3.58.11 \(\int \frac {e^{x+\frac {5 e^x x^2}{6 e-6 x^2}} (-5 x^4+e (10 x+5 x^2))}{6 e^2-12 e x^2+6 x^4} \, dx\)

Optimal. Leaf size=21 \[ e^{\frac {5 e^x x}{6 \left (\frac {e}{x}-x\right )}} \]

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Rubi [F]  time = 2.51, 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 {e^{x+\frac {5 e^x x^2}{6 e-6 x^2}} \left (-5 x^4+e \left (10 x+5 x^2\right )\right )}{6 e^2-12 e x^2+6 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(x + (5*E^x*x^2)/(6*E - 6*x^2))*(-5*x^4 + E*(10*x + 5*x^2)))/(6*E^2 - 12*E*x^2 + 6*x^4),x]

[Out]

(-5*Defer[Int][E^(x + (5*E^x*x^2)/(6*E - 6*x^2)), x])/6 + (5*Defer[Int][E^(1/2 + x + (5*E^x*x^2)/(6*E - 6*x^2)
)/(Sqrt[E] - x), x])/12 + (5*Defer[Int][E^(1/2 + x + (5*E^x*x^2)/(6*E - 6*x^2))/(Sqrt[E] + x), x])/12 + (5*Def
er[Int][(E^(1 + x + (5*E^x*x^2)/(6*E - 6*x^2))*x)/(E - x^2)^2, x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=6 \int \frac {e^{x+\frac {5 e^x x^2}{6 e-6 x^2}} \left (-5 x^4+e \left (10 x+5 x^2\right )\right )}{\left (-6 e+6 x^2\right )^2} \, dx\\ &=6 \int \frac {e^{x+\frac {5 e^x x^2}{6 e-6 x^2}} x \left (10 e+5 e x-5 x^3\right )}{\left (6 e-6 x^2\right )^2} \, dx\\ &=6 \int \left (-\frac {5}{36} e^{x+\frac {5 e^x x^2}{6 e-6 x^2}}+\frac {5 e^{1+x+\frac {5 e^x x^2}{6 e-6 x^2}} x}{18 \left (e-x^2\right )^2}+\frac {5 e^{1+x+\frac {5 e^x x^2}{6 e-6 x^2}}}{36 \left (e-x^2\right )}\right ) \, dx\\ &=-\left (\frac {5}{6} \int e^{x+\frac {5 e^x x^2}{6 e-6 x^2}} \, dx\right )+\frac {5}{6} \int \frac {e^{1+x+\frac {5 e^x x^2}{6 e-6 x^2}}}{e-x^2} \, dx+\frac {5}{3} \int \frac {e^{1+x+\frac {5 e^x x^2}{6 e-6 x^2}} x}{\left (e-x^2\right )^2} \, dx\\ &=-\left (\frac {5}{6} \int e^{x+\frac {5 e^x x^2}{6 e-6 x^2}} \, dx\right )+\frac {5}{6} \int \left (\frac {e^{\frac {1}{2}+x+\frac {5 e^x x^2}{6 e-6 x^2}}}{2 \left (\sqrt {e}-x\right )}+\frac {e^{\frac {1}{2}+x+\frac {5 e^x x^2}{6 e-6 x^2}}}{2 \left (\sqrt {e}+x\right )}\right ) \, dx+\frac {5}{3} \int \frac {e^{1+x+\frac {5 e^x x^2}{6 e-6 x^2}} x}{\left (e-x^2\right )^2} \, dx\\ &=\frac {5}{12} \int \frac {e^{\frac {1}{2}+x+\frac {5 e^x x^2}{6 e-6 x^2}}}{\sqrt {e}-x} \, dx+\frac {5}{12} \int \frac {e^{\frac {1}{2}+x+\frac {5 e^x x^2}{6 e-6 x^2}}}{\sqrt {e}+x} \, dx-\frac {5}{6} \int e^{x+\frac {5 e^x x^2}{6 e-6 x^2}} \, dx+\frac {5}{3} \int \frac {e^{1+x+\frac {5 e^x x^2}{6 e-6 x^2}} x}{\left (e-x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.47, size = 21, normalized size = 1.00 \begin {gather*} e^{-\frac {5 e^x x^2}{6 \left (-e+x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x + (5*E^x*x^2)/(6*E - 6*x^2))*(-5*x^4 + E*(10*x + 5*x^2)))/(6*E^2 - 12*E*x^2 + 6*x^4),x]

