3.37.39 \(\int \frac {e^{\frac {-150+10 e^{2 e^5 x^2}+2 x}{-15+e^{2 e^5 x^2}}} (-30+e^{2 e^5 x^2} (2-8 e^5 x^2))}{225-30 e^{2 e^5 x^2}+e^{4 e^5 x^2}} \, dx\)

Optimal. Leaf size=24 \[ 2+e^{2 \left (5+\frac {x}{-15+e^{2 e^5 x^2}}\right )} \]

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Rubi [F]  time = 8.89, 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 {\exp \left (\frac {-150+10 e^{2 e^5 x^2}+2 x}{-15+e^{2 e^5 x^2}}\right ) \left (-30+e^{2 e^5 x^2} \left (2-8 e^5 x^2\right )\right )}{225-30 e^{2 e^5 x^2}+e^{4 e^5 x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-150 + 10*E^(2*E^5*x^2) + 2*x)/(-15 + E^(2*E^5*x^2)))*(-30 + E^(2*E^5*x^2)*(2 - 8*E^5*x^2)))/(225 - 3
0*E^(2*E^5*x^2) + E^(4*E^5*x^2)),x]

[Out]

2*Defer[Int][E^((2*(-75 + 5*E^(2*E^5*x^2) + x))/(-15 + E^(2*E^5*x^2)))/(-15 + E^(2*E^5*x^2)), x] - 120*Defer[I
nt][(E^((-225 + 15*E^(2*E^5*x^2) + 2*x)/(-15 + E^(2*E^5*x^2)))*x^2)/(15 - E^(2*E^5*x^2))^2, x] - 8*Defer[Int][
(E^((-225 + 15*E^(2*E^5*x^2) + 2*x)/(-15 + E^(2*E^5*x^2)))*x^2)/(-15 + E^(2*E^5*x^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {2 \left (-75+5 e^{2 e^5 x^2}+x\right )}{-15+e^{2 e^5 x^2}}\right ) \left (-30+e^{2 e^5 x^2} \left (2-8 e^5 x^2\right )\right )}{\left (15-e^{2 e^5 x^2}\right )^2} \, dx\\ &=\int \left (-\frac {120 \exp \left (5+\frac {2 \left (-75+5 e^{2 e^5 x^2}+x\right )}{-15+e^{2 e^5 x^2}}\right ) x^2}{\left (-15+e^{2 e^5 x^2}\right )^2}-\frac {2 \exp \left (\frac {2 \left (-75+5 e^{2 e^5 x^2}+x\right )}{-15+e^{2 e^5 x^2}}\right ) \left (-1+4 e^5 x^2\right )}{-15+e^{2 e^5 x^2}}\right ) \, dx\\ &=-\left (2 \int \frac {\exp \left (\frac {2 \left (-75+5 e^{2 e^5 x^2}+x\right )}{-15+e^{2 e^5 x^2}}\right ) \left (-1+4 e^5 x^2\right )}{-15+e^{2 e^5 x^2}} \, dx\right )-120 \int \frac {\exp \left (5+\frac {2 \left (-75+5 e^{2 e^5 x^2}+x\right )}{-15+e^{2 e^5 x^2}}\right ) x^2}{\left (-15+e^{2 e^5 x^2}\right )^2} \, dx\\ &=-\left (2 \int \left (-\frac {\exp \left (\frac {2 \left (-75+5 e^{2 e^5 x^2}+x\right )}{-15+e^{2 e^5 x^2}}\right )}{-15+e^{2 e^5 x^2}}+\frac {4 \exp \left (5+\frac {2 \left (-75+5 e^{2 e^5 x^2}+x\right )}{-15+e^{2 e^5 x^2}}\right ) x^2}{-15+e^{2 e^5 x^2}}\right ) \, dx\right )-120 \int \frac {\exp \left (\frac {-225+15 e^{2 e^5 x^2}+2 x}{-15+e^{2 e^5 x^2}}\right ) x^2}{\left (15-e^{2 e^5 x^2}\right )^2} \, dx\\ &=2 \int \frac {\exp \left (\frac {2 \left (-75+5 e^{2 e^5 x^2}+x\right )}{-15+e^{2 e^5 x^2}}\right )}{-15+e^{2 e^5 x^2}} \, dx-8 \int \frac {\exp \left (5+\frac {2 \left (-75+5 e^{2 e^5 x^2}+x\right )}{-15+e^{2 e^5 x^2}}\right ) x^2}{-15+e^{2 e^5 x^2}} \, dx-120 \int \frac {\exp \left (\frac {-225+15 e^{2 e^5 x^2}+2 x}{-15+e^{2 e^5 x^2}}\right ) x^2}{\left (15-e^{2 e^5 x^2}\right )^2} \, dx\\ &=2 \int \frac {\exp \left (\frac {2 \left (-75+5 e^{2 e^5 x^2}+x\right )}{-15+e^{2 e^5 x^2}}\right )}{-15+e^{2 e^5 x^2}} \, dx-8 \int \frac {\exp \left (\frac {-225+15 e^{2 e^5 x^2}+2 x}{-15+e^{2 e^5 x^2}}\right ) x^2}{-15+e^{2 e^5 x^2}} \, dx-120 \int \frac {\exp \left (\frac {-225+15 e^{2 e^5 x^2}+2 x}{-15+e^{2 e^5 x^2}}\right ) x^2}{\left (15-e^{2 e^5 x^2}\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.16, size = 21, normalized size = 0.88 \begin {gather*} e^{10+\frac {2 x}{-15+e^{2 e^5 x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-150 + 10*E^(2*E^5*x^2) + 2*x)/(-15 + E^(2*E^5*x^2)))*(-30 + E^(2*E^5*x^2)*(2 - 8*E^5*x^2)))/(2
25 - 30*E^(2*E^5*x^2) + E^(4*E^5*x^2)),x]

