3.10.58 \(\int \frac {-6 e^{5+\frac {1}{x^2}}+45 e^5 x+e^x (e^5 (45 x-45 x^2)+e^{5+\frac {1}{x^2}} (-6+3 x^3))}{225 x-30 e^{\frac {1}{x^2}} x^2+e^{\frac {2}{x^2}} x^3+e^{2 x} (225 x-30 e^{\frac {1}{x^2}} x^2+e^{\frac {2}{x^2}} x^3)+e^x (450 x-60 e^{\frac {1}{x^2}} x^2+2 e^{\frac {2}{x^2}} x^3)} \, dx\)

Optimal. Leaf size=33 \[ \frac {e^5 x}{\left (1+e^x\right ) \left (5-x+\frac {1}{3} \left (3-e^{\frac {1}{x^2}}\right ) x\right )} \]

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Rubi [F]  time = 3.02, 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 {-6 e^{5+\frac {1}{x^2}}+45 e^5 x+e^x \left (e^5 \left (45 x-45 x^2\right )+e^{5+\frac {1}{x^2}} \left (-6+3 x^3\right )\right )}{225 x-30 e^{\frac {1}{x^2}} x^2+e^{\frac {2}{x^2}} x^3+e^{2 x} \left (225 x-30 e^{\frac {1}{x^2}} x^2+e^{\frac {2}{x^2}} x^3\right )+e^x \left (450 x-60 e^{\frac {1}{x^2}} x^2+2 e^{\frac {2}{x^2}} x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-6*E^(5 + x^(-2)) + 45*E^5*x + E^x*(E^5*(45*x - 45*x^2) + E^(5 + x^(-2))*(-6 + 3*x^3)))/(225*x - 30*E^x^(
-2)*x^2 + E^(2/x^2)*x^3 + E^(2*x)*(225*x - 30*E^x^(-2)*x^2 + E^(2/x^2)*x^3) + E^x*(450*x - 60*E^x^(-2)*x^2 + 2
*E^(2/x^2)*x^3)),x]

[Out]

