3.72.4 \(\int \frac {e^{-\frac {2 (5+(-3+x) \log (\frac {6 x}{5}))}{-3+x}} (9 x^2+e^{\frac {2 (5+(-3+x) \log (\frac {6 x}{5}))}{-3+x}} (-360+240 x-40 x^2))}{360-240 x+40 x^2} \, dx\)

Optimal. Leaf size=18 \[ -25+\frac {1}{64} e^{-\frac {10}{-3+x}}-x \]

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Rubi [A]  time = 0.50, antiderivative size = 19, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 5, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {27, 12, 6688, 14, 2209} \begin {gather*} \frac {1}{64} e^{\frac {10}{3-x}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(9*x^2 + E^((2*(5 + (-3 + x)*Log[(6*x)/5]))/(-3 + x))*(-360 + 240*x - 40*x^2))/(E^((2*(5 + (-3 + x)*Log[(6
*x)/5]))/(-3 + x))*(360 - 240*x + 40*x^2)),x]

[Out]

E^(10/(3 - x))/64 - x

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 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 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 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} &=\int \frac {\exp \left (-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}\right ) \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{40 (-3+x)^2} \, dx\\ &=\frac {1}{40} \int \frac {\exp \left (-\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}\right ) \left (9 x^2+e^{\frac {2 \left (5+(-3+x) \log \left (\frac {6 x}{5}\right )\right )}{-3+x}} \left (-360+240 x-40 x^2\right )\right )}{(-3+x)^2} \, dx\\ &=\frac {1}{40} \int \frac {5 \left (5 e^{-\frac {10}{-3+x}}-32 (-3+x)^2\right )}{4 (3-x)^2} \, dx\\ &=\frac {1}{32} \int \frac {5 e^{-\frac {10}{-3+x}}-32 (-3+x)^2}{(3-x)^2} \, dx\\ &=\frac {1}{32} \operatorname {Subst}\left (\int \frac {5 e^{-10/x}-32 x^2}{x^2} \, dx,x,-3+x\right )\\ &=\frac {1}{32} \operatorname {Subst}\left (\int \left (-32+\frac {5 e^{-10/x}}{x^2}\right ) \, dx,x,-3+x\right )\\ &=-x+\frac {5}{32} \operatorname {Subst}\left (\int \frac {e^{-10/x}}{x^2} \, dx,x,-3+x\right )\\ &=\frac {1}{64} e^{\frac {10}{3-x}}-x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 23, normalized size = 1.28 \begin {gather*} \frac {1}{32} \left (\frac {1}{2} e^{\frac {10}{3-x}}-32 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9*x^2 + E^((2*(5 + (-3 + x)*Log[(6*x)/5]))/(-3 + x))*(-360 + 240*x - 40*x^2))/(E^((2*(5 + (-3 + x)*
Log[(6*x)/5]))/(-3 + x))*(360 - 240*x + 40*x^2)),x]

[Out]

(E^(10/(3 - x))/2 - 32*x)/32

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fricas [A]  time = 1.02, size = 47, normalized size = 2.61 \begin {gather*} \frac {1}{400} \, {\left (9 \, x^{2} - 400 \, x e^{\left (\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )}\right )} e^{\left (-\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x^2+240*x-360)*exp(((x-3)*log(6/5*x)+5)/(x-3))^2+9*x^2)/(40*x^2-240*x+360)/exp(((x-3)*log(6/5*
x)+5)/(x-3))^2,x, algorithm="fricas")

[Out]

1/400*(9*x^2 - 400*x*e^(2*((x - 3)*log(6/5*x) + 5)/(x - 3)))*e^(-2*((x - 3)*log(6/5*x) + 5)/(x - 3))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (9 \, x^{2} - 40 \, {\left (x^{2} - 6 \, x + 9\right )} e^{\left (\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )}\right )} e^{\left (-\frac {2 \, {\left ({\left (x - 3\right )} \log \left (\frac {6}{5} \, x\right ) + 5\right )}}{x - 3}\right )}}{40 \, {\left (x^{2} - 6 \, x + 9\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x^2+240*x-360)*exp(((x-3)*log(6/5*x)+5)/(x-3))^2+9*x^2)/(40*x^2-240*x+360)/exp(((x-3)*log(6/5*
x)+5)/(x-3))^2,x, algorithm="giac")

