3.64.75 \(\int \frac {-36 e^{\frac {1}{x}}+352 x^2+8 x^3+e^{.\frac {1}{2}/x} (528+12 x-24 x^2)}{45 x^2} \, dx\)

Optimal. Leaf size=26 \[ \frac {4}{5} \left (16-e^{\left .\frac {1}{2}\right /x}+\frac {1}{3} (-4+x)\right )^2 \]

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Rubi [A]  time = 0.07, antiderivative size = 35, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {12, 14, 2209, 2288} \begin {gather*} \frac {4}{45} (x+44)^2-\frac {8}{15} e^{\left .\frac {1}{2}\right /x} (x+44)+\frac {4 e^{\frac {1}{x}}}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-36*E^x^(-1) + 352*x^2 + 8*x^3 + E^(1/(2*x))*(528 + 12*x - 24*x^2))/(45*x^2),x]

[Out]

(4*E^x^(-1))/5 - (8*E^(1/(2*x))*(44 + x))/15 + (4*(44 + x)^2)/45

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

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} &=\frac {1}{45} \int \frac {-36 e^{\frac {1}{x}}+352 x^2+8 x^3+e^{\left .\frac {1}{2}\right /x} \left (528+12 x-24 x^2\right )}{x^2} \, dx\\ &=\frac {1}{45} \int \left (-\frac {36 e^{\frac {1}{x}}}{x^2}+8 (44+x)-\frac {12 e^{\left .\frac {1}{2}\right /x} \left (-44-x+2 x^2\right )}{x^2}\right ) \, dx\\ &=\frac {4}{45} (44+x)^2-\frac {4}{15} \int \frac {e^{\left .\frac {1}{2}\right /x} \left (-44-x+2 x^2\right )}{x^2} \, dx-\frac {4}{5} \int \frac {e^{\frac {1}{x}}}{x^2} \, dx\\ &=\frac {4 e^{\frac {1}{x}}}{5}-\frac {8}{15} e^{\left .\frac {1}{2}\right /x} (44+x)+\frac {4}{45} (44+x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 0.77 \begin {gather*} \frac {4}{45} \left (44-3 e^{\left .\frac {1}{2}\right /x}+x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-36*E^x^(-1) + 352*x^2 + 8*x^3 + E^(1/(2*x))*(528 + 12*x - 24*x^2))/(45*x^2),x]

[Out]

(4*(44 - 3*E^(1/(2*x)) + x)^2)/45

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fricas [A]  time = 0.51, size = 26, normalized size = 1.00 \begin {gather*} \frac {4}{45} \, x^{2} - \frac {8}{15} \, {\left (x + 44\right )} e^{\left (\frac {1}{2 \, x}\right )} + \frac {352}{45} \, x + \frac {4}{5} \, e^{\frac {1}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/45*(-36*exp(1/2/x)^2+(-24*x^2+12*x+528)*exp(1/2/x)+8*x^3+352*x^2)/x^2,x, algorithm="fricas")

[Out]

4/45*x^2 - 8/15*(x + 44)*e^(1/2/x) + 352/45*x + 4/5*e^(1/x)

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giac [B]  time = 0.20, size = 32, normalized size = 1.23 \begin {gather*} \frac {4}{45} \, x^{2} - \frac {8}{15} \, x e^{\left (\frac {1}{2 \, x}\right )} + \frac {352}{45} \, x - \frac {352}{15} \, e^{\left (\frac {1}{2 \, x}\right )} + \frac {4}{5} \, e^{\frac {1}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/45*(-36*exp(1/2/x)^2+(-24*x^2+12*x+528)*exp(1/2/x)+8*x^3+352*x^2)/x^2,x, algorithm="giac")

[Out]

4/45*x^2 - 8/15*x*e^(1/2/x) + 352/45*x - 352/15*e^(1/2/x) + 4/5*e^(1/x)

