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]

-(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

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Rubi [A]  time = 0.0418457, antiderivative size = 48, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.002122, 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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Maple [C]  time = 0.187, size = 266, normalized size = 5.5 \begin{align*}{\frac{b{x}^{3} \left ( 3\,e{x}^{2}+5\,d \right ) \ln \left ({x}^{n} \right ) }{15}}+{\frac{i}{10}}\pi \,be{x}^{5}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-{\frac{i}{10}}\pi \,be{x}^{5}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -{\frac{i}{10}}\pi \,be{x}^{5} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+{\frac{i}{10}}\pi \,be{x}^{5} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{\ln \left ( c \right ) be{x}^{5}}{5}}-{\frac{ben{x}^{5}}{25}}+{\frac{ae{x}^{5}}{5}}+{\frac{i}{6}}\pi \,bd{x}^{3}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-{\frac{i}{6}}\pi \,bd{x}^{3}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -{\frac{i}{6}}\pi \,bd{x}^{3} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+{\frac{i}{6}}\pi \,bd{x}^{3} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{\ln \left ( c \right ) bd{x}^{3}}{3}}-{\frac{bdn{x}^{3}}{9}}+{\frac{ad{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(e*x^2+d)*(a+b*ln(c*x^n)),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*ln(c)
*b*e*x^5-1/25*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*ln(c)*
b*d*x^3-1/9*b*d*n*x^3+1/3*a*d*x^3

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Maxima [A]  time = 1.06058, size = 77, normalized size = 1.6 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 1.23287, size = 180, normalized size = 3.75 \begin{align*} -\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 \left (c\right ) + \frac{1}{15} \,{\left (3 \, b e n x^{5} + 5 \, b d n x^{3}\right )} \log \left (x\right ) \end{align*}

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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Sympy [B]  time = 3.4239, size = 87, normalized size = 1.81 \begin{align*} \frac{a d x^{3}}{3} + \frac{a e x^{5}}{5} + \frac{b d n x^{3} \log{\left (x \right )}}{3} - \frac{b d n x^{3}}{9} + \frac{b d x^{3} \log{\left (c \right )}}{3} + \frac{b e n x^{5} \log{\left (x \right )}}{5} - \frac{b e n x^{5}}{25} + \frac{b e x^{5} \log{\left (c \right )}}{5} \end{align*}

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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Giac [A]  time = 1.24387, size = 99, normalized size = 2.06 \begin{align*} \frac{1}{5} \, b n x^{5} e \log \left (x\right ) - \frac{1}{25} \, b n x^{5} e + \frac{1}{5} \, b x^{5} e \log \left (c\right ) + \frac{1}{5} \, a x^{5} e + \frac{1}{3} \, b d n x^{3} \log \left (x\right ) - \frac{1}{9} \, b d n x^{3} + \frac{1}{3} \, b d x^{3} \log \left (c\right ) + \frac{1}{3} \, a d x^{3} \end{align*}

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