3.83.24 \(\int \frac {e^{5/x} (-75+5 e^3)+30 x^3-2 e^3 x^3}{5 x^2} \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{5} \left (15-e^3\right ) x \left (\frac {e^{5/x}}{x}+x\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6, 12, 14, 2209} \begin {gather*} \frac {1}{5} \left (15-e^3\right ) x^2+\frac {1}{5} \left (15-e^3\right ) e^{5/x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5/x)*(-75 + 5*E^3) + 30*x^3 - 2*E^3*x^3)/(5*x^2),x]

[Out]

(E^(5/x)*(15 - E^3))/5 + ((15 - E^3)*x^2)/5

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 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} &=\int \frac {e^{5/x} \left (-75+5 e^3\right )+\left (30-2 e^3\right ) x^3}{5 x^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{5/x} \left (-75+5 e^3\right )+\left (30-2 e^3\right ) x^3}{x^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {5 e^{5/x} \left (-15+e^3\right )}{x^2}-2 \left (-15+e^3\right ) x\right ) \, dx\\ &=\frac {1}{5} \left (15-e^3\right ) x^2+\left (-15+e^3\right ) \int \frac {e^{5/x}}{x^2} \, dx\\ &=\frac {1}{5} e^{5/x} \left (15-e^3\right )+\frac {1}{5} \left (15-e^3\right ) x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 0.96 \begin {gather*} \frac {1}{5} \left (-15+e^3\right ) \left (-e^{5/x}-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5/x)*(-75 + 5*E^3) + 30*x^3 - 2*E^3*x^3)/(5*x^2),x]

[Out]

((-15 + E^3)*(-E^(5/x) - x^2))/5

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fricas [A]  time = 0.73, size = 25, normalized size = 1.00 \begin {gather*} -\frac {1}{5} \, x^{2} e^{3} + 3 \, x^{2} - \frac {1}{5} \, {\left (e^{3} - 15\right )} e^{\frac {5}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((5*exp(3)-75)*exp(5/x)-2*x^3*exp(3)+30*x^3)/x^2,x, algorithm="fricas")

[Out]

-1/5*x^2*e^3 + 3*x^2 - 1/5*(e^3 - 15)*e^(5/x)

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giac [A]  time = 0.22, size = 35, normalized size = 1.40 \begin {gather*} \frac {1}{5} \, x^{2} {\left (\frac {15 \, e^{\frac {5}{x}}}{x^{2}} - \frac {e^{\left (\frac {5}{x} + 3\right )}}{x^{2}} - e^{3} + 15\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((5*exp(3)-75)*exp(5/x)-2*x^3*exp(3)+30*x^3)/x^2,x, algorithm="giac")

[Out]

1/5*x^2*(15*e^(5/x)/x^2 - e^(5/x + 3)/x^2 - e^3 + 15)

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maple [A]  time = 0.06, size = 28, normalized size = 1.12

method	result	size
derivativedivides	\(3 x^{2}-\frac {x^{2} {\mathrm e}^{3}}{5}+\frac {\left (15-{\mathrm e}^{3}\right ) {\mathrm e}^{\frac {5}{x}}}{5}\)	\(28\)
default	\(3 x^{2}-\frac {x^{2} {\mathrm e}^{3}}{5}+\frac {\left (75-5 \,{\mathrm e}^{3}\right ) {\mathrm e}^{\frac {5}{x}}}{25}\)	\(28\)
norman	\(\frac {\left (3-\frac {{\mathrm e}^{3}}{5}\right ) x^{3}+\left (3-\frac {{\mathrm e}^{3}}{5}\right ) x \,{\mathrm e}^{\frac {5}{x}}}{x}\)	\(30\)
risch	\(3 x^{2}-\frac {x^{2} {\mathrm e}^{3}}{5}+3 \,{\mathrm e}^{\frac {5}{x}}-\frac {{\mathrm e}^{\frac {5}{x}} {\mathrm e}^{3}}{5}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((5*exp(3)-75)*exp(5/x)-2*x^3*exp(3)+30*x^3)/x^2,x,method=_RETURNVERBOSE)

[Out]

3*x^2-1/5*x^2*exp(3)+1/5*(15-exp(3))*exp(5/x)

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maxima [A]  time = 0.37, size = 31, normalized size = 1.24 \begin {gather*} -\frac {1}{5} \, x^{2} e^{3} + 3 \, x^{2} + 3 \, e^{\frac {5}{x}} - \frac {1}{5} \, e^{\left (\frac {5}{x} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((5*exp(3)-75)*exp(5/x)-2*x^3*exp(3)+30*x^3)/x^2,x, algorithm="maxima")

[Out]

-1/5*x^2*e^3 + 3*x^2 + 3*e^(5/x) - 1/5*e^(5/x + 3)

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mupad [B]  time = 6.10, size = 18, normalized size = 0.72 \begin {gather*} -\left (\frac {{\mathrm {e}}^3}{5}-3\right )\,\left ({\mathrm {e}}^{5/x}+x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(5/x)*(5*exp(3) - 75))/5 - (2*x^3*exp(3))/5 + 6*x^3)/x^2,x)

[Out]

-(exp(3)/5 - 3)*(exp(5/x) + x^2)

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sympy [A]  time = 0.13, size = 20, normalized size = 0.80 \begin {gather*} x^{2} \left (3 - \frac {e^{3}}{5}\right ) + \frac {\left (15 - e^{3}\right ) e^{\frac {5}{x}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((5*exp(3)-75)*exp(5/x)-2*x**3*exp(3)+30*x**3)/x**2,x)

[Out]

x**2*(3 - exp(3)/5) + (15 - exp(3))*exp(5/x)/5

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