3.24.64 \(\int \frac {-2 x^2+e^{2 x \log ^2(2)+2 (-5-x) \log ^2(2) \log (x)} (-10 e^2 \log ^2(2)-2 e^2 x \log ^2(2) \log (x))}{x} \, dx\)

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

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Rubi [F]  time = 0.41, 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^2+e^{2 x \log ^2(2)+2 (-5-x) \log ^2(2) \log (x)} \left (-10 e^2 \log ^2(2)-2 e^2 x \log ^2(2) \log (x)\right )}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x^2 + E^(2*x*Log[2]^2 + 2*(-5 - x)*Log[2]^2*Log[x])*(-10*E^2*Log[2]^2 - 2*E^2*x*Log[2]^2*Log[x]))/x,x]

[Out]

-x^2 - 10*Log[2]^2*Defer[Int][E^(2 + 2*x*Log[2]^2)*x^(-1 - 10*Log[2]^2 - 2*x*Log[2]^2), x] - 2*Log[2]^2*Log[x]
*Defer[Int][E^(2 + 2*x*Log[2]^2)*x^(-10*Log[2]^2 - 2*x*Log[2]^2), x] + 2*Log[2]^2*Defer[Int][Defer[Int][E^(2 +
2*x*Log[2]^2)*x^(-10*Log[2]^2 - 2*x*Log[2]^2), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-2 x+2 e^{2+2 x \log ^2(2)} x^{-1-10 \log ^2(2)-2 x \log ^2(2)} \log ^2(2) (-5-x \log (x))\right ) \, dx\\ &=-x^2+\left (2 \log ^2(2)\right ) \int e^{2+2 x \log ^2(2)} x^{-1-10 \log ^2(2)-2 x \log ^2(2)} (-5-x \log (x)) \, dx\\ &=-x^2+\left (2 \log ^2(2)\right ) \int \left (-5 e^{2+2 x \log ^2(2)} x^{-1-10 \log ^2(2)-2 x \log ^2(2)}-e^{2+2 x \log ^2(2)} x^{-10 \log ^2(2)-2 x \log ^2(2)} \log (x)\right ) \, dx\\ &=-x^2-\left (2 \log ^2(2)\right ) \int e^{2+2 x \log ^2(2)} x^{-10 \log ^2(2)-2 x \log ^2(2)} \log (x) \, dx-\left (10 \log ^2(2)\right ) \int e^{2+2 x \log ^2(2)} x^{-1-10 \log ^2(2)-2 x \log ^2(2)} \, dx\\ &=-x^2+\left (2 \log ^2(2)\right ) \int \frac {\int e^{2+2 x \log ^2(2)} x^{-10 \log ^2(2)-2 x \log ^2(2)} \, dx}{x} \, dx-\left (10 \log ^2(2)\right ) \int e^{2+2 x \log ^2(2)} x^{-1-10 \log ^2(2)-2 x \log ^2(2)} \, dx-\left (2 \log ^2(2) \log (x)\right ) \int e^{2+2 x \log ^2(2)} x^{-10 \log ^2(2)-2 x \log ^2(2)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 29, normalized size = 1.12 \begin {gather*} -x^2+e^{2+2 x \log ^2(2)} x^{-2 (5+x) \log ^2(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^2 + E^(2*x*Log[2]^2 + 2*(-5 - x)*Log[2]^2*Log[x])*(-10*E^2*Log[2]^2 - 2*E^2*x*Log[2]^2*Log[x])
)/x,x]

[Out]

-x^2 + E^(2 + 2*x*Log[2]^2)/x^(2*(5 + x)*Log[2]^2)

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fricas [A]  time = 0.50, size = 27, normalized size = 1.04 \begin {gather*} -x^{2} + e^{\left (-2 \, {\left (x + 5\right )} \log \relax (2)^{2} \log \relax (x) + 2 \, x \log \relax (2)^{2} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)^2*log(2)^2*log(x)-10*exp(1)^2*log(2)^2)*exp((-x-5)*log(2)^2*log(x)+x*log(2)^2)^2-2*x^2
)/x,x, algorithm="fricas")

