3.59.36 \(\int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} (-12 x^2-8 x^3)}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx\)

Optimal. Leaf size=24 \[ e^{\frac {4 x}{5 \left (x+e^{2 x} x^4\right )}}-2 x \]

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Rubi [F]  time = 2.61, 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 {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5+10 e^{2 x} x^3+5 e^{4 x} x^6} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10 - 20*E^(2*x)*x^3 - 10*E^(4*x)*x^6 + E^(2*x + 4/(5 + 5*E^(2*x)*x^3))*(-12*x^2 - 8*x^3))/(5 + 10*E^(2*x
)*x^3 + 5*E^(4*x)*x^6),x]

[Out]

-2*x - (12*Defer[Int][(E^(2*x + 4/(5 + 5*E^(2*x)*x^3))*x^2)/(1 + E^(2*x)*x^3)^2, x])/5 - (8*Defer[Int][(E^(2*x
+ 4/(5 + 5*E^(2*x)*x^3))*x^3)/(1 + E^(2*x)*x^3)^2, x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{5 \left (1+e^{2 x} x^3\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {-10-20 e^{2 x} x^3-10 e^{4 x} x^6+e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} \left (-12 x^2-8 x^3\right )}{\left (1+e^{2 x} x^3\right )^2} \, dx\\ &=\frac {1}{5} \int \left (-10-\frac {4 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2 (3+2 x)}{\left (1+e^{2 x} x^3\right )^2}\right ) \, dx\\ &=-2 x-\frac {4}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2 (3+2 x)}{\left (1+e^{2 x} x^3\right )^2} \, dx\\ &=-2 x-\frac {4}{5} \int \left (\frac {3 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2}{\left (1+e^{2 x} x^3\right )^2}+\frac {2 e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^3}{\left (1+e^{2 x} x^3\right )^2}\right ) \, dx\\ &=-2 x-\frac {8}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^3}{\left (1+e^{2 x} x^3\right )^2} \, dx-\frac {12}{5} \int \frac {e^{2 x+\frac {4}{5+5 e^{2 x} x^3}} x^2}{\left (1+e^{2 x} x^3\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 31, normalized size = 1.29 \begin {gather*} -\frac {2}{5} \left (-\frac {5}{2} e^{\frac {4}{5 \left (1+e^{2 x} x^3\right )}}+5 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 20*E^(2*x)*x^3 - 10*E^(4*x)*x^6 + E^(2*x + 4/(5 + 5*E^(2*x)*x^3))*(-12*x^2 - 8*x^3))/(5 + 10*
E^(2*x)*x^3 + 5*E^(4*x)*x^6),x]

[Out]

(-2*((-5*E^(4/(5*(1 + E^(2*x)*x^3))))/2 + 5*x))/5

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fricas [B]  time = 0.72, size = 45, normalized size = 1.88 \begin {gather*} -{\left (2 \, x e^{\left (2 \, x\right )} - e^{\left (\frac {2 \, {\left (5 \, x^{4} e^{\left (2 \, x\right )} + 5 \, x + 2\right )}}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )}\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp(x)^4-20*exp(x)^2*x^3-10)/(5*x^6*exp
(x)^4+10*exp(x)^2*x^3+5),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (5 \, x^{6} e^{\left (4 \, x\right )} + 10 \, x^{3} e^{\left (2 \, x\right )} + 2 \, {\left (2 \, x^{3} + 3 \, x^{2}\right )} e^{\left (2 \, x + \frac {4}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )} + 5\right )}}{5 \, {\left (x^{6} e^{\left (4 \, x\right )} + 2 \, x^{3} e^{\left (2 \, x\right )} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp(x)^4-20*exp(x)^2*x^3-10)/(5*x^6*exp
(x)^4+10*exp(x)^2*x^3+5),x, algorithm="giac")

[Out]

integrate(-2/5*(5*x^6*e^(4*x) + 10*x^3*e^(2*x) + 2*(2*x^3 + 3*x^2)*e^(2*x + 4/5/(x^3*e^(2*x) + 1)) + 5)/(x^6*e
^(4*x) + 2*x^3*e^(2*x) + 1), x)

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maple [A]  time = 0.05, size = 20, normalized size = 0.83

method	result	size
risch	\({\mathrm e}^{\frac {4}{5 \left ({\mathrm e}^{2 x} x^{3}+1\right )}}-2 x\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp(x)^4-20*exp(x)^2*x^3-10)/(5*x^6*exp(x)^4+
10*exp(x)^2*x^3+5),x,method=_RETURNVERBOSE)

[Out]

exp(4/5/(exp(2*x)*x^3+1))-2*x

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maxima [A]  time = 0.39, size = 19, normalized size = 0.79 \begin {gather*} -2 \, x + e^{\left (\frac {4}{5 \, {\left (x^{3} e^{\left (2 \, x\right )} + 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3-12*x^2)*exp(x)^2*exp(2/(5*exp(x)^2*x^3+5))^2-10*x^6*exp(x)^4-20*exp(x)^2*x^3-10)/(5*x^6*exp
(x)^4+10*exp(x)^2*x^3+5),x, algorithm="maxima")

[Out]

-2*x + e^(4/5/(x^3*e^(2*x) + 1))

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mupad [B]  time = 0.21, size = 20, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^{\frac {4}{5\,x^3\,{\mathrm {e}}^{2\,x}+5}}-2\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x^3*exp(2*x) + 10*x^6*exp(4*x) + exp(2*x)*exp(4/(5*x^3*exp(2*x) + 5))*(12*x^2 + 8*x^3) + 10)/(10*x^3*
exp(2*x) + 5*x^6*exp(4*x) + 5),x)

[Out]

exp(4/(5*x^3*exp(2*x) + 5)) - 2*x

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sympy [A]  time = 0.29, size = 17, normalized size = 0.71 \begin {gather*} - 2 x + e^{\frac {4}{5 x^{3} e^{2 x} + 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x**3-12*x**2)*exp(x)**2*exp(2/(5*exp(x)**2*x**3+5))**2-10*x**6*exp(x)**4-20*exp(x)**2*x**3-10)/
(5*x**6*exp(x)**4+10*exp(x)**2*x**3+5),x)

[Out]

-2*x + exp(4/(5*x**3*exp(2*x) + 5))

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