[Out]

E^(10 + (2*x)/(-15 + E^(2*E^5*x^2)))

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fricas [A]  time = 0.69, size = 28, normalized size = 1.17 \begin {gather*} e^{\left (\frac {2 \, {\left (x + 5 \, e^{\left (2 \, x^{2} e^{5}\right )} - 75\right )}}{e^{\left (2 \, x^{2} e^{5}\right )} - 15}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2*exp(5)+2)*exp(2*x^2*exp(5))-30)*exp((10*exp(2*x^2*exp(5))+2*x-150)/(exp(2*x^2*exp(5))-15))/
(exp(2*x^2*exp(5))^2-30*exp(2*x^2*exp(5))+225),x, algorithm="fricas")

[Out]

e^(2*(x + 5*e^(2*x^2*e^5) - 75)/(e^(2*x^2*e^5) - 15))

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giac [B]  time = 0.41, size = 53, normalized size = 2.21 \begin {gather*} e^{\left (\frac {2 \, x}{e^{\left (2 \, x^{2} e^{5}\right )} - 15} + \frac {10 \, e^{\left (2 \, x^{2} e^{5}\right )}}{e^{\left (2 \, x^{2} e^{5}\right )} - 15} - \frac {150}{e^{\left (2 \, x^{2} e^{5}\right )} - 15}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2*exp(5)+2)*exp(2*x^2*exp(5))-30)*exp((10*exp(2*x^2*exp(5))+2*x-150)/(exp(2*x^2*exp(5))-15))/
(exp(2*x^2*exp(5))^2-30*exp(2*x^2*exp(5))+225),x, algorithm="giac")

[Out]

e^(2*x/(e^(2*x^2*e^5) - 15) + 10*e^(2*x^2*e^5)/(e^(2*x^2*e^5) - 15) - 150/(e^(2*x^2*e^5) - 15))