45*E^5*Defer[Int][1/((1 + E^x)*(-15 + E^x^(-2)*x)^2), x] - 6*E^5*Defer[Int][E^x^(-2)/((1 + E^x)*x*(-15 + E^x^(
-2)*x)^2), x] - 45*E^5*Defer[Int][x/((1 + E^x)*(-15 + E^x^(-2)*x)^2), x] + 3*E^5*Defer[Int][(E^x^(-2)*x^2)/((1
+ E^x)*(-15 + E^x^(-2)*x)^2), x] - 3*E^5*Defer[Int][x/((1 + E^x)^2*(-15 + E^x^(-2)*x)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^5 \left (-2 e^{\frac {1}{x^2}}+15 x-15 e^x (-1+x) x+e^{\frac {1}{x^2}+x} \left (-2+x^3\right )\right )}{\left (1+e^x\right )^2 x \left (15-e^{\frac {1}{x^2}} x\right )^2} \, dx\\ &=\left (3 e^5\right ) \int \frac {-2 e^{\frac {1}{x^2}}+15 x-15 e^x (-1+x) x+e^{\frac {1}{x^2}+x} \left (-2+x^3\right )}{\left (1+e^x\right )^2 x \left (15-e^{\frac {1}{x^2}} x\right )^2} \, dx\\ &=\left (3 e^5\right ) \int \left (-\frac {x}{\left (1+e^x\right )^2 \left (-15+e^{\frac {1}{x^2}} x\right )}+\frac {-2 e^{\frac {1}{x^2}}+15 x-15 x^2+e^{\frac {1}{x^2}} x^3}{\left (1+e^x\right ) x \left (-15+e^{\frac {1}{x^2}} x\right )^2}\right ) \, dx\\ &=-\left (\left (3 e^5\right ) \int \frac {x}{\left (1+e^x\right )^2 \left (-15+e^{\frac {1}{x^2}} x\right )} \, dx\right )+\left (3 e^5\right ) \int \frac {-2 e^{\frac {1}{x^2}}+15 x-15 x^2+e^{\frac {1}{x^2}} x^3}{\left (1+e^x\right ) x \left (-15+e^{\frac {1}{x^2}} x\right )^2} \, dx\\ &=-\left (\left (3 e^5\right ) \int \frac {x}{\left (1+e^x\right )^2 \left (-15+e^{\frac {1}{x^2}} x\right )} \, dx\right )+\left (3 e^5\right ) \int \left (\frac {15}{\left (1+e^x\right ) \left (-15+e^{\frac {1}{x^2}} x\right )^2}-\frac {2 e^{\frac {1}{x^2}}}{\left (1+e^x\right ) x \left (-15+e^{\frac {1}{x^2}} x\right )^2}-\frac {15 x}{\left (1+e^x\right ) \left (-15+e^{\frac {1}{x^2}} x\right )^2}+\frac {e^{\frac {1}{x^2}} x^2}{\left (1+e^x\right ) \left (-15+e^{\frac {1}{x^2}} x\right )^2}\right ) \, dx\\ &=\left (3 e^5\right ) \int \frac {e^{\frac {1}{x^2}} x^2}{\left (1+e^x\right ) \left (-15+e^{\frac {1}{x^2}} x\right )^2} \, dx-\left (3 e^5\right ) \int \frac {x}{\left (1+e^x\right )^2 \left (-15+e^{\frac {1}{x^2}} x\right )} \, dx-\left (6 e^5\right ) \int \frac {e^{\frac {1}{x^2}}}{\left (1+e^x\right ) x \left (-15+e^{\frac {1}{x^2}} x\right )^2} \, dx+\left (45 e^5\right ) \int \frac {1}{\left (1+e^x\right ) \left (-15+e^{\frac {1}{x^2}} x\right )^2} \, dx-\left (45 e^5\right ) \int \frac {x}{\left (1+e^x\right ) \left (-15+e^{\frac {1}{x^2}} x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 24, normalized size = 0.73 \begin {gather*} -\frac {3 e^5 x}{\left (1+e^x\right ) \left (-15+e^{\frac {1}{x^2}} x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*E^(5 + x^(-2)) + 45*E^5*x + E^x*(E^5*(45*x - 45*x^2) + E^(5 + x^(-2))*(-6 + 3*x^3)))/(225*x - 30
*E^x^(-2)*x^2 + E^(2/x^2)*x^3 + E^(2*x)*(225*x - 30*E^x^(-2)*x^2 + E^(2/x^2)*x^3) + E^x*(450*x - 60*E^x^(-2)*x
^2 + 2*E^(2/x^2)*x^3)),x]

[Out]

(-3*E^5*x)/((1 + E^x)*(-15 + E^x^(-2)*x))

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fricas [A]  time = 0.53, size = 48, normalized size = 1.45 \begin {gather*} -\frac {3 \, x e^{10}}{{\left (x e^{\left (\frac {5 \, x^{2} + 1}{x^{2}}\right )} - 15 \, e^{5}\right )} e^{x} + x e^{\left (\frac {5 \, x^{2} + 1}{x^{2}}\right )} - 15 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^3-6)*exp(5)*exp(1/x^2)+(-45*x^2+45*x)*exp(5))*exp(x)-6*exp(5)*exp(1/x^2)+45*x*exp(5))/((x^3*e
xp(1/x^2)^2-30*x^2*exp(1/x^2)+225*x)*exp(x)^2+(2*x^3*exp(1/x^2)^2-60*x^2*exp(1/x^2)+450*x)*exp(x)+x^3*exp(1/x^
2)^2-30*x^2*exp(1/x^2)+225*x),x, algorithm="fricas")

[Out]

