3.92.49 \(\int \frac {e^{8-x} (100+100 x+100 x^2)+e^5 (2 x^2+4 x^3)+e^{3 x} (100 e^{8-x} x^2+e^5 (-2 x^3-3 x^4))}{25 x^2} \, dx\)

Optimal. Leaf size=38 \[ e^5 \left (4+\left (2-e^{3 x}+\frac {2}{x}\right ) \left (-2 e^{3-x}+\frac {x^2}{25}\right )\right ) \]

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Rubi [A]  time = 0.20, antiderivative size = 59, normalized size of antiderivative = 1.55, number of steps used = 17, number of rules used = 8, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 14, 2194, 2196, 2176, 2199, 2177, 2178} \begin {gather*} -\frac {1}{25} e^{3 x+5} x^2-4 e^{8-x}+2 e^{2 x+8}+\frac {1}{50} e^5 (2 x+1)^2-\frac {4 e^{8-x}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(8 - x)*(100 + 100*x + 100*x^2) + E^5*(2*x^2 + 4*x^3) + E^(3*x)*(100*E^(8 - x)*x^2 + E^5*(-2*x^3 - 3*x^
4)))/(25*x^2),x]

[Out]

-4*E^(8 - x) + 2*E^(8 + 2*x) - (4*E^(8 - x))/x - (E^(5 + 3*x)*x^2)/25 + (E^5*(1 + 2*x)^2)/50

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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 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 2196

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

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} &=\frac {1}{25} \int \frac {e^{8-x} \left (100+100 x+100 x^2\right )+e^5 \left (2 x^2+4 x^3\right )+e^{3 x} \left (100 e^{8-x} x^2+e^5 \left (-2 x^3-3 x^4\right )\right )}{x^2} \, dx\\ &=\frac {1}{25} \int \left (100 e^{8+2 x}+2 e^5 (1+2 x)-e^{5+3 x} x (2+3 x)+\frac {100 e^{8-x} \left (1+x+x^2\right )}{x^2}\right ) \, dx\\ &=\frac {1}{50} e^5 (1+2 x)^2-\frac {1}{25} \int e^{5+3 x} x (2+3 x) \, dx+4 \int e^{8+2 x} \, dx+4 \int \frac {e^{8-x} \left (1+x+x^2\right )}{x^2} \, dx\\ &=2 e^{8+2 x}+\frac {1}{50} e^5 (1+2 x)^2-\frac {1}{25} \int \left (2 e^{5+3 x} x+3 e^{5+3 x} x^2\right ) \, dx+4 \int \left (e^{8-x}+\frac {e^{8-x}}{x^2}+\frac {e^{8-x}}{x}\right ) \, dx\\ &=2 e^{8+2 x}+\frac {1}{50} e^5 (1+2 x)^2-\frac {2}{25} \int e^{5+3 x} x \, dx-\frac {3}{25} \int e^{5+3 x} x^2 \, dx+4 \int e^{8-x} \, dx+4 \int \frac {e^{8-x}}{x^2} \, dx+4 \int \frac {e^{8-x}}{x} \, dx\\ &=-4 e^{8-x}+2 e^{8+2 x}-\frac {4 e^{8-x}}{x}-\frac {2}{75} e^{5+3 x} x-\frac {1}{25} e^{5+3 x} x^2+\frac {1}{50} e^5 (1+2 x)^2+4 e^8 \text {Ei}(-x)+\frac {2}{75} \int e^{5+3 x} \, dx+\frac {2}{25} \int e^{5+3 x} x \, dx-4 \int \frac {e^{8-x}}{x} \, dx\\ &=-4 e^{8-x}+2 e^{8+2 x}+\frac {2}{225} e^{5+3 x}-\frac {4 e^{8-x}}{x}-\frac {1}{25} e^{5+3 x} x^2+\frac {1}{50} e^5 (1+2 x)^2-\frac {2}{75} \int e^{5+3 x} \, dx\\ &=-4 e^{8-x}+2 e^{8+2 x}-\frac {4 e^{8-x}}{x}-\frac {1}{25} e^{5+3 x} x^2+\frac {1}{50} e^5 (1+2 x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 38, normalized size = 1.00 \begin {gather*} -\frac {e^{5-x} \left (-2+\left (-2+e^{3 x}\right ) x\right ) \left (-50 e^3+e^x x^2\right )}{25 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(8 - x)*(100 + 100*x + 100*x^2) + E^5*(2*x^2 + 4*x^3) + E^(3*x)*(100*E^(8 - x)*x^2 + E^5*(-2*x^3
- 3*x^4)))/(25*x^2),x]

