3.66.57 \(\int \frac {-112-28 x+e^5 x^3}{e^5 x^3} \, dx\)

Optimal. Leaf size=18 \[ x+\frac {28 \left (2+x-x^2\right )}{e^5 x^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 14} \begin {gather*} \frac {56}{e^5 x^2}+x+\frac {28}{e^5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-112 - 28*x + E^5*x^3)/(E^5*x^3),x]

[Out]

56/(E^5*x^2) + 28/(E^5*x) + x

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 {-112-28 x+e^5 x^3}{x^3} \, dx}{e^5}\\ &=\frac {\int \left (e^5-\frac {112}{x^3}-\frac {28}{x^2}\right ) \, dx}{e^5}\\ &=\frac {56}{e^5 x^2}+\frac {28}{e^5 x}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.11 \begin {gather*} \frac {\frac {56}{x^2}+\frac {28}{x}+e^5 x}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-112 - 28*x + E^5*x^3)/(E^5*x^3),x]

[Out]

(56/x^2 + 28/x + E^5*x)/E^5

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fricas [A]  time = 0.54, size = 17, normalized size = 0.94 \begin {gather*} \frac {{\left (x^{3} e^{5} + 28 \, x + 56\right )} e^{\left (-5\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3*exp(5)-28*x-112)/x^3/exp(5),x, algorithm="fricas")

[Out]

(x^3*e^5 + 28*x + 56)*e^(-5)/x^2

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giac [A]  time = 0.19, size = 16, normalized size = 0.89 \begin {gather*} {\left (x e^{5} + \frac {28 \, {\left (x + 2\right )}}{x^{2}}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3*exp(5)-28*x-112)/x^3/exp(5),x, algorithm="giac")

[Out]

(x*e^5 + 28*(x + 2)/x^2)*e^(-5)

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maple [A]  time = 0.06, size = 14, normalized size = 0.78

method	result	size
risch	\(x +\frac {{\mathrm e}^{-5} \left (28 x +56\right )}{x^{2}}\)	\(14\)
gosper	\(\frac {\left (x^{3} {\mathrm e}^{5}+56+28 x \right ) {\mathrm e}^{-5}}{x^{2}}\)	\(20\)
default	\({\mathrm e}^{-5} \left (x \,{\mathrm e}^{5}+\frac {28}{x}+\frac {56}{x^{2}}\right )\)	\(21\)
norman	\(\frac {x^{3}+56 \,{\mathrm e}^{-5}+28 x \,{\mathrm e}^{-5}}{x^{2}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*exp(5)-28*x-112)/x^3/exp(5),x,method=_RETURNVERBOSE)

[Out]

x+exp(-5)*(28*x+56)/x^2

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maxima [A]  time = 0.40, size = 16, normalized size = 0.89 \begin {gather*} {\left (x e^{5} + \frac {28 \, {\left (x + 2\right )}}{x^{2}}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3*exp(5)-28*x-112)/x^3/exp(5),x, algorithm="maxima")

[Out]

(x*e^5 + 28*(x + 2)/x^2)*e^(-5)

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mupad [B]  time = 0.04, size = 13, normalized size = 0.72 \begin {gather*} x+\frac {{\mathrm {e}}^{-5}\,\left (28\,x+56\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-5)*(28*x - x^3*exp(5) + 112))/x^3,x)

[Out]

x + (exp(-5)*(28*x + 56))/x^2

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sympy [A]  time = 0.08, size = 15, normalized size = 0.83 \begin {gather*} \frac {x e^{5} + \frac {28 x + 56}{x^{2}}}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3*exp(5)-28*x-112)/x**3/exp(5),x)

[Out]

(x*exp(5) + (28*x + 56)/x**2)*exp(-5)

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