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

Optimal. Leaf size=47 \[ \frac{1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{4} b d n x^2-\frac{1}{16} b e n x^4 \]

[Out]

-(b*d*n*x^2)/4 - (b*e*n*x^4)/16 + ((2*d*x^2 + e*x^4)*(a + b*Log[c*x^n]))/4

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Rubi [A]  time = 0.0366205, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {14, 2334, 12} \[ \frac{1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{4} b d n x^2-\frac{1}{16} b e n x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d*n*x^2)/4 - (b*e*n*x^4)/16 + ((2*d*x^2 + e*x^4)*(a + b*Log[c*x^n]))/4

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])

Rule 12

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

Rubi steps

\begin{align*} \int x \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac{1}{4} x \left (2 d+e x^2\right ) \, dx\\ &=\frac{1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{4} (b n) \int x \left (2 d+e x^2\right ) \, dx\\ &=\frac{1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{4} (b n) \int \left (2 d x+e x^3\right ) \, dx\\ &=-\frac{1}{4} b d n x^2-\frac{1}{16} b e n x^4+\frac{1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0021318, size = 69, normalized size = 1.47 \[ \frac{1}{2} a d x^2+\frac{1}{4} a e x^4+\frac{1}{2} b d x^2 \log \left (c x^n\right )+\frac{1}{4} b e x^4 \log \left (c x^n\right )-\frac{1}{4} b d n x^2-\frac{1}{16} b e n x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*d*x^2)/2 - (b*d*n*x^2)/4 + (a*e*x^4)/4 - (b*e*n*x^4)/16 + (b*d*x^2*Log[c*x^n])/2 + (b*e*x^4*Log[c*x^n])/4

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Maple [C]  time = 0.202, size = 265, normalized size = 5.6 \begin{align*}{\frac{b{x}^{2} \left ( e{x}^{2}+2\,d \right ) \ln \left ({x}^{n} \right ) }{4}}+{\frac{i}{8}}\pi \,be{x}^{4}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-{\frac{i}{8}}\pi \,be{x}^{4}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -{\frac{i}{8}}\pi \,be{x}^{4} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+{\frac{i}{8}}\pi \,be{x}^{4} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{\ln \left ( c \right ) be{x}^{4}}{4}}-{\frac{ben{x}^{4}}{16}}+{\frac{ae{x}^{4}}{4}}+{\frac{i}{4}}\pi \,bd{x}^{2}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-{\frac{i}{4}}\pi \,bd{x}^{2}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -{\frac{i}{4}}\pi \,bd{x}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+{\frac{i}{4}}\pi \,bd{x}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{\ln \left ( c \right ) bd{x}^{2}}{2}}-{\frac{bdn{x}^{2}}{4}}+{\frac{ad{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^2*(e*x^2+2*d)*ln(x^n)+1/8*I*Pi*b*e*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2-1/8*I*Pi*b*e*x^4*csgn(I*x^n)*csgn(I
*c*x^n)*csgn(I*c)-1/8*I*Pi*b*e*x^4*csgn(I*c*x^n)^3+1/8*I*Pi*b*e*x^4*csgn(I*c*x^n)^2*csgn(I*c)+1/4*ln(c)*b*e*x^
4-1/16*b*e*n*x^4+1/4*a*e*x^4+1/4*I*Pi*b*d*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-1/4*I*Pi*b*d*x^2*csgn(I*x^n)*csgn(I*
c*x^n)*csgn(I*c)-1/4*I*Pi*b*d*x^2*csgn(I*c*x^n)^3+1/4*I*Pi*b*d*x^2*csgn(I*c*x^n)^2*csgn(I*c)+1/2*ln(c)*b*d*x^2
-1/4*b*d*n*x^2+1/2*a*d*x^2

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Maxima [A]  time = 1.04448, size = 77, normalized size = 1.64 \begin{align*} -\frac{1}{16} \, b e n x^{4} + \frac{1}{4} \, b e x^{4} \log \left (c x^{n}\right ) + \frac{1}{4} \, a e x^{4} - \frac{1}{4} \, b d n x^{2} + \frac{1}{2} \, b d x^{2} \log \left (c x^{n}\right ) + \frac{1}{2} \, a d x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*b*e*n*x^4 + 1/4*b*e*x^4*log(c*x^n) + 1/4*a*e*x^4 - 1/4*b*d*n*x^2 + 1/2*b*d*x^2*log(c*x^n) + 1/2*a*d*x^2

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Fricas [A]  time = 1.25674, size = 171, normalized size = 3.64 \begin{align*} -\frac{1}{16} \,{\left (b e n - 4 \, a e\right )} x^{4} - \frac{1}{4} \,{\left (b d n - 2 \, a d\right )} x^{2} + \frac{1}{4} \,{\left (b e x^{4} + 2 \, b d x^{2}\right )} \log \left (c\right ) + \frac{1}{4} \,{\left (b e n x^{4} + 2 \, b d n x^{2}\right )} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(b*e*n - 4*a*e)*x^4 - 1/4*(b*d*n - 2*a*d)*x^2 + 1/4*(b*e*x^4 + 2*b*d*x^2)*log(c) + 1/4*(b*e*n*x^4 + 2*b*
d*n*x^2)*log(x)

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Sympy [B]  time = 1.85302, size = 87, normalized size = 1.85 \begin{align*} \frac{a d x^{2}}{2} + \frac{a e x^{4}}{4} + \frac{b d n x^{2} \log{\left (x \right )}}{2} - \frac{b d n x^{2}}{4} + \frac{b d x^{2} \log{\left (c \right )}}{2} + \frac{b e n x^{4} \log{\left (x \right )}}{4} - \frac{b e n x^{4}}{16} + \frac{b e x^{4} \log{\left (c \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d*x**2/2 + a*e*x**4/4 + b*d*n*x**2*log(x)/2 - b*d*n*x**2/4 + b*d*x**2*log(c)/2 + b*e*n*x**4*log(x)/4 - b*e*n
*x**4/16 + b*e*x**4*log(c)/4

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Giac [A]  time = 1.27323, size = 99, normalized size = 2.11 \begin{align*} \frac{1}{4} \, b n x^{4} e \log \left (x\right ) - \frac{1}{16} \, b n x^{4} e + \frac{1}{4} \, b x^{4} e \log \left (c\right ) + \frac{1}{4} \, a x^{4} e + \frac{1}{2} \, b d n x^{2} \log \left (x\right ) - \frac{1}{4} \, b d n x^{2} + \frac{1}{2} \, b d x^{2} \log \left (c\right ) + \frac{1}{2} \, a d x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*n*x^4*e*log(x) - 1/16*b*n*x^4*e + 1/4*b*x^4*e*log(c) + 1/4*a*x^4*e + 1/2*b*d*n*x^2*log(x) - 1/4*b*d*n*x^
2 + 1/2*b*d*x^2*log(c) + 1/2*a*d*x^2