3.55.17 \(\int \frac {-50 e^2-5 x+e^x x^3}{x^3} \, dx\)

Optimal. Leaf size=25 \[ -e+e^x+\frac {5+25 e^2 \left (\frac {1}{x}-x\right )}{x} \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {14, 2194, 37} \begin {gather*} \frac {\left (x+10 e^2\right )^2}{4 e^2 x^2}+e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50*E^2 - 5*x + E^x*x^3)/x^3,x]

[Out]

E^x + (10*E^2 + x)^2/(4*E^2*x^2)

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 (e^x-\frac {5 \left (10 e^2+x\right )}{x^3}\right ) \, dx\\ &=-\left (5 \int \frac {10 e^2+x}{x^3} \, dx\right )+\int e^x \, dx\\ &=e^x+\frac {\left (10 e^2+x\right )^2}{4 e^2 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.68 \begin {gather*} e^x+\frac {25 e^2}{x^2}+\frac {5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50*E^2 - 5*x + E^x*x^3)/x^3,x]

[Out]

E^x + (25*E^2)/x^2 + 5/x

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fricas [A]  time = 0.53, size = 18, normalized size = 0.72 \begin {gather*} \frac {x^{2} e^{x} + 5 \, x + 25 \, e^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x^3-50*exp(2)-5*x)/x^3,x, algorithm="fricas")

[Out]

(x^2*e^x + 5*x + 25*e^2)/x^2

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giac [A]  time = 0.21, size = 18, normalized size = 0.72 \begin {gather*} \frac {x^{2} e^{x} + 5 \, x + 25 \, e^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x^3-50*exp(2)-5*x)/x^3,x, algorithm="giac")

[Out]

(x^2*e^x + 5*x + 25*e^2)/x^2

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maple [A]  time = 0.03, size = 16, normalized size = 0.64

method	result	size
default	\(\frac {5}{x}+\frac {25 \,{\mathrm e}^{2}}{x^{2}}+{\mathrm e}^{x}\)	\(16\)
risch	\(\frac {25 \,{\mathrm e}^{2}+5 x}{x^{2}}+{\mathrm e}^{x}\)	\(16\)
norman	\(\frac {{\mathrm e}^{x} x^{2}+5 x +25 \,{\mathrm e}^{2}}{x^{2}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*x^3-50*exp(2)-5*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

5/x+25*exp(2)/x^2+exp(x)

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maxima [A]  time = 0.36, size = 15, normalized size = 0.60 \begin {gather*} \frac {5}{x} + \frac {25 \, e^{2}}{x^{2}} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x^3-50*exp(2)-5*x)/x^3,x, algorithm="maxima")

[Out]

5/x + 25*e^2/x^2 + e^x

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mupad [B]  time = 3.49, size = 15, normalized size = 0.60 \begin {gather*} {\mathrm {e}}^x+\frac {5\,x+25\,{\mathrm {e}}^2}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + 50*exp(2) - x^3*exp(x))/x^3,x)

[Out]

exp(x) + (5*x + 25*exp(2))/x^2

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sympy [A]  time = 0.11, size = 15, normalized size = 0.60 \begin {gather*} e^{x} - \frac {- 5 x - 25 e^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x**3-50*exp(2)-5*x)/x**3,x)

[Out]

exp(x) - (-5*x - 25*exp(2))/x**2

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