3.33.76 \(\int \frac {-3125-250 x-5 x^2+e^{3/x} (-1125+705 x+15 x^2)}{625+50 x+x^2} \, dx\)

Optimal. Leaf size=24 \[ 5 \left (1-x+\frac {3 e^{3/x} x^2}{25+x}\right ) \]

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Rubi [A]  time = 0.64, antiderivative size = 37, normalized size of antiderivative = 1.54, number of steps used = 20, number of rules used = 10, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {27, 6688, 6742, 2206, 2210, 2223, 2209, 2222, 2228, 2178} \begin {gather*} 15 e^{3/x} x-5 x-375 e^{3/x}+\frac {9375 e^{3/x}}{x+25} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3125 - 250*x - 5*x^2 + E^(3/x)*(-1125 + 705*x + 15*x^2))/(625 + 50*x + x^2),x]

[Out]

-375*E^(3/x) - 5*x + 15*E^(3/x)*x + (9375*E^(3/x))/(25 + x)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 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 2222

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

Rule 2223

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(m + 1)*
F^(a + b/(c + d*x)))/(f*(m + 1)), x] + Dist[(b*d*Log[F])/(f*(m + 1)), Int[((e + f*x)^(m + 1)*F^(a + b/(c + d*x
)))/(c + d*x)^2, x], x] /; FreeQ[{F, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && ILtQ[m, -1]

Rule 2228

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/(((e_.) + (f_.)*(x_))*((g_.) + (h_.)*(x_))), x_Symbol] :> -Dist[
d/(f*(d*g - c*h)), Subst[Int[F^(a - (b*h)/(d*g - c*h) + (d*b*x)/(d*g - c*h))/x, x], x, (g + h*x)/(c + d*x)], x
] /; FreeQ[{F, a, b, c, d, e, f}, x] && EqQ[d*e - c*f, 0]

Rule 6688

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

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 \frac {-3125-250 x-5 x^2+e^{3/x} \left (-1125+705 x+15 x^2\right )}{(25+x)^2} \, dx\\ &=\int \left (-5+\frac {15 e^{3/x} \left (-75+47 x+x^2\right )}{(25+x)^2}\right ) \, dx\\ &=-5 x+15 \int \frac {e^{3/x} \left (-75+47 x+x^2\right )}{(25+x)^2} \, dx\\ &=-5 x+15 \int \left (e^{3/x}-\frac {625 e^{3/x}}{(25+x)^2}-\frac {3 e^{3/x}}{25+x}\right ) \, dx\\ &=-5 x+15 \int e^{3/x} \, dx-45 \int \frac {e^{3/x}}{25+x} \, dx-9375 \int \frac {e^{3/x}}{(25+x)^2} \, dx\\ &=-5 x+15 e^{3/x} x+\frac {9375 e^{3/x}}{25+x}+1125 \int \frac {e^{3/x}}{x (25+x)} \, dx+28125 \int \frac {e^{3/x}}{x^2 (25+x)} \, dx\\ &=-5 x+15 e^{3/x} x+\frac {9375 e^{3/x}}{25+x}-45 \operatorname {Subst}\left (\int \frac {e^{-\frac {3}{25}+\frac {3 x}{25}}}{x} \, dx,x,\frac {25+x}{x}\right )+28125 \int \left (\frac {e^{3/x}}{25 x^2}-\frac {e^{3/x}}{625 x}+\frac {e^{3/x}}{625 (25+x)}\right ) \, dx\\ &=-5 x+15 e^{3/x} x+\frac {9375 e^{3/x}}{25+x}-\frac {45 \text {Ei}\left (\frac {3}{25}+\frac {3}{x}\right )}{e^{3/25}}-45 \int \frac {e^{3/x}}{x} \, dx+45 \int \frac {e^{3/x}}{25+x} \, dx+1125 \int \frac {e^{3/x}}{x^2} \, dx\\ &=-375 e^{3/x}-5 x+15 e^{3/x} x+\frac {9375 e^{3/x}}{25+x}-\frac {45 \text {Ei}\left (\frac {3}{25}+\frac {3}{x}\right )}{e^{3/25}}+45 \text {Ei}\left (\frac {3}{x}\right )+45 \int \frac {e^{3/x}}{x} \, dx-1125 \int \frac {e^{3/x}}{x (25+x)} \, dx\\ &=-375 e^{3/x}-5 x+15 e^{3/x} x+\frac {9375 e^{3/x}}{25+x}-\frac {45 \text {Ei}\left (\frac {3}{25}+\frac {3}{x}\right )}{e^{3/25}}+45 \operatorname {Subst}\left (\int \frac {e^{-\frac {3}{25}+\frac {3 x}{25}}}{x} \, dx,x,\frac {25+x}{x}\right )\\ &=-375 e^{3/x}-5 x+15 e^{3/x} x+\frac {9375 e^{3/x}}{25+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 23, normalized size = 0.96 \begin {gather*} -5 x+15 e^{3/x} \left (-25+x+\frac {625}{25+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3125 - 250*x - 5*x^2 + E^(3/x)*(-1125 + 705*x + 15*x^2))/(625 + 50*x + x^2),x]