[Out]

-1/25*(E^(5 - x)*(-2 + (-2 + E^(3*x))*x)*(-50*E^3 + E^x*x^2))/x

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fricas [A]  time = 0.52, size = 53, normalized size = 1.39 \begin {gather*} -\frac {{\left (x^{3} e^{29} - 50 \, x e^{\left (-x + 32\right )} - 2 \, {\left (x^{3} + x^{2}\right )} e^{\left (-3 \, x + 29\right )} + 100 \, {\left (x + 1\right )} e^{\left (-4 \, x + 32\right )}\right )} e^{\left (3 \, x - 24\right )}}{25 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((100*x^2*exp(5)*exp(3-x)+(-3*x^4-2*x^3)*exp(5))*exp(3*x)+(100*x^2+100*x+100)*exp(5)*exp(3-x)+(
4*x^3+2*x^2)*exp(5))/x^2,x, algorithm="fricas")

[Out]

-1/25*(x^3*e^29 - 50*x*e^(-x + 32) - 2*(x^3 + x^2)*e^(-3*x + 29) + 100*(x + 1)*e^(-4*x + 32))*e^(3*x - 24)/x

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giac [A]  time = 0.16, size = 57, normalized size = 1.50 \begin {gather*} \frac {2 \, x^{3} e^{5} - x^{3} e^{\left (3 \, x + 5\right )} + 2 \, x^{2} e^{5} + 50 \, x e^{\left (2 \, x + 8\right )} - 100 \, x e^{\left (-x + 8\right )} - 100 \, e^{\left (-x + 8\right )}}{25 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((100*x^2*exp(5)*exp(3-x)+(-3*x^4-2*x^3)*exp(5))*exp(3*x)+(100*x^2+100*x+100)*exp(5)*exp(3-x)+(
4*x^3+2*x^2)*exp(5))/x^2,x, algorithm="giac")

[Out]

1/25*(2*x^3*e^5 - x^3*e^(3*x + 5) + 2*x^2*e^5 + 50*x*e^(2*x + 8) - 100*x*e^(-x + 8) - 100*e^(-x + 8))/x

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maple [A]  time = 0.11, size = 47, normalized size = 1.24

method	result	size
risch	\(\frac {2 x^{2} {\mathrm e}^{5}}{25}+\frac {2 x \,{\mathrm e}^{5}}{25}-\frac {x^{2} {\mathrm e}^{3 x +5}}{25}+2 \,{\mathrm e}^{2 x +8}-\frac {4 \left (x +1\right ) {\mathrm e}^{8-x}}{x}\)	\(47\)
norman	\(\frac {\left (-4 \,{\mathrm e}^{3} {\mathrm e}^{5}-4 x \,{\mathrm e}^{3} {\mathrm e}^{5}+\frac {2 x^{2} {\mathrm e}^{5} {\mathrm e}^{x}}{25}+\frac {2 x^{3} {\mathrm e}^{5} {\mathrm e}^{x}}{25}-\frac {x^{3} {\mathrm e}^{5} {\mathrm e}^{4 x}}{25}+2 x \,{\mathrm e}^{3} {\mathrm e}^{5} {\mathrm e}^{3 x}\right ) {\mathrm e}^{-x}}{x}\)	\(63\)
default	\(\frac {2 x^{2} {\mathrm e}^{5}}{25}-\frac {2 \,{\mathrm e}^{5} \left (\frac {x \,{\mathrm e}^{3 x}}{3}-\frac {{\mathrm e}^{3 x}}{9}\right )}{25}-\frac {3 \,{\mathrm e}^{5} \left (\frac {x^{2} {\mathrm e}^{3 x}}{3}-\frac {2 x \,{\mathrm e}^{3 x}}{9}+\frac {2 \,{\mathrm e}^{3 x}}{27}\right )}{25}-4 \,{\mathrm e}^{5} {\mathrm e}^{3} {\mathrm e}^{-x}+2 \,{\mathrm e}^{2 x} {\mathrm e}^{5} {\mathrm e}^{3}+4 \,{\mathrm e}^{3} {\mathrm e}^{5} \left (-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )\right )-4 \,{\mathrm e}^{3} {\mathrm e}^{5} \expIntegralEi \left (1, x\right )+\frac {2 x \,{\mathrm e}^{5}}{25}\)	\(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*((100*x^2*exp(5)*exp(3-x)+(-3*x^4-2*x^3)*exp(5))*exp(3*x)+(100*x^2+100*x+100)*exp(5)*exp(3-x)+(4*x^3+
2*x^2)*exp(5))/x^2,x,method=_RETURNVERBOSE)

