3.53.93 \(\int \frac {e^{162-36 x+2 x^2} (-17 x^2+180 x^3-20 x^4+e^x (-24 x+350 x^2-40 x^3))+e^{162-36 x+2 x^2} (3 x^2-36 x^3+4 x^4+e^x (4 x-70 x^2+8 x^3)) \log (x)}{-125+75 \log (x)-15 \log ^2(x)+\log ^3(x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {e^{2 (-9+x)^2} x^2 \left (2 e^x+x\right )}{(-5+\log (x))^2} \]

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Rubi [B]  time = 1.40, antiderivative size = 107, normalized size of antiderivative = 4.12, number of steps used = 5, number of rules used = 3, integrand size = 116, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6688, 6742, 2288} \begin {gather*} \frac {e^{2 (9-x)^2} x^2 \left (-5 x^2+x^2 \log (x)+45 x-9 x \log (x)\right )}{(9-x) (5-\log (x))^3}-\frac {2 e^{2 (9-x)^2+x} x \left (-20 x^2+4 x^2 \log (x)+175 x-35 x \log (x)\right )}{(1-4 (9-x)) (5-\log (x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(162 - 36*x + 2*x^2)*(-17*x^2 + 180*x^3 - 20*x^4 + E^x*(-24*x + 350*x^2 - 40*x^3)) + E^(162 - 36*x + 2*
x^2)*(3*x^2 - 36*x^3 + 4*x^4 + E^x*(4*x - 70*x^2 + 8*x^3))*Log[x])/(-125 + 75*Log[x] - 15*Log[x]^2 + Log[x]^3)
,x]

[Out]

(E^(2*(9 - x)^2)*x^2*(45*x - 5*x^2 - 9*x*Log[x] + x^2*Log[x]))/((9 - x)*(5 - Log[x])^3) - (2*E^(2*(9 - x)^2 +
x)*x*(175*x - 20*x^2 - 35*x*Log[x] + 4*x^2*Log[x]))/((1 - 4*(9 - x))*(5 - Log[x])^3)

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

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^{2 (-9+x)^2} x \left (-e^x \left (-24+350 x-40 x^2\right )-x \left (-17+180 x-20 x^2\right )-\left (x \left (3-36 x+4 x^2\right )+e^x \left (4-70 x+8 x^2\right )\right ) \log (x)\right )}{(5-\log (x))^3} \, dx\\ &=\int \left (\frac {e^{2 (-9+x)^2} x^2 \left (-17+180 x-20 x^2+3 \log (x)-36 x \log (x)+4 x^2 \log (x)\right )}{(-5+\log (x))^3}+\frac {2 e^{2 (-9+x)^2+x} x \left (-12+175 x-20 x^2+2 \log (x)-35 x \log (x)+4 x^2 \log (x)\right )}{(-5+\log (x))^3}\right ) \, dx\\ &=2 \int \frac {e^{2 (-9+x)^2+x} x \left (-12+175 x-20 x^2+2 \log (x)-35 x \log (x)+4 x^2 \log (x)\right )}{(-5+\log (x))^3} \, dx+\int \frac {e^{2 (-9+x)^2} x^2 \left (-17+180 x-20 x^2+3 \log (x)-36 x \log (x)+4 x^2 \log (x)\right )}{(-5+\log (x))^3} \, dx\\ &=\frac {e^{2 (9-x)^2} x^2 \left (45 x-5 x^2-9 x \log (x)+x^2 \log (x)\right )}{(9-x) (5-\log (x))^3}-\frac {2 e^{2 (9-x)^2+x} x \left (175 x-20 x^2-35 x \log (x)+4 x^2 \log (x)\right )}{(1-4 (9-x)) (5-\log (x))^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.48, size = 26, normalized size = 1.00 \begin {gather*} \frac {e^{2 (-9+x)^2} x^2 \left (2 e^x+x\right )}{(-5+\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(162 - 36*x + 2*x^2)*(-17*x^2 + 180*x^3 - 20*x^4 + E^x*(-24*x + 350*x^2 - 40*x^3)) + E^(162 - 36*
x + 2*x^2)*(3*x^2 - 36*x^3 + 4*x^4 + E^x*(4*x - 70*x^2 + 8*x^3))*Log[x])/(-125 + 75*Log[x] - 15*Log[x]^2 + Log
[x]^3),x]

[Out]

(E^(2*(-9 + x)^2)*x^2*(2*E^x + x))/(-5 + Log[x])^2

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fricas [A]  time = 0.55, size = 35, normalized size = 1.35 \begin {gather*} \frac {{\left (x^{3} + 2 \, x^{2} e^{x}\right )} e^{\left (2 \, x^{2} - 36 \, x + 162\right )}}{\log \relax (x)^{2} - 10 \, \log \relax (x) + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^3-70*x^2+4*x)*exp(x)+4*x^4-36*x^3+3*x^2)*exp(x^2-18*x+81)^2*log(x)+((-40*x^3+350*x^2-24*x)*ex
p(x)-20*x^4+180*x^3-17*x^2)*exp(x^2-18*x+81)^2)/(log(x)^3-15*log(x)^2+75*log(x)-125),x, algorithm="fricas")

