3.22.61 \(\int \frac {e^x (-3 x^2-5 x^3-2 x^4+e^{10/x} (20-2 x^2)+e^{5/x} (-20 x+4 x^2+4 x^3))}{x^2} \, dx\)

Optimal. Leaf size=26 \[ 1+e^x \left (x-2 \left (1+\left (e^{5/x}-x\right )^2+x\right )\right ) \]

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Rubi [B]  time = 0.61, antiderivative size = 61, normalized size of antiderivative = 2.35, number of steps used = 10, number of rules used = 5, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6742, 2194, 2176, 6706, 2288} \begin {gather*} -2 e^x x^2-\frac {4 e^{x+\frac {5}{x}} \left (5-x^2\right )}{\left (1-\frac {5}{x^2}\right ) x}-e^x x-2 e^x-2 e^{x+\frac {10}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*E^x - 2*E^(10/x + x) - E^x*x - 2*E^x*x^2 - (4*E^(5/x + x)*(5 - x^2))/((1 - 5/x^2)*x)

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

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]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 32, normalized size = 1.23 \begin {gather*} -e^x \left (2+2 e^{10/x}+x-4 e^{5/x} x+2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 29, normalized size = 1.12 \begin {gather*} -{\left (2 \, x^{2} - 4 \, x e^{\frac {5}{x}} + x + 2 \, e^{\frac {10}{x}} + 2\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+20)*exp(5/x)^2+(4*x^3+4*x^2-20*x)*exp(5/x)-2*x^4-5*x^3-3*x^2)*exp(x)/x^2,x, algorithm="fric
as")

[Out]

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

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giac [A]  time = 0.44, size = 42, normalized size = 1.62 \begin {gather*} -2 \, x^{2} e^{x} - x e^{x} + 4 \, x e^{\left (\frac {x^{2} + 5}{x}\right )} - 2 \, e^{x} - 2 \, e^{\left (\frac {x^{2} + 10}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 40, normalized size = 1.54

method	result	size
risch	\(\left (-2 x^{2}-x -2\right ) {\mathrm e}^{x}-2 \,{\mathrm e}^{\frac {x^{2}+10}{x}}+4 x \,{\mathrm e}^{\frac {x^{2}+5}{x}}\)	\(40\)
norman	\(\frac {-2 \,{\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x^{3}-2 \,{\mathrm e}^{x} x \,{\mathrm e}^{\frac {10}{x}}+4 \,{\mathrm e}^{x} x^{2} {\mathrm e}^{\frac {5}{x}}}{x}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*x^2-x-2)*exp(x)-2*exp((x^2+10)/x)+4*x*exp((x^2+5)/x)

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maxima [A]  time = 0.49, size = 45, normalized size = 1.73 \begin {gather*} 4 \, x e^{\left (x + \frac {5}{x}\right )} - 2 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} - 5 \, {\left (x - 1\right )} e^{x} - 2 \, e^{\left (x + \frac {10}{x}\right )} - 3 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2+20)*exp(5/x)^2+(4*x^3+4*x^2-20*x)*exp(5/x)-2*x^4-5*x^3-3*x^2)*exp(x)/x^2,x, algorithm="maxi
ma")

[Out]

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

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mupad [B]  time = 1.32, size = 29, normalized size = 1.12 \begin {gather*} -{\mathrm {e}}^x\,\left (x+2\,{\mathrm {e}}^{10/x}-4\,x\,{\mathrm {e}}^{5/x}+2\,x^2+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(x)*(x + 2*exp(10/x) - 4*x*exp(5/x) + 2*x^2 + 2)

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sympy [A]  time = 5.61, size = 26, normalized size = 1.00 \begin {gather*} \left (- 2 x^{2} + 4 x e^{\frac {5}{x}} - x - 2 e^{\frac {10}{x}} - 2\right ) e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2+20)*exp(5/x)**2+(4*x**3+4*x**2-20*x)*exp(5/x)-2*x**4-5*x**3-3*x**2)*exp(x)/x**2,x)

[Out]

(-2*x**2 + 4*x*exp(5/x) - x - 2*exp(10/x) - 2)*exp(x)

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