3.63.40 \(\int \frac {e^2 x^2+e^{25+e^{2 x}+10 x+11 x^2+2 x^3+x^4+e^x (-10-2 x-2 x^2)} (-4 e^{2+2 x} x+e^{2+x} (24 x+12 x^2+4 x^3)+e^2 (2-20 x-44 x^2-12 x^3-8 x^4))}{x^2} \, dx\)

Optimal. Leaf size=28 \[ e^2 \left (-\frac {2 e^{\left (-5+e^x-x-x^2\right )^2}}{x}+x\right ) \]

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Rubi [B]  time = 11.39, antiderivative size = 138, normalized size of antiderivative = 4.93, number of steps used = 3, number of rules used = 2, integrand size = 107, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {14, 2288} \begin {gather*} e^2 x-\frac {2 \left (2 x^4-e^x x^3+3 x^3-3 e^x x^2+11 x^2-6 e^x x+e^{2 x} x+5 x\right ) \exp \left (x^4+2 x^3+11 x^2-2 e^x \left (x^2+x+5\right )+10 x+e^{2 x}+27\right )}{x^2 \left (2 x^3+3 x^2-e^x \left (x^2+x+5\right )+11 x+e^{2 x}-e^x (2 x+1)+5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*x^2 + E^(25 + E^(2*x) + 10*x + 11*x^2 + 2*x^3 + x^4 + E^x*(-10 - 2*x - 2*x^2))*(-4*E^(2 + 2*x)*x + E^
(2 + x)*(24*x + 12*x^2 + 4*x^3) + E^2*(2 - 20*x - 44*x^2 - 12*x^3 - 8*x^4)))/x^2,x]

[Out]

E^2*x - (2*E^(27 + E^(2*x) + 10*x + 11*x^2 + 2*x^3 + x^4 - 2*E^x*(5 + x + x^2))*(5*x - 6*E^x*x + E^(2*x)*x + 1
1*x^2 - 3*E^x*x^2 + 3*x^3 - E^x*x^3 + 2*x^4))/(x^2*(5 + E^(2*x) + 11*x + 3*x^2 + 2*x^3 - E^x*(1 + 2*x) - E^x*(
5 + x + x^2)))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^2-\frac {2 \exp \left (27+e^{2 x}+10 x+11 x^2+2 x^3+x^4-2 e^x \left (5+x+x^2\right )\right ) \left (-1+10 x-12 e^x x+2 e^{2 x} x+22 x^2-6 e^x x^2+6 x^3-2 e^x x^3+4 x^4\right )}{x^2}\right ) \, dx\\ &=e^2 x-2 \int \frac {\exp \left (27+e^{2 x}+10 x+11 x^2+2 x^3+x^4-2 e^x \left (5+x+x^2\right )\right ) \left (-1+10 x-12 e^x x+2 e^{2 x} x+22 x^2-6 e^x x^2+6 x^3-2 e^x x^3+4 x^4\right )}{x^2} \, dx\\ &=e^2 x-\frac {2 \exp \left (27+e^{2 x}+10 x+11 x^2+2 x^3+x^4-2 e^x \left (5+x+x^2\right )\right ) \left (5 x-6 e^x x+e^{2 x} x+11 x^2-3 e^x x^2+3 x^3-e^x x^3+2 x^4\right )}{x^2 \left (5+e^{2 x}+11 x+3 x^2+2 x^3-e^x (1+2 x)-e^x \left (5+x+x^2\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 47, normalized size = 1.68 \begin {gather*} -\frac {2 e^{27+e^{2 x}+10 x+11 x^2+2 x^3+x^4-2 e^x \left (5+x+x^2\right )}}{x}+e^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*x^2 + E^(25 + E^(2*x) + 10*x + 11*x^2 + 2*x^3 + x^4 + E^x*(-10 - 2*x - 2*x^2))*(-4*E^(2 + 2*x)*
x + E^(2 + x)*(24*x + 12*x^2 + 4*x^3) + E^2*(2 - 20*x - 44*x^2 - 12*x^3 - 8*x^4)))/x^2,x]

[Out]

(-2*E^(27 + E^(2*x) + 10*x + 11*x^2 + 2*x^3 + x^4 - 2*E^x*(5 + x + x^2)))/x + E^2*x

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fricas [B]  time = 0.68, size = 59, normalized size = 2.11 \begin {gather*} \frac {x^{2} e^{2} - 2 \, e^{\left ({\left ({\left (x^{4} + 2 \, x^{3} + 11 \, x^{2} + 10 \, x + 25\right )} e^{4} - 2 \, {\left (x^{2} + x + 5\right )} e^{\left (x + 4\right )} + e^{\left (2 \, x + 4\right )}\right )} e^{\left (-4\right )} + 2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(2)*exp(x)^2+(4*x^3+12*x^2+24*x)*exp(2)*exp(x)+(-8*x^4-12*x^3-44*x^2-20*x+2)*exp(2))*exp(e
xp(x)^2+(-2*x^2-2*x-10)*exp(x)+x^4+2*x^3+11*x^2+10*x+25)+x^2*exp(2))/x^2,x, algorithm="fricas")

[Out]

(x^2*e^2 - 2*e^(((x^4 + 2*x^3 + 11*x^2 + 10*x + 25)*e^4 - 2*(x^2 + x + 5)*e^(x + 4) + e^(2*x + 4))*e^(-4) + 2)
)/x