[Out]

2/25*x^2*exp(5)+2/25*x*exp(5)-1/25*x^2*exp(3*x+5)+2*exp(2*x+8)-4*(x+1)/x*exp(8-x)

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maxima [C]  time = 0.39, size = 83, normalized size = 2.18 \begin {gather*} \frac {2}{25} \, x^{2} e^{5} + 4 \, {\rm Ei}\left (-x\right ) e^{8} + \frac {2}{25} \, x e^{5} - \frac {1}{225} \, {\left (9 \, x^{2} e^{5} - 6 \, x e^{5} + 2 \, e^{5}\right )} e^{\left (3 \, x\right )} - \frac {2}{225} \, {\left (3 \, x e^{5} - e^{5}\right )} e^{\left (3 \, x\right )} - 4 \, e^{8} \Gamma \left (-1, x\right ) + 2 \, e^{\left (2 \, x + 8\right )} - 4 \, e^{\left (-x + 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((100*x^2*exp(5)*exp(3-x)+(-3*x^4-2*x^3)*exp(5))*exp(3*x)+(100*x^2+100*x+100)*exp(5)*exp(3-x)+(
4*x^3+2*x^2)*exp(5))/x^2,x, algorithm="maxima")

[Out]

2/25*x^2*e^5 + 4*Ei(-x)*e^8 + 2/25*x*e^5 - 1/225*(9*x^2*e^5 - 6*x*e^5 + 2*e^5)*e^(3*x) - 2/225*(3*x*e^5 - e^5)
*e^(3*x) - 4*e^8*gamma(-1, x) + 2*e^(2*x + 8) - 4*e^(-x + 8)

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mupad [B]  time = 6.94, size = 53, normalized size = 1.39 \begin {gather*} 2\,{\mathrm {e}}^{2\,x+8}+\frac {2\,x\,{\mathrm {e}}^5}{25}+\frac {2\,x^2\,{\mathrm {e}}^5}{25}-\frac {x^2\,{\mathrm {e}}^{3\,x+5}}{25}-\frac {{\mathrm {e}}^{3-x}\,\left (4\,{\mathrm {e}}^5+4\,x\,{\mathrm {e}}^5\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(5)*(2*x^2 + 4*x^3))/25 - (exp(3*x)*(exp(5)*(2*x^3 + 3*x^4) - 100*x^2*exp(5)*exp(3 - x)))/25 + (exp(5
)*exp(3 - x)*(100*x + 100*x^2 + 100))/25)/x^2,x)

[Out]

2*exp(2*x + 8) + (2*x*exp(5))/25 + (2*x^2*exp(5))/25 - (x^2*exp(3*x + 5))/25 - (exp(3 - x)*(4*exp(5) + 4*x*exp
(5)))/x

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sympy [B]  time = 0.24, size = 70, normalized size = 1.84 \begin {gather*} \frac {2 x^{2} e^{5}}{25} + \frac {2 x e^{5}}{25} + \frac {- x^{3} e^{5} e^{3 x} + 50 x e^{8} \left (e^{3 x}\right )^{\frac {2}{3}} + \frac {- 100 x e^{8} - 100 e^{8}}{\sqrt [3]{e^{3 x}}}}{25 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*((100*x**2*exp(5)*exp(3-x)+(-3*x**4-2*x**3)*exp(5))*exp(3*x)+(100*x**2+100*x+100)*exp(5)*exp(3-
x)+(4*x**3+2*x**2)*exp(5))/x**2,x)

[Out]

2*x**2*exp(5)/25 + 2*x*exp(5)/25 + (-x**3*exp(5)*exp(3*x) + 50*x*exp(8)*exp(3*x)**(2/3) + (-100*x*exp(8) - 100
*exp(8))/exp(3*x)**(1/3))/(25*x)

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