3.18.68 \(\int \frac {(-128+320 x) \log ^2(9)+e^{x^2} (-20 x \log (9)+(64 x^3-160 x^4) \log ^2(9))+e^{x^2} (-8+20 x+16 x^2-40 x^3) \log (9) \log (2-5 x)}{(-512+1280 x) \log ^2(9)+e^{x^2} (-256 x+640 x^2) \log ^2(9)+e^{2 x^2} (-32 x^2+80 x^3) \log ^2(9)+(e^{x^2} (-64+160 x) \log (9)+e^{2 x^2} (-16 x+40 x^2) \log (9)) \log (2-5 x)+e^{2 x^2} (-2+5 x) \log ^2(2-5 x)} \, dx\)

Optimal. Leaf size=28 \[ \frac {x}{4+e^{x^2} \left (x+\frac {\log (2-5 x)}{4 \log (9)}\right )} \]

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Rubi [F]  time = 57.45, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {(-128+320 x) \log ^2(9)+e^{x^2} \left (-20 x \log (9)+\left (64 x^3-160 x^4\right ) \log ^2(9)\right )+e^{x^2} \left (-8+20 x+16 x^2-40 x^3\right ) \log (9) \log (2-5 x)}{(-512+1280 x) \log ^2(9)+e^{x^2} \left (-256 x+640 x^2\right ) \log ^2(9)+e^{2 x^2} \left (-32 x^2+80 x^3\right ) \log ^2(9)+\left (e^{x^2} (-64+160 x) \log (9)+e^{2 x^2} \left (-16 x+40 x^2\right ) \log (9)\right ) \log (2-5 x)+e^{2 x^2} (-2+5 x) \log ^2(2-5 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-128 + 320*x)*Log[9]^2 + E^x^2*(-20*x*Log[9] + (64*x^3 - 160*x^4)*Log[9]^2) + E^x^2*(-8 + 20*x + 16*x^2
- 40*x^3)*Log[9]*Log[2 - 5*x])/((-512 + 1280*x)*Log[9]^2 + E^x^2*(-256*x + 640*x^2)*Log[9]^2 + E^(2*x^2)*(-32*
x^2 + 80*x^3)*Log[9]^2 + (E^x^2*(-64 + 160*x)*Log[9] + E^(2*x^2)*(-16*x + 40*x^2)*Log[9])*Log[2 - 5*x] + E^(2*
x^2)*(-2 + 5*x)*Log[2 - 5*x]^2),x]

[Out]

