3.14.1 \(\int \frac {2 x^3+e^{\frac {2 (3 x^2+(16+4 x) \log (x))}{x^2}} (-32-8 x-x^2+(64+8 x) \log (x))}{x^2} \, dx\)

Optimal. Leaf size=21 \[ -e^{6+\frac {8 (4+x) \log (x)}{x^2}} x+x^2 \]

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Rubi [F]  time = 0.27, 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 {2 x^3+e^{\frac {2 \left (3 x^2+(16+4 x) \log (x)\right )}{x^2}} \left (-32-8 x-x^2+(64+8 x) \log (x)\right )}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*x^3 + E^((2*(3*x^2 + (16 + 4*x)*Log[x]))/x^2)*(-32 - 8*x - x^2 + (64 + 8*x)*Log[x]))/x^2,x]

[Out]

x^2 - 32*E^6*Defer[Int][x^(-2 + 32/x^2 + 8/x), x] + 64*E^6*Log[x]*Defer[Int][x^(-2 + 32/x^2 + 8/x), x] - 8*E^6
*Defer[Int][x^(-1 + 32/x^2 + 8/x), x] + 8*E^6*Log[x]*Defer[Int][x^(-1 + 32/x^2 + 8/x), x] - E^6*Defer[Int][x^(
(32 + 8*x)/x^2), x] - 64*E^6*Defer[Int][Defer[Int][x^(-2 + 32/x^2 + 8/x), x]/x, x] - 8*E^6*Defer[Int][Defer[In
t][x^(-1 + 32/x^2 + 8/x), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 x-e^6 x^{-2+\frac {32}{x^2}+\frac {8}{x}} \left (32+8 x+x^2-64 \log (x)-8 x \log (x)\right )\right ) \, dx\\ &=x^2-e^6 \int x^{-2+\frac {32}{x^2}+\frac {8}{x}} \left (32+8 x+x^2-64 \log (x)-8 x \log (x)\right ) \, dx\\ &=x^2-e^6 \int \left (32 x^{-2+\frac {32}{x^2}+\frac {8}{x}}+8 x^{-1+\frac {32}{x^2}+\frac {8}{x}}+x^{\frac {32}{x^2}+\frac {8}{x}}-64 x^{-2+\frac {32}{x^2}+\frac {8}{x}} \log (x)-8 x^{-1+\frac {32}{x^2}+\frac {8}{x}} \log (x)\right ) \, dx\\ &=x^2-e^6 \int x^{\frac {32}{x^2}+\frac {8}{x}} \, dx-\left (8 e^6\right ) \int x^{-1+\frac {32}{x^2}+\frac {8}{x}} \, dx+\left (8 e^6\right ) \int x^{-1+\frac {32}{x^2}+\frac {8}{x}} \log (x) \, dx-\left (32 e^6\right ) \int x^{-2+\frac {32}{x^2}+\frac {8}{x}} \, dx+\left (64 e^6\right ) \int x^{-2+\frac {32}{x^2}+\frac {8}{x}} \log (x) \, dx\\ &=x^2-e^6 \int x^{\frac {32+8 x}{x^2}} \, dx-\left (8 e^6\right ) \int x^{-1+\frac {32}{x^2}+\frac {8}{x}} \, dx-\left (8 e^6\right ) \int \frac {\int x^{-1+\frac {32}{x^2}+\frac {8}{x}} \, dx}{x} \, dx-\left (32 e^6\right ) \int x^{-2+\frac {32}{x^2}+\frac {8}{x}} \, dx-\left (64 e^6\right ) \int \frac {\int x^{-2+\frac {32}{x^2}+\frac {8}{x}} \, dx}{x} \, dx+\left (8 e^6 \log (x)\right ) \int x^{-1+\frac {32}{x^2}+\frac {8}{x}} \, dx+\left (64 e^6 \log (x)\right ) \int x^{-2+\frac {32}{x^2}+\frac {8}{x}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 21, normalized size = 1.00 \begin {gather*} x^2-e^6 x^{1+\frac {8 (4+x)}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^3 + E^((2*(3*x^2 + (16 + 4*x)*Log[x]))/x^2)*(-32 - 8*x - x^2 + (64 + 8*x)*Log[x]))/x^2,x]

[Out]

x^2 - E^6*x^(1 + (8*(4 + x))/x^2)

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fricas [A]  time = 1.17, size = 26, normalized size = 1.24 \begin {gather*} x^{2} - x e^{\left (\frac {2 \, {\left (3 \, x^{2} + 4 \, {\left (x + 4\right )} \log \relax (x)\right )}}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x+64)*log(x)-x^2-8*x-32)*exp(((4*x+16)*log(x)+3*x^2)/x^2)^2+2*x^3)/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.47, size = 28, normalized size = 1.33 \begin {gather*} x^{2} - x e^{\left (\frac {2 \, {\left (3 \, x^{2} + 4 \, x \log \relax (x) + 16 \, \log \relax (x)\right )}}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x+64)*log(x)-x^2-8*x-32)*exp(((4*x+16)*log(x)+3*x^2)/x^2)^2+2*x^3)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 28, normalized size = 1.33

method	result	size
risch	\(-x \,x^{\frac {32}{x^{2}}} x^{\frac {8}{x}} {\mathrm e}^{6}+x^{2}\)	\(28\)
default	\(-x \,{\mathrm e}^{\frac {2 \left (4 x +16\right ) \ln \relax (x )+6 x^{2}}{x^{2}}}+x^{2}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x+64)*ln(x)-x^2-8*x-32)*exp(((4*x+16)*ln(x)+3*x^2)/x^2)^2+2*x^3)/x^2,x,method=_RETURNVERBOSE)

[Out]

-x*(x^(16/x^2))^2*(x^(4/x))^2*exp(6)+x^2

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maxima [A]  time = 0.46, size = 24, normalized size = 1.14 \begin {gather*} x^{2} - x e^{\left (\frac {8 \, \log \relax (x)}{x} + \frac {32 \, \log \relax (x)}{x^{2}} + 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x+64)*log(x)-x^2-8*x-32)*exp(((4*x+16)*log(x)+3*x^2)/x^2)^2+2*x^3)/x^2,x, algorithm="maxima")

[Out]

x^2 - x*e^(8*log(x)/x + 32*log(x)/x^2 + 6)

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mupad [B]  time = 1.02, size = 22, normalized size = 1.05 \begin {gather*} x^2-x\,x^{\frac {8}{x}+\frac {32}{x^2}}\,{\mathrm {e}}^6 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*(log(x)*(4*x + 16) + 3*x^2))/x^2)*(8*x - log(x)*(8*x + 64) + x^2 + 32) - 2*x^3)/x^2,x)

[Out]

x^2 - x*x^(8/x + 32/x^2)*exp(6)

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sympy [A]  time = 0.37, size = 24, normalized size = 1.14 \begin {gather*} x^{2} - x e^{\frac {2 \left (3 x^{2} + \left (4 x + 16\right ) \log {\relax (x )}\right )}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x+64)*ln(x)-x**2-8*x-32)*exp(((4*x+16)*ln(x)+3*x**2)/x**2)**2+2*x**3)/x**2,x)

[Out]

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

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