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giac [B]  time = 0.24, size = 52, normalized size = 1.86 \begin {gather*} \frac {x^{2} e^{2} - 2 \, e^{\left (x^{4} + 2 \, x^{3} - 2 \, x^{2} e^{x} + 11 \, x^{2} - 2 \, x e^{x} + 10 \, x + e^{\left (2 \, x\right )} - 10 \, e^{x} + 27\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(2)*exp(x)^2+(4*x^3+12*x^2+24*x)*exp(2)*exp(x)+(-8*x^4-12*x^3-44*x^2-20*x+2)*exp(2))*exp(e
xp(x)^2+(-2*x^2-2*x-10)*exp(x)+x^4+2*x^3+11*x^2+10*x+25)+x^2*exp(2))/x^2,x, algorithm="giac")

[Out]

(x^2*e^2 - 2*e^(x^4 + 2*x^3 - 2*x^2*e^x + 11*x^2 - 2*x*e^x + 10*x + e^(2*x) - 10*e^x + 27))/x

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maple [A]  time = 0.15, size = 50, normalized size = 1.79

method	result	size
risch	\({\mathrm e}^{2} x -\frac {2 \,{\mathrm e}^{27+x^{4}-2 \,{\mathrm e}^{x} x^{2}+2 x^{3}-2 \,{\mathrm e}^{x} x +11 x^{2}-10 \,{\mathrm e}^{x}+{\mathrm e}^{2 x}+10 x}}{x}\)	\(50\)
norman	\(\frac {x^{2} {\mathrm e}^{2}-2 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{2 x}+\left (-2 x^{2}-2 x -10\right ) {\mathrm e}^{x}+x^{4}+2 x^{3}+11 x^{2}+10 x +25}}{x}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*exp(2)*exp(x)^2+(4*x^3+12*x^2+24*x)*exp(2)*exp(x)+(-8*x^4-12*x^3-44*x^2-20*x+2)*exp(2))*exp(exp(x)^
2+(-2*x^2-2*x-10)*exp(x)+x^4+2*x^3+11*x^2+10*x+25)+x^2*exp(2))/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(2)*x-2/x*exp(27+x^4-2*exp(x)*x^2+2*x^3-2*exp(x)*x+11*x^2-10*exp(x)+exp(2*x)+10*x)

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maxima [B]  time = 0.51, size = 49, normalized size = 1.75 \begin {gather*} x e^{2} - \frac {2 \, e^{\left (x^{4} + 2 \, x^{3} - 2 \, x^{2} e^{x} + 11 \, x^{2} - 2 \, x e^{x} + 10 \, x + e^{\left (2 \, x\right )} - 10 \, e^{x} + 27\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(2)*exp(x)^2+(4*x^3+12*x^2+24*x)*exp(2)*exp(x)+(-8*x^4-12*x^3-44*x^2-20*x+2)*exp(2))*exp(e
xp(x)^2+(-2*x^2-2*x-10)*exp(x)+x^4+2*x^3+11*x^2+10*x+25)+x^2*exp(2))/x^2,x, algorithm="maxima")

[Out]

x*e^2 - 2*e^(x^4 + 2*x^3 - 2*x^2*e^x + 11*x^2 - 2*x*e^x + 10*x + e^(2*x) - 10*e^x + 27)/x

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mupad [B]  time = 0.36, size = 56, normalized size = 2.00 \begin {gather*} x\,{\mathrm {e}}^2-\frac {2\,{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{10\,x}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{27}\,{\mathrm {e}}^{-2\,x^2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x^3}\,{\mathrm {e}}^{11\,x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{-10\,{\mathrm {e}}^x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(2) - exp(10*x + exp(2*x) - exp(x)*(2*x + 2*x^2 + 10) + 11*x^2 + 2*x^3 + x^4 + 25)*(exp(2)*(20*x +
44*x^2 + 12*x^3 + 8*x^4 - 2) - exp(2)*exp(x)*(24*x + 12*x^2 + 4*x^3) + 4*x*exp(2*x)*exp(2)))/x^2,x)

[Out]

x*exp(2) - (2*exp(-2*x*exp(x))*exp(10*x)*exp(x^4)*exp(27)*exp(-2*x^2*exp(x))*exp(2*x^3)*exp(11*x^2)*exp(exp(2*
x))*exp(-10*exp(x)))/x

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sympy [B]  time = 0.28, size = 51, normalized size = 1.82 \begin {gather*} x e^{2} - \frac {2 e^{2} e^{x^{4} + 2 x^{3} + 11 x^{2} + 10 x + \left (- 2 x^{2} - 2 x - 10\right ) e^{x} + e^{2 x} + 25}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(2)*exp(x)**2+(4*x**3+12*x**2+24*x)*exp(2)*exp(x)+(-8*x**4-12*x**3-44*x**2-20*x+2)*exp(2))
*exp(exp(x)**2+(-2*x**2-2*x-10)*exp(x)+x**4+2*x**3+11*x**2+10*x+25)+x**2*exp(2))/x**2,x)

[Out]

x*exp(2) - 2*exp(2)*exp(x**4 + 2*x**3 + 11*x**2 + 10*x + (-2*x**2 - 2*x - 10)*exp(x) + exp(2*x) + 25)/x

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