3.77.13 \(\int \frac {e^{-3/x} (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} (16 x^2+8 x^3+x^4)+e^{\frac {3}{x}+x^2} (32 x^3+16 x^4+2 x^5))}{16 x^2+8 x^3+x^4} \, dx\)

Optimal. Leaf size=31 \[ 4+e^x+e^{x^2}-\frac {\left (-4+5 e^{-3/x}\right ) (-3+x)}{4+x} \]

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Rubi [A]  time = 2.87, antiderivative size = 39, normalized size of antiderivative = 1.26, number of steps used = 48, number of rules used = 10, integrand size = 98, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {1594, 27, 6742, 2194, 2209, 2223, 2210, 2222, 2228, 2178} \begin {gather*} e^{x^2}-5 e^{-3/x}+e^x+\frac {35 e^{-3/x}}{x+4}-\frac {28}{x+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(180 - 15*x - 50*x^2 + 28*E^(3/x)*x^2 + E^(3/x + x)*(16*x^2 + 8*x^3 + x^4) + E^(3/x + x^2)*(32*x^3 + 16*x^
4 + 2*x^5))/(E^(3/x)*(16*x^2 + 8*x^3 + x^4)),x]

[Out]

-5/E^(3/x) + E^x + E^x^2 - 28/(4 + x) + 35/(E^(3/x)*(4 + 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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - 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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

