3.54.11 \(\int \frac {-12 x+60 e^x x^3+e^{e^3 x} (2-e^3 x)}{60 x^3} \, dx\)

Optimal. Leaf size=26 \[ e^x+\frac {-\frac {1}{6} e^{e^3 x}+2 x}{10 x^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 25, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {12, 14, 2197, 2194} \begin {gather*} -\frac {e^{e^3 x}}{60 x^2}+e^x+\frac {1}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-12*x + 60*E^x*x^3 + E^(E^3*x)*(2 - E^3*x))/(60*x^3),x]

[Out]

E^x - E^(E^3*x)/(60*x^2) + 1/(5*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]

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{60} \int \frac {-12 x+60 e^x x^3+e^{e^3 x} \left (2-e^3 x\right )}{x^3} \, dx\\ &=\frac {1}{60} \int \left (-\frac {e^{e^3 x} \left (-2+e^3 x\right )}{x^3}+\frac {12 \left (-1+5 e^x x^2\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{60} \int \frac {e^{e^3 x} \left (-2+e^3 x\right )}{x^3} \, dx\right )+\frac {1}{5} \int \frac {-1+5 e^x x^2}{x^2} \, dx\\ &=-\frac {e^{e^3 x}}{60 x^2}+\frac {1}{5} \int \left (5 e^x-\frac {1}{x^2}\right ) \, dx\\ &=-\frac {e^{e^3 x}}{60 x^2}+\frac {1}{5 x}+\int e^x \, dx\\ &=e^x-\frac {e^{e^3 x}}{60 x^2}+\frac {1}{5 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 0.96 \begin {gather*} e^x-\frac {e^{e^3 x}}{60 x^2}+\frac {1}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12*x + 60*E^x*x^3 + E^(E^3*x)*(2 - E^3*x))/(60*x^3),x]

[Out]

E^x - E^(E^3*x)/(60*x^2) + 1/(5*x)

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fricas [A]  time = 0.87, size = 23, normalized size = 0.88 \begin {gather*} \frac {60 \, x^{2} e^{x} + 12 \, x - e^{\left (x e^{3}\right )}}{60 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/60*((-x*exp(3)+2)*exp(x*exp(3))+60*exp(x)*x^3-12*x)/x^3,x, algorithm="fricas")

[Out]

1/60*(60*x^2*e^x + 12*x - e^(x*e^3))/x^2

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giac [A]  time = 0.13, size = 23, normalized size = 0.88 \begin {gather*} \frac {60 \, x^{2} e^{x} + 12 \, x - e^{\left (x e^{3}\right )}}{60 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/60*((-x*exp(3)+2)*exp(x*exp(3))+60*exp(x)*x^3-12*x)/x^3,x, algorithm="giac")

[Out]

1/60*(60*x^2*e^x + 12*x - e^(x*e^3))/x^2

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maple [A]  time = 0.14, size = 19, normalized size = 0.73

method	result	size
risch	\(\frac {1}{5 x}+{\mathrm e}^{x}-\frac {{\mathrm e}^{x \,{\mathrm e}^{3}}}{60 x^{2}}\)	\(19\)
norman	\(\frac {{\mathrm e}^{x} x^{2}+\frac {x}{5}-\frac {{\mathrm e}^{x \,{\mathrm e}^{3}}}{60}}{x^{2}}\)	\(22\)
default	\(\frac {1}{5 x}+\frac {{\mathrm e}^{6} \left (-\frac {{\mathrm e}^{x \,{\mathrm e}^{3}} {\mathrm e}^{-6}}{2 x^{2}}-\frac {{\mathrm e}^{x \,{\mathrm e}^{3}} {\mathrm e}^{-3}}{2 x}-\frac {\expIntegralEi \left (1, -x \,{\mathrm e}^{3}\right )}{2}\right )}{30}-\frac {{\mathrm e}^{6} \left (-\frac {{\mathrm e}^{x \,{\mathrm e}^{3}} {\mathrm e}^{-3}}{x}-\expIntegralEi \left (1, -x \,{\mathrm e}^{3}\right )\right )}{60}+{\mathrm e}^{x}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/60*((-x*exp(3)+2)*exp(x*exp(3))+60*exp(x)*x^3-12*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

1/5/x+exp(x)-1/60/x^2*exp(x*exp(3))

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maxima [C]  time = 0.51, size = 30, normalized size = 1.15 \begin {gather*} -\frac {1}{60} \, e^{6} \Gamma \left (-1, -x e^{3}\right ) - \frac {1}{30} \, e^{6} \Gamma \left (-2, -x e^{3}\right ) + \frac {1}{5 \, x} + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/60*((-x*exp(3)+2)*exp(x*exp(3))+60*exp(x)*x^3-12*x)/x^3,x, algorithm="maxima")

[Out]

-1/60*e^6*gamma(-1, -x*e^3) - 1/30*e^6*gamma(-2, -x*e^3) + 1/5/x + e^x

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mupad [B]  time = 3.49, size = 18, normalized size = 0.69 \begin {gather*} {\mathrm {e}}^x+\frac {\frac {x}{5}-\frac {{\mathrm {e}}^{x\,{\mathrm {e}}^3}}{60}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x/5 - x^3*exp(x) + (exp(x*exp(3))*(x*exp(3) - 2))/60)/x^3,x)

[Out]

exp(x) + (x/5 - exp(x*exp(3))/60)/x^2

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sympy [A]  time = 0.70, size = 19, normalized size = 0.73 \begin {gather*} e^{x} + \frac {1}{5 x} - \frac {e^{x e^{3}}}{60 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/60*((-x*exp(3)+2)*exp(x*exp(3))+60*exp(x)*x**3-12*x)/x**3,x)

[Out]

exp(x) + 1/(5*x) - exp(x*exp(3))/(60*x**2)

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