3.48.29 \(\int \frac {-2 x^2+60 x^3+e^{\frac {-1-3 x+2 x^2}{2 x}} (5+10 x^2)}{10 e^{\frac {-1-3 x+2 x^2}{2 x}} x^2-2 x^3+30 x^4} \, dx\)

Optimal. Leaf size=27 \[ \log \left (e^{-1+x-\frac {1+x}{2 x}}-\frac {x}{5}+3 x^2\right ) \]

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Rubi [F]  time = 5.80, 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 {-2 x^2+60 x^3+e^{\frac {-1-3 x+2 x^2}{2 x}} \left (5+10 x^2\right )}{10 e^{\frac {-1-3 x+2 x^2}{2 x}} x^2-2 x^3+30 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x^2 + 60*x^3 + E^((-1 - 3*x + 2*x^2)/(2*x))*(5 + 10*x^2))/(10*E^((-1 - 3*x + 2*x^2)/(2*x))*x^2 - 2*x^3
+ 30*x^4),x]

[Out]

-1/2*1/x + x - (17*Defer[Int][E^(3/2 + 1/(2*x))/(5*E^x + E^((3 + x^(-1))/2)*x*(-1 + 15*x)), x])/2 + 31*Defer[I
nt][(E^(3/2 + 1/(2*x))*x)/(5*E^x + E^((3 + x^(-1))/2)*x*(-1 + 15*x)), x] - 15*Defer[Int][(E^(3/2 + 1/(2*x))*x^
2)/(5*E^x + E^((3 + x^(-1))/2)*x*(-1 + 15*x)), x] - Defer[Subst][Defer[Int][E^((15*E^x + x + E^((3 + x^(-1))/2
)*(1 + 45*x^2))/(10*(E^x + 3*E^((3 + x^(-1))/2)*x^2)))/x, x], x, (5*E^x - E^(3/2 + 1/(2*x))*x + 15*E^(3/2 + 1/
(2*x))*x^2)/x]/(10*(E^x + 3*E^(3/2 + 1/(2*x))*x^2))

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1+2 x^2}{2 x^2}-\frac {e^{\frac {3}{2}+\frac {1}{2 x}} \left (-1+17 x-62 x^2+30 x^3\right )}{2 x \left (5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2\right )}\right ) \, dx\\ &=\frac {1}{2} \int \frac {1+2 x^2}{x^2} \, dx-\frac {1}{2} \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}} \left (-1+17 x-62 x^2+30 x^3\right )}{x \left (5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2\right )} \, dx\\ &=\frac {1}{2} \int \left (2+\frac {1}{x^2}\right ) \, dx-\frac {1}{2} \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}} \left (-1+17 x-62 x^2+30 x^3\right )}{x \left (5 e^x+e^{\frac {1}{2} \left (3+\frac {1}{x}\right )} x (-1+15 x)\right )} \, dx\\ &=-\frac {1}{2 x}+x-\frac {1}{2} \int \left (\frac {17 e^{\frac {3}{2}+\frac {1}{2 x}}}{5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2}-\frac {e^{\frac {3}{2}+\frac {1}{2 x}}}{x \left (5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2\right )}-\frac {62 e^{\frac {3}{2}+\frac {1}{2 x}} x}{5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2}+\frac {30 e^{\frac {3}{2}+\frac {1}{2 x}} x^2}{5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2}\right ) \, dx\\ &=-\frac {1}{2 x}+x+\frac {1}{2} \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}}}{x \left (5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2\right )} \, dx-\frac {17}{2} \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}}}{5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2} \, dx-15 \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}} x^2}{5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2} \, dx+31 \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}} x}{5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2} \, dx\\ &=-\frac {1}{2 x}+x-\frac {17}{2} \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}}}{5 e^x+e^{\frac {1}{2} \left (3+\frac {1}{x}\right )} x (-1+15 x)} \, dx-15 \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}} x^2}{5 e^x+e^{\frac {1}{2} \left (3+\frac {1}{x}\right )} x (-1+15 x)} \, dx+31 \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}} x}{5 e^x+e^{\frac {1}{2} \left (3+\frac {1}{x}\right )} x (-1+15 x)} \, dx-\frac {\operatorname {Subst}\left (\int \frac {\exp \left (\frac {3}{2}+\frac {e^{\frac {3}{2}+\frac {1}{2 x}}}{2 \left (5 e^x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2\right )}+\frac {x}{2 \left (5 e^x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2\right )}\right )}{x} \, dx,x,\frac {5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2}{x}\right )}{2 \left (5 e^x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2\right )}\\ &=-\frac {1}{2 x}+x-\frac {17}{2} \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}}}{5 e^x+e^{\frac {1}{2} \left (3+\frac {1}{x}\right )} x (-1+15 x)} \, dx-15 \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}} x^2}{5 e^x+e^{\frac {1}{2} \left (3+\frac {1}{x}\right )} x (-1+15 x)} \, dx+31 \int \frac {e^{\frac {3}{2}+\frac {1}{2 x}} x}{5 e^x+e^{\frac {1}{2} \left (3+\frac {1}{x}\right )} x (-1+15 x)} \, dx-\frac {\operatorname {Subst}\left (\int \frac {\exp \left (\frac {15 e^x+x+e^{\frac {1}{2} \left (3+\frac {1}{x}\right )} \left (1+45 x^2\right )}{10 \left (e^x+3 e^{\frac {1}{2} \left (3+\frac {1}{x}\right )} x^2\right )}\right )}{x} \, dx,x,\frac {5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2}{x}\right )}{2 \left (5 e^x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.06, size = 53, normalized size = 1.96 \begin {gather*} \frac {1}{2} \left (-\frac {1}{x}+2 \log \left (5 e^x-e^{\frac {3}{2}+\frac {1}{2 x}} x+15 e^{\frac {3}{2}+\frac {1}{2 x}} x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^2 + 60*x^3 + E^((-1 - 3*x + 2*x^2)/(2*x))*(5 + 10*x^2))/(10*E^((-1 - 3*x + 2*x^2)/(2*x))*x^2 -
2*x^3 + 30*x^4),x]

