3.90.58 \(\int \frac {e^{e^5-x} (e^3 (-x^2+x^3)+e^{-e^5+x} (3+9 x^4))}{3 x^2} \, dx\)

Optimal. Leaf size=27 \[ 3+\frac {-1+x}{x}-\frac {1}{3} e^{3+e^5-x} x+x^3 \]

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Rubi [A]  time = 0.21, antiderivative size = 42, normalized size of antiderivative = 1.56, number of steps used = 7, number of rules used = 5, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {12, 6688, 2176, 2194, 14} \begin {gather*} x^3-\frac {1}{3} e^{-x+e^5+3}+\frac {1}{3} e^{-x+e^5+3} (1-x)-\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^5 - x)*(E^3*(-x^2 + x^3) + E^(-E^5 + x)*(3 + 9*x^4)))/(3*x^2),x]

[Out]

-1/3*E^(3 + E^5 - x) + (E^(3 + E^5 - x)*(1 - x))/3 - x^(-1) + x^3

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {e^{e^5-x} \left (e^3 \left (-x^2+x^3\right )+e^{-e^5+x} \left (3+9 x^4\right )\right )}{x^2} \, dx\\ &=\frac {1}{3} \int \left (e^{3+e^5-x} (-1+x)+\frac {3+9 x^4}{x^2}\right ) \, dx\\ &=\frac {1}{3} \int e^{3+e^5-x} (-1+x) \, dx+\frac {1}{3} \int \frac {3+9 x^4}{x^2} \, dx\\ &=\frac {1}{3} e^{3+e^5-x} (1-x)+\frac {1}{3} \int e^{3+e^5-x} \, dx+\frac {1}{3} \int \left (\frac {3}{x^2}+9 x^2\right ) \, dx\\ &=-\frac {1}{3} e^{3+e^5-x}+\frac {1}{3} e^{3+e^5-x} (1-x)-\frac {1}{x}+x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 28, normalized size = 1.04 \begin {gather*} \frac {1}{3} \left (-\frac {3}{x}-e^{3+e^5-x} x+3 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^5 - x)*(E^3*(-x^2 + x^3) + E^(-E^5 + x)*(3 + 9*x^4)))/(3*x^2),x]

[Out]

(-3/x - E^(3 + E^5 - x)*x + 3*x^3)/3

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fricas [A]  time = 0.71, size = 25, normalized size = 0.93 \begin {gather*} \frac {3 \, x^{4} - x^{2} e^{\left (-x + e^{5} + 3\right )} - 3}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x^4+3)*exp(-exp(5)+x)+(x^3-x^2)*exp(3))/x^2/exp(-exp(5)+x),x, algorithm="fricas")

[Out]

1/3*(3*x^4 - x^2*e^(-x + e^5 + 3) - 3)/x

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giac [A]  time = 0.18, size = 25, normalized size = 0.93 \begin {gather*} \frac {3 \, x^{4} - x^{2} e^{\left (-x + e^{5} + 3\right )} - 3}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x^4+3)*exp(-exp(5)+x)+(x^3-x^2)*exp(3))/x^2/exp(-exp(5)+x),x, algorithm="giac")

[Out]

1/3*(3*x^4 - x^2*e^(-x + e^5 + 3) - 3)/x

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maple [A]  time = 0.15, size = 21, normalized size = 0.78

