3.9.56 \(\int \frac {e^{\frac {-x^2+x \log (x^2)+\log (\frac {16-\log (3+x)}{e})}{x}} (95 x-16 x^2-16 x^3+(-6 x+x^2+x^3) \log (3+x)+(-48-16 x+(3+x) \log (3+x)) \log (\frac {16-\log (3+x)}{e}))}{-48 x^2-16 x^3+(3 x^2+x^3) \log (3+x)} \, dx\)

Optimal. Leaf size=28 \[ -e^{-x+\frac {\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} x^2 \]

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Rubi [F]  time = 18.43, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{\frac {-x^2+x \log \left (x^2\right )+\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} \left (95 x-16 x^2-16 x^3+\left (-6 x+x^2+x^3\right ) \log (3+x)+(-48-16 x+(3+x) \log (3+x)) \log \left (\frac {16-\log (3+x)}{e}\right )\right )}{-48 x^2-16 x^3+\left (3 x^2+x^3\right ) \log (3+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-x^2 + x*Log[x^2] + Log[(16 - Log[3 + x])/E])/x)*(95*x - 16*x^2 - 16*x^3 + (-6*x + x^2 + x^3)*Log[3 +
x] + (-48 - 16*x + (3 + x)*Log[3 + x])*Log[(16 - Log[3 + x])/E]))/(-48*x^2 - 16*x^3 + (3*x^2 + x^3)*Log[3 + x
]),x]

[Out]