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 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 {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{x^2 \left (16+8 x+x^2\right )} \, dx\\ &=\int \frac {e^{-3/x} \left (180-15 x-50 x^2+28 e^{3/x} x^2+e^{\frac {3}{x}+x} \left (16 x^2+8 x^3+x^4\right )+e^{\frac {3}{x}+x^2} \left (32 x^3+16 x^4+2 x^5\right )\right )}{x^2 (4+x)^2} \, dx\\ &=\int \left (e^x+2 e^{x^2} x+\frac {28}{(4+x)^2}-\frac {50 e^{-3/x}}{(4+x)^2}+\frac {180 e^{-3/x}}{x^2 (4+x)^2}-\frac {15 e^{-3/x}}{x (4+x)^2}\right ) \, dx\\ &=-\frac {28}{4+x}+2 \int e^{x^2} x \, dx-15 \int \frac {e^{-3/x}}{x (4+x)^2} \, dx-50 \int \frac {e^{-3/x}}{(4+x)^2} \, dx+180 \int \frac {e^{-3/x}}{x^2 (4+x)^2} \, dx+\int e^x \, dx\\ &=e^x+e^{x^2}-\frac {28}{4+x}+\frac {50 e^{-3/x}}{4+x}-15 \int \left (\frac {e^{-3/x}}{16 x}-\frac {e^{-3/x}}{4 (4+x)^2}-\frac {e^{-3/x}}{16 (4+x)}\right ) \, dx-150 \int \frac {e^{-3/x}}{x^2 (4+x)} \, dx+180 \int \left (\frac {e^{-3/x}}{16 x^2}-\frac {e^{-3/x}}{32 x}+\frac {e^{-3/x}}{16 (4+x)^2}+\frac {e^{-3/x}}{32 (4+x)}\right ) \, dx\\ &=e^x+e^{x^2}-\frac {28}{4+x}+\frac {50 e^{-3/x}}{4+x}-\frac {15}{16} \int \frac {e^{-3/x}}{x} \, dx+\frac {15}{16} \int \frac {e^{-3/x}}{4+x} \, dx+\frac {15}{4} \int \frac {e^{-3/x}}{(4+x)^2} \, dx-\frac {45}{8} \int \frac {e^{-3/x}}{x} \, dx+\frac {45}{8} \int \frac {e^{-3/x}}{4+x} \, dx+\frac {45}{4} \int \frac {e^{-3/x}}{x^2} \, dx+\frac {45}{4} \int \frac {e^{-3/x}}{(4+x)^2} \, dx-150 \int \left (\frac {e^{-3/x}}{4 x^2}-\frac {e^{-3/x}}{16 x}+\frac {e^{-3/x}}{16 (4+x)}\right ) \, dx\\ &=\frac {15 e^{-3/x}}{4}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x}+\frac {105 \text {Ei}\left (-\frac {3}{x}\right )}{16}+\frac {15}{16} \int \frac {e^{-3/x}}{x} \, dx-\frac {15}{4} \int \frac {e^{-3/x}}{x (4+x)} \, dx+\frac {45}{8} \int \frac {e^{-3/x}}{x} \, dx+\frac {75}{8} \int \frac {e^{-3/x}}{x} \, dx-\frac {75}{8} \int \frac {e^{-3/x}}{4+x} \, dx+\frac {45}{4} \int \frac {e^{-3/x}}{x^2 (4+x)} \, dx-\frac {45}{2} \int \frac {e^{-3/x}}{x (4+x)} \, dx+\frac {135}{4} \int \frac {e^{-3/x}}{x^2 (4+x)} \, dx-\frac {75}{2} \int \frac {e^{-3/x}}{x^2} \, dx\\ &=-\frac {35}{4} e^{-3/x}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x}-\frac {75 \text {Ei}\left (-\frac {3}{x}\right )}{8}+\frac {15}{16} \operatorname {Subst}\left (\int \frac {e^{\frac {3}{4}-\frac {3 x}{4}}}{x} \, dx,x,\frac {4+x}{x}\right )+\frac {45}{8} \operatorname {Subst}\left (\int \frac {e^{\frac {3}{4}-\frac {3 x}{4}}}{x} \, dx,x,\frac {4+x}{x}\right )-\frac {75}{8} \int \frac {e^{-3/x}}{x} \, dx+\frac {45}{4} \int \left (\frac {e^{-3/x}}{4 x^2}-\frac {e^{-3/x}}{16 x}+\frac {e^{-3/x}}{16 (4+x)}\right ) \, dx+\frac {135}{4} \int \left (\frac {e^{-3/x}}{4 x^2}-\frac {e^{-3/x}}{16 x}+\frac {e^{-3/x}}{16 (4+x)}\right ) \, dx+\frac {75}{2} \int \frac {e^{-3/x}}{x (4+x)} \, dx\\ &=-\frac {35}{4} e^{-3/x}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x}+\frac {105}{16} e^{3/4} \text {Ei}\left (-\frac {3 (4+x)}{4 x}\right )-\frac {45}{64} \int \frac {e^{-3/x}}{x} \, dx+\frac {45}{64} \int \frac {e^{-3/x}}{4+x} \, dx-\frac {135}{64} \int \frac {e^{-3/x}}{x} \, dx+\frac {135}{64} \int \frac {e^{-3/x}}{4+x} \, dx+\frac {45}{16} \int \frac {e^{-3/x}}{x^2} \, dx+\frac {135}{16} \int \frac {e^{-3/x}}{x^2} \, dx-\frac {75}{8} \operatorname {Subst}\left (\int \frac {e^{\frac {3}{4}-\frac {3 x}{4}}}{x} \, dx,x,\frac {4+x}{x}\right )\\ &=-5 e^{-3/x}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x}+\frac {45 \text {Ei}\left (-\frac {3}{x}\right )}{16}-\frac {45}{16} e^{3/4} \text {Ei}\left (-\frac {3 (4+x)}{4 x}\right )+\frac {45}{64} \int \frac {e^{-3/x}}{x} \, dx+\frac {135}{64} \int \frac {e^{-3/x}}{x} \, dx-\frac {45}{16} \int \frac {e^{-3/x}}{x (4+x)} \, dx-\frac {135}{16} \int \frac {e^{-3/x}}{x (4+x)} \, dx\\ &=-5 e^{-3/x}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x}-\frac {45}{16} e^{3/4} \text {Ei}\left (-\frac {3 (4+x)}{4 x}\right )+\frac {45}{64} \operatorname {Subst}\left (\int \frac {e^{\frac {3}{4}-\frac {3 x}{4}}}{x} \, dx,x,\frac {4+x}{x}\right )+\frac {135}{64} \operatorname {Subst}\left (\int \frac {e^{\frac {3}{4}-\frac {3 x}{4}}}{x} \, dx,x,\frac {4+x}{x}\right )\\ &=-5 e^{-3/x}+e^x+e^{x^2}-\frac {28}{4+x}+\frac {35 e^{-3/x}}{4+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.26, size = 33, normalized size = 1.06 \begin {gather*} e^x+e^{x^2}-\frac {28}{4+x}-\frac {5 e^{-3/x} (-3+x)}{4+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(180 - 15*x - 50*x^2 + 28*E^(3/x)*x^2 + E^(3/x + x)*(16*x^2 + 8*x^3 + x^4) + E^(3/x + x^2)*(32*x^3 +
16*x^4 + 2*x^5))/(E^(3/x)*(16*x^2 + 8*x^3 + x^4)),x]

[Out]

E^x + E^x^2 - 28/(4 + x) - (5*(-3 + x))/(E^(3/x)*(4 + x))

