3.91.13 \(\int \frac {e^{-x} (e^8 (-10-20 x-5 x^2)+e^{5+2 x} (-20-20 x+10 x^2)+e^{2+4 x} (-10+15 x^2))}{x^3+2 x^4+x^5} \, dx\) [9013]

Optimal. Leaf size=30 \[ \frac {5 e^{-x} \left (e^4+e^{1+2 x}\right )^2}{x \left (x+x^2\right )} \]

[Out]

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

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(109\) vs. \(2(30)=60\).
time = 1.74, antiderivative size = 109, normalized size of antiderivative = 3.63, number of steps used = 35, number of rules used = 5, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1608, 27, 6874, 2208, 2209} \begin {gather*} \frac {5 e^{8-x}}{x^2}+\frac {10 e^{x+5}}{x^2}+\frac {5 e^{3 x+2}}{x^2}+\frac {5 e^{8-x}}{x+1}+\frac {10 e^{x+5}}{x+1}+\frac {5 e^{3 x+2}}{x+1}-\frac {5 e^{8-x}}{x}-\frac {10 e^{x+5}}{x}-\frac {5 e^{3 x+2}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^8*(-10 - 20*x - 5*x^2) + E^(5 + 2*x)*(-20 - 20*x + 10*x^2) + E^(2 + 4*x)*(-10 + 15*x^2))/(E^x*(x^3 + 2*
x^4 + x^5)),x]

[Out]

(5*E^(8 - x))/x^2 + (10*E^(5 + x))/x^2 + (5*E^(2 + 3*x))/x^2 - (5*E^(8 - x))/x - (10*E^(5 + x))/x - (5*E^(2 +
3*x))/x + (5*E^(8 - x))/(1 + x) + (10*E^(5 + x))/(1 + x) + (5*E^(2 + 3*x))/(1 + 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 1608

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 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2209