(64*(5 - 8*Log[9])*Log[9]^2*Defer[Int][1/((4*x*Log[9] + Log[2 - 5*x])*(16*Log[9] + 4*E^x^2*x*Log[9] + E^x^2*Lo
g[2 - 5*x])^2), x])/5 + (512*Log[9]^3*Defer[Int][1/((4*x*Log[9] + Log[2 - 5*x])*(16*Log[9] + 4*E^x^2*x*Log[9]
+ E^x^2*Log[2 - 5*x])^2), x])/5 + 256*Log[9]^3*Defer[Int][x/((4*x*Log[9] + Log[2 - 5*x])*(16*Log[9] + 4*E^x^2*
x*Log[9] + E^x^2*Log[2 - 5*x])^2), x] + 512*Log[9]^3*Defer[Int][x^3/((4*x*Log[9] + Log[2 - 5*x])*(16*Log[9] +
4*E^x^2*x*Log[9] + E^x^2*Log[2 - 5*x])^2), x] + (128*(5 - 8*Log[9])*Log[9]^2*Defer[Int][1/((-2 + 5*x)*(4*x*Log
[9] + Log[2 - 5*x])*(16*Log[9] + 4*E^x^2*x*Log[9] + E^x^2*Log[2 - 5*x])^2), x])/5 + (1024*Log[9]^3*Defer[Int][
1/((-2 + 5*x)*(4*x*Log[9] + Log[2 - 5*x])*(16*Log[9] + 4*E^x^2*x*Log[9] + E^x^2*Log[2 - 5*x])^2), x])/5 + 128*
Log[9]^2*Defer[Int][(x^2*Log[2 - 5*x])/((4*x*Log[9] + Log[2 - 5*x])*(16*Log[9] + 4*E^x^2*x*Log[9] + E^x^2*Log[
2 - 5*x])^2), x] - 4*Log[9]*Defer[Int][1/((4*x*Log[9] + Log[2 - 5*x])*(16*Log[9] + 4*E^x^2*x*Log[9] + E^x^2*Lo
g[2 - 5*x])), x] - 32*Log[9]^2*Defer[Int][x^3/((4*x*Log[9] + Log[2 - 5*x])*(16*Log[9] + 4*E^x^2*x*Log[9] + E^x
^2*Log[2 - 5*x])), x] - 8*Log[9]*Defer[Int][1/((-2 + 5*x)*(4*x*Log[9] + Log[2 - 5*x])*(16*Log[9] + 4*E^x^2*x*L
og[9] + E^x^2*Log[2 - 5*x])), x] + 4*Log[9]*Defer[Int][Log[2 - 5*x]/((4*x*Log[9] + Log[2 - 5*x])*(16*Log[9] +
4*E^x^2*x*Log[9] + E^x^2*Log[2 - 5*x])), x] - 8*Log[9]*Defer[Int][(x^2*Log[2 - 5*x])/((4*x*Log[9] + Log[2 - 5*
x])*(16*Log[9] + 4*E^x^2*x*Log[9] + E^x^2*Log[2 - 5*x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \log (9) \left (16 (2-5 x) \log (9)+e^{x^2} x \left (5-16 x^2 \log (9)+40 x^3 \log (9)\right )+e^{x^2} \left (2-5 x-4 x^2+10 x^3\right ) \log (2-5 x)\right )}{(2-5 x) \left (4 \left (4+e^{x^2} x\right ) \log (9)+e^{x^2} \log (2-5 x)\right )^2} \, dx\\ &=(4 \log (9)) \int \frac {16 (2-5 x) \log (9)+e^{x^2} x \left (5-16 x^2 \log (9)+40 x^3 \log (9)\right )+e^{x^2} \left (2-5 x-4 x^2+10 x^3\right ) \log (2-5 x)}{(2-5 x) \left (4 \left (4+e^{x^2} x\right ) \log (9)+e^{x^2} \log (2-5 x)\right )^2} \, dx\\ &=(4 \log (9)) \int \left (\frac {16 x \log (9) \left (-5 \left (1-\frac {8 \log (9)}{5}\right )-20 x \log (9)+16 x^2 \log (9)-40 x^3 \log (9)+4 x \log (2-5 x)-10 x^2 \log (2-5 x)\right )}{(2-5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )^2}-\frac {5 x-16 x^3 \log (9)+40 x^4 \log (9)+2 \log (2-5 x)-5 x \log (2-5 x)-4 x^2 \log (2-5 x)+10 x^3 \log (2-5 x)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )}\right ) \, dx\\ &=-\left ((4 \log (9)) \int \frac {5 x-16 x^3 \log (9)+40 x^4 \log (9)+2 \log (2-5 x)-5 x \log (2-5 x)-4 x^2 \log (2-5 x)+10 x^3 \log (2-5 x)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )} \, dx\right )+\left (64 \log ^2(9)\right ) \int \frac {x \left (-5 \left (1-\frac {8 \log (9)}{5}\right )-20 x \log (9)+16 x^2 \log (9)-40 x^3 \log (9)+4 x \log (2-5 x)-10 x^2 \log (2-5 x)\right )}{(2-5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )^2} \, dx\\ &=-\left ((4 \log (9)) \int \left (\frac {5 x}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )}-\frac {16 x^3 \log (9)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )}+\frac {40 x^4 \log (9)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )}+\frac {2 \log (2-5 x)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )}-\frac {5 x \log (2-5 x)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )}-\frac {4 x^2 \log (2-5 x)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )}+\frac {10 x^3 \log (2-5 x)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )}\right ) \, dx\right )+\left (64 \log ^2(9)\right ) \int \left (\frac {2 \left (-5 \left (1-\frac {8 \log (9)}{5}\right )-20 x \log (9)+16 x^2 \log (9)-40 x^3 \log (9)+4 x \log (2-5 x)-10 x^2 \log (2-5 x)\right )}{5 (2-5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )^2}+\frac {5 \left (1-\frac {8 \log (9)}{5}\right )+20 x \log (9)-16 x^2 \log (9)+40 x^3 \log (9)-4 x \log (2-5 x)+10 x^2 \log (2-5 x)}{5 (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )^2}\right ) \, dx\\ &=-\left ((8 \log (9)) \int \frac {\log (2-5 x)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )} \, dx\right )+(16 \log (9)) \int \frac {x^2 \log (2-5 x)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )} \, dx-(20 \log (9)) \int \frac {x}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )} \, dx+(20 \log (9)) \int \frac {x \log (2-5 x)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )} \, dx-(40 \log (9)) \int \frac {x^3 \log (2-5 x)}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )} \, dx+\frac {1}{5} \left (64 \log ^2(9)\right ) \int \frac {5 \left (1-\frac {8 \log (9)}{5}\right )+20 x \log (9)-16 x^2 \log (9)+40 x^3 \log (9)-4 x \log (2-5 x)+10 x^2 \log (2-5 x)}{(4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )^2} \, dx+\frac {1}{5} \left (128 \log ^2(9)\right ) \int \frac {-5 \left (1-\frac {8 \log (9)}{5}\right )-20 x \log (9)+16 x^2 \log (9)-40 x^3 \log (9)+4 x \log (2-5 x)-10 x^2 \log (2-5 x)}{(2-5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )^2} \, dx+\left (64 \log ^2(9)\right ) \int \frac {x^3}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )} \, dx-\left (160 \log ^2(9)\right ) \int \frac {x^4}{(-2+5 x) (4 x \log (9)+\log (2-5 x)) \left (16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 34, normalized size = 1.21 \begin {gather*} \frac {4 x \log (9)}{16 \log (9)+4 e^{x^2} x \log (9)+e^{x^2} \log (2-5 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-128 + 320*x)*Log[9]^2 + E^x^2*(-20*x*Log[9] + (64*x^3 - 160*x^4)*Log[9]^2) + E^x^2*(-8 + 20*x + 1
6*x^2 - 40*x^3)*Log[9]*Log[2 - 5*x])/((-512 + 1280*x)*Log[9]^2 + E^x^2*(-256*x + 640*x^2)*Log[9]^2 + E^(2*x^2)
*(-32*x^2 + 80*x^3)*Log[9]^2 + (E^x^2*(-64 + 160*x)*Log[9] + E^(2*x^2)*(-16*x + 40*x^2)*Log[9])*Log[2 - 5*x] +
E^(2*x^2)*(-2 + 5*x)*Log[2 - 5*x]^2),x]