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maple [A]  time = 0.25, size = 29, normalized size = 1.12

method	result	size
risch	\(\frac {4 x^{2}}{45}+\frac {4 \,{\mathrm e}^{\frac {1}{x}}}{5}+\frac {352 x}{45}+\frac {\left (-1056-24 x \right ) {\mathrm e}^{\frac {1}{2 x}}}{45}\)	\(29\)
derivativedivides	\(\frac {4 x^{2}}{45}+\frac {352 x}{45}+\frac {4 \,{\mathrm e}^{\frac {1}{x}}}{5}-\frac {8 \,{\mathrm e}^{\frac {1}{2 x}} x}{15}-\frac {352 \,{\mathrm e}^{\frac {1}{2 x}}}{15}\)	\(37\)
default	\(\frac {4 x^{2}}{45}+\frac {352 x}{45}+\frac {4 \,{\mathrm e}^{\frac {1}{x}}}{5}-\frac {8 \,{\mathrm e}^{\frac {1}{2 x}} x}{15}-\frac {352 \,{\mathrm e}^{\frac {1}{2 x}}}{15}\)	\(37\)
norman	\(\frac {\frac {352 x^{2}}{45}+\frac {4 x^{3}}{45}-\frac {352 \,{\mathrm e}^{\frac {1}{2 x}} x}{15}-\frac {8 \,{\mathrm e}^{\frac {1}{2 x}} x^{2}}{15}+\frac {4 x \,{\mathrm e}^{\frac {1}{x}}}{5}}{x}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/45*(-36*exp(1/2/x)^2+(-24*x^2+12*x+528)*exp(1/2/x)+8*x^3+352*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

4/45*x^2+4/5*exp(1/x)+352/45*x+1/45*(-1056-24*x)*exp(1/2/x)

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maxima [C]  time = 0.39, size = 40, normalized size = 1.54 \begin {gather*} \frac {4}{45} \, x^{2} + \frac {352}{45} \, x - \frac {4}{15} \, {\rm Ei}\left (\frac {1}{2 \, x}\right ) - \frac {352}{15} \, e^{\left (\frac {1}{2 \, x}\right )} + \frac {4}{5} \, e^{\frac {1}{x}} + \frac {4}{15} \, \Gamma \left (-1, -\frac {1}{2 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/45*(-36*exp(1/2/x)^2+(-24*x^2+12*x+528)*exp(1/2/x)+8*x^3+352*x^2)/x^2,x, algorithm="maxima")

[Out]

4/45*x^2 + 352/45*x - 4/15*Ei(1/2/x) - 352/15*e^(1/2/x) + 4/5*e^(1/x) + 4/15*gamma(-1, -1/2/x)

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mupad [B]  time = 4.16, size = 33, normalized size = 1.27 \begin {gather*} \frac {4\,{\mathrm {e}}^{1/x}}{5}-\frac {352\,{\mathrm {e}}^{\frac {1}{2\,x}}}{15}-x\,\left (\frac {8\,{\mathrm {e}}^{\frac {1}{2\,x}}}{15}-\frac {352}{45}\right )+\frac {4\,x^2}{45} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(1/(2*x))*(12*x - 24*x^2 + 528))/45 - (4*exp(1/x))/5 + (352*x^2)/45 + (8*x^3)/45)/x^2,x)

[Out]

(4*exp(1/x))/5 - (352*exp(1/(2*x)))/15 - x*((8*exp(1/(2*x)))/15 - 352/45) + (4*x^2)/45

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sympy [A]  time = 0.16, size = 34, normalized size = 1.31 \begin {gather*} \frac {4 x^{2}}{45} + \frac {352 x}{45} + \frac {\left (- 40 x - 1760\right ) e^{\frac {1}{2 x}}}{75} + \frac {4 e^{\frac {1}{x}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/45*(-36*exp(1/2/x)**2+(-24*x**2+12*x+528)*exp(1/2/x)+8*x**3+352*x**2)/x**2,x)

[Out]

4*x**2/45 + 352*x/45 + (-40*x - 1760)*exp(1/(2*x))/75 + 4*exp(1/x)/5

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