3.44.52 \(\int \frac {24+e^{2 x^2} (6+24 x^2-32 \log (4))-64 \log (4)-4 e^{4 x^2} \log (4)}{16 x^2+8 e^{2 x^2} x^2+e^{4 x^2} x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {x-4 \left (\frac {3}{2 \left (4+e^{2 x^2}\right )}-\log (4)\right )}{x} \]

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Rubi [F]  time = 0.89, 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 {24+e^{2 x^2} \left (6+24 x^2-32 \log (4)\right )-64 \log (4)-4 e^{4 x^2} \log (4)}{16 x^2+8 e^{2 x^2} x^2+e^{4 x^2} x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(24 + E^(2*x^2)*(6 + 24*x^2 - 32*Log[4]) - 64*Log[4] - 4*E^(4*x^2)*Log[4])/(16*x^2 + 8*E^(2*x^2)*x^2 + E^(
4*x^2)*x^2),x]

[Out]

(4*Log[4])/x - 96*Defer[Int][(4 + E^(2*x^2))^(-2), x] + 24*Defer[Int][(4 + E^(2*x^2))^(-1), x] + 6*Defer[Int][
1/((4 + E^(2*x^2))*x^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x^2} \left (6+24 x^2-32 \log (4)\right )+24 \left (1-\frac {8 \log (4)}{3}\right )-4 e^{4 x^2} \log (4)}{\left (4+e^{2 x^2}\right )^2 x^2} \, dx\\ &=\int \left (-\frac {96}{\left (4+e^{2 x^2}\right )^2}+\frac {6 \left (1+4 x^2\right )}{\left (4+e^{2 x^2}\right ) x^2}-\frac {4 \log (4)}{x^2}\right ) \, dx\\ &=\frac {4 \log (4)}{x}+6 \int \frac {1+4 x^2}{\left (4+e^{2 x^2}\right ) x^2} \, dx-96 \int \frac {1}{\left (4+e^{2 x^2}\right )^2} \, dx\\ &=\frac {4 \log (4)}{x}+6 \int \left (\frac {4}{4+e^{2 x^2}}+\frac {1}{\left (4+e^{2 x^2}\right ) x^2}\right ) \, dx-96 \int \frac {1}{\left (4+e^{2 x^2}\right )^2} \, dx\\ &=\frac {4 \log (4)}{x}+6 \int \frac {1}{\left (4+e^{2 x^2}\right ) x^2} \, dx+24 \int \frac {1}{4+e^{2 x^2}} \, dx-96 \int \frac {1}{\left (4+e^{2 x^2}\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 21, normalized size = 0.75 \begin {gather*} \frac {2 \left (-\frac {3}{4+e^{2 x^2}}+\log (16)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24 + E^(2*x^2)*(6 + 24*x^2 - 32*Log[4]) - 64*Log[4] - 4*E^(4*x^2)*Log[4])/(16*x^2 + 8*E^(2*x^2)*x^2
+ E^(4*x^2)*x^2),x]

[Out]

(2*(-3/(4 + E^(2*x^2)) + Log[16]))/x

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fricas [A]  time = 0.53, size = 32, normalized size = 1.14 \begin {gather*} \frac {2 \, {\left (4 \, e^{\left (2 \, x^{2}\right )} \log \relax (2) + 16 \, \log \relax (2) - 3\right )}}{x e^{\left (2 \, x^{2}\right )} + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*log(2)*exp(x^2)^4+(-64*log(2)+24*x^2+6)*exp(x^2)^2-128*log(2)+24)/(x^2*exp(x^2)^4+8*x^2*exp(x^2)
^2+16*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 32, normalized size = 1.14 \begin {gather*} \frac {2 \, {\left (4 \, e^{\left (2 \, x^{2}\right )} \log \relax (2) + 16 \, \log \relax (2) - 3\right )}}{x e^{\left (2 \, x^{2}\right )} + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*log(2)*exp(x^2)^4+(-64*log(2)+24*x^2+6)*exp(x^2)^2-128*log(2)+24)/(x^2*exp(x^2)^4+8*x^2*exp(x^2)
^2+16*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 24, normalized size = 0.86

method	result	size
risch	\(\frac {8 \ln \relax (2)}{x}-\frac {6}{x \left ({\mathrm e}^{2 x^{2}}+4\right )}\)	\(24\)
norman	\(\frac {-6+8 \ln \relax (2) {\mathrm e}^{2 x^{2}}+32 \ln \relax (2)}{x \left ({\mathrm e}^{2 x^{2}}+4\right )}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*ln(2)*exp(x^2)^4+(-64*ln(2)+24*x^2+6)*exp(x^2)^2-128*ln(2)+24)/(x^2*exp(x^2)^4+8*x^2*exp(x^2)^2+16*x^2
),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.49, size = 32, normalized size = 1.14 \begin {gather*} \frac {2 \, {\left (4 \, e^{\left (2 \, x^{2}\right )} \log \relax (2) + 16 \, \log \relax (2) - 3\right )}}{x e^{\left (2 \, x^{2}\right )} + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*log(2)*exp(x^2)^4+(-64*log(2)+24*x^2+6)*exp(x^2)^2-128*log(2)+24)/(x^2*exp(x^2)^4+8*x^2*exp(x^2)
^2+16*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.14, size = 23, normalized size = 0.82 \begin {gather*} \frac {8\,\ln \relax (2)}{x}-\frac {6}{x\,\left ({\mathrm {e}}^{2\,x^2}+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(128*log(2) - exp(2*x^2)*(24*x^2 - 64*log(2) + 6) + 8*exp(4*x^2)*log(2) - 24)/(8*x^2*exp(2*x^2) + x^2*exp
(4*x^2) + 16*x^2),x)

[Out]

(8*log(2))/x - 6/(x*(exp(2*x^2) + 4))

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sympy [A]  time = 0.13, size = 19, normalized size = 0.68 \begin {gather*} - \frac {6}{x e^{2 x^{2}} + 4 x} + \frac {8 \log {\relax (2 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*ln(2)*exp(x**2)**4+(-64*ln(2)+24*x**2+6)*exp(x**2)**2-128*ln(2)+24)/(x**2*exp(x**2)**4+8*x**2*ex
p(x**2)**2+16*x**2),x)

[Out]

-6/(x*exp(2*x**2) + 4*x) + 8*log(2)/x

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