3.40.57 \(\int \frac {15+e^x (-10 x-10 x^2)}{-3+e^x x^2} \, dx\)

Optimal. Leaf size=32 \[ 5 \left (4-x-\log \left (4 \sqrt {e}\right )+\log \left (\frac {4}{3-e^x x^2}\right )\right ) \]

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Rubi [F]  time = 0.31, 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 {15+e^x \left (-10 x-10 x^2\right )}{-3+e^x x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(15 + E^x*(-10*x - 10*x^2))/(-3 + E^x*x^2),x]

[Out]

-10*x - 10*Log[x] - 15*Defer[Int][(-3 + E^x*x^2)^(-1), x] - 30*Defer[Int][1/(x*(-3 + E^x*x^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {10 (1+x)}{x}-\frac {15 (2+x)}{x \left (-3+e^x x^2\right )}\right ) \, dx\\ &=-\left (10 \int \frac {1+x}{x} \, dx\right )-15 \int \frac {2+x}{x \left (-3+e^x x^2\right )} \, dx\\ &=-\left (10 \int \left (1+\frac {1}{x}\right ) \, dx\right )-15 \int \left (\frac {1}{-3+e^x x^2}+\frac {2}{x \left (-3+e^x x^2\right )}\right ) \, dx\\ &=-10 x-10 \log (x)-15 \int \frac {1}{-3+e^x x^2} \, dx-30 \int \frac {1}{x \left (-3+e^x x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 17, normalized size = 0.53 \begin {gather*} -5 x-5 \log \left (3-e^x x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 + E^x*(-10*x - 10*x^2))/(-3 + E^x*x^2),x]

[Out]

-5*x - 5*Log[3 - E^x*x^2]

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fricas [A]  time = 0.78, size = 23, normalized size = 0.72 \begin {gather*} -5 \, x - 10 \, \log \relax (x) - 5 \, \log \left (\frac {x^{2} e^{x} - 3}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2-10*x)*exp(x)+15)/(exp(x)*x^2-3),x, algorithm="fricas")

[Out]

-5*x - 10*log(x) - 5*log((x^2*e^x - 3)/x^2)

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giac [A]  time = 0.23, size = 15, normalized size = 0.47 \begin {gather*} -5 \, x - 5 \, \log \left (x^{2} e^{x} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2-10*x)*exp(x)+15)/(exp(x)*x^2-3),x, algorithm="giac")

[Out]

-5*x - 5*log(x^2*e^x - 3)

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maple [A]  time = 0.02, size = 16, normalized size = 0.50

method	result	size
norman	\(-5 x -5 \ln \left ({\mathrm e}^{x} x^{2}-3\right )\)	\(16\)
risch	\(-5 x -10 \ln \relax (x )-5 \ln \left ({\mathrm e}^{x}-\frac {3}{x^{2}}\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-10*x^2-10*x)*exp(x)+15)/(exp(x)*x^2-3),x,method=_RETURNVERBOSE)

[Out]

-5*x-5*ln(exp(x)*x^2-3)

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maxima [A]  time = 0.67, size = 23, normalized size = 0.72 \begin {gather*} -5 \, x - 10 \, \log \relax (x) - 5 \, \log \left (\frac {x^{2} e^{x} - 3}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x^2-10*x)*exp(x)+15)/(exp(x)*x^2-3),x, algorithm="maxima")

[Out]

-5*x - 10*log(x) - 5*log((x^2*e^x - 3)/x^2)

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mupad [B]  time = 0.08, size = 15, normalized size = 0.47 \begin {gather*} -5\,x-5\,\ln \left (x^2\,{\mathrm {e}}^x-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(10*x + 10*x^2) - 15)/(x^2*exp(x) - 3),x)

[Out]

- 5*x - 5*log(x^2*exp(x) - 3)

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sympy [A]  time = 0.16, size = 20, normalized size = 0.62 \begin {gather*} - 5 x - 10 \log {\relax (x )} - 5 \log {\left (e^{x} - \frac {3}{x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x**2-10*x)*exp(x)+15)/(exp(x)*x**2-3),x)

[Out]

-5*x - 10*log(x) - 5*log(exp(x) - 3/x**2)

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