3.67.50 \(\int \frac {-8 x^4+e^{\frac {2 (-3+2 x)}{x}} (3750+3125 x+600 x^2+100 x^3)}{x^3} \, dx\)

Optimal. Leaf size=28 \[ 5-4 x^2+25 e^{4-\frac {6}{x}} \left (4+\frac {5}{x}\right ) (5+x) \]

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Rubi [A]  time = 0.23, antiderivative size = 43, normalized size of antiderivative = 1.54, number of steps used = 10, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {14, 6742, 2206, 2210, 2212, 2209} \begin {gather*} -4 x^2+100 e^{4-\frac {6}{x}} x+625 e^{4-\frac {6}{x}}+\frac {625 e^{4-\frac {6}{x}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*x^4 + E^((2*(-3 + 2*x))/x)*(3750 + 3125*x + 600*x^2 + 100*x^3))/x^3,x]

[Out]

625*E^(4 - 6/x) + (625*E^(4 - 6/x))/x + 100*E^(4 - 6/x)*x - 4*x^2

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 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
 && ILtQ[n, 0]

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 2210

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

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
 - n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-8 x+\frac {25 e^{4-\frac {6}{x}} \left (150+125 x+24 x^2+4 x^3\right )}{x^3}\right ) \, dx\\ &=-4 x^2+25 \int \frac {e^{4-\frac {6}{x}} \left (150+125 x+24 x^2+4 x^3\right )}{x^3} \, dx\\ &=-4 x^2+25 \int \left (4 e^{4-\frac {6}{x}}+\frac {150 e^{4-\frac {6}{x}}}{x^3}+\frac {125 e^{4-\frac {6}{x}}}{x^2}+\frac {24 e^{4-\frac {6}{x}}}{x}\right ) \, dx\\ &=-4 x^2+100 \int e^{4-\frac {6}{x}} \, dx+600 \int \frac {e^{4-\frac {6}{x}}}{x} \, dx+3125 \int \frac {e^{4-\frac {6}{x}}}{x^2} \, dx+3750 \int \frac {e^{4-\frac {6}{x}}}{x^3} \, dx\\ &=\frac {3125}{6} e^{4-\frac {6}{x}}+\frac {625 e^{4-\frac {6}{x}}}{x}+100 e^{4-\frac {6}{x}} x-4 x^2-600 e^4 \text {Ei}\left (-\frac {6}{x}\right )-600 \int \frac {e^{4-\frac {6}{x}}}{x} \, dx+625 \int \frac {e^{4-\frac {6}{x}}}{x^2} \, dx\\ &=625 e^{4-\frac {6}{x}}+\frac {625 e^{4-\frac {6}{x}}}{x}+100 e^{4-\frac {6}{x}} x-4 x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 35, normalized size = 1.25 \begin {gather*} -4 x^2+25 e^{-6/x} \left (25 e^4+\frac {25 e^4}{x}+4 e^4 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*x^4 + E^((2*(-3 + 2*x))/x)*(3750 + 3125*x + 600*x^2 + 100*x^3))/x^3,x]

[Out]

-4*x^2 + (25*(25*E^4 + (25*E^4)/x + 4*E^4*x))/E^(6/x)

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fricas [A]  time = 0.52, size = 34, normalized size = 1.21 \begin {gather*} -\frac {4 \, x^{3} - 25 \, {\left (4 \, x^{2} + 25 \, x + 25\right )} e^{\left (\frac {2 \, {\left (2 \, x - 3\right )}}{x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x^3+600*x^2+3125*x+3750)*exp((2*x-3)/x)^2-8*x^4)/x^3,x, algorithm="fricas")

[Out]

-(4*x^3 - 25*(4*x^2 + 25*x + 25)*e^(2*(2*x - 3)/x))/x

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giac [B]  time = 0.23, size = 109, normalized size = 3.89 \begin {gather*} -\frac {\frac {625 \, {\left (2 \, x - 3\right )}^{3} e^{\left (\frac {2 \, {\left (2 \, x - 3\right )}}{x}\right )}}{x^{3}} - \frac {5625 \, {\left (2 \, x - 3\right )}^{2} e^{\left (\frac {2 \, {\left (2 \, x - 3\right )}}{x}\right )}}{x^{2}} + \frac {15900 \, {\left (2 \, x - 3\right )} e^{\left (\frac {2 \, {\left (2 \, x - 3\right )}}{x}\right )}}{x} - 14300 \, e^{\left (\frac {2 \, {\left (2 \, x - 3\right )}}{x}\right )} + 108}{3 \, {\left (\frac {{\left (2 \, x - 3\right )}^{2}}{x^{2}} - \frac {4 \, {\left (2 \, x - 3\right )}}{x} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x^3+600*x^2+3125*x+3750)*exp((2*x-3)/x)^2-8*x^4)/x^3,x, algorithm="giac")

