3.20.35 \(\int \frac {-1+48 x^2+16 e^x x^2}{48 x^2} \, dx\)

Optimal. Leaf size=17 \[ \frac {1}{3} \left (e^x+\frac {1}{16 x}\right )+x \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {12, 14, 2194} \begin {gather*} x+\frac {e^x}{3}+\frac {1}{48 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 48*x^2 + 16*E^x*x^2)/(48*x^2),x]

[Out]

E^x/3 + 1/(48*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 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} &=\frac {1}{48} \int \frac {-1+48 x^2+16 e^x x^2}{x^2} \, dx\\ &=\frac {1}{48} \int \left (16 e^x+\frac {-1+48 x^2}{x^2}\right ) \, dx\\ &=\frac {1}{48} \int \frac {-1+48 x^2}{x^2} \, dx+\frac {\int e^x \, dx}{3}\\ &=\frac {e^x}{3}+\frac {1}{48} \int \left (48-\frac {1}{x^2}\right ) \, dx\\ &=\frac {e^x}{3}+\frac {1}{48 x}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.94 \begin {gather*} \frac {e^x}{3}+\frac {1}{48 x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 48*x^2 + 16*E^x*x^2)/(48*x^2),x]

[Out]

E^x/3 + 1/(48*x) + x

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fricas [A]  time = 0.79, size = 17, normalized size = 1.00 \begin {gather*} \frac {48 \, x^{2} + 16 \, x e^{x} + 1}{48 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/48*(16*exp(x)*x^2+48*x^2-1)/x^2,x, algorithm="fricas")

[Out]

1/48*(48*x^2 + 16*x*e^x + 1)/x

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giac [A]  time = 0.93, size = 17, normalized size = 1.00 \begin {gather*} \frac {48 \, x^{2} + 16 \, x e^{x} + 1}{48 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/48*(16*exp(x)*x^2+48*x^2-1)/x^2,x, algorithm="giac")

[Out]

1/48*(48*x^2 + 16*x*e^x + 1)/x

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maple [A]  time = 0.02, size = 12, normalized size = 0.71

method	result	size
default	\(\frac {1}{48 x}+\frac {{\mathrm e}^{x}}{3}+x\)	\(12\)
risch	\(\frac {1}{48 x}+\frac {{\mathrm e}^{x}}{3}+x\)	\(12\)
norman	\(\frac {\frac {1}{48}+x^{2}+\frac {{\mathrm e}^{x} x}{3}}{x}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/48*(16*exp(x)*x^2+48*x^2-1)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/48/x+1/3*exp(x)+x

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maxima [A]  time = 0.56, size = 11, normalized size = 0.65 \begin {gather*} x + \frac {1}{48 \, x} + \frac {1}{3} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/48*(16*exp(x)*x^2+48*x^2-1)/x^2,x, algorithm="maxima")

[Out]

x + 1/48/x + 1/3*e^x

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mupad [B]  time = 1.12, size = 11, normalized size = 0.65 \begin {gather*} x+\frac {{\mathrm {e}}^x}{3}+\frac {1}{48\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2*exp(x))/3 + x^2 - 1/48)/x^2,x)

[Out]

x + exp(x)/3 + 1/(48*x)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.59 \begin {gather*} x + \frac {e^{x}}{3} + \frac {1}{48 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/48*(16*exp(x)*x**2+48*x**2-1)/x**2,x)

[Out]

x + exp(x)/3 + 1/(48*x)

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