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

Rule 6874

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^{-x} \left (e^8 \left (-10-20 x-5 x^2\right )+e^{5+2 x} \left (-20-20 x+10 x^2\right )+e^{2+4 x} \left (-10+15 x^2\right )\right )}{x^3 \left (1+2 x+x^2\right )} \, dx\\ &=\int \frac {e^{-x} \left (e^8 \left (-10-20 x-5 x^2\right )+e^{5+2 x} \left (-20-20 x+10 x^2\right )+e^{2+4 x} \left (-10+15 x^2\right )\right )}{x^3 (1+x)^2} \, dx\\ &=\int \left (\frac {10 e^{5+x} \left (-2-2 x+x^2\right )}{x^3 (1+x)^2}-\frac {5 e^{8-x} \left (2+4 x+x^2\right )}{x^3 (1+x)^2}+\frac {5 e^{2+3 x} \left (-2+3 x^2\right )}{x^3 (1+x)^2}\right ) \, dx\\ &=-\left (5 \int \frac {e^{8-x} \left (2+4 x+x^2\right )}{x^3 (1+x)^2} \, dx\right )+5 \int \frac {e^{2+3 x} \left (-2+3 x^2\right )}{x^3 (1+x)^2} \, dx+10 \int \frac {e^{5+x} \left (-2-2 x+x^2\right )}{x^3 (1+x)^2} \, dx\\ &=-\left (5 \int \left (\frac {2 e^{8-x}}{x^3}-\frac {e^{8-x}}{x}+\frac {e^{8-x}}{(1+x)^2}+\frac {e^{8-x}}{1+x}\right ) \, dx\right )+5 \int \left (-\frac {2 e^{2+3 x}}{x^3}+\frac {4 e^{2+3 x}}{x^2}-\frac {3 e^{2+3 x}}{x}-\frac {e^{2+3 x}}{(1+x)^2}+\frac {3 e^{2+3 x}}{1+x}\right ) \, dx+10 \int \left (-\frac {2 e^{5+x}}{x^3}+\frac {2 e^{5+x}}{x^2}-\frac {e^{5+x}}{x}-\frac {e^{5+x}}{(1+x)^2}+\frac {e^{5+x}}{1+x}\right ) \, dx\\ &=5 \int \frac {e^{8-x}}{x} \, dx-5 \int \frac {e^{8-x}}{(1+x)^2} \, dx-5 \int \frac {e^{2+3 x}}{(1+x)^2} \, dx-5 \int \frac {e^{8-x}}{1+x} \, dx-10 \int \frac {e^{8-x}}{x^3} \, dx-10 \int \frac {e^{2+3 x}}{x^3} \, dx-10 \int \frac {e^{5+x}}{x} \, dx-10 \int \frac {e^{5+x}}{(1+x)^2} \, dx+10 \int \frac {e^{5+x}}{1+x} \, dx-15 \int \frac {e^{2+3 x}}{x} \, dx+15 \int \frac {e^{2+3 x}}{1+x} \, dx-20 \int \frac {e^{5+x}}{x^3} \, dx+20 \int \frac {e^{5+x}}{x^2} \, dx+20 \int \frac {e^{2+3 x}}{x^2} \, dx\\ &=\frac {5 e^{8-x}}{x^2}+\frac {10 e^{5+x}}{x^2}+\frac {5 e^{2+3 x}}{x^2}-\frac {20 e^{5+x}}{x}-\frac {20 e^{2+3 x}}{x}+\frac {5 e^{8-x}}{1+x}+\frac {10 e^{5+x}}{1+x}+\frac {5 e^{2+3 x}}{1+x}-5 e^9 \text {Ei}(-1-x)+5 e^8 \text {Ei}(-x)-10 e^5 \text {Ei}(x)-15 e^2 \text {Ei}(3 x)+10 e^4 \text {Ei}(1+x)+\frac {15 \text {Ei}(3 (1+x))}{e}+5 \int \frac {e^{8-x}}{x^2} \, dx+5 \int \frac {e^{8-x}}{1+x} \, dx-10 \int \frac {e^{5+x}}{x^2} \, dx-10 \int \frac {e^{5+x}}{1+x} \, dx-15 \int \frac {e^{2+3 x}}{x^2} \, dx-15 \int \frac {e^{2+3 x}}{1+x} \, dx+20 \int \frac {e^{5+x}}{x} \, dx+60 \int \frac {e^{2+3 x}}{x} \, dx\\ &=\frac {5 e^{8-x}}{x^2}+\frac {10 e^{5+x}}{x^2}+\frac {5 e^{2+3 x}}{x^2}-\frac {5 e^{8-x}}{x}-\frac {10 e^{5+x}}{x}-\frac {5 e^{2+3 x}}{x}+\frac {5 e^{8-x}}{1+x}+\frac {10 e^{5+x}}{1+x}+\frac {5 e^{2+3 x}}{1+x}+5 e^8 \text {Ei}(-x)+10 e^5 \text {Ei}(x)+45 e^2 \text {Ei}(3 x)-5 \int \frac {e^{8-x}}{x} \, dx-10 \int \frac {e^{5+x}}{x} \, dx-45 \int \frac {e^{2+3 x}}{x} \, dx\\ &=\frac {5 e^{8-x}}{x^2}+\frac {10 e^{5+x}}{x^2}+\frac {5 e^{2+3 x}}{x^2}-\frac {5 e^{8-x}}{x}-\frac {10 e^{5+x}}{x}-\frac {5 e^{2+3 x}}{x}+\frac {5 e^{8-x}}{1+x}+\frac {10 e^{5+x}}{1+x}+\frac {5 e^{2+3 x}}{1+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 1.57, size = 28, normalized size = 0.93 \begin {gather*} \frac {5 e^{2-x} \left (e^3+e^{2 x}\right )^2}{x^2 (1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^8*(-10 - 20*x - 5*x^2) + E^(5 + 2*x)*(-20 - 20*x + 10*x^2) + E^(2 + 4*x)*(-10 + 15*x^2))/(E^x*(x^
3 + 2*x^4 + x^5)),x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.46, size = 316, normalized size = 10.53

