3.53.74 \(\int \frac {-10 x^2-2 x^3+e^{4/x} (20+23 x+4 x^2)}{10 x^5} \, dx\)

Optimal. Leaf size=27 \[ 1-\frac {\left (e^{4/x}-x\right ) \left (2+\frac {4 x}{5}\right )}{4 x^3} \]

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Rubi [A]  time = 0.22, antiderivative size = 41, normalized size of antiderivative = 1.52, number of steps used = 15, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 14, 37, 6742, 2212, 2209} \begin {gather*} -\frac {e^{4/x}}{2 x^3}+\frac {(x+5)^2}{50 x^2}-\frac {e^{4/x}}{5 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*x^2 - 2*x^3 + E^(4/x)*(20 + 23*x + 4*x^2))/(10*x^5),x]

[Out]

-1/2*E^(4/x)/x^3 - E^(4/x)/(5*x^2) + (5 + x)^2/(50*x^2)

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 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} &=\frac {1}{10} \int \frac {-10 x^2-2 x^3+e^{4/x} \left (20+23 x+4 x^2\right )}{x^5} \, dx\\ &=\frac {1}{10} \int \left (-\frac {2 (5+x)}{x^3}+\frac {e^{4/x} \left (20+23 x+4 x^2\right )}{x^5}\right ) \, dx\\ &=\frac {1}{10} \int \frac {e^{4/x} \left (20+23 x+4 x^2\right )}{x^5} \, dx-\frac {1}{5} \int \frac {5+x}{x^3} \, dx\\ &=\frac {(5+x)^2}{50 x^2}+\frac {1}{10} \int \left (\frac {20 e^{4/x}}{x^5}+\frac {23 e^{4/x}}{x^4}+\frac {4 e^{4/x}}{x^3}\right ) \, dx\\ &=\frac {(5+x)^2}{50 x^2}+\frac {2}{5} \int \frac {e^{4/x}}{x^3} \, dx+2 \int \frac {e^{4/x}}{x^5} \, dx+\frac {23}{10} \int \frac {e^{4/x}}{x^4} \, dx\\ &=-\frac {e^{4/x}}{2 x^3}-\frac {23 e^{4/x}}{40 x^2}-\frac {e^{4/x}}{10 x}+\frac {(5+x)^2}{50 x^2}-\frac {1}{10} \int \frac {e^{4/x}}{x^2} \, dx-\frac {23}{20} \int \frac {e^{4/x}}{x^3} \, dx-\frac {3}{2} \int \frac {e^{4/x}}{x^4} \, dx\\ &=\frac {e^{4/x}}{40}-\frac {e^{4/x}}{2 x^3}-\frac {e^{4/x}}{5 x^2}+\frac {3 e^{4/x}}{16 x}+\frac {(5+x)^2}{50 x^2}+\frac {23}{80} \int \frac {e^{4/x}}{x^2} \, dx+\frac {3}{4} \int \frac {e^{4/x}}{x^3} \, dx\\ &=-\frac {3 e^{4/x}}{64}-\frac {e^{4/x}}{2 x^3}-\frac {e^{4/x}}{5 x^2}+\frac {(5+x)^2}{50 x^2}-\frac {3}{16} \int \frac {e^{4/x}}{x^2} \, dx\\ &=-\frac {e^{4/x}}{2 x^3}-\frac {e^{4/x}}{5 x^2}+\frac {(5+x)^2}{50 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 23, normalized size = 0.85 \begin {gather*} \frac {\left (-e^{4/x}+x\right ) (5+2 x)}{10 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x^2 - 2*x^3 + E^(4/x)*(20 + 23*x + 4*x^2))/(10*x^5),x]

[Out]

((-E^(4/x) + x)*(5 + 2*x))/(10*x^3)

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fricas [A]  time = 0.72, size = 27, normalized size = 1.00 \begin {gather*} \frac {2 \, x^{2} - {\left (2 \, x + 5\right )} e^{\frac {4}{x}} + 5 \, x}{10 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((4*x^2+23*x+20)*exp(2/x)^2-2*x^3-10*x^2)/x^5,x, algorithm="fricas")

