3.99.77 \(\int (-6+e^{5 x-x^3} (-20+12 x^2)-\log (x)) \, dx\)

Optimal. Leaf size=26 \[ -4 e^{\left (\frac {5}{x}-x\right ) x^2}-5 x-x \log (x) \]

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Rubi [A]  time = 0.04, antiderivative size = 22, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6706, 2295} \begin {gather*} -4 e^{5 x-x^3}-5 x+x (-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-6 + E^(5*x - x^3)*(-20 + 12*x^2) - Log[x],x]

[Out]

-4*E^(5*x - x^3) - 5*x - x*Log[x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-6 x+\int e^{5 x-x^3} \left (-20+12 x^2\right ) \, dx-\int \log (x) \, dx\\ &=-4 e^{5 x-x^3}-5 x-x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 22, normalized size = 0.85 \begin {gather*} -4 e^{5 x-x^3}-5 x-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-6 + E^(5*x - x^3)*(-20 + 12*x^2) - Log[x],x]

[Out]

-4*E^(5*x - x^3) - 5*x - x*Log[x]

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fricas [A]  time = 0.71, size = 21, normalized size = 0.81 \begin {gather*} -x \log \relax (x) - 5 \, x - 4 \, e^{\left (-x^{3} + 5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)+(12*x^2-20)*exp(-x^3+5*x)-6,x, algorithm="fricas")

[Out]

-x*log(x) - 5*x - 4*e^(-x^3 + 5*x)

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giac [A]  time = 0.27, size = 21, normalized size = 0.81 \begin {gather*} -x \log \relax (x) - 5 \, x - 4 \, e^{\left (-x^{3} + 5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)+(12*x^2-20)*exp(-x^3+5*x)-6,x, algorithm="giac")

[Out]

-x*log(x) - 5*x - 4*e^(-x^3 + 5*x)

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maple [A]  time = 0.04, size = 21, normalized size = 0.81

method	result	size
risch	\(-5 x -4 \,{\mathrm e}^{-x \left (x^{2}-5\right )}-x \ln \relax (x )\)	\(21\)
default	\(-5 x -4 \,{\mathrm e}^{-x^{3}+5 x}-x \ln \relax (x )\)	\(22\)
norman	\(-5 x -4 \,{\mathrm e}^{-x^{3}+5 x}-x \ln \relax (x )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-ln(x)+(12*x^2-20)*exp(-x^3+5*x)-6,x,method=_RETURNVERBOSE)

[Out]

-5*x-4*exp(-x*(x^2-5))-x*ln(x)

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maxima [A]  time = 0.35, size = 21, normalized size = 0.81 \begin {gather*} -x \log \relax (x) - 5 \, x - 4 \, e^{\left (-x^{3} + 5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)+(12*x^2-20)*exp(-x^3+5*x)-6,x, algorithm="maxima")

[Out]

-x*log(x) - 5*x - 4*e^(-x^3 + 5*x)

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mupad [B]  time = 5.55, size = 21, normalized size = 0.81 \begin {gather*} -5\,x-4\,{\mathrm {e}}^{5\,x-x^3}-x\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(5*x - x^3)*(12*x^2 - 20) - log(x) - 6,x)

[Out]

- 5*x - 4*exp(5*x - x^3) - x*log(x)

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sympy [A]  time = 0.29, size = 19, normalized size = 0.73 \begin {gather*} - x \log {\relax (x )} - 5 x - 4 e^{- x^{3} + 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-ln(x)+(12*x**2-20)*exp(-x**3+5*x)-6,x)

[Out]

-x*log(x) - 5*x - 4*exp(-x**3 + 5*x)

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