3.49.5 \(\int \frac {e^{\frac {1}{x}} (112-112 x)-336 x^3}{x} \, dx\)

Optimal. Leaf size=21 \[ 112 x \left (-e^{\frac {1}{x}}+\frac {25}{x}-x^2\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 14, normalized size of antiderivative = 0.67, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {14, 2288} \begin {gather*} -112 x^3-112 e^{\frac {1}{x}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x^(-1)*(112 - 112*x) - 336*x^3)/x,x]

[Out]

-112*E^x^(-1)*x - 112*x^3

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {112 e^{\frac {1}{x}} (-1+x)}{x}-336 x^2\right ) \, dx\\ &=-112 x^3-112 \int \frac {e^{\frac {1}{x}} (-1+x)}{x} \, dx\\ &=-112 e^{\frac {1}{x}} x-112 x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.62 \begin {gather*} -112 \left (e^{\frac {1}{x}} x+x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x^(-1)*(112 - 112*x) - 336*x^3)/x,x]

[Out]

-112*(E^x^(-1)*x + x^3)

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fricas [A]  time = 0.88, size = 13, normalized size = 0.62 \begin {gather*} -112 \, x^{3} - 112 \, x e^{\frac {1}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-112*x+112)*exp(1/x)-336*x^3)/x,x, algorithm="fricas")

[Out]

-112*x^3 - 112*x*e^(1/x)

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giac [A]  time = 0.13, size = 15, normalized size = 0.71 \begin {gather*} -112 \, x^{3} {\left (\frac {e^{\frac {1}{x}}}{x^{2}} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-112*x+112)*exp(1/x)-336*x^3)/x,x, algorithm="giac")

[Out]

-112*x^3*(e^(1/x)/x^2 + 1)

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maple [A]  time = 0.09, size = 14, normalized size = 0.67

method	result	size
derivativedivides	\(-112 x^{3}-112 x \,{\mathrm e}^{\frac {1}{x}}\)	\(14\)
default	\(-112 x^{3}-112 x \,{\mathrm e}^{\frac {1}{x}}\)	\(14\)
norman	\(-112 x^{3}-112 x \,{\mathrm e}^{\frac {1}{x}}\)	\(14\)
risch	\(-112 x^{3}-112 x \,{\mathrm e}^{\frac {1}{x}}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-112*x+112)*exp(1/x)-336*x^3)/x,x,method=_RETURNVERBOSE)

[Out]

-112*x^3-112*x*exp(1/x)

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maxima [C]  time = 0.37, size = 21, normalized size = 1.00 \begin {gather*} -112 \, x^{3} - 112 \, {\rm Ei}\left (\frac {1}{x}\right ) + 112 \, \Gamma \left (-1, -\frac {1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-112*x+112)*exp(1/x)-336*x^3)/x,x, algorithm="maxima")

[Out]

-112*x^3 - 112*Ei(1/x) + 112*gamma(-1, -1/x)

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mupad [B]  time = 3.25, size = 11, normalized size = 0.52 \begin {gather*} -112\,x\,\left ({\mathrm {e}}^{1/x}+x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1/x)*(112*x - 112) + 336*x^3)/x,x)

[Out]

-112*x*(exp(1/x) + x^2)

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sympy [A]  time = 0.11, size = 14, normalized size = 0.67 \begin {gather*} - 112 x^{3} - 112 x e^{\frac {1}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-112*x+112)*exp(1/x)-336*x**3)/x,x)

[Out]

-112*x**3 - 112*x*exp(1/x)

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