3.32.47 \(\int \frac {e^{-\log ^2(3)} (75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} (260-160 x+20 x^2))}{50-20 x+2 x^2} \, dx\)

Optimal. Leaf size=32 \[ (3-x) \left (\frac {10 e^x}{5-x}+\frac {1}{2} e^{-\log ^2(3)} x\right ) \]

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Rubi [A]  time = 0.28, antiderivative size = 46, normalized size of antiderivative = 1.44, number of steps used = 11, number of rules used = 7, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.127, Rules used = {12, 27, 6688, 2199, 2194, 2177, 2178} \begin {gather*} -\frac {1}{2} x^2 e^{-\log ^2(3)}+10 e^x-\frac {20 e^x}{5-x}+\frac {3}{2} x e^{-\log ^2(3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75 - 80*x + 23*x^2 - 2*x^3 + E^(x + Log[3]^2)*(260 - 160*x + 20*x^2))/(E^Log[3]^2*(50 - 20*x + 2*x^2)),x]

[Out]

10*E^x - (20*E^x)/(5 - x) + (3*x)/(2*E^Log[3]^2) - x^2/(2*E^Log[3]^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 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

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{-\log ^2(3)} \int \frac {75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )}{50-20 x+2 x^2} \, dx\\ &=e^{-\log ^2(3)} \int \frac {75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )}{2 (-5+x)^2} \, dx\\ &=\frac {1}{2} e^{-\log ^2(3)} \int \frac {75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )}{(-5+x)^2} \, dx\\ &=\frac {1}{2} e^{-\log ^2(3)} \int \left (3-2 x+\frac {20 e^{x+\log ^2(3)} \left (13-8 x+x^2\right )}{(-5+x)^2}\right ) \, dx\\ &=\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2+\left (10 e^{-\log ^2(3)}\right ) \int \frac {e^{x+\log ^2(3)} \left (13-8 x+x^2\right )}{(-5+x)^2} \, dx\\ &=\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2+\left (10 e^{-\log ^2(3)}\right ) \int \left (e^{x+\log ^2(3)}-\frac {2 e^{x+\log ^2(3)}}{(-5+x)^2}+\frac {2 e^{x+\log ^2(3)}}{-5+x}\right ) \, dx\\ &=\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2+\left (10 e^{-\log ^2(3)}\right ) \int e^{x+\log ^2(3)} \, dx-\left (20 e^{-\log ^2(3)}\right ) \int \frac {e^{x+\log ^2(3)}}{(-5+x)^2} \, dx+\left (20 e^{-\log ^2(3)}\right ) \int \frac {e^{x+\log ^2(3)}}{-5+x} \, dx\\ &=10 e^x-\frac {20 e^x}{5-x}+\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2+20 e^5 \text {Ei}(-5+x)-\left (20 e^{-\log ^2(3)}\right ) \int \frac {e^{x+\log ^2(3)}}{-5+x} \, dx\\ &=10 e^x-\frac {20 e^x}{5-x}+\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 46, normalized size = 1.44 \begin {gather*} 10 e^x-\frac {20 e^x}{5-x}+\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75 - 80*x + 23*x^2 - 2*x^3 + E^(x + Log[3]^2)*(260 - 160*x + 20*x^2))/(E^Log[3]^2*(50 - 20*x + 2*x^
2)),x]

[Out]

10*E^x - (20*E^x)/(5 - x) + (3*x)/(2*E^Log[3]^2) - x^2/(2*E^Log[3]^2)

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fricas [A]  time = 0.50, size = 38, normalized size = 1.19 \begin {gather*} -\frac {{\left (x^{3} - 8 \, x^{2} - 20 \, {\left (x - 3\right )} e^{\left (\log \relax (3)^{2} + x\right )} + 15 \, x\right )} e^{\left (-\log \relax (3)^{2}\right )}}{2 \, {\left (x - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x^2-160*x+260)*exp(x)*exp(log(3)^2)-2*x^3+23*x^2-80*x+75)/(2*x^2-20*x+50)/exp(log(3)^2),x, algo
rithm="fricas")

[Out]

-1/2*(x^3 - 8*x^2 - 20*(x - 3)*e^(log(3)^2 + x) + 15*x)*e^(-log(3)^2)/(x - 5)

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giac [A]  time = 0.28, size = 45, normalized size = 1.41 \begin {gather*} -\frac {{\left (x^{3} - 8 \, x^{2} - 20 \, x e^{\left (\log \relax (3)^{2} + x\right )} + 15 \, x + 60 \, e^{\left (\log \relax (3)^{2} + x\right )}\right )} e^{\left (-\log \relax (3)^{2}\right )}}{2 \, {\left (x - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x^2-160*x+260)*exp(x)*exp(log(3)^2)-2*x^3+23*x^2-80*x+75)/(2*x^2-20*x+50)/exp(log(3)^2),x, algo
rithm="giac")