method	result	size
risch	\(x^{3}-\frac {1}{x}-\frac {x \,{\mathrm e}^{3+{\mathrm e}^{5}-x}}{3}\)	\(21\)
norman	\(\frac {\left ({\mathrm e}^{-{\mathrm e}^{5}+x} x^{4}-\frac {x^{2} {\mathrm e}^{3}}{3}-{\mathrm e}^{-{\mathrm e}^{5}+x}\right ) {\mathrm e}^{{\mathrm e}^{5}-x}}{x}\)	\(42\)
derivativedivides	\(\frac {{\mathrm e}^{3} \left ({\mathrm e}^{{\mathrm e}^{5}-x} \left (3 \,{\mathrm e}^{5}-x -1\right )+\frac {{\mathrm e}^{{\mathrm e}^{5}-x} {\mathrm e}^{15}}{x}-\left ({\mathrm e}^{15}+3 \,{\mathrm e}^{10}\right ) {\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )}{3}+\frac {{\mathrm e}^{15} {\mathrm e}^{3} \left (-\frac {{\mathrm e}^{{\mathrm e}^{5}-x}}{x}+{\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )}{3}-\frac {1}{x}+\left (-{\mathrm e}^{5}+x \right )^{3}+3 \,{\mathrm e}^{5} \left (-{\mathrm e}^{5}+x \right )^{2}+3 \,{\mathrm e}^{10} \left (-{\mathrm e}^{5}+x \right )-36 \,{\mathrm e}^{15} \ln \relax (x )-\frac {36 \,{\mathrm e}^{15} {\mathrm e}^{5}}{x}-\frac {18 \,{\mathrm e}^{20}}{x}+36 \,{\mathrm e}^{5} {\mathrm e}^{10} \ln \relax (x )+\frac {54 \left ({\mathrm e}^{10}\right )^{2}}{x}-\frac {{\mathrm e}^{3} \left (-{\mathrm e}^{{\mathrm e}^{5}-x}-\frac {{\mathrm e}^{{\mathrm e}^{5}-x} {\mathrm e}^{10}}{x}-\left (-{\mathrm e}^{10}-2 \,{\mathrm e}^{5}\right ) {\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )}{3}-\frac {{\mathrm e}^{10} {\mathrm e}^{3} \left (-\frac {{\mathrm e}^{{\mathrm e}^{5}-x}}{x}+{\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )}{3}-\frac {2 \,{\mathrm e}^{3} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}-x} {\mathrm e}^{5}}{x}-\left ({\mathrm e}^{5}+1\right ) {\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )}{3}+{\mathrm e}^{3} {\mathrm e}^{5} \left (-{\mathrm e}^{{\mathrm e}^{5}-x}-\frac {{\mathrm e}^{{\mathrm e}^{5}-x} {\mathrm e}^{10}}{x}-\left (-{\mathrm e}^{10}-2 \,{\mathrm e}^{5}\right ) {\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )+{\mathrm e}^{10} {\mathrm e}^{3} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}-x} {\mathrm e}^{5}}{x}-\left ({\mathrm e}^{5}+1\right ) {\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )\)	\(365\)
default	\(\frac {{\mathrm e}^{3} \left ({\mathrm e}^{{\mathrm e}^{5}-x} \left (3 \,{\mathrm e}^{5}-x -1\right )+\frac {{\mathrm e}^{{\mathrm e}^{5}-x} {\mathrm e}^{15}}{x}-\left ({\mathrm e}^{15}+3 \,{\mathrm e}^{10}\right ) {\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )}{3}+\frac {{\mathrm e}^{15} {\mathrm e}^{3} \left (-\frac {{\mathrm e}^{{\mathrm e}^{5}-x}}{x}+{\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )}{3}-\frac {1}{x}+\left (-{\mathrm e}^{5}+x \right )^{3}+3 \,{\mathrm e}^{5} \left (-{\mathrm e}^{5}+x \right )^{2}+3 \,{\mathrm e}^{10} \left (-{\mathrm e}^{5}+x \right )-36 \,{\mathrm e}^{15} \ln \relax (x )-\frac {36 \,{\mathrm e}^{15} {\mathrm e}^{5}}{x}-\frac {18 \,{\mathrm e}^{20}}{x}+36 \,{\mathrm e}^{5} {\mathrm e}^{10} \ln \relax (x )+\frac {54 \left ({\mathrm e}^{10}\right )^{2}}{x}-\frac {{\mathrm e}^{3} \left (-{\mathrm e}^{{\mathrm e}^{5}-x}-\frac {{\mathrm e}^{{\mathrm e}^{5}-x} {\mathrm e}^{10}}{x}-\left (-{\mathrm e}^{10}-2 \,{\mathrm e}^{5}\right ) {\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )}{3}-\frac {{\mathrm e}^{10} {\mathrm e}^{3} \left (-\frac {{\mathrm e}^{{\mathrm e}^{5}-x}}{x}+{\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )}{3}-\frac {2 \,{\mathrm e}^{3} {\mathrm e}^{5} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}-x} {\mathrm e}^{5}}{x}-\left ({\mathrm e}^{5}+1\right ) {\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )}{3}+{\mathrm e}^{3} {\mathrm e}^{5} \left (-{\mathrm e}^{{\mathrm e}^{5}-x}-\frac {{\mathrm e}^{{\mathrm e}^{5}-x} {\mathrm e}^{10}}{x}-\left (-{\mathrm e}^{10}-2 \,{\mathrm e}^{5}\right ) {\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )+{\mathrm e}^{10} {\mathrm e}^{3} \left (\frac {{\mathrm e}^{{\mathrm e}^{5}-x} {\mathrm e}^{5}}{x}-\left ({\mathrm e}^{5}+1\right ) {\mathrm e}^{{\mathrm e}^{5}} \expIntegralEi \left (1, x\right )\right )\)	\(365\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((9*x^4+3)*exp(-exp(5)+x)+(x^3-x^2)*exp(3))/x^2/exp(-exp(5)+x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.36, size = 38, normalized size = 1.41 \begin {gather*} x^{3} - \frac {1}{3} \, {\left (x e^{\left (e^{5} + 3\right )} + e^{\left (e^{5} + 3\right )}\right )} e^{\left (-x\right )} - \frac {1}{x} + \frac {1}{3} \, e^{\left (-x + e^{5} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x^4+3)*exp(-exp(5)+x)+(x^3-x^2)*exp(3))/x^2/exp(-exp(5)+x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.43, size = 21, normalized size = 0.78 \begin {gather*} x^3-\frac {1}{x}-\frac {x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3\,{\mathrm {e}}^{{\mathrm {e}}^5}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(5) - x)*((exp(3)*(x^2 - x^3))/3 - (exp(x - exp(5))*(9*x^4 + 3))/3))/x^2,x)

[Out]

x^3 - 1/x - (x*exp(-x)*exp(3)*exp(exp(5)))/3

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sympy [A]  time = 0.14, size = 19, normalized size = 0.70 \begin {gather*} x^{3} - \frac {x e^{3} e^{- x + e^{5}}}{3} - \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x**4+3)*exp(-exp(5)+x)+(x**3-x**2)*exp(3))/x**2/exp(-exp(5)+x),x)

[Out]

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

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