3.39.69 \(\int \frac {80 e^{2 x}+160 e^x x+80 x^2+(e^{2 x} (40-160 x)+40 x-160 x^2+e^x (40-120 x-160 x^2)) \log (-3+12 x)+(-3+12 x) \log ^2(-3+12 x)}{(-2+8 x) \log ^2(-3+12 x)} \, dx\)

Optimal. Leaf size=25 \[ x+\frac {1}{2} \left (x-\frac {20 \left (e^x+x\right )^2}{\log (-3+12 x)}\right ) \]

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Rubi [F]  time = 1.42, 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 {80 e^{2 x}+160 e^x x+80 x^2+\left (e^{2 x} (40-160 x)+40 x-160 x^2+e^x \left (40-120 x-160 x^2\right )\right ) \log (-3+12 x)+(-3+12 x) \log ^2(-3+12 x)}{(-2+8 x) \log ^2(-3+12 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(80*E^(2*x) + 160*E^x*x + 80*x^2 + (E^(2*x)*(40 - 160*x) + 40*x - 160*x^2 + E^x*(40 - 120*x - 160*x^2))*Lo
g[-3 + 12*x] + (-3 + 12*x)*Log[-3 + 12*x]^2)/((-2 + 8*x)*Log[-3 + 12*x]^2),x]

[Out]