[Out]

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

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fricas [A]  time = 0.60, size = 28, normalized size = 1.04 \begin {gather*} \log \left (15 \, x^{2} - x + 5 \, e^{\left (\frac {2 \, x^{2} - 3 \, x - 1}{2 \, x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+5)*exp(1/2*(2*x^2-3*x-1)/x)+60*x^3-2*x^2)/(10*x^2*exp(1/2*(2*x^2-3*x-1)/x)+30*x^4-2*x^3),x,
algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 28, normalized size = 1.04 \begin {gather*} \log \left (15 \, x^{2} - x + 5 \, e^{\left (\frac {2 \, x^{2} - 3 \, x - 1}{2 \, x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+5)*exp(1/2*(2*x^2-3*x-1)/x)+60*x^3-2*x^2)/(10*x^2*exp(1/2*(2*x^2-3*x-1)/x)+30*x^4-2*x^3),x,
algorithm="giac")

[Out]

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

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maple [A]  time = 0.11, size = 29, normalized size = 1.07

method	result	size
norman	\(\ln \left (15 x^{2}+5 \,{\mathrm e}^{\frac {2 x^{2}-3 x -1}{2 x}}-x \right )\)	\(29\)
risch	\(x -\frac {1}{2 x}-\frac {2 x^{2}-3 x -1}{2 x}+\ln \left (3 x^{2}-\frac {x}{5}+{\mathrm e}^{\frac {2 x^{2}-3 x -1}{2 x}}\right )\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^2+5)*exp(1/2*(2*x^2-3*x-1)/x)+60*x^3-2*x^2)/(10*x^2*exp(1/2*(2*x^2-3*x-1)/x)+30*x^4-2*x^3),x,method
=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.42, size = 56, normalized size = 2.07 \begin {gather*} -\frac {1}{2 \, x} + \log \left (15 \, x - 1\right ) + \log \relax (x) + \log \left (\frac {{\left (15 \, x^{2} e^{\frac {3}{2}} - x e^{\frac {3}{2}}\right )} e^{\left (\frac {1}{2 \, x}\right )} + 5 \, e^{x}}{15 \, x^{2} e^{\frac {3}{2}} - x e^{\frac {3}{2}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+5)*exp(1/2*(2*x^2-3*x-1)/x)+60*x^3-2*x^2)/(10*x^2*exp(1/2*(2*x^2-3*x-1)/x)+30*x^4-2*x^3),x,
algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {60\,x^3-2\,x^2+{\mathrm {e}}^{-\frac {-x^2+\frac {3\,x}{2}+\frac {1}{2}}{x}}\,\left (10\,x^2+5\right )}{10\,x^2\,{\mathrm {e}}^{-\frac {-x^2+\frac {3\,x}{2}+\frac {1}{2}}{x}}-2\,x^3+30\,x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((60*x^3 - 2*x^2 + exp(-((3*x)/2 - x^2 + 1/2)/x)*(10*x^2 + 5))/(10*x^2*exp(-((3*x)/2 - x^2 + 1/2)/x) - 2*x^
3 + 30*x^4),x)

[Out]

int((60*x^3 - 2*x^2 + exp(-((3*x)/2 - x^2 + 1/2)/x)*(10*x^2 + 5))/(10*x^2*exp(-((3*x)/2 - x^2 + 1/2)/x) - 2*x^
3 + 30*x^4), x)

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sympy [A]  time = 0.24, size = 24, normalized size = 0.89 \begin {gather*} \log {\left (3 x^{2} - \frac {x}{5} + e^{\frac {x^{2} - \frac {3 x}{2} - \frac {1}{2}}{x}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**2+5)*exp(1/2*(2*x**2-3*x-1)/x)+60*x**3-2*x**2)/(10*x**2*exp(1/2*(2*x**2-3*x-1)/x)+30*x**4-2*
x**3),x)

[Out]

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

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