3.102.45 \(\int \frac {x^2+e^{\frac {-25-9 x-e^2 x-e^x x+3 x^2+x \log (x)}{x}} (25+x+3 x^2-e^x x^2)}{x^2} \, dx\)

Optimal. Leaf size=26 \[ x+e^{-9-e^2-e^x-\frac {25}{x}+3 x} x \]

________________________________________________________________________________________

Rubi [B]  time = 0.25, antiderivative size = 57, normalized size of antiderivative = 2.19, number of steps used = 3, number of rules used = 2, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {14, 2288} \begin {gather*} \frac {e^{3 x-e^x-\frac {25}{x}-e^2-9} \left (-e^x x^2+3 x^2+25\right )}{\left (\frac {25}{x^2}-e^x+3\right ) x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + E^((-25 - 9*x - E^2*x - E^x*x + 3*x^2 + x*Log[x])/x)*(25 + x + 3*x^2 - E^x*x^2))/x^2,x]

[Out]

x + (E^(-9 - E^2 - E^x - 25/x + 3*x)*(25 + 3*x^2 - E^x*x^2))/((3 - E^x + 25/x^2)*x)

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 (1+\frac {e^{-e^x-9 \left (1+\frac {e^2}{9}\right )-\frac {25}{x}+3 x} \left (25+x+3 x^2-e^x x^2\right )}{x}\right ) \, dx\\ &=x+\int \frac {e^{-e^x-9 \left (1+\frac {e^2}{9}\right )-\frac {25}{x}+3 x} \left (25+x+3 x^2-e^x x^2\right )}{x} \, dx\\ &=x+\frac {e^{-9-e^2-e^x-\frac {25}{x}+3 x} \left (25+3 x^2-e^x x^2\right )}{\left (3-e^x+\frac {25}{x^2}\right ) x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 26, normalized size = 1.00 \begin {gather*} \left (1+e^{-9-e^2-e^x-\frac {25}{x}+3 x}\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + E^((-25 - 9*x - E^2*x - E^x*x + 3*x^2 + x*Log[x])/x)*(25 + x + 3*x^2 - E^x*x^2))/x^2,x]

[Out]

(1 + E^(-9 - E^2 - E^x - 25/x + 3*x))*x

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 31, normalized size = 1.19 \begin {gather*} x + e^{\left (\frac {3 \, x^{2} - x e^{2} - x e^{x} + x \log \relax (x) - 9 \, x - 25}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+3*x^2+x+25)*exp((x*log(x)-exp(x)*x-exp(2)*x+3*x^2-9*x-25)/x)+x^2)/x^2,x, algorithm="fr
icas")

[Out]

x + e^((3*x^2 - x*e^2 - x*e^x + x*log(x) - 9*x - 25)/x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} - {\left (x^{2} e^{x} - 3 \, x^{2} - x - 25\right )} e^{\left (\frac {3 \, x^{2} - x e^{2} - x e^{x} + x \log \relax (x) - 9 \, x - 25}{x}\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+3*x^2+x+25)*exp((x*log(x)-exp(x)*x-exp(2)*x+3*x^2-9*x-25)/x)+x^2)/x^2,x, algorithm="gi
ac")

[Out]

integrate((x^2 - (x^2*e^x - 3*x^2 - x - 25)*e^((3*x^2 - x*e^2 - x*e^x + x*log(x) - 9*x - 25)/x))/x^2, x)

________________________________________________________________________________________

maple [A]  time = 0.22, size = 29, normalized size = 1.12

method	result	size
risch	\(x +x \,{\mathrm e}^{-\frac {{\mathrm e}^{2} x +{\mathrm e}^{x} x -3 x^{2}+9 x +25}{x}}\)	\(29\)
norman	\(\frac {x^{2}+x \,{\mathrm e}^{\frac {x \ln \relax (x )-{\mathrm e}^{x} x -{\mathrm e}^{2} x +3 x^{2}-9 x -25}{x}}}{x}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)*x^2+3*x^2+x+25)*exp((x*ln(x)-exp(x)*x-exp(2)*x+3*x^2-9*x-25)/x)+x^2)/x^2,x,method=_RETURNVERBOSE
)

[Out]

x+x*exp(-(exp(2)*x+exp(x)*x-3*x^2+9*x+25)/x)

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 23, normalized size = 0.88 \begin {gather*} x e^{\left (3 \, x - \frac {25}{x} - e^{2} - e^{x} - 9\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x^2+3*x^2+x+25)*exp((x*log(x)-exp(x)*x-exp(2)*x+3*x^2-9*x-25)/x)+x^2)/x^2,x, algorithm="ma
xima")

[Out]

x*e^(3*x - 25/x - e^2 - e^x - 9) + x

________________________________________________________________________________________

mupad [B]  time = 7.93, size = 26, normalized size = 1.00 \begin {gather*} x+x\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{-9}\,{\mathrm {e}}^{-\frac {25}{x}}\,{\mathrm {e}}^{-{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(9*x + x*exp(2) + x*exp(x) - x*log(x) - 3*x^2 + 25)/x)*(x - x^2*exp(x) + 3*x^2 + 25) + x^2)/x^2,x)

[Out]

x + x*exp(-exp(2))*exp(3*x)*exp(-9)*exp(-25/x)*exp(-exp(x))

________________________________________________________________________________________

sympy [A]  time = 0.41, size = 29, normalized size = 1.12 \begin {gather*} x + e^{\frac {3 x^{2} - x e^{x} + x \log {\relax (x )} - 9 x - x e^{2} - 25}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x**2+3*x**2+x+25)*exp((x*ln(x)-exp(x)*x-exp(2)*x+3*x**2-9*x-25)/x)+x**2)/x**2,x)

[Out]

x + exp((3*x**2 - x*exp(x) + x*log(x) - 9*x - x*exp(2) - 25)/x)

________________________________________________________________________________________