[Out]

E^((-5*E^x*x^2)/(6*(-E + x^2)))

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fricas [B]  time = 1.21, size = 35, normalized size = 1.67 \begin {gather*} e^{\left (-x + \frac {6 \, x^{3} - 5 \, x^{2} e^{x} - 6 \, x e}{6 \, {\left (x^{2} - e\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+10*x)*exp(1)-5*x^4)*exp(x)*exp(5*x^2*exp(x)/(6*exp(1)-6*x^2))/(6*exp(1)^2-12*x^2*exp(1)+6*x^
4),x, algorithm="fricas")

[Out]

e^(-x + 1/6*(6*x^3 - 5*x^2*e^x - 6*x*e)/(x^2 - e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {5 \, {\left (x^{4} - {\left (x^{2} + 2 \, x\right )} e\right )} e^{\left (-\frac {5 \, x^{2} e^{x}}{6 \, {\left (x^{2} - e\right )}} + x\right )}}{6 \, {\left (x^{4} - 2 \, x^{2} e + e^{2}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+10*x)*exp(1)-5*x^4)*exp(x)*exp(5*x^2*exp(x)/(6*exp(1)-6*x^2))/(6*exp(1)^2-12*x^2*exp(1)+6*x^
4),x, algorithm="giac")

[Out]

integrate(-5/6*(x^4 - (x^2 + 2*x)*e)*e^(-5/6*x^2*e^x/(x^2 - e) + x)/(x^4 - 2*x^2*e + e^2), x)

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maple [A]  time = 0.20, size = 19, normalized size = 0.90

method	result	size
risch	\({\mathrm e}^{\frac {5 \,{\mathrm e}^{x} x^{2}}{6 \left ({\mathrm e}-x^{2}\right )}}\)	\(19\)
norman	\(\frac {{\mathrm e} \,{\mathrm e}^{\frac {5 x^{2} {\mathrm e}^{x}}{6 \,{\mathrm e}-6 x^{2}}}-x^{2} {\mathrm e}^{\frac {5 x^{2} {\mathrm e}^{x}}{6 \,{\mathrm e}-6 x^{2}}}}{{\mathrm e}-x^{2}}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^2+10*x)*exp(1)-5*x^4)*exp(x)*exp(5*x^2*exp(x)/(6*exp(1)-6*x^2))/(6*exp(1)^2-12*x^2*exp(1)+6*x^4),x,m
ethod=_RETURNVERBOSE)

[Out]

exp(5/6*exp(x)*x^2/(exp(1)-x^2))

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maxima [A]  time = 0.50, size = 22, normalized size = 1.05 \begin {gather*} e^{\left (-\frac {5 \, e^{\left (x + 1\right )}}{6 \, {\left (x^{2} - e\right )}} - \frac {5}{6} \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+10*x)*exp(1)-5*x^4)*exp(x)*exp(5*x^2*exp(x)/(6*exp(1)-6*x^2))/(6*exp(1)^2-12*x^2*exp(1)+6*x^
4),x, algorithm="maxima")

[Out]

e^(-5/6*e^(x + 1)/(x^2 - e) - 5/6*e^x)

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mupad [B]  time = 4.06, size = 20, normalized size = 0.95 \begin {gather*} {\mathrm {e}}^{\frac {5\,x^2\,{\mathrm {e}}^x}{6\,\mathrm {e}-6\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((5*x^2*exp(x))/(6*exp(1) - 6*x^2))*exp(x)*(exp(1)*(10*x + 5*x^2) - 5*x^4))/(6*exp(2) - 12*x^2*exp(1)
+ 6*x^4),x)

[Out]

exp((5*x^2*exp(x))/(6*exp(1) - 6*x^2))

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sympy [A]  time = 0.28, size = 19, normalized size = 0.90 \begin {gather*} e^{\frac {5 x^{2} e^{x}}{- 6 x^{2} + 6 e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**2+10*x)*exp(1)-5*x**4)*exp(x)*exp(5*x**2*exp(x)/(6*exp(1)-6*x**2))/(6*exp(1)**2-12*x**2*exp(1
)+6*x**4),x)

[Out]

exp(5*x**2*exp(x)/(-6*x**2 + 6*E))

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