(3*x)/2 - 5/(8*Log[-3 + 12*x]) + (5*(1 - 4*x))/(4*Log[-3 + 12*x]) - (5*(1 - 4*x)^2)/(8*Log[-3 + 12*x]) + 20*De
fer[Int][E^x/Log[-3 + 12*x]^2, x] + 20*Defer[Int][E^x/((-1 + 4*x)*Log[-3 + 12*x]^2), x] + 40*Defer[Int][E^(2*x
)/((-1 + 4*x)*Log[-3 + 12*x]^2), x] - 25*Defer[Int][E^x/Log[-3 + 12*x], x] - 20*Defer[Int][E^(2*x)/Log[-3 + 12
*x], x] - (5*Defer[Int][(E^x*(-3 + 12*x))/Log[-3 + 12*x], x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {3}{2}+\frac {40 \left (e^x+x\right )^2}{(-1+4 x) \log ^2(-3+12 x)}-\frac {20 \left (1+e^x\right ) \left (e^x+x\right )}{\log (-3+12 x)}\right ) \, dx\\ &=\frac {3 x}{2}-20 \int \frac {\left (1+e^x\right ) \left (e^x+x\right )}{\log (-3+12 x)} \, dx+40 \int \frac {\left (e^x+x\right )^2}{(-1+4 x) \log ^2(-3+12 x)} \, dx\\ &=\frac {3 x}{2}-20 \int \left (\frac {e^{2 x}}{\log (-3+12 x)}+\frac {x}{\log (-3+12 x)}+\frac {e^x (1+x)}{\log (-3+12 x)}\right ) \, dx+40 \int \left (\frac {e^{2 x}}{(-1+4 x) \log ^2(-3+12 x)}+\frac {2 e^x x}{(-1+4 x) \log ^2(-3+12 x)}+\frac {x^2}{(-1+4 x) \log ^2(-3+12 x)}\right ) \, dx\\ &=\frac {3 x}{2}-20 \int \frac {e^{2 x}}{\log (-3+12 x)} \, dx-20 \int \frac {x}{\log (-3+12 x)} \, dx-20 \int \frac {e^x (1+x)}{\log (-3+12 x)} \, dx+40 \int \frac {e^{2 x}}{(-1+4 x) \log ^2(-3+12 x)} \, dx+40 \int \frac {x^2}{(-1+4 x) \log ^2(-3+12 x)} \, dx+80 \int \frac {e^x x}{(-1+4 x) \log ^2(-3+12 x)} \, dx\\ &=\frac {3 x}{2}+\frac {10}{3} \operatorname {Subst}\left (\int \frac {3 \left (\frac {1}{4}+\frac {x}{12}\right )^2}{x \log ^2(x)} \, dx,x,-3+12 x\right )-20 \int \left (\frac {1}{4 \log (-3+12 x)}+\frac {-3+12 x}{12 \log (-3+12 x)}\right ) \, dx-20 \int \left (\frac {5 e^x}{4 \log (-3+12 x)}+\frac {e^x (-3+12 x)}{12 \log (-3+12 x)}\right ) \, dx-20 \int \frac {e^{2 x}}{\log (-3+12 x)} \, dx+40 \int \frac {e^{2 x}}{(-1+4 x) \log ^2(-3+12 x)} \, dx+80 \int \left (\frac {e^x}{4 \log ^2(-3+12 x)}+\frac {e^x}{4 (-1+4 x) \log ^2(-3+12 x)}\right ) \, dx\\ &=\frac {3 x}{2}-\frac {5}{3} \int \frac {-3+12 x}{\log (-3+12 x)} \, dx-\frac {5}{3} \int \frac {e^x (-3+12 x)}{\log (-3+12 x)} \, dx-5 \int \frac {1}{\log (-3+12 x)} \, dx+10 \operatorname {Subst}\left (\int \frac {\left (\frac {1}{4}+\frac {x}{12}\right )^2}{x \log ^2(x)} \, dx,x,-3+12 x\right )+20 \int \frac {e^x}{\log ^2(-3+12 x)} \, dx+20 \int \frac {e^x}{(-1+4 x) \log ^2(-3+12 x)} \, dx-20 \int \frac {e^{2 x}}{\log (-3+12 x)} \, dx-25 \int \frac {e^x}{\log (-3+12 x)} \, dx+40 \int \frac {e^{2 x}}{(-1+4 x) \log ^2(-3+12 x)} \, dx\\ &=\frac {3 x}{2}-\frac {5}{36} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-3+12 x\right )-\frac {5}{12} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-3+12 x\right )-\frac {5}{3} \int \frac {e^x (-3+12 x)}{\log (-3+12 x)} \, dx+10 \operatorname {Subst}\left (\int \left (\frac {1}{24 \log ^2(x)}+\frac {1}{16 x \log ^2(x)}+\frac {x}{144 \log ^2(x)}\right ) \, dx,x,-3+12 x\right )+20 \int \frac {e^x}{\log ^2(-3+12 x)} \, dx+20 \int \frac {e^x}{(-1+4 x) \log ^2(-3+12 x)} \, dx-20 \int \frac {e^{2 x}}{\log (-3+12 x)} \, dx-25 \int \frac {e^x}{\log (-3+12 x)} \, dx+40 \int \frac {e^{2 x}}{(-1+4 x) \log ^2(-3+12 x)} \, dx\\ &=\frac {3 x}{2}-\frac {5}{12} \text {li}(-3+12 x)+\frac {5}{72} \operatorname {Subst}\left (\int \frac {x}{\log ^2(x)} \, dx,x,-3+12 x\right )-\frac {5}{36} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-3+12 x)\right )+\frac {5}{12} \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-3+12 x\right )+\frac {5}{8} \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-3+12 x\right )-\frac {5}{3} \int \frac {e^x (-3+12 x)}{\log (-3+12 x)} \, dx+20 \int \frac {e^x}{\log ^2(-3+12 x)} \, dx+20 \int \frac {e^x}{(-1+4 x) \log ^2(-3+12 x)} \, dx-20 \int \frac {e^{2 x}}{\log (-3+12 x)} \, dx-25 \int \frac {e^x}{\log (-3+12 x)} \, dx+40 \int \frac {e^{2 x}}{(-1+4 x) \log ^2(-3+12 x)} \, dx\\ &=\frac {3 x}{2}-\frac {5}{36} \text {Ei}(2 \log (-3+12 x))+\frac {5 (1-4 x)}{4 \log (-3+12 x)}-\frac {5 (1-4 x)^2}{8 \log (-3+12 x)}-\frac {5}{12} \text {li}(-3+12 x)+\frac {5}{36} \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-3+12 x\right )+\frac {5}{12} \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-3+12 x\right )+\frac {5}{8} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-3+12 x)\right )-\frac {5}{3} \int \frac {e^x (-3+12 x)}{\log (-3+12 x)} \, dx+20 \int \frac {e^x}{\log ^2(-3+12 x)} \, dx+20 \int \frac {e^x}{(-1+4 x) \log ^2(-3+12 x)} \, dx-20 \int \frac {e^{2 x}}{\log (-3+12 x)} \, dx-25 \int \frac {e^x}{\log (-3+12 x)} \, dx+40 \int \frac {e^{2 x}}{(-1+4 x) \log ^2(-3+12 x)} \, dx\\ &=\frac {3 x}{2}-\frac {5}{36} \text {Ei}(2 \log (-3+12 x))-\frac {5}{8 \log (-3+12 x)}+\frac {5 (1-4 x)}{4 \log (-3+12 x)}-\frac {5 (1-4 x)^2}{8 \log (-3+12 x)}+\frac {5}{36} \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-3+12 x)\right )-\frac {5}{3} \int \frac {e^x (-3+12 x)}{\log (-3+12 x)} \, dx+20 \int \frac {e^x}{\log ^2(-3+12 x)} \, dx+20 \int \frac {e^x}{(-1+4 x) \log ^2(-3+12 x)} \, dx-20 \int \frac {e^{2 x}}{\log (-3+12 x)} \, dx-25 \int \frac {e^x}{\log (-3+12 x)} \, dx+40 \int \frac {e^{2 x}}{(-1+4 x) \log ^2(-3+12 x)} \, dx\\ &=\frac {3 x}{2}-\frac {5}{8 \log (-3+12 x)}+\frac {5 (1-4 x)}{4 \log (-3+12 x)}-\frac {5 (1-4 x)^2}{8 \log (-3+12 x)}-\frac {5}{3} \int \frac {e^x (-3+12 x)}{\log (-3+12 x)} \, dx+20 \int \frac {e^x}{\log ^2(-3+12 x)} \, dx+20 \int \frac {e^x}{(-1+4 x) \log ^2(-3+12 x)} \, dx-20 \int \frac {e^{2 x}}{\log (-3+12 x)} \, dx-25 \int \frac {e^x}{\log (-3+12 x)} \, dx+40 \int \frac {e^{2 x}}{(-1+4 x) \log ^2(-3+12 x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.54, size = 23, normalized size = 0.92 \begin {gather*} \frac {3 x}{2}-\frac {10 \left (e^x+x\right )^2}{\log (-3+12 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(80*E^(2*x) + 160*E^x*x + 80*x^2 + (E^(2*x)*(40 - 160*x) + 40*x - 160*x^2 + E^x*(40 - 120*x - 160*x^
2))*Log[-3 + 12*x] + (-3 + 12*x)*Log[-3 + 12*x]^2)/((-2 + 8*x)*Log[-3 + 12*x]^2),x]

