3.9.12 \(\int \frac {16-24 x^3+(4-6 x^3) \log (\frac {3}{4} e^{-x^3} x^2)}{x} \, dx\)

Optimal. Leaf size=19 \[ \left (4+\log \left (\frac {3}{4} e^{-x^3} x^2\right )\right )^2 \]

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Rubi [A]  time = 0.06, antiderivative size = 27, normalized size of antiderivative = 1.42, number of steps used = 5, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14, 6686} \begin {gather*} -8 x^3+\log ^2\left (\frac {3}{4} e^{-x^3} x^2\right )+16 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 - 24*x^3 + (4 - 6*x^3)*Log[(3*x^2)/(4*E^x^3)])/x,x]

[Out]

-8*x^3 + 16*Log[x] + Log[(3*x^2)/(4*E^x^3)]^2

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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {8 \left (-2+3 x^3\right )}{x}-\frac {2 \left (-2+3 x^3\right ) \log \left (\frac {3}{4} e^{-x^3} x^2\right )}{x}\right ) \, dx\\ &=-\left (2 \int \frac {\left (-2+3 x^3\right ) \log \left (\frac {3}{4} e^{-x^3} x^2\right )}{x} \, dx\right )-8 \int \frac {-2+3 x^3}{x} \, dx\\ &=\log ^2\left (\frac {3}{4} e^{-x^3} x^2\right )-8 \int \left (-\frac {2}{x}+3 x^2\right ) \, dx\\ &=-8 x^3+16 \log (x)+\log ^2\left (\frac {3}{4} e^{-x^3} x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 70, normalized size = 3.68 \begin {gather*} -8 x^3-x^6+16 \log (x)+\log ^2\left (x^2\right )-2 x^3 \log \left (\frac {3}{4} e^{-x^3} x^2\right )+4 \log (x) \left (x^3-\log \left (x^2\right )+\log \left (\frac {3}{4} e^{-x^3} x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 - 24*x^3 + (4 - 6*x^3)*Log[(3*x^2)/(4*E^x^3)])/x,x]

[Out]

-8*x^3 - x^6 + 16*Log[x] + Log[x^2]^2 - 2*x^3*Log[(3*x^2)/(4*E^x^3)] + 4*Log[x]*(x^3 - Log[x^2] + Log[(3*x^2)/
(4*E^x^3)])

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fricas [A]  time = 1.01, size = 29, normalized size = 1.53 \begin {gather*} \log \left (\frac {3}{4} \, x^{2} e^{\left (-x^{3}\right )}\right )^{2} + 8 \, \log \left (\frac {3}{4} \, x^{2} e^{\left (-x^{3}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^3+4)*log(3/4*x^2/exp(x^3))-24*x^3+16)/x,x, algorithm="fricas")

[Out]

log(3/4*x^2*e^(-x^3))^2 + 8*log(3/4*x^2*e^(-x^3))

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giac [A]  time = 0.52, size = 32, normalized size = 1.68 \begin {gather*} x^{6} - 2 \, x^{3} \log \left (\frac {3}{4} \, x^{2}\right ) - 8 \, x^{3} + \log \left (\frac {3}{4} \, x^{2}\right )^{2} + 16 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^3+4)*log(3/4*x^2/exp(x^3))-24*x^3+16)/x,x, algorithm="giac")

[Out]

x^6 - 2*x^3*log(3/4*x^2) - 8*x^3 + log(3/4*x^2)^2 + 16*log(x)

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maple [A]  time = 0.16, size = 25, normalized size = 1.32

method	result	size
norman	\(\ln \left (\frac {3 x^{2} {\mathrm e}^{-x^{3}}}{4}\right )^{2}-8 x^{3}+16 \ln \relax (x )\)	\(25\)
default	\(-8 x^{3}+16 \ln \relax (x )-2 x^{3} \ln \left (\frac {3 x^{2} {\mathrm e}^{-x^{3}}}{4}\right )+4 \ln \relax (x ) \ln \left (\frac {3 x^{2} {\mathrm e}^{-x^{3}}}{4}\right )-x^{6}+4 x^{3} \ln \relax (x )-4 \ln \relax (x )^{2}\)	\(62\)
risch	\(16 \ln \relax (x )+4 \ln \relax (x )^{2}-x^{6}-8 x^{3}-8 \ln \relax (2) \ln \relax (x )-2 x^{3} \ln \relax (3)+4 \ln \relax (3) \ln \relax (x )+4 x^{3} \ln \relax (2)+i \pi \,x^{3} \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \pi \,x^{3} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i \pi \,x^{3} \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x^{3}}\right )^{2}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x^{3}}\right )^{3}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+4 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\left (2 x^{3}-4 \ln \relax (x )\right ) \ln \left ({\mathrm e}^{x^{3}}\right )-i \pi \,x^{3} \mathrm {csgn}\left (i {\mathrm e}^{-x^{3}}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x^{3}}\right )^{2}+i \pi \,x^{3} \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x^{3}}\right )^{3}+i \pi \,x^{3} \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x^{3}}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x^{3}}\right )+2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x^{3}}\right )^{2}-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x^{3}}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x^{3}}\right )-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x^{2}\right )^{3}+i \pi \,x^{3} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i {\mathrm e}^{-x^{3}}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{-x^{3}}\right )^{2}\)	\(413\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6*x^3+4)*ln(3/4*x^2/exp(x^3))-24*x^3+16)/x,x,method=_RETURNVERBOSE)

[Out]

ln(3/4*x^2/exp(x^3))^2-8*x^3+16*ln(x)

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maxima [B]  time = 0.79, size = 66, normalized size = 3.47 \begin {gather*} -x^{6} - 2 \, x^{3} \log \left (\frac {3}{4} \, x^{2} e^{\left (-x^{3}\right )}\right ) - 8 \, x^{3} + 4 \, {\left (x^{3} - 2 \, \log \relax (x)\right )} \log \relax (x) + 4 \, \log \left (\frac {3}{4} \, x^{2} e^{\left (-x^{3}\right )}\right ) \log \relax (x) + 4 \, \log \relax (x)^{2} + 16 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^3+4)*log(3/4*x^2/exp(x^3))-24*x^3+16)/x,x, algorithm="maxima")

[Out]

-x^6 - 2*x^3*log(3/4*x^2*e^(-x^3)) - 8*x^3 + 4*(x^3 - 2*log(x))*log(x) + 4*log(3/4*x^2*e^(-x^3))*log(x) + 4*lo
g(x)^2 + 16*log(x)

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mupad [B]  time = 0.67, size = 26, normalized size = 1.37 \begin {gather*} \left (\ln \left (\frac {3\,x^2}{4}\right )-x^3\right )\,\left (\ln \left (\frac {3\,x^2}{4}\right )-x^3+8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((3*x^2*exp(-x^3))/4)*(6*x^3 - 4) + 24*x^3 - 16)/x,x)

[Out]

(log((3*x^2)/4) - x^3)*(log((3*x^2)/4) - x^3 + 8)

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sympy [A]  time = 0.29, size = 24, normalized size = 1.26 \begin {gather*} - 8 x^{3} + 16 \log {\relax (x )} + \log {\left (\frac {3 x^{2} e^{- x^{3}}}{4} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x**3+4)*ln(3/4*x**2/exp(x**3))-24*x**3+16)/x,x)

[Out]

-8*x**3 + 16*log(x) + log(3*x**2*exp(-x**3)/4)**2

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