3.68.95 \(\int \frac {-90+49 x-4 x^3}{9 e x^3} \, dx\)

Optimal. Leaf size=31 \[ 1+\frac {-\frac {1}{9} (7-2 x)^2+\frac {5}{x}+\frac {x}{3}}{e x} \]

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 14} \begin {gather*} \frac {5}{e x^2}-\frac {4 x}{9 e}-\frac {49}{9 e x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-90 + 49*x - 4*x^3)/(9*E*x^3),x]

[Out]

5/(E*x^2) - 49/(9*E*x) - (4*x)/(9*E)

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-90+49 x-4 x^3}{x^3} \, dx}{9 e}\\ &=\frac {\int \left (-4-\frac {90}{x^3}+\frac {49}{x^2}\right ) \, dx}{9 e}\\ &=\frac {5}{e x^2}-\frac {49}{9 e x}-\frac {4 x}{9 e}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 0.68 \begin {gather*} -\frac {-\frac {45}{x^2}+\frac {49}{x}+4 x}{9 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-90 + 49*x - 4*x^3)/(9*E*x^3),x]

[Out]

-1/9*(-45/x^2 + 49/x + 4*x)/E

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fricas [A]  time = 0.53, size = 17, normalized size = 0.55 \begin {gather*} -\frac {{\left (4 \, x^{3} + 49 \, x - 45\right )} e^{\left (-1\right )}}{9 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-4*x^3+49*x-90)/x^3/exp(1),x, algorithm="fricas")

[Out]

-1/9*(4*x^3 + 49*x - 45)*e^(-1)/x^2

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giac [A]  time = 0.17, size = 17, normalized size = 0.55 \begin {gather*} -\frac {1}{9} \, {\left (4 \, x + \frac {49 \, x - 45}{x^{2}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-4*x^3+49*x-90)/x^3/exp(1),x, algorithm="giac")

[Out]

-1/9*(4*x + (49*x - 45)/x^2)*e^(-1)

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maple [A]  time = 0.03, size = 19, normalized size = 0.61

method	result	size
risch	\(-\frac {4 \,{\mathrm e}^{-1} x}{9}+\frac {{\mathrm e}^{-1} \left (-49 x +45\right )}{9 x^{2}}\)	\(19\)
gosper	\(-\frac {\left (4 x^{3}+49 x -45\right ) {\mathrm e}^{-1}}{9 x^{2}}\)	\(20\)
default	\(\frac {{\mathrm e}^{-1} \left (-4 x +\frac {45}{x^{2}}-\frac {49}{x}\right )}{9}\)	\(21\)
norman	\(\frac {5 \,{\mathrm e}^{-1}-\frac {49 \,{\mathrm e}^{-1} x}{9}-\frac {4 \,{\mathrm e}^{-1} x^{3}}{9}}{x^{2}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(-4*x^3+49*x-90)/x^3/exp(1),x,method=_RETURNVERBOSE)

[Out]

-4/9*exp(-1)*x+1/9*exp(-1)*(-49*x+45)/x^2

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maxima [A]  time = 0.36, size = 17, normalized size = 0.55 \begin {gather*} -\frac {1}{9} \, {\left (4 \, x + \frac {49 \, x - 45}{x^{2}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-4*x^3+49*x-90)/x^3/exp(1),x, algorithm="maxima")

[Out]

-1/9*(4*x + (49*x - 45)/x^2)*e^(-1)

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mupad [B]  time = 4.11, size = 17, normalized size = 0.55 \begin {gather*} -\frac {{\mathrm {e}}^{-1}\,\left (4\,x^3+49\,x-45\right )}{9\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-1)*((4*x^3)/9 - (49*x)/9 + 10))/x^3,x)

[Out]

-(exp(-1)*(49*x + 4*x^3 - 45))/(9*x^2)

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sympy [A]  time = 0.07, size = 17, normalized size = 0.55 \begin {gather*} \frac {- 4 x - \frac {49 x - 45}{x^{2}}}{9 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-4*x**3+49*x-90)/x**3/exp(1),x)

[Out]

(-4*x - (49*x - 45)/x**2)*exp(-1)/9

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