[Out]

-x^2 + e^(-2*(x + 5)*log(2)^2*log(x) + 2*x*log(2)^2 + 2)

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giac [A]  time = 0.31, size = 33, normalized size = 1.27 \begin {gather*} -x^{2} + e^{\left (-2 \, x \log \relax (2)^{2} \log \relax (x) + 2 \, x \log \relax (2)^{2} - 10 \, \log \relax (2)^{2} \log \relax (x) + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)^2*log(2)^2*log(x)-10*exp(1)^2*log(2)^2)*exp((-x-5)*log(2)^2*log(x)+x*log(2)^2)^2-2*x^2
)/x,x, algorithm="giac")

[Out]

-x^2 + e^(-2*x*log(2)^2*log(x) + 2*x*log(2)^2 - 10*log(2)^2*log(x) + 2)

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maple [A]  time = 0.24, size = 31, normalized size = 1.19

method	result	size
risch	\(x^{-2 \left (5+x \right ) \ln \relax (2)^{2}} {\mathrm e}^{2+2 x \ln \relax (2)^{2}}-x^{2}\)	\(31\)
default	\({\mathrm e}^{2} {\mathrm e}^{\left (-2 x -10\right ) \ln \relax (2)^{2} \ln \relax (x )+2 x \ln \relax (2)^{2}}-x^{2}\)	\(34\)
norman	\({\mathrm e}^{2} {\mathrm e}^{\left (-2 x -10\right ) \ln \relax (2)^{2} \ln \relax (x )+2 x \ln \relax (2)^{2}}-x^{2}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(1)^2*ln(2)^2*ln(x)-10*exp(1)^2*ln(2)^2)*exp((-x-5)*ln(2)^2*ln(x)+x*ln(2)^2)^2-2*x^2)/x,x,method
=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.72, size = 33, normalized size = 1.27 \begin {gather*} -x^{2} + e^{\left (-2 \, x \log \relax (2)^{2} \log \relax (x) + 2 \, x \log \relax (2)^{2} - 10 \, \log \relax (2)^{2} \log \relax (x) + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)^2*log(2)^2*log(x)-10*exp(1)^2*log(2)^2)*exp((-x-5)*log(2)^2*log(x)+x*log(2)^2)^2-2*x^2
)/x,x, algorithm="maxima")

[Out]

-x^2 + e^(-2*x*log(2)^2*log(x) + 2*x*log(2)^2 - 10*log(2)^2*log(x) + 2)

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mupad [B]  time = 1.49, size = 38, normalized size = 1.46 \begin {gather*} \frac {{\mathrm {e}}^{2\,x\,{\ln \relax (2)}^2}\,{\mathrm {e}}^2}{x^{10\,{\ln \relax (2)}^2}\,x^{2\,x\,{\ln \relax (2)}^2}}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x*log(2)^2 - 2*log(2)^2*log(x)*(x + 5))*(10*exp(2)*log(2)^2 + 2*x*exp(2)*log(2)^2*log(x)) + 2*x^2)
/x,x)

[Out]

(exp(2*x*log(2)^2)*exp(2))/(x^(10*log(2)^2)*x^(2*x*log(2)^2)) - x^2

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sympy [A]  time = 0.41, size = 31, normalized size = 1.19 \begin {gather*} - x^{2} + e^{2} e^{2 x \log {\relax (2 )}^{2} + 2 \left (- x - 5\right ) \log {\relax (2 )}^{2} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)**2*ln(2)**2*ln(x)-10*exp(1)**2*ln(2)**2)*exp((-x-5)*ln(2)**2*ln(x)+x*ln(2)**2)**2-2*x*
*2)/x,x)

[Out]

-x**2 + exp(2)*exp(2*x*log(2)**2 + 2*(-x - 5)*log(2)**2*log(x))

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