-15*Defer[Int][E^(-x^(-1) - x)*(16 - Log[3 + x])^(-1 + x^(-1)), x] - 32*Defer[Int][E^(-x^(-1) - x)*x*(16 - Log
[3 + x])^(-1 + x^(-1)), x] + 16*Defer[Int][E^(-x^(-1) - x)*x^2*(16 - Log[3 + x])^(-1 + x^(-1)), x] - 3*Defer[I
nt][(E^(-x^(-1) - x)*(16 - Log[3 + x])^(-1 + x^(-1)))/(3 + x), x] + Defer[Int][E^(-x^(-1) - x)*(16 - Log[3 + x
])^(-1 + x^(-1))*Log[3 + x], x] + 2*Defer[Int][E^(-x^(-1) - x)*x*(16 - Log[3 + x])^(-1 + x^(-1))*Log[3 + x], x
] - Defer[Int][E^(-x^(-1) - x)*x^2*(16 - Log[3 + x])^(-1 + x^(-1))*Log[3 + x], x] + 16*Defer[Int][E^(-x^(-1) -
x)*(16 - Log[3 + x])^(-1 + x^(-1))*Log[16 - Log[3 + x]], x] - Defer[Int][E^(-x^(-1) - x)*(16 - Log[3 + x])^(-
1 + x^(-1))*Log[3 + x]*Log[16 - Log[3 + x]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \left (-48-111 x+16 x^2+16 x^3+16 (3+x) \log (16-\log (3+x))-(3+x) \log (3+x) \left (-1-2 x+x^2+\log (16-\log (3+x))\right )\right )}{3+x} \, dx\\ &=\int \left (\frac {e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \left (-48-111 x+16 x^2+16 x^3+3 \log (3+x)+7 x \log (3+x)-x^2 \log (3+x)-x^3 \log (3+x)\right )}{3+x}-e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} (-16+\log (3+x)) \log (16-\log (3+x))\right ) \, dx\\ &=\int \frac {e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \left (-48-111 x+16 x^2+16 x^3+3 \log (3+x)+7 x \log (3+x)-x^2 \log (3+x)-x^3 \log (3+x)\right )}{3+x} \, dx-\int e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} (-16+\log (3+x)) \log (16-\log (3+x)) \, dx\\ &=\int \frac {e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \left (-48-111 x+16 x^2+16 x^3-\left (-3-7 x+x^2+x^3\right ) \log (3+x)\right )}{3+x} \, dx-\int \left (-16 e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (16-\log (3+x))+e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x) \log (16-\log (3+x))\right ) \, dx\\ &=16 \int e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (16-\log (3+x)) \, dx+\int \left (\frac {e^{-\frac {1}{x}-x} \left (-48-111 x+16 x^2+16 x^3\right ) (16-\log (3+x))^{-1+\frac {1}{x}}}{3+x}-e^{-\frac {1}{x}-x} \left (-1-2 x+x^2\right ) (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x)\right ) \, dx-\int e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x) \log (16-\log (3+x)) \, dx\\ &=16 \int e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (16-\log (3+x)) \, dx+\int \frac {e^{-\frac {1}{x}-x} \left (-48-111 x+16 x^2+16 x^3\right ) (16-\log (3+x))^{-1+\frac {1}{x}}}{3+x} \, dx-\int e^{-\frac {1}{x}-x} \left (-1-2 x+x^2\right ) (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x) \, dx-\int e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x) \log (16-\log (3+x)) \, dx\\ &=16 \int e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (16-\log (3+x)) \, dx+\int \left (-15 e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}}-32 e^{-\frac {1}{x}-x} x (16-\log (3+x))^{-1+\frac {1}{x}}+16 e^{-\frac {1}{x}-x} x^2 (16-\log (3+x))^{-1+\frac {1}{x}}-\frac {3 e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}}}{3+x}\right ) \, dx-\int \left (-e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x)-2 e^{-\frac {1}{x}-x} x (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x)+e^{-\frac {1}{x}-x} x^2 (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x)\right ) \, dx-\int e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x) \log (16-\log (3+x)) \, dx\\ &=2 \int e^{-\frac {1}{x}-x} x (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x) \, dx-3 \int \frac {e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}}}{3+x} \, dx-15 \int e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \, dx+16 \int e^{-\frac {1}{x}-x} x^2 (16-\log (3+x))^{-1+\frac {1}{x}} \, dx+16 \int e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (16-\log (3+x)) \, dx-32 \int e^{-\frac {1}{x}-x} x (16-\log (3+x))^{-1+\frac {1}{x}} \, dx+\int e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x) \, dx-\int e^{-\frac {1}{x}-x} x^2 (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x) \, dx-\int e^{-\frac {1}{x}-x} (16-\log (3+x))^{-1+\frac {1}{x}} \log (3+x) \log (16-\log (3+x)) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 28, normalized size = 1.00 \begin {gather*} -e^{-\frac {1}{x}-x} x^2 (16-\log (3+x))^{\frac {1}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-x^2 + x*Log[x^2] + Log[(16 - Log[3 + x])/E])/x)*(95*x - 16*x^2 - 16*x^3 + (-6*x + x^2 + x^3)*L
og[3 + x] + (-48 - 16*x + (3 + x)*Log[3 + x])*Log[(16 - Log[3 + x])/E]))/(-48*x^2 - 16*x^3 + (3*x^2 + x^3)*Log
[3 + x]),x]

[Out]

-(E^(-x^(-1) - x)*x^2*(16 - Log[3 + x])^x^(-1))

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fricas [A]  time = 0.74, size = 32, normalized size = 1.14 \begin {gather*} -e^{\left (-\frac {x^{2} - x \log \left (x^{2}\right ) - \log \left (-{\left (\log \left (x + 3\right ) - 16\right )} e^{\left (-1\right )}\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3+x)*log(3+x)-16*x-48)*log((-log(3+x)+16)/exp(1))+(x^3+x^2-6*x)*log(3+x)-16*x^3-16*x^2+95*x)*exp(
(log((-log(3+x)+16)/exp(1))+x*log(x^2)-x^2)/x)/((x^3+3*x^2)*log(3+x)-16*x^3-48*x^2),x, algorithm="fricas")

[Out]

-e^(-(x^2 - x*log(x^2) - log(-(log(x + 3) - 16)*e^(-1)))/x)

