3.87.28 \(\int \frac {5 e^{.\frac {5}{4}/x}+4 x^2}{4 x^2} \, dx\)

Optimal. Leaf size=21 \[ 4+\frac {e^{25}}{3}-e^{\left .\frac {5}{4}\right /x}+x \]

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Rubi [A]  time = 0.02, antiderivative size = 13, normalized size of antiderivative = 0.62, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14, 2209} \begin {gather*} x-e^{\left .\frac {5}{4}\right /x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^(5/(4*x)) + 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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.62 \begin {gather*} -e^{\left .\frac {5}{4}\right /x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^(5/(4*x)) + x

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fricas [A]  time = 0.66, size = 10, normalized size = 0.48 \begin {gather*} x - e^{\left (\frac {5}{4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(5*exp(5/4/x)+4*x^2)/x^2,x, algorithm="fricas")

[Out]

x - e^(5/4/x)

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giac [A]  time = 0.19, size = 15, normalized size = 0.71 \begin {gather*} -x {\left (\frac {e^{\left (\frac {5}{4 \, x}\right )}}{x} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(5*exp(5/4/x)+4*x^2)/x^2,x, algorithm="giac")

[Out]

-x*(e^(5/4/x)/x - 1)

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maple [A]  time = 0.04, size = 11, normalized size = 0.52

method	result	size
derivativedivides	\(x -{\mathrm e}^{\frac {5}{4 x}}\)	\(11\)
default	\(x -{\mathrm e}^{\frac {5}{4 x}}\)	\(11\)
risch	\(x -{\mathrm e}^{\frac {5}{4 x}}\)	\(11\)
norman	\(\frac {x^{2}-x \,{\mathrm e}^{\frac {5}{4 x}}}{x}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(5*exp(5/4/x)+4*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

x-exp(5/4/x)

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maxima [A]  time = 0.36, size = 10, normalized size = 0.48 \begin {gather*} x - e^{\left (\frac {5}{4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(5*exp(5/4/x)+4*x^2)/x^2,x, algorithm="maxima")

[Out]

x - e^(5/4/x)

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mupad [B]  time = 5.19, size = 10, normalized size = 0.48 \begin {gather*} x-{\mathrm {e}}^{\frac {5}{4\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*exp(5/(4*x)))/4 + x^2)/x^2,x)

[Out]

x - exp(5/(4*x))

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sympy [A]  time = 0.09, size = 7, normalized size = 0.33 \begin {gather*} x - e^{\frac {5}{4 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(5*exp(5/4/x)+4*x**2)/x**2,x)

[Out]

x - exp(5/(4*x))

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