}{5} - \frac {b e n x^{5}}{25} + \frac {b e x^{5} \log {\relax (c )}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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