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giac [A]  time = 1.32, size = 29, normalized size = 1.04 \begin {gather*} -e^{\left (-x + \frac {\log \left (-e^{\left (-1\right )} \log \left (x + 3\right ) + 16 \, e^{\left (-1\right )}\right )}{x} + \log \left (x^{2}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3+x)*log(3+x)-16*x-48)*log((-log(3+x)+16)/exp(1))+(x^3+x^2-6*x)*log(3+x)-16*x^3-16*x^2+95*x)*exp(
(log((-log(3+x)+16)/exp(1))+x*log(x^2)-x^2)/x)/((x^3+3*x^2)*log(3+x)-16*x^3-48*x^2),x, algorithm="giac")

[Out]

-e^(-x + log(-e^(-1)*log(x + 3) + 16*e^(-1))/x + log(x^2))

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maple [C]  time = 0.11, size = 75, normalized size = 2.68

method	result	size
risch	\(-x^{2} \left (\left (-\ln \left (3+x \right )+16\right ) {\mathrm e}^{-1}\right )^{\frac {1}{x}} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}}{2}-x}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3+x)*ln(3+x)-16*x-48)*ln((-ln(3+x)+16)/exp(1))+(x^3+x^2-6*x)*ln(3+x)-16*x^3-16*x^2+95*x)*exp((ln((-ln(3
+x)+16)/exp(1))+x*ln(x^2)-x^2)/x)/((x^3+3*x^2)*ln(3+x)-16*x^3-48*x^2),x,method=_RETURNVERBOSE)

[Out]

-x^2*((-ln(3+x)+16)*exp(-1))^(1/x)*exp(-1/2*I*Pi*csgn(I*x^2)^3+I*Pi*csgn(I*x^2)^2*csgn(I*x)-1/2*I*Pi*csgn(I*x^
2)*csgn(I*x)^2-x)

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maxima [A]  time = 0.60, size = 28, normalized size = 1.00 \begin {gather*} -x^{2} e^{\left (-x + \frac {\log \left (-\log \left (x + 3\right ) + 16\right )}{x} - \frac {1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3+x)*log(3+x)-16*x-48)*log((-log(3+x)+16)/exp(1))+(x^3+x^2-6*x)*log(3+x)-16*x^3-16*x^2+95*x)*exp(
(log((-log(3+x)+16)/exp(1))+x*log(x^2)-x^2)/x)/((x^3+3*x^2)*log(3+x)-16*x^3-48*x^2),x, algorithm="maxima")

[Out]

-x^2*e^(-x + log(-log(x + 3) + 16)/x - 1/x)

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mupad [B]  time = 0.92, size = 26, normalized size = 0.93 \begin {gather*} -x^2\,{\mathrm {e}}^{-x}\,{\left (16\,{\mathrm {e}}^{-1}-\ln \left (x+3\right )\,{\mathrm {e}}^{-1}\right )}^{1/x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((log(-exp(-1)*(log(x + 3) - 16)) + x*log(x^2) - x^2)/x)*(16*x^2 - log(x + 3)*(x^2 - 6*x + x^3) - 95*x
+ 16*x^3 + log(-exp(-1)*(log(x + 3) - 16))*(16*x - log(x + 3)*(x + 3) + 48)))/(48*x^2 - log(x + 3)*(3*x^2 + x
^3) + 16*x^3),x)

[Out]

-x^2*exp(-x)*(16*exp(-1) - log(x + 3)*exp(-1))^(1/x)

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sympy [A]  time = 2.60, size = 26, normalized size = 0.93 \begin {gather*} - e^{\frac {- x^{2} + x \log {\left (x^{2} \right )} + \log {\left (\frac {16 - \log {\left (x + 3 \right )}}{e} \right )}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3+x)*ln(3+x)-16*x-48)*ln((-ln(3+x)+16)/exp(1))+(x**3+x**2-6*x)*ln(3+x)-16*x**3-16*x**2+95*x)*exp(
(ln((-ln(3+x)+16)/exp(1))+x*ln(x**2)-x**2)/x)/((x**3+3*x**2)*ln(3+x)-16*x**3-48*x**2),x)

[Out]

-exp((-x**2 + x*log(x**2) + log((16 - log(x + 3))*exp(-1)))/x)

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