3.75.4 \(\int \frac {-1-5 e^x x^2-10 x^3}{5 x^2} \, dx\)

Optimal. Leaf size=19 \[ 3-e^x+\frac {1}{5 x}-x^2 \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {12, 14, 2194} \begin {gather*} -x^2-e^x+\frac {1}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 5*E^x*x^2 - 10*x^3)/(5*x^2),x]

[Out]

-E^x + 1/(5*x) - x^2

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

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.95 \begin {gather*} -e^x+\frac {1}{5 x}-x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 5*E^x*x^2 - 10*x^3)/(5*x^2),x]

[Out]

-E^x + 1/(5*x) - x^2

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fricas [A]  time = 0.54, size = 17, normalized size = 0.89 \begin {gather*} -\frac {5 \, x^{3} + 5 \, x e^{x} - 1}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-5*exp(x)*x^2-10*x^3-1)/x^2,x, algorithm="fricas")

[Out]

-1/5*(5*x^3 + 5*x*e^x - 1)/x

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giac [A]  time = 0.17, size = 17, normalized size = 0.89 \begin {gather*} -\frac {5 \, x^{3} + 5 \, x e^{x} - 1}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-5*exp(x)*x^2-10*x^3-1)/x^2,x, algorithm="giac")

[Out]

-1/5*(5*x^3 + 5*x*e^x - 1)/x

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maple [A]  time = 0.02, size = 16, normalized size = 0.84

method	result	size
default	\(-x^{2}+\frac {1}{5 x}-{\mathrm e}^{x}\)	\(16\)
risch	\(-x^{2}+\frac {1}{5 x}-{\mathrm e}^{x}\)	\(16\)
norman	\(\frac {\frac {1}{5}-x^{3}-{\mathrm e}^{x} x}{x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-5*exp(x)*x^2-10*x^3-1)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.36, size = 15, normalized size = 0.79 \begin {gather*} -x^{2} + \frac {1}{5 \, x} - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-5*exp(x)*x^2-10*x^3-1)/x^2,x, algorithm="maxima")

[Out]

-x^2 + 1/5/x - e^x

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mupad [B]  time = 0.06, size = 15, normalized size = 0.79 \begin {gather*} \frac {1}{5\,x}-{\mathrm {e}}^x-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(5*x) - exp(x) - x^2

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sympy [A]  time = 0.12, size = 10, normalized size = 0.53 \begin {gather*} - x^{2} - e^{x} + \frac {1}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-5*exp(x)*x**2-10*x**3-1)/x**2,x)

[Out]

-x**2 - exp(x) + 1/(5*x)

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