3.99.97 \(\int \frac {1}{5} e^{\frac {4 e^x+20 x}{5 x}} (120 x+60 x^2+e^{2 x} (-12+8 x+4 x^2)+e^x (-48+62 x+46 x^2+5 x^3)) \, dx\)

Optimal. Leaf size=26 \[ e^{4+\frac {4 e^x}{5 x}} \left (4+e^x\right ) x^2 (3+x) \]

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Rubi [F]  time = 1.31, 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 {1}{5} e^{\frac {4 e^x+20 x}{5 x}} \left (120 x+60 x^2+e^{2 x} \left (-12+8 x+4 x^2\right )+e^x \left (-48+62 x+46 x^2+5 x^3\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((4*E^x + 20*x)/(5*x))*(120*x + 60*x^2 + E^(2*x)*(-12 + 8*x + 4*x^2) + E^x*(-48 + 62*x + 46*x^2 + 5*x^3
)))/5,x]

[Out]

(-48*Defer[Int][E^(x + (4*(E^x + 5*x))/(5*x)), x])/5 - (12*Defer[Int][E^(2*x + (4*(E^x + 5*x))/(5*x)), x])/5 +
24*Defer[Int][E^((4*(E^x + 5*x))/(5*x))*x, x] + (62*Defer[Int][E^(x + (4*(E^x + 5*x))/(5*x))*x, x])/5 + (8*De
fer[Int][E^(2*x + (4*(E^x + 5*x))/(5*x))*x, x])/5 + 12*Defer[Int][E^((4*(E^x + 5*x))/(5*x))*x^2, x] + (46*Defe
r[Int][E^(x + (4*(E^x + 5*x))/(5*x))*x^2, x])/5 + (4*Defer[Int][E^(2*x + (4*(E^x + 5*x))/(5*x))*x^2, x])/5 + D
efer[Int][E^(x + (4*(E^x + 5*x))/(5*x))*x^3, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int e^{\frac {4 e^x+20 x}{5 x}} \left (120 x+60 x^2+e^{2 x} \left (-12+8 x+4 x^2\right )+e^x \left (-48+62 x+46 x^2+5 x^3\right )\right ) \, dx\\ &=\frac {1}{5} \int e^{\frac {4 \left (e^x+5 x\right )}{5 x}} \left (120 x+60 x^2+e^{2 x} \left (-12+8 x+4 x^2\right )+e^x \left (-48+62 x+46 x^2+5 x^3\right )\right ) \, dx\\ &=\frac {1}{5} \int \left (120 e^{\frac {4 \left (e^x+5 x\right )}{5 x}} x+60 e^{\frac {4 \left (e^x+5 x\right )}{5 x}} x^2+4 e^{2 x+\frac {4 \left (e^x+5 x\right )}{5 x}} \left (-3+2 x+x^2\right )+e^{x+\frac {4 \left (e^x+5 x\right )}{5 x}} \left (-48+62 x+46 x^2+5 x^3\right )\right ) \, dx\\ &=\frac {1}{5} \int e^{x+\frac {4 \left (e^x+5 x\right )}{5 x}} \left (-48+62 x+46 x^2+5 x^3\right ) \, dx+\frac {4}{5} \int e^{2 x+\frac {4 \left (e^x+5 x\right )}{5 x}} \left (-3+2 x+x^2\right ) \, dx+12 \int e^{\frac {4 \left (e^x+5 x\right )}{5 x}} x^2 \, dx+24 \int e^{\frac {4 \left (e^x+5 x\right )}{5 x}} x \, dx\\ &=\frac {1}{5} \int \left (-48 e^{x+\frac {4 \left (e^x+5 x\right )}{5 x}}+62 e^{x+\frac {4 \left (e^x+5 x\right )}{5 x}} x+46 e^{x+\frac {4 \left (e^x+5 x\right )}{5 x}} x^2+5 e^{x+\frac {4 \left (e^x+5 x\right )}{5 x}} x^3\right ) \, dx+\frac {4}{5} \int \left (-3 e^{2 x+\frac {4 \left (e^x+5 x\right )}{5 x}}+2 e^{2 x+\frac {4 \left (e^x+5 x\right )}{5 x}} x+e^{2 x+\frac {4 \left (e^x+5 x\right )}{5 x}} x^2\right ) \, dx+12 \int e^{\frac {4 \left (e^x+5 x\right )}{5 x}} x^2 \, dx+24 \int e^{\frac {4 \left (e^x+5 x\right )}{5 x}} x \, dx\\ &=\frac {4}{5} \int e^{2 x+\frac {4 \left (e^x+5 x\right )}{5 x}} x^2 \, dx+\frac {8}{5} \int e^{2 x+\frac {4 \left (e^x+5 x\right )}{5 x}} x \, dx-\frac {12}{5} \int e^{2 x+\frac {4 \left (e^x+5 x\right )}{5 x}} \, dx+\frac {46}{5} \int e^{x+\frac {4 \left (e^x+5 x\right )}{5 x}} x^2 \, dx-\frac {48}{5} \int e^{x+\frac {4 \left (e^x+5 x\right )}{5 x}} \, dx+12 \int e^{\frac {4 \left (e^x+5 x\right )}{5 x}} x^2 \, dx+\frac {62}{5} \int e^{x+\frac {4 \left (e^x+5 x\right )}{5 x}} x \, dx+24 \int e^{\frac {4 \left (e^x+5 x\right )}{5 x}} x \, dx+\int e^{x+\frac {4 \left (e^x+5 x\right )}{5 x}} x^3 \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.59, size = 26, normalized size = 1.00 \begin {gather*} e^{4+\frac {4 e^x}{5 x}} \left (4+e^x\right ) x^2 (3+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((4*E^x + 20*x)/(5*x))*(120*x + 60*x^2 + E^(2*x)*(-12 + 8*x + 4*x^2) + E^x*(-48 + 62*x + 46*x^2 +
5*x^3)))/5,x]

