3.3.30 \(\int x \log (x+x^3) \, dx\) [230]

Optimal. Leaf size=31 \[ -\frac {3 x^2}{4}+\frac {1}{2} \log \left (1+x^2\right )+\frac {1}{2} x^2 \log \left (x+x^3\right ) \]

[Out]

-3/4*x^2+1/2*ln(x^2+1)+1/2*x^2*ln(x^3+x)

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Rubi [A]
time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2605, 455, 45} \begin {gather*} -\frac {3 x^2}{4}+\frac {1}{2} \log \left (x^2+1\right )+\frac {1}{2} x^2 \log \left (x^3+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Log[x + x^3],x]

[Out]

(-3*x^2)/4 + Log[1 + x^2]/2 + (x^2*Log[x + x^3])/2

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 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 2605

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[(d + e*x)^(m +
1)*((a + b*Log[c*RFx^p])^n/(e*(m + 1))), x] - Dist[b*n*(p/(e*(m + 1))), Int[SimplifyIntegrand[(d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rubi steps

\begin {align*} \int x \log \left (x+x^3\right ) \, dx &=\frac {1}{2} x^2 \log \left (x+x^3\right )-\frac {1}{2} \int \frac {x \left (1+3 x^2\right )}{1+x^2} \, dx\\ &=\frac {1}{2} x^2 \log \left (x+x^3\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1+3 x}{1+x} \, dx,x,x^2\right )\\ &=\frac {1}{2} x^2 \log \left (x+x^3\right )-\frac {1}{4} \text {Subst}\left (\int \left (3-\frac {2}{1+x}\right ) \, dx,x,x^2\right )\\ &=-\frac {3 x^2}{4}+\frac {1}{2} \log \left (1+x^2\right )+\frac {1}{2} x^2 \log \left (x+x^3\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 31, normalized size = 1.00 \begin {gather*} -\frac {3 x^2}{4}+\frac {1}{2} \log \left (1+x^2\right )+\frac {1}{2} x^2 \log \left (x+x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Log[x + x^3],x]

[Out]

(-3*x^2)/4 + Log[1 + x^2]/2 + (x^2*Log[x + x^3])/2

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Maple [A]
time = 0.02, size = 26, normalized size = 0.84

method	result	size
default	\(-\frac {3 x^{2}}{4}+\frac {\ln \left (x^{2}+1\right )}{2}+\frac {x^{2} \ln \left (x^{3}+x \right )}{2}\)	\(26\)
risch	\(-\frac {3 x^{2}}{4}+\frac {\ln \left (x^{2}+1\right )}{2}+\frac {x^{2} \ln \left (x^{3}+x \right )}{2}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*ln(x^3+x),x,method=_RETURNVERBOSE)

[Out]

-3/4*x^2+1/2*ln(x^2+1)+1/2*x^2*ln(x^3+x)

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Maxima [A]
time = 0.48, size = 25, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, x^{2} \log \left (x^{3} + x\right ) - \frac {3}{4} \, x^{2} + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(x^3+x),x, algorithm="maxima")

[Out]

1/2*x^2*log(x^3 + x) - 3/4*x^2 + 1/2*log(x^2 + 1)

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Fricas [A]
time = 0.40, size = 25, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, x^{2} \log \left (x^{3} + x\right ) - \frac {3}{4} \, x^{2} + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(x^3+x),x, algorithm="fricas")

[Out]

1/2*x^2*log(x^3 + x) - 3/4*x^2 + 1/2*log(x^2 + 1)

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Sympy [A]
time = 0.11, size = 26, normalized size = 0.84 \begin {gather*} \frac {x^{2} \log {\left (x^{3} + x \right )}}{2} - \frac {3 x^{2}}{4} + \frac {\log {\left (x^{2} + 1 \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*ln(x**3+x),x)

[Out]

x**2*log(x**3 + x)/2 - 3*x**2/4 + log(x**2 + 1)/2

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Giac [A]
time = 6.01, size = 25, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, x^{2} \log \left (x^{3} + x\right ) - \frac {3}{4} \, x^{2} + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(x^3+x),x, algorithm="giac")

[Out]

1/2*x^2*log(x^3 + x) - 3/4*x^2 + 1/2*log(x^2 + 1)

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Mupad [B]
time = 0.41, size = 25, normalized size = 0.81 \begin {gather*} \frac {\ln \left (x^2+1\right )}{2}+\frac {x^2\,\ln \left (x^3+x\right )}{2}-\frac {3\,x^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(x + x^3),x)

[Out]

log(x^2 + 1)/2 + (x^2*log(x + x^3))/2 - (3*x^2)/4

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