3.6.92 \(\int \frac {100-100 x^2+e^{6-2 e^3-2 x} (50 x^2+50 x^3)+e^{3-e^3-x} (150 x+50 x^2-50 x^3-50 x^4)}{x^5} \, dx\)

Optimal. Leaf size=37 \[ -\frac {\left (5-5 \left (-e^{3-e^3-x}-\frac {1-x}{x}+x\right )\right )^2}{x^2} \]

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Rubi [A]  time = 0.33, antiderivative size = 65, normalized size of antiderivative = 1.76, number of steps used = 17, number of rules used = 5, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {14, 2197, 2199, 2177, 2178} \begin {gather*} -\frac {25}{x^4}-\frac {50 e^{-x-e^3+3}}{x^3}-\frac {25 e^{2 \left (3-e^3\right )-2 x}}{x^2}+\frac {50}{x^2}+\frac {50 e^{-x-e^3+3}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(100 - 100*x^2 + E^(6 - 2*E^3 - 2*x)*(50*x^2 + 50*x^3) + E^(3 - E^3 - x)*(150*x + 50*x^2 - 50*x^3 - 50*x^4
))/x^5,x]

[Out]

-25/x^4 - (50*E^(3 - E^3 - x))/x^3 + 50/x^2 - (25*E^(2*(3 - E^3) - 2*x))/x^2 + (50*E^(3 - E^3 - x))/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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {50 e^{6-2 e^3-2 x} (1+x)}{x^3}-\frac {100 \left (-1+x^2\right )}{x^5}-\frac {50 e^{3-e^3-x} \left (-3-x+x^2+x^3\right )}{x^4}\right ) \, dx\\ &=50 \int \frac {e^{6-2 e^3-2 x} (1+x)}{x^3} \, dx-50 \int \frac {e^{3-e^3-x} \left (-3-x+x^2+x^3\right )}{x^4} \, dx-100 \int \frac {-1+x^2}{x^5} \, dx\\ &=-\frac {25 e^{2 \left (3-e^3\right )-2 x}}{x^2}-50 \int \left (-\frac {3 e^{3-e^3-x}}{x^4}-\frac {e^{3-e^3-x}}{x^3}+\frac {e^{3-e^3-x}}{x^2}+\frac {e^{3-e^3-x}}{x}\right ) \, dx-100 \int \left (-\frac {1}{x^5}+\frac {1}{x^3}\right ) \, dx\\ &=-\frac {25}{x^4}+\frac {50}{x^2}-\frac {25 e^{2 \left (3-e^3\right )-2 x}}{x^2}+50 \int \frac {e^{3-e^3-x}}{x^3} \, dx-50 \int \frac {e^{3-e^3-x}}{x^2} \, dx-50 \int \frac {e^{3-e^3-x}}{x} \, dx+150 \int \frac {e^{3-e^3-x}}{x^4} \, dx\\ &=-\frac {25}{x^4}-\frac {50 e^{3-e^3-x}}{x^3}+\frac {50}{x^2}-\frac {25 e^{2 \left (3-e^3\right )-2 x}}{x^2}-\frac {25 e^{3-e^3-x}}{x^2}+\frac {50 e^{3-e^3-x}}{x}-50 e^{3-e^3} \text {Ei}(-x)-25 \int \frac {e^{3-e^3-x}}{x^2} \, dx-50 \int \frac {e^{3-e^3-x}}{x^3} \, dx+50 \int \frac {e^{3-e^3-x}}{x} \, dx\\ &=-\frac {25}{x^4}-\frac {50 e^{3-e^3-x}}{x^3}+\frac {50}{x^2}-\frac {25 e^{2 \left (3-e^3\right )-2 x}}{x^2}+\frac {75 e^{3-e^3-x}}{x}+25 \int \frac {e^{3-e^3-x}}{x^2} \, dx+25 \int \frac {e^{3-e^3-x}}{x} \, dx\\ &=-\frac {25}{x^4}-\frac {50 e^{3-e^3-x}}{x^3}+\frac {50}{x^2}-\frac {25 e^{2 \left (3-e^3\right )-2 x}}{x^2}+\frac {50 e^{3-e^3-x}}{x}+25 e^{3-e^3} \text {Ei}(-x)-25 \int \frac {e^{3-e^3-x}}{x} \, dx\\ &=-\frac {25}{x^4}-\frac {50 e^{3-e^3-x}}{x^3}+\frac {50}{x^2}-\frac {25 e^{2 \left (3-e^3\right )-2 x}}{x^2}+\frac {50 e^{3-e^3-x}}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 58, normalized size = 1.57 \begin {gather*} -50 \left (\frac {1}{2 x^4}-\frac {1}{x^2}+\frac {e^{6-2 e^3-2 x}}{2 x^2}-\frac {e^{3-e^3-x} \left (-x+x^3\right )}{x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100 - 100*x^2 + E^(6 - 2*E^3 - 2*x)*(50*x^2 + 50*x^3) + E^(3 - E^3 - x)*(150*x + 50*x^2 - 50*x^3 -
50*x^4))/x^5,x]

[Out]

-50*(1/(2*x^4) - x^(-2) + E^(6 - 2*E^3 - 2*x)/(2*x^2) - (E^(3 - E^3 - x)*(-x + x^3))/x^4)

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fricas [A]  time = 0.65, size = 45, normalized size = 1.22 \begin {gather*} -\frac {25 \, {\left (x^{2} e^{\left (-2 \, x - 2 \, e^{3} + 6\right )} - 2 \, x^{2} - 2 \, {\left (x^{3} - x\right )} e^{\left (-x - e^{3} + 3\right )} + 1\right )}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^3+50*x^2)*exp(-exp(3)+3-x)^2+(-50*x^4-50*x^3+50*x^2+150*x)*exp(-exp(3)+3-x)-100*x^2+100)/x^5,
x, algorithm="fricas")