[Out]

(3*x)/2 - (10*(E^x + x)^2)/Log[-3 + 12*x]

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fricas [A]  time = 1.01, size = 36, normalized size = 1.44 \begin {gather*} -\frac {20 \, x^{2} + 40 \, x e^{x} - 3 \, x \log \left (12 \, x - 3\right ) + 20 \, e^{\left (2 \, x\right )}}{2 \, \log \left (12 \, x - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x-3)*log(12*x-3)^2+((-160*x+40)*exp(x)^2+(-160*x^2-120*x+40)*exp(x)-160*x^2+40*x)*log(12*x-3)+8
0*exp(x)^2+160*exp(x)*x+80*x^2)/(8*x-2)/log(12*x-3)^2,x, algorithm="fricas")

[Out]

-1/2*(20*x^2 + 40*x*e^x - 3*x*log(12*x - 3) + 20*e^(2*x))/log(12*x - 3)

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giac [B]  time = 0.21, size = 44, normalized size = 1.76 \begin {gather*} -\frac {20 \, x^{2} + 40 \, x e^{x} - 3 \, x \log \relax (3) - 3 \, x \log \left (4 \, x - 1\right ) + 20 \, e^{\left (2 \, x\right )}}{2 \, {\left (\log \relax (3) + \log \left (4 \, x - 1\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x-3)*log(12*x-3)^2+((-160*x+40)*exp(x)^2+(-160*x^2-120*x+40)*exp(x)-160*x^2+40*x)*log(12*x-3)+8
0*exp(x)^2+160*exp(x)*x+80*x^2)/(8*x-2)/log(12*x-3)^2,x, algorithm="giac")

