3.54.58 \(\int \frac {-5+32 e^x x^2}{x^2} \, dx\)

Optimal. Leaf size=14 \[ -\frac {17}{6}+32 e^x+\frac {5}{x} \]

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Rubi [A]  time = 0.01, antiderivative size = 11, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {14, 2194} \begin {gather*} 32 e^x+\frac {5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + 32*E^x*x^2)/x^2,x]

[Out]

32*E^x + 5/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 \left (32 e^x-\frac {5}{x^2}\right ) \, dx\\ &=\frac {5}{x}+32 \int e^x \, dx\\ &=32 e^x+\frac {5}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 0.79 \begin {gather*} 32 e^x+\frac {5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 32*E^x*x^2)/x^2,x]

[Out]

32*E^x + 5/x

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fricas [A]  time = 0.77, size = 15, normalized size = 1.07 \begin {gather*} \frac {x e^{\left (x + 5 \, \log \relax (2)\right )} + 5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*e^(x + 5*log(2)) + 5)/x

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giac [A]  time = 0.14, size = 11, normalized size = 0.79 \begin {gather*} \frac {32 \, x e^{x} + 5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(32*x*e^x + 5)/x

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maple [A]  time = 0.10, size = 11, normalized size = 0.79

method	result	size
risch	\(32 \,{\mathrm e}^{x}+\frac {5}{x}\)	\(11\)
derivativedivides	\({\mathrm e}^{5 \ln \relax (2)+x}+\frac {5}{x}\)	\(14\)
default	\({\mathrm e}^{5 \ln \relax (2)+x}+\frac {5}{x}\)	\(14\)
norman	\(\frac {5+x \,{\mathrm e}^{5 \ln \relax (2)+x}}{x}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32*exp(x)+5/x

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maxima [A]  time = 0.47, size = 10, normalized size = 0.71 \begin {gather*} \frac {5}{x} + 32 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/x + 32*e^x

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mupad [B]  time = 0.05, size = 10, normalized size = 0.71 \begin {gather*} 32\,{\mathrm {e}}^x+\frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32*exp(x) + 5/x

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sympy [A]  time = 0.08, size = 7, normalized size = 0.50 \begin {gather*} 32 e^{x} + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32*exp(x) + 5/x

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