[Out]

-1/3*(625*(2*x - 3)^3*e^(2*(2*x - 3)/x)/x^3 - 5625*(2*x - 3)^2*e^(2*(2*x - 3)/x)/x^2 + 15900*(2*x - 3)*e^(2*(2
*x - 3)/x)/x - 14300*e^(2*(2*x - 3)/x) + 108)/((2*x - 3)^2/x^2 - 4*(2*x - 3)/x + 4)

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maple [A]  time = 0.10, size = 33, normalized size = 1.18




method result size



risch \(-4 x^{2}+\frac {25 \left (4 x^{2}+25 x +25\right ) {\mathrm e}^{\frac {4 x -6}{x}}}{x}\) \(33\)
derivativedivides \(\frac {3125 \,{\mathrm e}^{-\frac {6}{x}+4}}{3}-4 x^{2}-\frac {625 \,{\mathrm e}^{-\frac {6}{x}+4} \left (-\frac {3}{x}+2\right )}{3}+100 \,{\mathrm e}^{-\frac {6}{x}+4} x\) \(51\)
default \(\frac {3125 \,{\mathrm e}^{-\frac {6}{x}+4}}{3}-4 x^{2}-\frac {625 \,{\mathrm e}^{-\frac {6}{x}+4} \left (-\frac {3}{x}+2\right )}{3}+100 \,{\mathrm e}^{-\frac {6}{x}+4} x\) \(51\)
norman \(\frac {-4 x^{4}+625 \,{\mathrm e}^{\frac {4 x -6}{x}} x +625 \,{\mathrm e}^{\frac {4 x -6}{x}} x^{2}+100 \,{\mathrm e}^{\frac {4 x -6}{x}} x^{3}}{x^{2}}\) \(60\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((100*x^3+600*x^2+3125*x+3750)*exp((2*x-3)/x)^2-8*x^4)/x^3,x,method=_RETURNVERBOSE)

[Out]

-4*x^2+25*(4*x^2+25*x+25)/x*exp(2*(2*x-3)/x)

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maxima [C]  time = 0.42, size = 48, normalized size = 1.71 \begin {gather*} -4 \, x^{2} - 600 \, {\rm Ei}\left (-\frac {6}{x}\right ) e^{4} + \frac {625}{6} \, e^{4} \Gamma \left (2, \frac {6}{x}\right ) + 600 \, e^{4} \Gamma \left (-1, \frac {6}{x}\right ) + \frac {3125}{6} \, e^{\left (-\frac {6}{x} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x^3+600*x^2+3125*x+3750)*exp((2*x-3)/x)^2-8*x^4)/x^3,x, algorithm="maxima")

[Out]

-4*x^2 - 600*Ei(-6/x)*e^4 + 625/6*e^4*gamma(2, 6/x) + 600*e^4*gamma(-1, 6/x) + 3125/6*e^(-6/x + 4)

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mupad [B]  time = 4.20, size = 40, normalized size = 1.43 \begin {gather*} 625\,{\mathrm {e}}^{4-\frac {6}{x}}+\frac {625\,{\mathrm {e}}^{4-\frac {6}{x}}}{x}+100\,x\,{\mathrm {e}}^{4-\frac {6}{x}}-4\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((2*(2*x - 3))/x)*(3125*x + 600*x^2 + 100*x^3 + 3750) - 8*x^4)/x^3,x)

[Out]

625*exp(4 - 6/x) + (625*exp(4 - 6/x))/x + 100*x*exp(4 - 6/x) - 4*x^2

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sympy [A]  time = 0.12, size = 24, normalized size = 0.86 \begin {gather*} - 4 x^{2} + \frac {\left (100 x^{2} + 625 x + 625\right ) e^{\frac {2 \left (2 x - 3\right )}{x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x**3+600*x**2+3125*x+3750)*exp((2*x-3)/x)**2-8*x**4)/x**3,x)

[Out]

-4*x**2 + (100*x**2 + 625*x + 625)*exp(2*(2*x - 3)/x)/x

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