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maple [A]  time = 0.31, size = 29, normalized size = 1.21

method	result	size
risch	\({\mathrm e}^{\frac {10 \,{\mathrm e}^{2 x^{2} {\mathrm e}^{5}}+2 x -150}{{\mathrm e}^{2 x^{2} {\mathrm e}^{5}}-15}}\)	\(29\)
norman	\(\frac {{\mathrm e}^{2 x^{2} {\mathrm e}^{5}} {\mathrm e}^{\frac {10 \,{\mathrm e}^{2 x^{2} {\mathrm e}^{5}}+2 x -150}{{\mathrm e}^{2 x^{2} {\mathrm e}^{5}}-15}}-15 \,{\mathrm e}^{\frac {10 \,{\mathrm e}^{2 x^{2} {\mathrm e}^{5}}+2 x -150}{{\mathrm e}^{2 x^{2} {\mathrm e}^{5}}-15}}}{{\mathrm e}^{2 x^{2} {\mathrm e}^{5}}-15}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x^2*exp(5)+2)*exp(2*x^2*exp(5))-30)*exp((10*exp(2*x^2*exp(5))+2*x-150)/(exp(2*x^2*exp(5))-15))/(exp(2
*x^2*exp(5))^2-30*exp(2*x^2*exp(5))+225),x,method=_RETURNVERBOSE)

[Out]

exp(2*(5*exp(2*x^2*exp(5))+x-75)/(exp(2*x^2*exp(5))-15))

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maxima [B]  time = 0.59, size = 53, normalized size = 2.21 \begin {gather*} e^{\left (\frac {2 \, x}{e^{\left (2 \, x^{2} e^{5}\right )} - 15} + \frac {10 \, e^{\left (2 \, x^{2} e^{5}\right )}}{e^{\left (2 \, x^{2} e^{5}\right )} - 15} - \frac {150}{e^{\left (2 \, x^{2} e^{5}\right )} - 15}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2*exp(5)+2)*exp(2*x^2*exp(5))-30)*exp((10*exp(2*x^2*exp(5))+2*x-150)/(exp(2*x^2*exp(5))-15))/
(exp(2*x^2*exp(5))^2-30*exp(2*x^2*exp(5))+225),x, algorithm="maxima")

[Out]

e^(2*x/(e^(2*x^2*e^5) - 15) + 10*e^(2*x^2*e^5)/(e^(2*x^2*e^5) - 15) - 150/(e^(2*x^2*e^5) - 15))

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mupad [B]  time = 2.41, size = 55, normalized size = 2.29 \begin {gather*} {\mathrm {e}}^{-\frac {150}{{\mathrm {e}}^{2\,x^2\,{\mathrm {e}}^5}-15}}\,{\mathrm {e}}^{\frac {10\,{\mathrm {e}}^{2\,x^2\,{\mathrm {e}}^5}}{{\mathrm {e}}^{2\,x^2\,{\mathrm {e}}^5}-15}}\,{\mathrm {e}}^{\frac {2\,x}{{\mathrm {e}}^{2\,x^2\,{\mathrm {e}}^5}-15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*x + 10*exp(2*x^2*exp(5)) - 150)/(exp(2*x^2*exp(5)) - 15))*(exp(2*x^2*exp(5))*(8*x^2*exp(5) - 2) +
30))/(exp(4*x^2*exp(5)) - 30*exp(2*x^2*exp(5)) + 225),x)

[Out]

exp(-150/(exp(2*x^2*exp(5)) - 15))*exp((10*exp(2*x^2*exp(5)))/(exp(2*x^2*exp(5)) - 15))*exp((2*x)/(exp(2*x^2*e
xp(5)) - 15))

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sympy [A]  time = 0.49, size = 29, normalized size = 1.21 \begin {gather*} e^{\frac {2 x + 10 e^{2 x^{2} e^{5}} - 150}{e^{2 x^{2} e^{5}} - 15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x**2*exp(5)+2)*exp(2*x**2*exp(5))-30)*exp((10*exp(2*x**2*exp(5))+2*x-150)/(exp(2*x**2*exp(5))-1
5))/(exp(2*x**2*exp(5))**2-30*exp(2*x**2*exp(5))+225),x)

[Out]

exp((2*x + 10*exp(2*x**2*exp(5)) - 150)/(exp(2*x**2*exp(5)) - 15))

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