method	result	size
norman	\(\frac {\left (5 \,{\mathrm e}^{8}+5 \,{\mathrm e}^{2} {\mathrm e}^{4 x}+10 \,{\mathrm e}^{4} {\mathrm e} \,{\mathrm e}^{2 x}\right ) {\mathrm e}^{-x}}{x^{2} \left (x +1\right )}\)	\(41\)
risch	\(\frac {5 \,{\mathrm e}^{3 x +2}}{x^{2} \left (x +1\right )}+\frac {10 \,{\mathrm e}^{5+x}}{x^{2} \left (x +1\right )}+\frac {5 \,{\mathrm e}^{8-x}}{x^{2} \left (x +1\right )}\)	\(48\)
default	\(-10 \,{\mathrm e}^{8} \left (\frac {{\mathrm e}^{-x} \left (7 x^{2}+4 x -1\right )}{2 x^{2} \left (x +1\right )}-\frac {11 \expIntegral \left (1, x\right )}{2}+2 \,{\mathrm e} \expIntegral \left (1, x +1\right )\right )-10 \,{\mathrm e}^{2} \left (\frac {{\mathrm e}^{3 x}}{2 x}-\frac {3 \expIntegral \left (1, -3 x \right )}{2}-\frac {{\mathrm e}^{3 x}}{2 x^{2}}+\frac {{\mathrm e}^{3 x}}{x +1}+6 \,{\mathrm e}^{-3} \expIntegral \left (1, -3 x -3\right )\right )-20 \,{\mathrm e}^{8} \left (-\frac {{\mathrm e}^{-x} \left (2 x +1\right )}{\left (x +1\right ) x}+3 \expIntegral \left (1, x\right )-{\mathrm e} \expIntegral \left (1, x +1\right )\right )-5 \,{\mathrm e}^{8} \left (\frac {{\mathrm e}^{-x}}{x +1}-\expIntegral \left (1, x\right )\right )-20 \,{\mathrm e} \,{\mathrm e}^{4} \left (\frac {3 \,{\mathrm e}^{x}}{2 x}-\frac {3 \expIntegral \left (1, -x \right )}{2}-\frac {{\mathrm e}^{x}}{2 x^{2}}+\frac {{\mathrm e}^{x}}{x +1}+4 \,{\mathrm e}^{-1} \expIntegral \left (1, -x -1\right )\right )+15 \,{\mathrm e}^{2} \left (-\expIntegral \left (1, -3 x \right )+\frac {{\mathrm e}^{3 x}}{x +1}+4 \,{\mathrm e}^{-3} \expIntegral \left (1, -3 x -3\right )\right )-20 \,{\mathrm e} \,{\mathrm e}^{4} \left (-\frac {{\mathrm e}^{x}}{x}+\expIntegral \left (1, -x \right )-\frac {{\mathrm e}^{x}}{x +1}-3 \,{\mathrm e}^{-1} \expIntegral \left (1, -x -1\right )\right )+10 \,{\mathrm e} \,{\mathrm e}^{4} \left (-\expIntegral \left (1, -x \right )+\frac {{\mathrm e}^{x}}{x +1}+2 \,{\mathrm e}^{-1} \expIntegral \left (1, -x -1\right )\right )\)	\(316\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((15*x^2-10)*exp(2*x+1)^2+(10*x^2-20*x-20)*exp(4)*exp(2*x+1)+(-5*x^2-20*x-10)*exp(4)^2)/(x^5+2*x^4+x^3)/ex
p(x),x,method=_RETURNVERBOSE)

[Out]

-10*exp(4)^2*(1/2*exp(-x)*(7*x^2+4*x-1)/x^2/(x+1)-11/2*Ei(1,x)+2*exp(1)*Ei(1,x+1))-10*exp(1)^2*(1/2/x*exp(x)^3
-3/2*Ei(1,-3*x)-1/2/x^2*exp(x)^3+exp(x)^3/(x+1)+6*exp(-3)*Ei(1,-3*x-3))-20*exp(4)^2*(-exp(-x)*(2*x+1)/(x+1)/x+
3*Ei(1,x)-exp(1)*Ei(1,x+1))-5*exp(4)^2*(exp(-x)/(x+1)-Ei(1,x))-20*exp(1)*exp(4)*(3/2*exp(x)/x-3/2*Ei(1,-x)-1/2
*exp(x)/x^2+exp(x)/(x+1)+4*exp(-1)*Ei(1,-x-1))+15*exp(1)^2*(-Ei(1,-3*x)+exp(x)^3/(x+1)+4*exp(-3)*Ei(1,-3*x-3))
-20*exp(1)*exp(4)*(-exp(x)/x+Ei(1,-x)-exp(x)/(x+1)-3*exp(-1)*Ei(1,-x-1))+10*exp(1)*exp(4)*(-Ei(1,-x)+exp(x)/(x
+1)+2*exp(-1)*Ei(1,-x-1))

