3.101.48 \(\int ((45-15 x^2) \log (2)-15 x^2 \log (2) \log (x^3)) \, dx\)

Optimal. Leaf size=18 \[ \frac {5}{2} x \log (2) \left (18-2 x^2 \log \left (x^3\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {2304} \begin {gather*} 45 x \log (2)-5 x^3 \log (2) \log \left (x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(45 - 15*x^2)*Log[2] - 15*x^2*Log[2]*Log[x^3],x]

[Out]

45*x*Log[2] - 5*x^3*Log[2]*Log[x^3]

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} &=\log (2) \int \left (45-15 x^2\right ) \, dx-(15 \log (2)) \int x^2 \log \left (x^3\right ) \, dx\\ &=45 x \log (2)-5 x^3 \log (2) \log \left (x^3\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 0.94 \begin {gather*} 45 x \log (2)-5 x^3 \log (2) \log \left (x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(45 - 15*x^2)*Log[2] - 15*x^2*Log[2]*Log[x^3],x]

[Out]

45*x*Log[2] - 5*x^3*Log[2]*Log[x^3]

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fricas [A]  time = 0.84, size = 17, normalized size = 0.94 \begin {gather*} -5 \, x^{3} \log \relax (2) \log \left (x^{3}\right ) + 45 \, x \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15*x^2*log(2)*log(x^3)+(-15*x^2+45)*log(2),x, algorithm="fricas")

[Out]

-5*x^3*log(2)*log(x^3) + 45*x*log(2)

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giac [A]  time = 0.15, size = 30, normalized size = 1.67 \begin {gather*} -5 \, {\left (x^{3} \log \left (x^{3}\right ) - x^{3}\right )} \log \relax (2) - 5 \, {\left (x^{3} - 9 \, x\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15*x^2*log(2)*log(x^3)+(-15*x^2+45)*log(2),x, algorithm="giac")

[Out]

-5*(x^3*log(x^3) - x^3)*log(2) - 5*(x^3 - 9*x)*log(2)

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maple [A]  time = 0.02, size = 18, normalized size = 1.00

method	result	size
default	\(45 x \ln \relax (2)-5 \ln \relax (2) x^{3} \ln \left (x^{3}\right )\)	\(18\)
norman	\(45 x \ln \relax (2)-5 \ln \relax (2) x^{3} \ln \left (x^{3}\right )\)	\(18\)
risch	\(45 x \ln \relax (2)-5 \ln \relax (2) x^{3} \ln \left (x^{3}\right )\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-15*x^2*ln(2)*ln(x^3)+(-15*x^2+45)*ln(2),x,method=_RETURNVERBOSE)

[Out]

45*x*ln(2)-5*ln(2)*x^3*ln(x^3)

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maxima [A]  time = 0.36, size = 30, normalized size = 1.67 \begin {gather*} -5 \, {\left (x^{3} \log \left (x^{3}\right ) - x^{3}\right )} \log \relax (2) - 5 \, {\left (x^{3} - 9 \, x\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15*x^2*log(2)*log(x^3)+(-15*x^2+45)*log(2),x, algorithm="maxima")

[Out]

-5*(x^3*log(x^3) - x^3)*log(2) - 5*(x^3 - 9*x)*log(2)

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mupad [B]  time = 7.62, size = 15, normalized size = 0.83 \begin {gather*} -5\,x\,\ln \relax (2)\,\left (x^2\,\ln \left (x^3\right )-9\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(- log(2)*(15*x^2 - 45) - 15*x^2*log(x^3)*log(2),x)

[Out]

-5*x*log(2)*(x^2*log(x^3) - 9)

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sympy [A]  time = 0.11, size = 19, normalized size = 1.06 \begin {gather*} - 5 x^{3} \log {\relax (2 )} \log {\left (x^{3} \right )} + 45 x \log {\relax (2 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15*x**2*ln(2)*ln(x**3)+(-15*x**2+45)*ln(2),x)

[Out]

-5*x**3*log(2)*log(x**3) + 45*x*log(2)

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