[Out]

1/10*(2*x^2 - (2*x + 5)*e^(4/x) + 5*x)/x^3

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giac [A]  time = 3.56, size = 33, normalized size = 1.22 \begin {gather*} \frac {1}{5 \, x} - \frac {e^{\frac {4}{x}}}{5 \, x^{2}} + \frac {1}{2 \, x^{2}} - \frac {e^{\frac {4}{x}}}{2 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((4*x^2+23*x+20)*exp(2/x)^2-2*x^3-10*x^2)/x^5,x, algorithm="giac")

[Out]

1/5/x - 1/5*e^(4/x)/x^2 + 1/2/x^2 - 1/2*e^(4/x)/x^3

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maple [A]  time = 0.09, size = 28, normalized size = 1.04

method	result	size
risch	\(\frac {5+2 x}{10 x^{2}}-\frac {\left (5+2 x \right ) {\mathrm e}^{\frac {4}{x}}}{10 x^{3}}\)	\(28\)
derivativedivides	\(\frac {1}{2 x^{2}}+\frac {1}{5 x}-\frac {{\mathrm e}^{\frac {4}{x}}}{5 x^{2}}-\frac {{\mathrm e}^{\frac {4}{x}}}{2 x^{3}}\)	\(38\)
default	\(\frac {1}{2 x^{2}}+\frac {1}{5 x}-\frac {{\mathrm e}^{\frac {4}{x}}}{5 x^{2}}-\frac {{\mathrm e}^{\frac {4}{x}}}{2 x^{3}}\)	\(38\)
norman	\(\frac {\frac {x^{2}}{2}+\frac {x^{3}}{5}-\frac {x^{2} {\mathrm e}^{\frac {4}{x}}}{5}-\frac {x \,{\mathrm e}^{\frac {4}{x}}}{2}}{x^{4}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*((4*x^2+23*x+20)*exp(2/x)^2-2*x^3-10*x^2)/x^5,x,method=_RETURNVERBOSE)

[Out]

1/10*(5+2*x)/x^2-1/10*(5+2*x)/x^3*exp(4/x)

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maxima [C]  time = 0.38, size = 38, normalized size = 1.41 \begin {gather*} \frac {1}{5 \, x} + \frac {1}{2 \, x^{2}} + \frac {1}{128} \, \Gamma \left (4, -\frac {4}{x}\right ) - \frac {23}{640} \, \Gamma \left (3, -\frac {4}{x}\right ) + \frac {1}{40} \, \Gamma \left (2, -\frac {4}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((4*x^2+23*x+20)*exp(2/x)^2-2*x^3-10*x^2)/x^5,x, algorithm="maxima")

[Out]

1/5/x + 1/2/x^2 + 1/128*gamma(4, -4/x) - 23/640*gamma(3, -4/x) + 1/40*gamma(2, -4/x)

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mupad [B]  time = 3.42, size = 36, normalized size = 1.33 \begin {gather*} -\frac {x^3\,\left (\frac {{\mathrm {e}}^{4/x}}{5}-\frac {1}{2}\right )+\frac {x^2\,{\mathrm {e}}^{4/x}}{2}-\frac {x^4}{5}}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2 - (exp(4/x)*(23*x + 4*x^2 + 20))/10 + x^3/5)/x^5,x)

[Out]

-(x^3*(exp(4/x)/5 - 1/2) + (x^2*exp(4/x))/2 - x^4/5)/x^5

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sympy [A]  time = 0.16, size = 27, normalized size = 1.00 \begin {gather*} - \frac {- 2 x - 5}{10 x^{2}} + \frac {\left (- 2 x - 5\right ) e^{\frac {4}{x}}}{10 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((4*x**2+23*x+20)*exp(2/x)**2-2*x**3-10*x**2)/x**5,x)

[Out]

-(-2*x - 5)/(10*x**2) + (-2*x - 5)*exp(4/x)/(10*x**3)

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