[Out]

(4*x*Log[9])/(16*Log[9] + 4*E^x^2*x*Log[9] + E^x^2*Log[2 - 5*x])

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fricas [A]  time = 0.83, size = 32, normalized size = 1.14 \begin {gather*} \frac {8 \, x \log \relax (3)}{8 \, x e^{\left (x^{2}\right )} \log \relax (3) + e^{\left (x^{2}\right )} \log \left (-5 \, x + 2\right ) + 32 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-40*x^3+16*x^2+20*x-8)*log(3)*exp(x^2)*log(-5*x+2)+(4*(-160*x^4+64*x^3)*log(3)^2-40*x*log(3))*ex
p(x^2)+4*(320*x-128)*log(3)^2)/((5*x-2)*exp(x^2)^2*log(-5*x+2)^2+(2*(40*x^2-16*x)*log(3)*exp(x^2)^2+2*(160*x-6
4)*log(3)*exp(x^2))*log(-5*x+2)+4*(80*x^3-32*x^2)*log(3)^2*exp(x^2)^2+4*(640*x^2-256*x)*log(3)^2*exp(x^2)+4*(1
280*x-512)*log(3)^2),x, algorithm="fricas")

[Out]

8*x*log(3)/(8*x*e^(x^2)*log(3) + e^(x^2)*log(-5*x + 2) + 32*log(3))

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giac [A]  time = 0.48, size = 32, normalized size = 1.14 \begin {gather*} \frac {8 \, x \log \relax (3)}{8 \, x e^{\left (x^{2}\right )} \log \relax (3) + e^{\left (x^{2}\right )} \log \left (-5 \, x + 2\right ) + 32 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-40*x^3+16*x^2+20*x-8)*log(3)*exp(x^2)*log(-5*x+2)+(4*(-160*x^4+64*x^3)*log(3)^2-40*x*log(3))*ex
p(x^2)+4*(320*x-128)*log(3)^2)/((5*x-2)*exp(x^2)^2*log(-5*x+2)^2+(2*(40*x^2-16*x)*log(3)*exp(x^2)^2+2*(160*x-6
4)*log(3)*exp(x^2))*log(-5*x+2)+4*(80*x^3-32*x^2)*log(3)^2*exp(x^2)^2+4*(640*x^2-256*x)*log(3)^2*exp(x^2)+4*(1
280*x-512)*log(3)^2),x, algorithm="giac")

[Out]

8*x*log(3)/(8*x*e^(x^2)*log(3) + e^(x^2)*log(-5*x + 2) + 32*log(3))

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maple [A]  time = 0.05, size = 33, normalized size = 1.18

method	result	size
risch	\(\frac {8 \ln \relax (3) x}{8 \ln \relax (3) {\mathrm e}^{x^{2}} x +{\mathrm e}^{x^{2}} \ln \left (-5 x +2\right )+32 \ln \relax (3)}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(-40*x^3+16*x^2+20*x-8)*ln(3)*exp(x^2)*ln(-5*x+2)+(4*(-160*x^4+64*x^3)*ln(3)^2-40*x*ln(3))*exp(x^2)+4*(
320*x-128)*ln(3)^2)/((5*x-2)*exp(x^2)^2*ln(-5*x+2)^2+(2*(40*x^2-16*x)*ln(3)*exp(x^2)^2+2*(160*x-64)*ln(3)*exp(
x^2))*ln(-5*x+2)+4*(80*x^3-32*x^2)*ln(3)^2*exp(x^2)^2+4*(640*x^2-256*x)*ln(3)^2*exp(x^2)+4*(1280*x-512)*ln(3)^
2),x,method=_RETURNVERBOSE)

[Out]

8*ln(3)*x/(8*ln(3)*exp(x^2)*x+exp(x^2)*ln(-5*x+2)+32*ln(3))

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maxima [A]  time = 0.72, size = 32, normalized size = 1.14 \begin {gather*} \frac {8 \, x \log \relax (3)}{8 \, x e^{\left (x^{2}\right )} \log \relax (3) + e^{\left (x^{2}\right )} \log \left (-5 \, x + 2\right ) + 32 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-40*x^3+16*x^2+20*x-8)*log(3)*exp(x^2)*log(-5*x+2)+(4*(-160*x^4+64*x^3)*log(3)^2-40*x*log(3))*ex
p(x^2)+4*(320*x-128)*log(3)^2)/((5*x-2)*exp(x^2)^2*log(-5*x+2)^2+(2*(40*x^2-16*x)*log(3)*exp(x^2)^2+2*(160*x-6
4)*log(3)*exp(x^2))*log(-5*x+2)+4*(80*x^3-32*x^2)*log(3)^2*exp(x^2)^2+4*(640*x^2-256*x)*log(3)^2*exp(x^2)+4*(1
280*x-512)*log(3)^2),x, algorithm="maxima")

[Out]