[Out]

-1/2*(x^3 - 8*x^2 - 20*x*e^(log(3)^2 + x) + 15*x + 60*e^(log(3)^2 + x))*e^(-log(3)^2)/(x - 5)

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maple [A]  time = 0.59, size = 36, normalized size = 1.12

method	result	size
risch	\(-\frac {{\mathrm e}^{-\ln \relax (3)^{2}} x^{2}}{2}+\frac {3 x \,{\mathrm e}^{-\ln \relax (3)^{2}}}{2}+\frac {10 \left (x -3\right ) {\mathrm e}^{x}}{x -5}\)	\(36\)
default	\(\frac {{\mathrm e}^{-\ln \relax (3)^{2}} \left (-x^{2}+3 x +\frac {40 \,{\mathrm e}^{\ln \relax (3)^{2}} {\mathrm e}^{x}}{x -5}+20 \,{\mathrm e}^{x} {\mathrm e}^{\ln \relax (3)^{2}}\right )}{2}\)	\(42\)
norman	\(\frac {10 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{-\ln \relax (3)^{2}} x^{2}-\frac {{\mathrm e}^{-\ln \relax (3)^{2}} x^{3}}{2}-30 \,{\mathrm e}^{x}-\frac {75 \,{\mathrm e}^{-\ln \relax (3)^{2}}}{2}}{x -5}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((20*x^2-160*x+260)*exp(x)*exp(ln(3)^2)-2*x^3+23*x^2-80*x+75)/(2*x^2-20*x+50)/exp(ln(3)^2),x,method=_RETUR
NVERBOSE)

[Out]

-1/2*exp(-ln(3)^2)*x^2+3/2*x*exp(-ln(3)^2)+10*(x-3)/(x-5)*exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{2} \, {\left (x^{2} - 3 \, x - \frac {20 \, {\left (x^{2} e^{\left (\log \relax (3)^{2}\right )} - 8 \, x e^{\left (\log \relax (3)^{2}\right )}\right )} e^{x}}{x^{2} - 10 \, x + 25} + \frac {260 \, e^{\left (\log \relax (3)^{2} + 5\right )} E_{2}\left (-x + 5\right )}{x - 5} - \int \frac {40 \, {\left (x e^{\left (\log \relax (3)^{2}\right )} - 20 \, e^{\left (\log \relax (3)^{2}\right )}\right )} e^{x}}{x^{3} - 15 \, x^{2} + 75 \, x - 125}\,{d x}\right )} e^{\left (-\log \relax (3)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x^2-160*x+260)*exp(x)*exp(log(3)^2)-2*x^3+23*x^2-80*x+75)/(2*x^2-20*x+50)/exp(log(3)^2),x, algo
rithm="maxima")

[Out]

-1/2*(x^2 - 3*x - 20*(x^2*e^(log(3)^2) - 8*x*e^(log(3)^2))*e^x/(x^2 - 10*x + 25) + 260*e^(log(3)^2 + 5)*exp_in
tegral_e(2, -x + 5)/(x - 5) - integrate(40*(x*e^(log(3)^2) - 20*e^(log(3)^2))*e^x/(x^3 - 15*x^2 + 75*x - 125),
x))*e^(-log(3)^2)

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mupad [B]  time = 1.93, size = 36, normalized size = 1.12 \begin {gather*} \frac {{\mathrm {e}}^{-{\ln \relax (3)}^2}\,\left (x-3\right )\,\left (5\,x+20\,{\mathrm {e}}^{x+{\ln \relax (3)}^2}-x^2\right )}{2\,\left (x-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-log(3)^2)*(23*x^2 - 80*x - 2*x^3 + exp(log(3)^2)*exp(x)*(20*x^2 - 160*x + 260) + 75))/(2*x^2 - 20*x
+ 50),x)

[Out]

(exp(-log(3)^2)*(x - 3)*(5*x + 20*exp(x + log(3)^2) - x^2))/(2*(x - 5))

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sympy [A]  time = 0.14, size = 34, normalized size = 1.06 \begin {gather*} - \frac {x^{2}}{2 e^{\log {\relax (3 )}^{2}}} + \frac {3 x}{2 e^{\log {\relax (3 )}^{2}}} + \frac {\left (10 x - 30\right ) e^{x}}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x**2-160*x+260)*exp(x)*exp(ln(3)**2)-2*x**3+23*x**2-80*x+75)/(2*x**2-20*x+50)/exp(ln(3)**2),x)

[Out]

-x**2*exp(-log(3)**2)/2 + 3*x*exp(-log(3)**2)/2 + (10*x - 30)*exp(x)/(x - 5)

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