[Out]

integrate(1/40*(9*x^2 - 40*(x^2 - 6*x + 9)*e^(2*((x - 3)*log(6/5*x) + 5)/(x - 3)))*e^(-2*((x - 3)*log(6/5*x) +
5)/(x - 3))/(x^2 - 6*x + 9), x)

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maple [A]  time = 0.15, size = 32, normalized size = 1.78

method	result	size
risch	\(-x +\frac {9 x^{2} {\mathrm e}^{-\frac {2 \left (\ln \left (\frac {6 x}{5}\right ) x -3 \ln \left (\frac {6 x}{5}\right )+5\right )}{x -3}}}{400}\)	\(32\)
default	\(-x +\frac {\left (-\frac {27}{10} x^{2}+\frac {9}{10} x^{3}\right ) {\mathrm e}^{-\frac {2 \left (\left (x -3\right ) \ln \left (\frac {6 x}{5}\right )+5\right )}{x -3}}}{40 x -120}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-40*x^2+240*x-360)*exp(((x-3)*ln(6/5*x)+5)/(x-3))^2+9*x^2)/(40*x^2-240*x+360)/exp(((x-3)*ln(6/5*x)+5)/(x
-3))^2,x,method=_RETURNVERBOSE)

[Out]

-x+9/400*x^2*exp(-2*(ln(6/5*x)*x-3*ln(6/5*x)+5)/(x-3))

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maxima [A]  time = 0.48, size = 14, normalized size = 0.78 \begin {gather*} -x + \frac {1}{64} \, e^{\left (-\frac {10}{x - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x^2+240*x-360)*exp(((x-3)*log(6/5*x)+5)/(x-3))^2+9*x^2)/(40*x^2-240*x+360)/exp(((x-3)*log(6/5*
x)+5)/(x-3))^2,x, algorithm="maxima")

[Out]

-x + 1/64*e^(-10/(x - 3))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} -\int \frac {{\mathrm {e}}^{-\frac {2\,\left (\ln \left (\frac {6\,x}{5}\right )\,\left (x-3\right )+5\right )}{x-3}}\,\left ({\mathrm {e}}^{\frac {2\,\left (\ln \left (\frac {6\,x}{5}\right )\,\left (x-3\right )+5\right )}{x-3}}\,\left (40\,x^2-240\,x+360\right )-9\,x^2\right )}{40\,x^2-240\,x+360} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(2*(log((6*x)/5)*(x - 3) + 5))/(x - 3))*(exp((2*(log((6*x)/5)*(x - 3) + 5))/(x - 3))*(40*x^2 - 240*
x + 360) - 9*x^2))/(40*x^2 - 240*x + 360),x)

[Out]

-int((exp(-(2*(log((6*x)/5)*(x - 3) + 5))/(x - 3))*(exp((2*(log((6*x)/5)*(x - 3) + 5))/(x - 3))*(40*x^2 - 240*
x + 360) - 9*x^2))/(40*x^2 - 240*x + 360), x)

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sympy [B]  time = 0.52, size = 26, normalized size = 1.44 \begin {gather*} \frac {9 x^{2} e^{- \frac {2 \left (\left (x - 3\right ) \log {\left (\frac {6 x}{5} \right )} + 5\right )}{x - 3}}}{400} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x**2+240*x-360)*exp(((x-3)*ln(6/5*x)+5)/(x-3))**2+9*x**2)/(40*x**2-240*x+360)/exp(((x-3)*ln(6/
5*x)+5)/(x-3))**2,x)

[Out]

9*x**2*exp(-2*((x - 3)*log(6*x/5) + 5)/(x - 3))/400 - x

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