3.92.86 \(\int \frac {10+5 e^{4-x} x^2}{4 x^2+e^2 x^2} \, dx\)

Optimal. Leaf size=27 \[ 4-\frac {5 \left (7+e^{4-x}+\frac {2}{x}+\log (4)\right )}{4+e^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6, 12, 14, 2194} \begin {gather*} -\frac {5 e^{4-x}}{4+e^2}-\frac {10}{\left (4+e^2\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 25, normalized size = 0.93 \begin {gather*} -\frac {10+5 e^{4-x} x}{4 x+e^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((10 + 5*E^(4 - x)*x)/(4*x + E^2*x))

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 22, normalized size = 0.81 \begin {gather*} -\frac {5 \, {\left (x e^{\left (-x + 4\right )} + 2\right )}}{x e^{2} + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*exp(-x+4)+10)/(x^2*exp(2)+4*x^2),x, algorithm="fricas")

[Out]

-5*(x*e^(-x + 4) + 2)/(x*e^2 + 4*x)

________________________________________________________________________________________

giac [A]  time = 0.12, size = 22, normalized size = 0.81 \begin {gather*} -\frac {5 \, {\left (x e^{\left (-x + 4\right )} + 2\right )}}{x e^{2} + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*exp(-x+4)+10)/(x^2*exp(2)+4*x^2),x, algorithm="giac")

[Out]

-5*(x*e^(-x + 4) + 2)/(x*e^2 + 4*x)

________________________________________________________________________________________

maple [A]  time = 0.54, size = 27, normalized size = 1.00

method	result	size
derivativedivides	\(-\frac {5 \,{\mathrm e}^{-x +4}}{4+{\mathrm e}^{2}}-\frac {10}{x \left (4+{\mathrm e}^{2}\right )}\)	\(27\)
default	\(-\frac {5 \,{\mathrm e}^{-x +4}}{4+{\mathrm e}^{2}}-\frac {10}{x \left (4+{\mathrm e}^{2}\right )}\)	\(27\)
risch	\(-\frac {5 \,{\mathrm e}^{-x +4}}{4+{\mathrm e}^{2}}-\frac {10}{x \left (4+{\mathrm e}^{2}\right )}\)	\(27\)
norman	\(\frac {-\frac {10}{4+{\mathrm e}^{2}}-\frac {5 x \,{\mathrm e}^{-x +4}}{4+{\mathrm e}^{2}}}{x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2*exp(-x+4)+10)/(x^2*exp(2)+4*x^2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 26, normalized size = 0.96 \begin {gather*} -\frac {5 \, e^{\left (-x + 4\right )}}{e^{2} + 4} - \frac {10}{x {\left (e^{2} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2*exp(-x+4)+10)/(x^2*exp(2)+4*x^2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.13, size = 22, normalized size = 0.81 \begin {gather*} -\frac {5\,x\,{\mathrm {e}}^{4-x}+10}{x\,\left ({\mathrm {e}}^2+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.11, size = 20, normalized size = 0.74 \begin {gather*} - \frac {5 e^{4 - x}}{4 + e^{2}} - \frac {10}{x \left (4 + e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________