-3*x*e^10/((x*e^((5*x^2 + 1)/x^2) - 15*e^5)*e^x + x*e^((5*x^2 + 1)/x^2) - 15*e^5)

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giac [B]  time = 0.54, size = 247, normalized size = 7.48 \begin {gather*} -\frac {3 \, {\left (x^{2} e^{\left (\frac {x^{3} + 2}{x^{2}} + \frac {1}{x^{2}} + 5\right )} + x^{2} e^{\left (\frac {3}{x^{2}} + 5\right )} - 15 \, x e^{\left (x + \frac {2}{x^{2}} + 5\right )} - 15 \, x e^{\left (\frac {2}{x^{2}} + 5\right )} + 15 \, e^{\left (x + \frac {2}{x^{2}} + 5\right )} - 15 \, e^{\left (\frac {x^{3} + 2}{x^{2}} + 5\right )}\right )}}{x^{2} e^{\left (x + \frac {x^{3} + 2}{x^{2}} + \frac {2}{x^{2}}\right )} + x^{2} e^{\left (x + \frac {4}{x^{2}}\right )} + x^{2} e^{\left (\frac {x^{3} + 2}{x^{2}} + \frac {2}{x^{2}}\right )} + x^{2} e^{\left (\frac {4}{x^{2}}\right )} - 30 \, x e^{\left (x + \frac {x^{3} + 2}{x^{2}} + \frac {1}{x^{2}}\right )} - 30 \, x e^{\left (x + \frac {3}{x^{2}}\right )} - 30 \, x e^{\left (\frac {x^{3} + 2}{x^{2}} + \frac {1}{x^{2}}\right )} - 30 \, x e^{\left (\frac {3}{x^{2}}\right )} + 225 \, e^{\left (x + \frac {x^{3} + 2}{x^{2}}\right )} + 225 \, e^{\left (x + \frac {2}{x^{2}}\right )} + 225 \, e^{\left (\frac {x^{3} + 2}{x^{2}}\right )} + 225 \, e^{\left (\frac {2}{x^{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^3-6)*exp(5)*exp(1/x^2)+(-45*x^2+45*x)*exp(5))*exp(x)-6*exp(5)*exp(1/x^2)+45*x*exp(5))/((x^3*e
xp(1/x^2)^2-30*x^2*exp(1/x^2)+225*x)*exp(x)^2+(2*x^3*exp(1/x^2)^2-60*x^2*exp(1/x^2)+450*x)*exp(x)+x^3*exp(1/x^
2)^2-30*x^2*exp(1/x^2)+225*x),x, algorithm="giac")

[Out]

-3*(x^2*e^((x^3 + 2)/x^2 + 1/x^2 + 5) + x^2*e^(3/x^2 + 5) - 15*x*e^(x + 2/x^2 + 5) - 15*x*e^(2/x^2 + 5) + 15*e
^(x + 2/x^2 + 5) - 15*e^((x^3 + 2)/x^2 + 5))/(x^2*e^(x + (x^3 + 2)/x^2 + 2/x^2) + x^2*e^(x + 4/x^2) + x^2*e^((
x^3 + 2)/x^2 + 2/x^2) + x^2*e^(4/x^2) - 30*x*e^(x + (x^3 + 2)/x^2 + 1/x^2) - 30*x*e^(x + 3/x^2) - 30*x*e^((x^3
+ 2)/x^2 + 1/x^2) - 30*x*e^(3/x^2) + 225*e^(x + (x^3 + 2)/x^2) + 225*e^(x + 2/x^2) + 225*e^((x^3 + 2)/x^2) +
225*e^(2/x^2))