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Maxima [A]
time = 0.30, size = 30, normalized size = 1.00 \begin {gather*} \frac {5 \, {\left (e^{\left (3 \, x + 2\right )} + 2 \, e^{\left (x + 5\right )} + e^{\left (-x + 8\right )}\right )}}{x^{3} + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^2-10)*exp(1+2*x)^2+(10*x^2-20*x-20)*exp(4)*exp(1+2*x)+(-5*x^2-20*x-10)*exp(4)^2)/(x^5+2*x^4+x
^3)/exp(x),x, algorithm="maxima")

[Out]

5*(e^(3*x + 2) + 2*e^(x + 5) + e^(-x + 8))/(x^3 + x^2)

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Fricas [A]
time = 0.40, size = 32, normalized size = 1.07 \begin {gather*} \frac {5 \, {\left (e^{2} + 2 \, e^{\left (-2 \, x + 5\right )} + e^{\left (-4 \, x + 8\right )}\right )} e^{\left (3 \, x\right )}}{x^{3} + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^2-10)*exp(1+2*x)^2+(10*x^2-20*x-20)*exp(4)*exp(1+2*x)+(-5*x^2-20*x-10)*exp(4)^2)/(x^5+2*x^4+x
^3)/exp(x),x, algorithm="fricas")

[Out]

5*(e^2 + 2*e^(-2*x + 5) + e^(-4*x + 8))*e^(3*x)/(x^3 + x^2)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (22) = 44\).
time = 0.16, size = 104, normalized size = 3.47 \begin {gather*} \frac {\left (5 x^{6} e^{2} + 10 x^{5} e^{2} + 5 x^{4} e^{2}\right ) e^{3 x} + \left (10 x^{6} e^{5} + 20 x^{5} e^{5} + 10 x^{4} e^{5}\right ) e^{x} + \left (5 x^{6} e^{8} + 10 x^{5} e^{8} + 5 x^{4} e^{8}\right ) e^{- x}}{x^{9} + 3 x^{8} + 3 x^{7} + x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x**2-10)*exp(1+2*x)**2+(10*x**2-20*x-20)*exp(4)*exp(1+2*x)+(-5*x**2-20*x-10)*exp(4)**2)/(x**5+2
*x**4+x**3)/exp(x),x)

[Out]

((5*x**6*exp(2) + 10*x**5*exp(2) + 5*x**4*exp(2))*exp(3*x) + (10*x**6*exp(5) + 20*x**5*exp(5) + 10*x**4*exp(5)
)*exp(x) + (5*x**6*exp(8) + 10*x**5*exp(8) + 5*x**4*exp(8))*exp(-x))/(x**9 + 3*x**8 + 3*x**7 + x**6)

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Giac [A]
time = 0.39, size = 30, normalized size = 1.00 \begin {gather*} \frac {5 \, {\left (e^{\left (3 \, x + 2\right )} + 2 \, e^{\left (x + 5\right )} + e^{\left (-x + 8\right )}\right )}}{x^{3} + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15*x^2-10)*exp(1+2*x)^2+(10*x^2-20*x-20)*exp(4)*exp(1+2*x)+(-5*x^2-20*x-10)*exp(4)^2)/(x^5+2*x^4+x
^3)/exp(x),x, algorithm="giac")

[Out]

5*(e^(3*x + 2) + 2*e^(x + 5) + e^(-x + 8))/(x^3 + x^2)

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Mupad [B]
time = 7.84, size = 50, normalized size = 1.67 \begin {gather*} \frac {10\,{\mathrm {e}}^5\,{\mathrm {e}}^x}{x^3+x^2}+\frac {5\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^2}{x^3+x^2}+\frac {5\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^8}{x^3+x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(exp(8)*(20*x + 5*x^2 + 10) - exp(4*x + 2)*(15*x^2 - 10) + exp(4)*exp(2*x + 1)*(20*x - 10*x^2 +
20)))/(x^3 + 2*x^4 + x^5),x)

[Out]

(10*exp(5)*exp(x))/(x^2 + x^3) + (5*exp(3*x)*exp(2))/(x^2 + x^3) + (5*exp(-x)*exp(8))/(x^2 + x^3)

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