3.73.7 \(\int \frac {25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} (-4-25 x^3)}{25 e^{\frac {2-50 x^2-25 x^3}{25 x^2}} x^3+25 x^4+225 x^3 \log (2)} \, dx\)

Optimal. Leaf size=24 \[ \log \left (9+\frac {e^{-2+\frac {2}{25 x^2}-x}+x}{\log (2)}\right ) \]

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Rubi [F]  time = 4.34, 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 {25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} \left (-4-25 x^3\right )}{25 e^{\frac {2-50 x^2-25 x^3}{25 x^2}} x^3+25 x^4+225 x^3 \log (2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25*x^3 + E^((2 - 50*x^2 - 25*x^3)/(25*x^2))*(-4 - 25*x^3))/(25*E^((2 - 50*x^2 - 25*x^3)/(25*x^2))*x^3 + 2
5*x^4 + 225*x^3*Log[2]),x]

[Out]

2/(25*x^2) - x + (1 + Log[512])*Defer[Int][E^(2 + x)/(E^(2/(25*x^2)) + E^(2 + x)*x + E^(2 + x)*Log[512]), x] +
(4*Log[512]*Defer[Int][E^(2 + x)/(x^3*(E^(2/(25*x^2)) + E^(2 + x)*x + E^(2 + x)*Log[512])), x])/25 + (4*Defer
[Int][E^(2 + x)/(x^2*(E^(2/(25*x^2)) + E^(2 + x)*x + E^(2 + x)*Log[512])), x])/25 + Defer[Int][(E^(2 + x)*x)/(
E^(2/(25*x^2)) + E^(2 + x)*x + E^(2 + x)*Log[512]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2+x} \left (25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} \left (-4-25 x^3\right )\right )}{25 x^3 \left (e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)\right )} \, dx\\ &=\frac {1}{25} \int \frac {e^{2+x} \left (25 x^3+e^{\frac {2-50 x^2-25 x^3}{25 x^2}} \left (-4-25 x^3\right )\right )}{x^3 \left (e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)\right )} \, dx\\ &=\frac {1}{25} \int \left (-\frac {4+25 x^3}{x^3}+\frac {e^{2+x} \left (4 x+25 x^4+4 \log (512)+25 x^3 (1+\log (512))\right )}{x^3 \left (e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)\right )}\right ) \, dx\\ &=-\left (\frac {1}{25} \int \frac {4+25 x^3}{x^3} \, dx\right )+\frac {1}{25} \int \frac {e^{2+x} \left (4 x+25 x^4+4 \log (512)+25 x^3 (1+\log (512))\right )}{x^3 \left (e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)\right )} \, dx\\ &=-\left (\frac {1}{25} \int \left (25+\frac {4}{x^3}\right ) \, dx\right )+\frac {1}{25} \int \left (\frac {4 e^{2+x}}{x^2 \left (e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)\right )}+\frac {25 e^{2+x} x}{e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)}+\frac {4 e^{2+x} \log (512)}{x^3 \left (e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)\right )}+\frac {25 e^{2+x} (1+\log (512))}{e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)}\right ) \, dx\\ &=\frac {2}{25 x^2}-x+\frac {4}{25} \int \frac {e^{2+x}}{x^2 \left (e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)\right )} \, dx+\frac {1}{25} (4 \log (512)) \int \frac {e^{2+x}}{x^3 \left (e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)\right )} \, dx+(1+\log (512)) \int \frac {e^{2+x}}{e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)} \, dx+\int \frac {e^{2+x} x}{e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.45, size = 36, normalized size = 1.50 \begin {gather*} \frac {1}{25} \left (-25 x+25 \log \left (e^{\frac {2}{25 x^2}}+e^{2+x} x+e^{2+x} \log (512)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*x^3 + E^((2 - 50*x^2 - 25*x^3)/(25*x^2))*(-4 - 25*x^3))/(25*E^((2 - 50*x^2 - 25*x^3)/(25*x^2))*x
^3 + 25*x^4 + 225*x^3*Log[2]),x]