[Out]

-25*(x^2*e^(-2*x - 2*e^3 + 6) - 2*x^2 - 2*(x^3 - x)*e^(-x - e^3 + 3) + 1)/x^4

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giac [A]  time = 0.32, size = 55, normalized size = 1.49 \begin {gather*} \frac {25 \, {\left (2 \, x^{3} e^{\left (-x - e^{3} + 3\right )} - x^{2} e^{\left (-2 \, x - 2 \, e^{3} + 6\right )} + 2 \, x^{2} - 2 \, x e^{\left (-x - e^{3} + 3\right )} - 1\right )}}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^3+50*x^2)*exp(-exp(3)+3-x)^2+(-50*x^4-50*x^3+50*x^2+150*x)*exp(-exp(3)+3-x)-100*x^2+100)/x^5,
x, algorithm="giac")

[Out]

25*(2*x^3*e^(-x - e^3 + 3) - x^2*e^(-2*x - 2*e^3 + 6) + 2*x^2 - 2*x*e^(-x - e^3 + 3) - 1)/x^4

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maple [A]  time = 0.19, size = 48, normalized size = 1.30

method	result	size
risch	\(\frac {50 x^{2}-25}{x^{4}}-\frac {25 \,{\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}}{x^{2}}+\frac {50 \left (x^{2}-1\right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}}{x^{3}}\)	\(48\)
norman	\(\frac {-25+50 x^{2}-50 \,{\mathrm e}^{-{\mathrm e}^{3}+3-x} x +50 \,{\mathrm e}^{-{\mathrm e}^{3}+3-x} x^{3}-25 \,{\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x} x^{2}}{x^{4}}\)	\(57\)
derivativedivides	\(\text {Expression too large to display}\)	\(3528\)
default	\(\text {Expression too large to display}\)	\(3528\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50*x^3+50*x^2)*exp(-exp(3)+3-x)^2+(-50*x^4-50*x^3+50*x^2+150*x)*exp(-exp(3)+3-x)-100*x^2+100)/x^5,x,meth
od=_RETURNVERBOSE)

[Out]

(50*x^2-25)/x^4-25*exp(-2*exp(3)+6-2*x)/x^2+50*(x^2-1)/x^3*exp(-exp(3)+3-x)

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maxima [C]  time = 1.05, size = 88, normalized size = 2.38 \begin {gather*} -50 \, {\rm Ei}\left (-x\right ) e^{\left (-e^{3} + 3\right )} - 100 \, e^{\left (-2 \, e^{3} + 6\right )} \Gamma \left (-1, 2 \, x\right ) + 50 \, e^{\left (-e^{3} + 3\right )} \Gamma \left (-1, x\right ) - 200 \, e^{\left (-2 \, e^{3} + 6\right )} \Gamma \left (-2, 2 \, x\right ) - 50 \, e^{\left (-e^{3} + 3\right )} \Gamma \left (-2, x\right ) - 150 \, e^{\left (-e^{3} + 3\right )} \Gamma \left (-3, x\right ) + \frac {50}{x^{2}} - \frac {25}{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^3+50*x^2)*exp(-exp(3)+3-x)^2+(-50*x^4-50*x^3+50*x^2+150*x)*exp(-exp(3)+3-x)-100*x^2+100)/x^5,
x, algorithm="maxima")

[Out]

-50*Ei(-x)*e^(-e^3 + 3) - 100*e^(-2*e^3 + 6)*gamma(-1, 2*x) + 50*e^(-e^3 + 3)*gamma(-1, x) - 200*e^(-2*e^3 + 6
)*gamma(-2, 2*x) - 50*e^(-e^3 + 3)*gamma(-2, x) - 150*e^(-e^3 + 3)*gamma(-3, x) + 50/x^2 - 25/x^4

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mupad [B]  time = 0.56, size = 59, normalized size = 1.59 \begin {gather*} \frac {50}{x^2}-\frac {25}{x^4}+\frac {50\,{\mathrm {e}}^{-{\mathrm {e}}^3}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3}{x}-\frac {50\,{\mathrm {e}}^{-{\mathrm {e}}^3}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3}{x^3}-\frac {25\,{\mathrm {e}}^{-2\,{\mathrm {e}}^3}\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^6}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(6 - 2*exp(3) - 2*x)*(50*x^2 + 50*x^3) + exp(3 - exp(3) - x)*(150*x + 50*x^2 - 50*x^3 - 50*x^4) - 100*
x^2 + 100)/x^5,x)

[Out]

50/x^2 - 25/x^4 + (50*exp(-exp(3))*exp(-x)*exp(3))/x - (50*exp(-exp(3))*exp(-x)*exp(3))/x^3 - (25*exp(-2*exp(3
))*exp(-2*x)*exp(6))/x^2

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sympy [A]  time = 0.17, size = 48, normalized size = 1.30 \begin {gather*} - \frac {25 - 50 x^{2}}{x^{4}} + \frac {- 25 x^{3} e^{- 2 x - 2 e^{3} + 6} + \left (50 x^{4} - 50 x^{2}\right ) e^{- x - e^{3} + 3}}{x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x**3+50*x**2)*exp(-exp(3)+3-x)**2+(-50*x**4-50*x**3+50*x**2+150*x)*exp(-exp(3)+3-x)-100*x**2+10
0)/x**5,x)

[Out]

-(25 - 50*x**2)/x**4 + (-25*x**3*exp(-2*x - 2*exp(3) + 6) + (50*x**4 - 50*x**2)*exp(-x - exp(3) + 3))/x**5

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