8*x*log(3)/(8*x*e^(x^2)*log(3) + e^(x^2)*log(-5*x + 2) + 32*log(3))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {4\,{\ln \relax (3)}^2\,\left (320\,x-128\right )-{\mathrm {e}}^{x^2}\,\left (40\,x\,\ln \relax (3)-4\,{\ln \relax (3)}^2\,\left (64\,x^3-160\,x^4\right )\right )+2\,{\mathrm {e}}^{x^2}\,\ln \relax (3)\,\ln \left (2-5\,x\right )\,\left (-40\,x^3+16\,x^2+20\,x-8\right )}{{\mathrm {e}}^{2\,x^2}\,\left (5\,x-2\right )\,{\ln \left (2-5\,x\right )}^2+\left (2\,{\mathrm {e}}^{x^2}\,\ln \relax (3)\,\left (160\,x-64\right )-2\,{\mathrm {e}}^{2\,x^2}\,\ln \relax (3)\,\left (16\,x-40\,x^2\right )\right )\,\ln \left (2-5\,x\right )+4\,{\ln \relax (3)}^2\,\left (1280\,x-512\right )-4\,{\mathrm {e}}^{2\,x^2}\,{\ln \relax (3)}^2\,\left (32\,x^2-80\,x^3\right )-4\,{\mathrm {e}}^{x^2}\,{\ln \relax (3)}^2\,\left (256\,x-640\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(3)^2*(320*x - 128) - exp(x^2)*(40*x*log(3) - 4*log(3)^2*(64*x^3 - 160*x^4)) + 2*exp(x^2)*log(3)*log
(2 - 5*x)*(20*x + 16*x^2 - 40*x^3 - 8))/(4*log(3)^2*(1280*x - 512) + log(2 - 5*x)*(2*exp(x^2)*log(3)*(160*x -
64) - 2*exp(2*x^2)*log(3)*(16*x - 40*x^2)) + exp(2*x^2)*log(2 - 5*x)^2*(5*x - 2) - 4*exp(2*x^2)*log(3)^2*(32*x
^2 - 80*x^3) - 4*exp(x^2)*log(3)^2*(256*x - 640*x^2)),x)

[Out]

int((4*log(3)^2*(320*x - 128) - exp(x^2)*(40*x*log(3) - 4*log(3)^2*(64*x^3 - 160*x^4)) + 2*exp(x^2)*log(3)*log
(2 - 5*x)*(20*x + 16*x^2 - 40*x^3 - 8))/(4*log(3)^2*(1280*x - 512) + log(2 - 5*x)*(2*exp(x^2)*log(3)*(160*x -
64) - 2*exp(2*x^2)*log(3)*(16*x - 40*x^2)) + exp(2*x^2)*log(2 - 5*x)^2*(5*x - 2) - 4*exp(2*x^2)*log(3)^2*(32*x
^2 - 80*x^3) - 4*exp(x^2)*log(3)^2*(256*x - 640*x^2)), x)

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sympy [A]  time = 0.47, size = 29, normalized size = 1.04 \begin {gather*} \frac {8 x \log {\relax (3 )}}{\left (8 x \log {\relax (3 )} + \log {\left (2 - 5 x \right )}\right ) e^{x^{2}} + 32 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-40*x**3+16*x**2+20*x-8)*ln(3)*exp(x**2)*ln(-5*x+2)+(4*(-160*x**4+64*x**3)*ln(3)**2-40*x*ln(3))*
exp(x**2)+4*(320*x-128)*ln(3)**2)/((5*x-2)*exp(x**2)**2*ln(-5*x+2)**2+(2*(40*x**2-16*x)*ln(3)*exp(x**2)**2+2*(
160*x-64)*ln(3)*exp(x**2))*ln(-5*x+2)+4*(80*x**3-32*x**2)*ln(3)**2*exp(x**2)**2+4*(640*x**2-256*x)*ln(3)**2*ex
p(x**2)+4*(1280*x-512)*ln(3)**2),x)

[Out]

8*x*log(3)/((8*x*log(3) + log(2 - 5*x))*exp(x**2) + 32*log(3))

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