[Out]

-5*x + 15*E^(3/x)*(-25 + x + 625/(25 + x))

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fricas [A]  time = 0.50, size = 27, normalized size = 1.12 \begin {gather*} \frac {5 \, {\left (3 \, x^{2} e^{\frac {3}{x}} - x^{2} - 25 \, x\right )}}{x + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^2+705*x-1125)*exp(3/x)-5*x^2-250*x-3125)/(x^2+50*x+625),x, algorithm="fricas")

[Out]

5*(3*x^2*e^(3/x) - x^2 - 25*x)/(x + 25)

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giac [A]  time = 0.18, size = 28, normalized size = 1.17 \begin {gather*} -\frac {5 \, {\left (\frac {25}{x} - 3 \, e^{\frac {3}{x}} + 1\right )}}{\frac {1}{x} + \frac {25}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^2+705*x-1125)*exp(3/x)-5*x^2-250*x-3125)/(x^2+50*x+625),x, algorithm="giac")

[Out]

-5*(25/x - 3*e^(3/x) + 1)/(1/x + 25/x^2)

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maple [A]  time = 0.46, size = 21, normalized size = 0.88

method	result	size
risch	\(-5 x +\frac {15 x^{2} {\mathrm e}^{\frac {3}{x}}}{x +25}\)	\(21\)
norman	\(\frac {-5 x^{2}+15 x^{2} {\mathrm e}^{\frac {3}{x}}+3125}{x +25}\)	\(25\)
derivativedivides	\(-5 x -\frac {45 \,{\mathrm e}^{\frac {3}{x}}}{\frac {3}{x}+\frac {3}{25}}+15 x \,{\mathrm e}^{\frac {3}{x}}\)	\(31\)
default	\(-5 x -\frac {45 \,{\mathrm e}^{\frac {3}{x}}}{\frac {3}{x}+\frac {3}{25}}+15 x \,{\mathrm e}^{\frac {3}{x}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((15*x^2+705*x-1125)*exp(3/x)-5*x^2-250*x-3125)/(x^2+50*x+625),x,method=_RETURNVERBOSE)

[Out]

-5*x+15*x^2*exp(3/x)/(x+25)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -5 \, x + \frac {15 \, {\left (x^{3} - 1175 \, x\right )} e^{\frac {3}{x}}}{x^{2} + 50 \, x + 625} + 5 \, \int \frac {75 \, {\left (959 \, x - 3525\right )} e^{\frac {3}{x}}}{x^{4} + 75 \, x^{3} + 1875 \, x^{2} + 15625 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^2+705*x-1125)*exp(3/x)-5*x^2-250*x-3125)/(x^2+50*x+625),x, algorithm="maxima")

[Out]

-5*x + 15*(x^3 - 1175*x)*e^(3/x)/(x^2 + 50*x + 625) + 5*integrate(75*(959*x - 3525)*e^(3/x)/(x^4 + 75*x^3 + 18
75*x^2 + 15625*x), x)

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mupad [B]  time = 1.98, size = 27, normalized size = 1.12 \begin {gather*} -\frac {125\,x-15\,x^2\,{\mathrm {e}}^{3/x}+5\,x^2}{x+25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(250*x - exp(3/x)*(705*x + 15*x^2 - 1125) + 5*x^2 + 3125)/(50*x + x^2 + 625),x)

[Out]

-(125*x - 15*x^2*exp(3/x) + 5*x^2)/(x + 25)

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sympy [A]  time = 0.13, size = 15, normalized size = 0.62 \begin {gather*} \frac {15 x^{2} e^{\frac {3}{x}}}{x + 25} - 5 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x**2+705*x-1125)*exp(3/x)-5*x**2-250*x-3125)/(x**2+50*x+625),x)

[Out]

15*x**2*exp(3/x)/(x + 25) - 5*x

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