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maple [A]  time = 0.06, size = 22, normalized size = 0.67

method	result	size
risch	\(-\frac {3 x \,{\mathrm e}^{5}}{\left ({\mathrm e}^{x}+1\right ) \left (x \,{\mathrm e}^{\frac {1}{x^{2}}}-15\right )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x^3-6)*exp(5)*exp(1/x^2)+(-45*x^2+45*x)*exp(5))*exp(x)-6*exp(5)*exp(1/x^2)+45*x*exp(5))/((x^3*exp(1/x
^2)^2-30*x^2*exp(1/x^2)+225*x)*exp(x)^2+(2*x^3*exp(1/x^2)^2-60*x^2*exp(1/x^2)+450*x)*exp(x)+x^3*exp(1/x^2)^2-3
0*x^2*exp(1/x^2)+225*x),x,method=_RETURNVERBOSE)

[Out]

-3*x*exp(5)/(exp(x)+1)/(x*exp(1/x^2)-15)

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maxima [A]  time = 0.50, size = 24, normalized size = 0.73 \begin {gather*} -\frac {3 \, x e^{5}}{{\left (x e^{x} + x\right )} e^{\left (\frac {1}{x^{2}}\right )} - 15 \, e^{x} - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^3-6)*exp(5)*exp(1/x^2)+(-45*x^2+45*x)*exp(5))*exp(x)-6*exp(5)*exp(1/x^2)+45*x*exp(5))/((x^3*e
xp(1/x^2)^2-30*x^2*exp(1/x^2)+225*x)*exp(x)^2+(2*x^3*exp(1/x^2)^2-60*x^2*exp(1/x^2)+450*x)*exp(x)+x^3*exp(1/x^
2)^2-30*x^2*exp(1/x^2)+225*x),x, algorithm="maxima")

[Out]

-3*x*e^5/((x*e^x + x)*e^(x^(-2)) - 15*e^x - 15)

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mupad [B]  time = 0.82, size = 54, normalized size = 1.64 \begin {gather*} \frac {3\,\left (2\,x\,{\mathrm {e}}^{x+5}+2\,x\,{\mathrm {e}}^5-x^3\,{\mathrm {e}}^{x+5}-x^3\,{\mathrm {e}}^5\right )}{\left (x^2-2\right )\,{\left ({\mathrm {e}}^x+1\right )}^2\,\left (x\,{\mathrm {e}}^{\frac {1}{x^2}}-15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((45*x*exp(5) - 6*exp(1/x^2)*exp(5) + exp(x)*(exp(5)*(45*x - 45*x^2) + exp(1/x^2)*exp(5)*(3*x^3 - 6)))/(225
*x + exp(2*x)*(225*x - 30*x^2*exp(1/x^2) + x^3*exp(2/x^2)) - 30*x^2*exp(1/x^2) + exp(x)*(450*x - 60*x^2*exp(1/
x^2) + 2*x^3*exp(2/x^2)) + x^3*exp(2/x^2)),x)

[Out]

(3*(2*x*exp(x + 5) + 2*x*exp(5) - x^3*exp(x + 5) - x^3*exp(5)))/((x^2 - 2)*(exp(x) + 1)^2*(x*exp(1/x^2) - 15))

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sympy [A]  time = 0.30, size = 31, normalized size = 0.94 \begin {gather*} - \frac {3 x e^{5}}{x e^{\frac {1}{x^{2}}} + \left (x e^{\frac {1}{x^{2}}} - 15\right ) e^{x} - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x**3-6)*exp(5)*exp(1/x**2)+(-45*x**2+45*x)*exp(5))*exp(x)-6*exp(5)*exp(1/x**2)+45*x*exp(5))/((x
**3*exp(1/x**2)**2-30*x**2*exp(1/x**2)+225*x)*exp(x)**2+(2*x**3*exp(1/x**2)**2-60*x**2*exp(1/x**2)+450*x)*exp(
x)+x**3*exp(1/x**2)**2-30*x**2*exp(1/x**2)+225*x),x)

[Out]

-3*x*exp(5)/(x*exp(x**(-2)) + (x*exp(x**(-2)) - 15)*exp(x) - 15)

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