[Out]

(-25*x + 25*Log[E^(2/(25*x^2)) + E^(2 + x)*x + E^(2 + x)*Log[512]])/25

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fricas [A]  time = 0.60, size = 25, normalized size = 1.04 \begin {gather*} \log \left (x + e^{\left (-\frac {25 \, x^{3} + 50 \, x^{2} - 2}{25 \, x^{2}}\right )} + 9 \, \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^3-4)*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+25*x^3)/(25*x^3*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+225*x
^3*log(2)+25*x^4),x, algorithm="fricas")

[Out]

log(x + e^(-1/25*(25*x^3 + 50*x^2 - 2)/x^2) + 9*log(2))

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giac [A]  time = 0.23, size = 25, normalized size = 1.04 \begin {gather*} \log \left (x + e^{\left (-\frac {25 \, x^{3} + 50 \, x^{2} - 2}{25 \, x^{2}}\right )} + 9 \, \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^3-4)*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+25*x^3)/(25*x^3*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+225*x
^3*log(2)+25*x^4),x, algorithm="giac")

[Out]

log(x + e^(-1/25*(25*x^3 + 50*x^2 - 2)/x^2) + 9*log(2))

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maple [A]  time = 0.11, size = 26, normalized size = 1.08

method	result	size
norman	\(\ln \left (9 \ln \relax (2)+x +{\mathrm e}^{\frac {-25 x^{3}-50 x^{2}+2}{25 x^{2}}}\right )\)	\(26\)
risch	\(-x +\frac {2}{25 x^{2}}-\frac {-25 x^{3}-50 x^{2}+2}{25 x^{2}}+\ln \left (9 \ln \relax (2)+{\mathrm e}^{-\frac {25 x^{3}+50 x^{2}-2}{25 x^{2}}}+x \right )\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-25*x^3-4)*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+25*x^3)/(25*x^3*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+225*x^3*ln(
2)+25*x^4),x,method=_RETURNVERBOSE)

[Out]

ln(9*ln(2)+x+exp(1/25*(-25*x^3-50*x^2+2)/x^2))

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maxima [A]  time = 0.50, size = 26, normalized size = 1.08 \begin {gather*} -x + \log \left ({\left (x e^{2} + 9 \, e^{2} \log \relax (2)\right )} e^{x} + e^{\left (\frac {2}{25 \, x^{2}}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^3-4)*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+25*x^3)/(25*x^3*exp(1/25*(-25*x^3-50*x^2+2)/x^2)+225*x
^3*log(2)+25*x^4),x, algorithm="maxima")

[Out]

-x + log((x*e^2 + 9*e^2*log(2))*e^x + e^(2/25/x^2))

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mupad [B]  time = 4.45, size = 16, normalized size = 0.67 \begin {gather*} \ln \left (x+\ln \left (512\right )+{\mathrm {e}}^{\frac {2}{25\,x^2}-x-2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x^3 - exp(-(2*x^2 + x^3 - 2/25)/x^2)*(25*x^3 + 4))/(25*x^3*exp(-(2*x^2 + x^3 - 2/25)/x^2) + 225*x^3*lo
g(2) + 25*x^4),x)

[Out]

log(x + log(512) + exp(2/(25*x^2) - x - 2))

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sympy [A]  time = 0.27, size = 24, normalized size = 1.00 \begin {gather*} \log {\left (x + e^{\frac {- x^{3} - 2 x^{2} + \frac {2}{25}}{x^{2}}} + 9 \log {\relax (2 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x**3-4)*exp(1/25*(-25*x**3-50*x**2+2)/x**2)+25*x**3)/(25*x**3*exp(1/25*(-25*x**3-50*x**2+2)/x*
*2)+225*x**3*ln(2)+25*x**4),x)

[Out]

log(x + exp((-x**3 - 2*x**2 + 2/25)/x**2) + 9*log(2))

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