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fricas [A]  time = 0.69, size = 53, normalized size = 1.71 \begin {gather*} \frac {{\left ({\left (x + 4\right )} e^{\left (\frac {x^{3} + 3}{x}\right )} + {\left (x + 4\right )} e^{\left (\frac {x^{2} + 3}{x}\right )} - 5 \, x - 28 \, e^{\frac {3}{x}} + 15\right )} e^{\left (-\frac {3}{x}\right )}}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+16*x^4+32*x^3)*exp(3/x)*exp(x^2)+(x^4+8*x^3+16*x^2)*exp(3/x)*exp(x)+28*x^2*exp(3/x)-50*x^2-1
5*x+180)/(x^4+8*x^3+16*x^2)/exp(3/x),x, algorithm="fricas")

[Out]

((x + 4)*e^((x^3 + 3)/x) + (x + 4)*e^((x^2 + 3)/x) - 5*x - 28*e^(3/x) + 15)*e^(-3/x)/(x + 4)

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giac [A]  time = 0.17, size = 45, normalized size = 1.45 \begin {gather*} \frac {x e^{\left (x^{2}\right )} + x e^{x} - 5 \, x e^{\left (-\frac {3}{x}\right )} + 4 \, e^{\left (x^{2}\right )} + 4 \, e^{x} + 15 \, e^{\left (-\frac {3}{x}\right )} - 28}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+16*x^4+32*x^3)*exp(3/x)*exp(x^2)+(x^4+8*x^3+16*x^2)*exp(3/x)*exp(x)+28*x^2*exp(3/x)-50*x^2-1
5*x+180)/(x^4+8*x^3+16*x^2)/exp(3/x),x, algorithm="giac")

[Out]

(x*e^(x^2) + x*e^x - 5*x*e^(-3/x) + 4*e^(x^2) + 4*e^x + 15*e^(-3/x) - 28)/(x + 4)

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maple [A]  time = 0.21, size = 31, normalized size = 1.00

method	result	size
risch	\(-\frac {28}{4+x}+{\mathrm e}^{x}+{\mathrm e}^{x^{2}}-\frac {5 \left (x -3\right ) {\mathrm e}^{-\frac {3}{x}}}{4+x}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^5+16*x^4+32*x^3)*exp(3/x)*exp(x^2)+(x^4+8*x^3+16*x^2)*exp(3/x)*exp(x)+28*x^2*exp(3/x)-50*x^2-15*x+18
0)/(x^4+8*x^3+16*x^2)/exp(3/x),x,method=_RETURNVERBOSE)

[Out]

-28/(4+x)+exp(x)+exp(x^2)-5*(x-3)/(4+x)*exp(-3/x)

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maxima [A]  time = 0.40, size = 33, normalized size = 1.06 \begin {gather*} \frac {{\left (x + 4\right )} e^{\left (x^{2}\right )} + {\left (x + 4\right )} e^{x} - 5 \, {\left (x - 3\right )} e^{\left (-\frac {3}{x}\right )} - 28}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+16*x^4+32*x^3)*exp(3/x)*exp(x^2)+(x^4+8*x^3+16*x^2)*exp(3/x)*exp(x)+28*x^2*exp(3/x)-50*x^2-1
5*x+180)/(x^4+8*x^3+16*x^2)/exp(3/x),x, algorithm="maxima")

[Out]

((x + 4)*e^(x^2) + (x + 4)*e^x - 5*(x - 3)*e^(-3/x) - 28)/(x + 4)

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mupad [B]  time = 5.53, size = 32, normalized size = 1.03 \begin {gather*} {\mathrm {e}}^{x^2}+{\mathrm {e}}^x-\frac {28}{x+4}-\frac {{\mathrm {e}}^{-\frac {3}{x}}\,\left (5\,x-15\right )}{x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3/x)*(28*x^2*exp(3/x) - 15*x - 50*x^2 + exp(x^2)*exp(3/x)*(32*x^3 + 16*x^4 + 2*x^5) + exp(3/x)*exp(x
)*(16*x^2 + 8*x^3 + x^4) + 180))/(16*x^2 + 8*x^3 + x^4),x)

[Out]

exp(x^2) + exp(x) - 28/(x + 4) - (exp(-3/x)*(5*x - 15))/(x + 4)

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sympy [A]  time = 0.61, size = 26, normalized size = 0.84 \begin {gather*} \frac {\left (15 - 5 x\right ) e^{- \frac {3}{x}}}{x + 4} + e^{x} + e^{x^{2}} - \frac {28}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**5+16*x**4+32*x**3)*exp(3/x)*exp(x**2)+(x**4+8*x**3+16*x**2)*exp(3/x)*exp(x)+28*x**2*exp(3/x)-
50*x**2-15*x+180)/(x**4+8*x**3+16*x**2)/exp(3/x),x)

[Out]

(15 - 5*x)*exp(-3/x)/(x + 4) + exp(x) + exp(x**2) - 28/(x + 4)

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