3.103.10 \(\int \frac {e^{-8 x-e^{.\frac {1}{4}/x} x+x \log (x^2)} (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log (x^2))}{4 x} \, dx\)

Optimal. Leaf size=23 \[ e^{3+x \left (-8-e^{\left .\frac {1}{4}\right /x}+\log \left (x^2\right )\right )} \]

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Rubi [A]  time = 0.56, antiderivative size = 25, normalized size of antiderivative = 1.09, number of steps used = 2, number of rules used = 2, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {12, 6706} \begin {gather*} e^{-e^{\left .\frac {1}{4}\right /x} x-8 x+3} \left (x^2\right )^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-8*x - E^(1/(4*x))*x + x*Log[x^2])*(E^(3 + 1/(4*x))*(1 - 4*x) - 24*E^3*x + 4*E^3*x*Log[x^2]))/(4*x),x]

[Out]

E^(3 - 8*x - E^(1/(4*x))*x)*(x^2)^x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\frac {1}{4} \int \frac {e^{-8 x-e^{\left .\frac {1}{4}\right /x} x+x \log \left (x^2\right )} \left (e^{3+\frac {1}{4 x}} (1-4 x)-24 e^3 x+4 e^3 x \log \left (x^2\right )\right )}{x} \, dx\\ &=e^{3-8 x-e^{\left .\frac {1}{4}\right /x} x} \left (x^2\right )^x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.55, size = 25, normalized size = 1.09 \begin {gather*} e^{3-8 x-e^{\left .\frac {1}{4}\right /x} x} \left (x^2\right )^x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-8*x - E^(1/(4*x))*x + x*Log[x^2])*(E^(3 + 1/(4*x))*(1 - 4*x) - 24*E^3*x + 4*E^3*x*Log[x^2]))/(4
*x),x]

[Out]

E^(3 - 8*x - E^(1/(4*x))*x)*(x^2)^x

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fricas [A]  time = 0.92, size = 34, normalized size = 1.48 \begin {gather*} e^{\left ({\left (x e^{3} \log \left (x^{2}\right ) - 8 \, x e^{3} - x e^{\left (\frac {12 \, x + 1}{4 \, x}\right )}\right )} e^{\left (-3\right )} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*x*exp(3)*log(x^2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3))*exp(x*log(x^2)-x*exp(1/4/x)-8*x)/x,
x, algorithm="fricas")

[Out]

e^((x*e^3*log(x^2) - 8*x*e^3 - x*e^(1/4*(12*x + 1)/x))*e^(-3) + 3)

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giac [A]  time = 0.31, size = 21, normalized size = 0.91 \begin {gather*} e^{\left (-x e^{\left (\frac {1}{4 \, x}\right )} + x \log \left (x^{2}\right ) - 8 \, x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*x*exp(3)*log(x^2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3))*exp(x*log(x^2)-x*exp(1/4/x)-8*x)/x,
x, algorithm="giac")

[Out]

e^(-x*e^(1/4/x) + x*log(x^2) - 8*x + 3)

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maple [A]  time = 0.34, size = 24, normalized size = 1.04

method	result	size
default	\({\mathrm e}^{3} {\mathrm e}^{x \ln \left (x^{2}\right )-x \,{\mathrm e}^{\frac {1}{4 x}}-8 x}\)	\(24\)
norman	\({\mathrm e}^{3} {\mathrm e}^{x \ln \left (x^{2}\right )-x \,{\mathrm e}^{\frac {1}{4 x}}-8 x}\)	\(24\)
risch	\({\mathrm e}^{3-\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3} x}{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right ) x -\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2} x}{2}+2 x \ln \relax (x )-x \,{\mathrm e}^{\frac {1}{4 x}}-8 x}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(4*x*exp(3)*ln(x^2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3))*exp(x*ln(x^2)-x*exp(1/4/x)-8*x)/x,x,method
=_RETURNVERBOSE)

[Out]

exp(3)*exp(x*ln(x^2)-x*exp(1/4/x)-8*x)

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maxima [A]  time = 0.40, size = 20, normalized size = 0.87 \begin {gather*} e^{\left (-x e^{\left (\frac {1}{4 \, x}\right )} + 2 \, x \log \relax (x) - 8 \, x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*x*exp(3)*log(x^2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3))*exp(x*log(x^2)-x*exp(1/4/x)-8*x)/x,
x, algorithm="maxima")

[Out]

e^(-x*e^(1/4/x) + 2*x*log(x) - 8*x + 3)

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mupad [B]  time = 6.89, size = 22, normalized size = 0.96 \begin {gather*} {\mathrm {e}}^{-8\,x}\,{\mathrm {e}}^3\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{\frac {1}{4\,x}}}\,{\left (x^2\right )}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x*log(x^2) - 8*x - x*exp(1/(4*x)))*(24*x*exp(3) - 4*x*log(x^2)*exp(3) + exp(3)*exp(1/(4*x))*(4*x - 1
)))/(4*x),x)

[Out]

exp(-8*x)*exp(3)*exp(-x*exp(1/(4*x)))*(x^2)^x

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sympy [A]  time = 0.57, size = 22, normalized size = 0.96 \begin {gather*} e^{3} e^{- x e^{\frac {1}{4 x}} + x \log {\left (x^{2} \right )} - 8 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4*x*exp(3)*ln(x**2)+(-4*x+1)*exp(3)*exp(1/4/x)-24*x*exp(3))*exp(x*ln(x**2)-x*exp(1/4/x)-8*x)/x,
x)

[Out]

exp(3)*exp(-x*exp(1/(4*x)) + x*log(x**2) - 8*x)

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