[Out]

(x^3 + 2*x^2*e^x)*e^(2*x^2 - 36*x + 162)/(log(x)^2 - 10*log(x) + 25)

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giac [A]  time = 0.24, size = 45, normalized size = 1.73 \begin {gather*} \frac {x^{3} e^{\left (2 \, x^{2} - 36 \, x + 162\right )} + 2 \, x^{2} e^{\left (2 \, x^{2} - 35 \, x + 162\right )}}{\log \relax (x)^{2} - 10 \, \log \relax (x) + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^3-70*x^2+4*x)*exp(x)+4*x^4-36*x^3+3*x^2)*exp(x^2-18*x+81)^2*log(x)+((-40*x^3+350*x^2-24*x)*ex
p(x)-20*x^4+180*x^3-17*x^2)*exp(x^2-18*x+81)^2)/(log(x)^3-15*log(x)^2+75*log(x)-125),x, algorithm="giac")

[Out]

(x^3*e^(2*x^2 - 36*x + 162) + 2*x^2*e^(2*x^2 - 35*x + 162))/(log(x)^2 - 10*log(x) + 25)

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maple [A]  time = 0.06, size = 25, normalized size = 0.96

method	result	size
risch	\(\frac {x^{2} {\mathrm e}^{2 \left (x -9\right )^{2}} \left (2 \,{\mathrm e}^{x}+x \right )}{\left (\ln \relax (x )-5\right )^{2}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x^3-70*x^2+4*x)*exp(x)+4*x^4-36*x^3+3*x^2)*exp(x^2-18*x+81)^2*ln(x)+((-40*x^3+350*x^2-24*x)*exp(x)-20
*x^4+180*x^3-17*x^2)*exp(x^2-18*x+81)^2)/(ln(x)^3-15*ln(x)^2+75*ln(x)-125),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.46, size = 39, normalized size = 1.50 \begin {gather*} \frac {{\left (x^{3} e^{162} + 2 \, x^{2} e^{\left (x + 162\right )}\right )} e^{\left (2 \, x^{2} - 36 \, x\right )}}{\log \relax (x)^{2} - 10 \, \log \relax (x) + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^3-70*x^2+4*x)*exp(x)+4*x^4-36*x^3+3*x^2)*exp(x^2-18*x+81)^2*log(x)+((-40*x^3+350*x^2-24*x)*ex
p(x)-20*x^4+180*x^3-17*x^2)*exp(x^2-18*x+81)^2)/(log(x)^3-15*log(x)^2+75*log(x)-125),x, algorithm="maxima")

[Out]

(x^3*e^162 + 2*x^2*e^(x + 162))*e^(2*x^2 - 36*x)/(log(x)^2 - 10*log(x) + 25)

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mupad [B]  time = 3.67, size = 27, normalized size = 1.04 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{2\,x^2-36\,x+162}\,\left (x+2\,{\mathrm {e}}^x\right )}{{\left (\ln \relax (x)-5\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x^2 - 36*x + 162)*(17*x^2 - 180*x^3 + 20*x^4 + exp(x)*(24*x - 350*x^2 + 40*x^3)) - exp(2*x^2 - 36*
x + 162)*log(x)*(3*x^2 - 36*x^3 + 4*x^4 + exp(x)*(4*x - 70*x^2 + 8*x^3)))/(75*log(x) - 15*log(x)^2 + log(x)^3
- 125),x)

[Out]

(x^2*exp(2*x^2 - 36*x + 162)*(x + 2*exp(x)))/(log(x) - 5)^2

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sympy [A]  time = 0.46, size = 34, normalized size = 1.31 \begin {gather*} \frac {\left (x^{3} + 2 x^{2} e^{x}\right ) e^{2 x^{2} - 36 x + 162}}{\log {\relax (x )}^{2} - 10 \log {\relax (x )} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x**3-70*x**2+4*x)*exp(x)+4*x**4-36*x**3+3*x**2)*exp(x**2-18*x+81)**2*ln(x)+((-40*x**3+350*x**2-
24*x)*exp(x)-20*x**4+180*x**3-17*x**2)*exp(x**2-18*x+81)**2)/(ln(x)**3-15*ln(x)**2+75*ln(x)-125),x)

[Out]

(x**3 + 2*x**2*exp(x))*exp(2*x**2 - 36*x + 162)/(log(x)**2 - 10*log(x) + 25)

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