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

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

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Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.36, number of steps used = 6, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 14, 2194} \begin {gather*} -\frac {1-e^5}{5 x^2}+\frac {x}{5}+\frac {e^x}{5}+\frac {2}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x/5 - (1 - E^5)/(5*x^2) + 2/(5*x) + x/5

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 26, normalized size = 1.04 \begin {gather*} \frac {1}{5} \left (e^x-\frac {1}{x^2}+\frac {e^5}{x^2}+\frac {2}{x}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^x - x^(-2) + E^5/x^2 + 2/x + x)/5

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fricas [A]  time = 0.71, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{3} + x^{2} e^{x} + 2 \, x + e^{5} - 1}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^{3} + x^{2} e^{x} + 2 \, x + e^{5} - 1}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 21, normalized size = 0.84

method	result	size
risch	\(\frac {x}{5}+\frac {2 x +{\mathrm e}^{5}-1}{5 x^{2}}+\frac {{\mathrm e}^{x}}{5}\)	\(21\)
default	\(\frac {x}{5}-\frac {1}{5 x^{2}}+\frac {2}{5 x}+\frac {{\mathrm e}^{5}}{5 x^{2}}+\frac {{\mathrm e}^{x}}{5}\)	\(26\)
norman	\(\frac {\frac {2 x}{5}+\frac {x^{3}}{5}+\frac {{\mathrm e}^{x} x^{2}}{5}+\frac {{\mathrm e}^{5}}{5}-\frac {1}{5}}{x^{2}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 25, normalized size = 1.00 \begin {gather*} \frac {1}{5} \, x + \frac {2}{5 \, x} + \frac {e^{5}}{5 \, x^{2}} - \frac {1}{5 \, x^{2}} + \frac {1}{5} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x + 2/5/x + 1/5*e^5/x^2 - 1/5/x^2 + 1/5*e^x

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mupad [B]  time = 6.59, size = 21, normalized size = 0.84 \begin {gather*} \frac {x}{5}+\frac {{\mathrm {e}}^x}{5}+\frac {\frac {2\,x}{5}+\frac {{\mathrm {e}}^5}{5}-\frac {1}{5}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.16, size = 20, normalized size = 0.80 \begin {gather*} \frac {x}{5} + \frac {e^{x}}{5} + \frac {2 x - 1 + e^{5}}{5 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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