[Out]

-1/2*(20*x^2 + 40*x*e^x - 3*x*log(3) - 3*x*log(4*x - 1) + 20*e^(2*x))/(log(3) + log(4*x - 1))

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maple [A]  time = 0.10, size = 28, normalized size = 1.12

method	result	size
risch	\(\frac {3 x}{2}-\frac {10 \left ({\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x +x^{2}\right )}{\ln \left (12 x -3\right )}\)	\(28\)
default	\(\frac {3 x}{2}-\frac {10 \,{\mathrm e}^{2 x}}{\ln \left (12 x -3\right )}-\frac {20 \,{\mathrm e}^{x} x}{\ln \left (12 x -3\right )}-\frac {5 \left (12 x -3\right )^{2}}{72 \ln \left (12 x -3\right )}-\frac {5 \left (12 x -3\right )}{12 \ln \left (12 x -3\right )}-\frac {5}{8 \ln \left (12 x -3\right )}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((12*x-3)*ln(12*x-3)^2+((-160*x+40)*exp(x)^2+(-160*x^2-120*x+40)*exp(x)-160*x^2+40*x)*ln(12*x-3)+80*exp(x)
^2+160*exp(x)*x+80*x^2)/(8*x-2)/ln(12*x-3)^2,x,method=_RETURNVERBOSE)

[Out]

3/2*x-10*(exp(2*x)+2*exp(x)*x+x^2)/ln(12*x-3)

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maxima [B]  time = 0.47, size = 44, normalized size = 1.76 \begin {gather*} -\frac {20 \, x^{2} + 40 \, x e^{x} - 3 \, x \log \relax (3) - 3 \, x \log \left (4 \, x - 1\right ) + 20 \, e^{\left (2 \, x\right )}}{2 \, {\left (\log \relax (3) + \log \left (4 \, x - 1\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x-3)*log(12*x-3)^2+((-160*x+40)*exp(x)^2+(-160*x^2-120*x+40)*exp(x)-160*x^2+40*x)*log(12*x-3)+8
0*exp(x)^2+160*exp(x)*x+80*x^2)/(8*x-2)/log(12*x-3)^2,x, algorithm="maxima")

[Out]

-1/2*(20*x^2 + 40*x*e^x - 3*x*log(3) - 3*x*log(4*x - 1) + 20*e^(2*x))/(log(3) + log(4*x - 1))

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mupad [B]  time = 2.47, size = 31, normalized size = 1.24 \begin {gather*} \frac {3\,x}{2}-\frac {10\,{\mathrm {e}}^{2\,x}+20\,x\,{\mathrm {e}}^x+10\,x^2}{\ln \left (12\,x-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((80*exp(2*x) + log(12*x - 3)^2*(12*x - 3) + 160*x*exp(x) + 80*x^2 - log(12*x - 3)*(exp(x)*(120*x + 160*x^2
- 40) - 40*x + exp(2*x)*(160*x - 40) + 160*x^2))/(log(12*x - 3)^2*(8*x - 2)),x)

[Out]

(3*x)/2 - (10*exp(2*x) + 20*x*exp(x) + 10*x^2)/log(12*x - 3)

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sympy [B]  time = 0.37, size = 53, normalized size = 2.12 \begin {gather*} - \frac {10 x^{2}}{\log {\left (12 x - 3 \right )}} + \frac {3 x}{2} + \frac {- 20 x e^{x} \log {\left (12 x - 3 \right )} - 10 e^{2 x} \log {\left (12 x - 3 \right )}}{\log {\left (12 x - 3 \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x-3)*ln(12*x-3)**2+((-160*x+40)*exp(x)**2+(-160*x**2-120*x+40)*exp(x)-160*x**2+40*x)*ln(12*x-3)
+80*exp(x)**2+160*exp(x)*x+80*x**2)/(8*x-2)/ln(12*x-3)**2,x)

[Out]

-10*x**2/log(12*x - 3) + 3*x/2 + (-20*x*exp(x)*log(12*x - 3) - 10*exp(2*x)*log(12*x - 3))/log(12*x - 3)**2

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