3.44.92 \(\int \frac {x^3-10 e^{\frac {2}{e^3}-5 x^2} x^4+e^x (-6-x+4 x^2)}{x^3} \, dx\)

Optimal. Leaf size=30 \[ -3+e^{\frac {2}{e^3}-5 x^2}+\frac {e^x \left (4+\frac {3}{x}\right )}{x}+x \]

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Rubi [A]  time = 0.14, antiderivative size = 31, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {14, 2209, 2199, 2177, 2178} \begin {gather*} e^{\frac {2}{e^3}-5 x^2}+\frac {3 e^x}{x^2}+x+\frac {4 e^x}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3 - 10*E^(2/E^3 - 5*x^2)*x^4 + E^x*(-6 - x + 4*x^2))/x^3,x]

[Out]

E^(2/E^3 - 5*x^2) + (3*E^x)/x^2 + (4*E^x)/x + 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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

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 \left (-10 e^{\frac {2}{e^3}-5 x^2} x+\frac {-6 e^x-e^x x+4 e^x x^2+x^3}{x^3}\right ) \, dx\\ &=-\left (10 \int e^{\frac {2}{e^3}-5 x^2} x \, dx\right )+\int \frac {-6 e^x-e^x x+4 e^x x^2+x^3}{x^3} \, dx\\ &=e^{\frac {2}{e^3}-5 x^2}+\int \left (1+\frac {e^x \left (-6-x+4 x^2\right )}{x^3}\right ) \, dx\\ &=e^{\frac {2}{e^3}-5 x^2}+x+\int \frac {e^x \left (-6-x+4 x^2\right )}{x^3} \, dx\\ &=e^{\frac {2}{e^3}-5 x^2}+x+\int \left (-\frac {6 e^x}{x^3}-\frac {e^x}{x^2}+\frac {4 e^x}{x}\right ) \, dx\\ &=e^{\frac {2}{e^3}-5 x^2}+x+4 \int \frac {e^x}{x} \, dx-6 \int \frac {e^x}{x^3} \, dx-\int \frac {e^x}{x^2} \, dx\\ &=e^{\frac {2}{e^3}-5 x^2}+\frac {3 e^x}{x^2}+\frac {e^x}{x}+x+4 \text {Ei}(x)-3 \int \frac {e^x}{x^2} \, dx-\int \frac {e^x}{x} \, dx\\ &=e^{\frac {2}{e^3}-5 x^2}+\frac {3 e^x}{x^2}+\frac {4 e^x}{x}+x+3 \text {Ei}(x)-3 \int \frac {e^x}{x} \, dx\\ &=e^{\frac {2}{e^3}-5 x^2}+\frac {3 e^x}{x^2}+\frac {4 e^x}{x}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 31, normalized size = 1.03 \begin {gather*} e^{\frac {2}{e^3}-5 x^2}+\frac {3 e^x}{x^2}+\frac {4 e^x}{x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3 - 10*E^(2/E^3 - 5*x^2)*x^4 + E^x*(-6 - x + 4*x^2))/x^3,x]

[Out]

E^(2/E^3 - 5*x^2) + (3*E^x)/x^2 + (4*E^x)/x + x

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fricas [A]  time = 0.67, size = 32, normalized size = 1.07 \begin {gather*} \frac {x^{3} + x^{2} e^{\left (-5 \, x^{2} + e^{\left (\log \relax (2) - 3\right )}\right )} + {\left (4 \, x + 3\right )} e^{x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x^4*exp(exp(log(2)-3)-5*x^2)+(4*x^2-x-6)*exp(x)+x^3)/x^3,x, algorithm="fricas")

[Out]

(x^3 + x^2*e^(-5*x^2 + e^(log(2) - 3)) + (4*x + 3)*e^x)/x^2

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giac [A]  time = 0.24, size = 32, normalized size = 1.07 \begin {gather*} \frac {x^{3} + x^{2} e^{\left (-5 \, x^{2} + 2 \, e^{\left (-3\right )}\right )} + 4 \, x e^{x} + 3 \, e^{x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x^4*exp(exp(log(2)-3)-5*x^2)+(4*x^2-x-6)*exp(x)+x^3)/x^3,x, algorithm="giac")

[Out]

(x^3 + x^2*e^(-5*x^2 + 2*e^(-3)) + 4*x*e^x + 3*e^x)/x^2

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maple [A]  time = 0.07, size = 25, normalized size = 0.83

method	result	size
risch	\(x +\frac {\left (3+4 x \right ) {\mathrm e}^{x}}{x^{2}}+{\mathrm e}^{2 \,{\mathrm e}^{-3}-5 x^{2}}\)	\(25\)
default	\(x +\frac {3 \,{\mathrm e}^{x}}{x^{2}}+\frac {4 \,{\mathrm e}^{x}}{x}+{\mathrm e}^{2 \,{\mathrm e}^{-3}} {\mathrm e}^{-5 x^{2}}\)	\(29\)
norman	\(\frac {x^{3}+x^{2} {\mathrm e}^{{\mathrm e}^{\ln \relax (2)-3}-5 x^{2}}+4 \,{\mathrm e}^{x} x +3 \,{\mathrm e}^{x}}{x^{2}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-10*x^4*exp(exp(ln(2)-3)-5*x^2)+(4*x^2-x-6)*exp(x)+x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

x+(3+4*x)/x^2*exp(x)+exp(2*exp(-3)-5*x^2)

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maxima [C]  time = 0.42, size = 31, normalized size = 1.03 \begin {gather*} x + 4 \, {\rm Ei}\relax (x) + e^{\left (-5 \, x^{2} + 2 \, e^{\left (-3\right )}\right )} - \Gamma \left (-1, -x\right ) + 6 \, \Gamma \left (-2, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x^4*exp(exp(log(2)-3)-5*x^2)+(4*x^2-x-6)*exp(x)+x^3)/x^3,x, algorithm="maxima")

[Out]

x + 4*Ei(x) + e^(-5*x^2 + 2*e^(-3)) - gamma(-1, -x) + 6*gamma(-2, -x)

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mupad [B]  time = 3.39, size = 28, normalized size = 0.93 \begin {gather*} x+\frac {4\,{\mathrm {e}}^x}{x}+\frac {3\,{\mathrm {e}}^x}{x^2}+{\mathrm {e}}^{2\,{\mathrm {e}}^{-3}}\,{\mathrm {e}}^{-5\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(x - 4*x^2 + 6) + 10*x^4*exp(exp(log(2) - 3) - 5*x^2) - x^3)/x^3,x)

[Out]

x + (4*exp(x))/x + (3*exp(x))/x^2 + exp(2*exp(-3))*exp(-5*x^2)

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sympy [A]  time = 0.16, size = 24, normalized size = 0.80 \begin {gather*} x + e^{- 5 x^{2} + \frac {2}{e^{3}}} + \frac {\left (4 x + 3\right ) e^{x}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*x**4*exp(exp(ln(2)-3)-5*x**2)+(4*x**2-x-6)*exp(x)+x**3)/x**3,x)

[Out]

x + exp(-5*x**2 + 2*exp(-3)) + (4*x + 3)*exp(x)/x**2

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