[Out]

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

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fricas [A]  time = 1.51, size = 36, normalized size = 1.38 \begin {gather*} {\left (4 \, x^{3} + 12 \, x^{2} + {\left (x^{3} + 3 \, x^{2}\right )} e^{x}\right )} e^{\left (\frac {4 \, {\left (5 \, x + e^{x}\right )}}{5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((4*x^2+8*x-12)*exp(x)^2+(5*x^3+46*x^2+62*x-48)*exp(x)+60*x^2+120*x)*exp(1/5*(4*exp(x)+20*x)/x),
x, algorithm="fricas")

[Out]

(4*x^3 + 12*x^2 + (x^3 + 3*x^2)*e^x)*e^(4/5*(5*x + e^x)/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{5} \, {\left (60 \, x^{2} + 4 \, {\left (x^{2} + 2 \, x - 3\right )} e^{\left (2 \, x\right )} + {\left (5 \, x^{3} + 46 \, x^{2} + 62 \, x - 48\right )} e^{x} + 120 \, x\right )} e^{\left (\frac {4 \, {\left (5 \, x + e^{x}\right )}}{5 \, x}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((4*x^2+8*x-12)*exp(x)^2+(5*x^3+46*x^2+62*x-48)*exp(x)+60*x^2+120*x)*exp(1/5*(4*exp(x)+20*x)/x),
x, algorithm="giac")

[Out]

integrate(1/5*(60*x^2 + 4*(x^2 + 2*x - 3)*e^(2*x) + (5*x^3 + 46*x^2 + 62*x - 48)*e^x + 120*x)*e^(4/5*(5*x + e^
x)/x), x)

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maple [A]  time = 0.06, size = 24, normalized size = 0.92

method	result	size
risch	\(\left ({\mathrm e}^{x}+4\right ) x^{2} \left (3+x \right ) {\mathrm e}^{\frac {4 x +\frac {4 \,{\mathrm e}^{x}}{5}}{x}}\)	\(24\)
norman	\({\mathrm e}^{x} x^{3} {\mathrm e}^{\frac {4 \,{\mathrm e}^{x}+20 x}{5 x}}+12 x^{2} {\mathrm e}^{\frac {4 \,{\mathrm e}^{x}+20 x}{5 x}}+4 x^{3} {\mathrm e}^{\frac {4 \,{\mathrm e}^{x}+20 x}{5 x}}+3 \,{\mathrm e}^{x} x^{2} {\mathrm e}^{\frac {4 \,{\mathrm e}^{x}+20 x}{5 x}}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((4*x^2+8*x-12)*exp(x)^2+(5*x^3+46*x^2+62*x-48)*exp(x)+60*x^2+120*x)*exp(1/5*(4*exp(x)+20*x)/x),x,meth
od=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.43, size = 41, normalized size = 1.58 \begin {gather*} {\left (4 \, x^{3} e^{4} + 12 \, x^{2} e^{4} + {\left (x^{3} e^{4} + 3 \, x^{2} e^{4}\right )} e^{x}\right )} e^{\left (\frac {4 \, e^{x}}{5 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((4*x^2+8*x-12)*exp(x)^2+(5*x^3+46*x^2+62*x-48)*exp(x)+60*x^2+120*x)*exp(1/5*(4*exp(x)+20*x)/x),
x, algorithm="maxima")

[Out]

(4*x^3*e^4 + 12*x^2*e^4 + (x^3*e^4 + 3*x^2*e^4)*e^x)*e^(4/5*e^x/x)

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mupad [B]  time = 5.69, size = 21, normalized size = 0.81 \begin {gather*} x^2\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^x}{5\,x}+4}\,\left ({\mathrm {e}}^x+4\right )\,\left (x+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((4*x + (4*exp(x))/5)/x)*(120*x + exp(2*x)*(8*x + 4*x^2 - 12) + 60*x^2 + exp(x)*(62*x + 46*x^2 + 5*x^3
- 48)))/5,x)

[Out]

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

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sympy [A]  time = 0.31, size = 37, normalized size = 1.42 \begin {gather*} \left (x^{3} e^{x} + 4 x^{3} + 3 x^{2} e^{x} + 12 x^{2}\right ) e^{\frac {4 x + \frac {4 e^{x}}{5}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((4*x**2+8*x-12)*exp(x)**2+(5*x**3+46*x**2+62*x-48)*exp(x)+60*x**2+120*x)*exp(1/5*(4*exp(x)+20*x
)/x),x)

[Out]

(x**3*exp(x) + 4*x**3 + 3*x**2*exp(x) + 12*x**2)*exp((4*x + 4*exp(x)/5)/x)

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