3.25.28 \(\int \frac {1}{8} (40-6 x^2-9 x^2 \log (x^2)) \, dx\)

Optimal. Leaf size=15 \[ x \left (5-\frac {3}{8} x^2 \log \left (x^2\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 2304} \begin {gather*} 5 x-\frac {3}{8} x^3 \log \left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40 - 6*x^2 - 9*x^2*Log[x^2])/8,x]

[Out]

5*x - (3*x^3*Log[x^2])/8

Rule 12

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

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \left (40-6 x^2-9 x^2 \log \left (x^2\right )\right ) \, dx\\ &=5 x-\frac {x^3}{4}-\frac {9}{8} \int x^2 \log \left (x^2\right ) \, dx\\ &=5 x-\frac {3}{8} x^3 \log \left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} 5 x-\frac {3}{8} x^3 \log \left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40 - 6*x^2 - 9*x^2*Log[x^2])/8,x]

[Out]

5*x - (3*x^3*Log[x^2])/8

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fricas [A]  time = 0.56, size = 13, normalized size = 0.87 \begin {gather*} -\frac {3}{8} \, x^{3} \log \left (x^{2}\right ) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9/8*x^2*log(x^2)-3/4*x^2+5,x, algorithm="fricas")

[Out]

-3/8*x^3*log(x^2) + 5*x

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giac [A]  time = 0.23, size = 13, normalized size = 0.87 \begin {gather*} -\frac {3}{8} \, x^{3} \log \left (x^{2}\right ) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9/8*x^2*log(x^2)-3/4*x^2+5,x, algorithm="giac")

[Out]

-3/8*x^3*log(x^2) + 5*x

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maple [A]  time = 0.02, size = 14, normalized size = 0.93




method result size



default \(5 x -\frac {3 x^{3} \ln \left (x^{2}\right )}{8}\) \(14\)
norman \(5 x -\frac {3 x^{3} \ln \left (x^{2}\right )}{8}\) \(14\)
risch \(5 x -\frac {3 x^{3} \ln \left (x^{2}\right )}{8}\) \(14\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-9/8*x^2*ln(x^2)-3/4*x^2+5,x,method=_RETURNVERBOSE)

[Out]

5*x-3/8*x^3*ln(x^2)

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maxima [A]  time = 0.49, size = 13, normalized size = 0.87 \begin {gather*} -\frac {3}{8} \, x^{3} \log \left (x^{2}\right ) + 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9/8*x^2*log(x^2)-3/4*x^2+5,x, algorithm="maxima")

[Out]

-3/8*x^3*log(x^2) + 5*x

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mupad [B]  time = 1.39, size = 13, normalized size = 0.87 \begin {gather*} 5\,x-\frac {3\,x^3\,\ln \left (x^2\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5 - (3*x^2)/4 - (9*x^2*log(x^2))/8,x)

[Out]

5*x - (3*x^3*log(x^2))/8

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sympy [A]  time = 0.10, size = 14, normalized size = 0.93 \begin {gather*} - \frac {3 x^{3} \log {\left (x^{2} \right )}}{8} + 5 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9/8*x**2*ln(x**2)-3/4*x**2+5,x)

[Out]

-3*x**3*log(x**2)/8 + 5*x

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