3.5.80 \(\int \frac {32 x-10 x^2+e^x (16 x-5 x^2)+(30 x-10 x^2+e^x (15 x+10 x^2-5 x^3)+(-6+2 x+e^x (-3-2 x+x^2)) \log (-12+4 x)) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{5} \left (2+e^x\right ) x \log (5 x-\log (4 (-3+x))) \]

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Rubi [F]  time = 5.27, 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 {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{75 x-25 x^2+(-15+5 x) \log (-12+4 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(32*x - 10*x^2 + E^x*(16*x - 5*x^2) + (30*x - 10*x^2 + E^x*(15*x + 10*x^2 - 5*x^3) + (-6 + 2*x + E^x*(-3 -
2*x + x^2))*Log[-12 + 4*x])*Log[5*x - Log[-12 + 4*x]])/(75*x - 25*x^2 + (-15 + 5*x)*Log[-12 + 4*x]),x]

[Out]

(-6*Defer[Int][1/((-3 + x)*(5*x - Log[4*(-3 + x)])), x])/5 - (3*Defer[Int][E^x/((-3 + x)*(5*x - Log[4*(-3 + x)
])), x])/5 + 2*Defer[Int][x/(5*x - Log[4*(-3 + x)]), x] + Defer[Int][(E^x*x)/(5*x - Log[4*(-3 + x)]), x] + (2*
Defer[Int][(-5*x + Log[4*(-3 + x)])^(-1), x])/5 + Defer[Int][E^x/(-5*x + Log[4*(-3 + x)]), x]/5 + (2*Defer[Int
][Log[5*x - Log[4*(-3 + x)]], x])/5 + Defer[Int][E^x*Log[5*x - Log[4*(-3 + x)]], x]/5 + Defer[Int][E^x*x*Log[5
*x - Log[4*(-3 + x)]], x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{5 (3-x) (5 x-\log (4 (-3+x)))} \, dx\\ &=\frac {1}{5} \int \frac {32 x-10 x^2+e^x \left (16 x-5 x^2\right )+\left (30 x-10 x^2+e^x \left (15 x+10 x^2-5 x^3\right )+\left (-6+2 x+e^x \left (-3-2 x+x^2\right )\right ) \log (-12+4 x)\right ) \log (5 x-\log (-12+4 x))}{(3-x) (5 x-\log (4 (-3+x)))} \, dx\\ &=\frac {1}{5} \int \frac {32 x-10 x^2-e^x x (-16+5 x)-(-3+x) \left (2+e^x (1+x)\right ) (5 x-\log (4 (-3+x))) \log (5 x-\log (4 (-3+x)))}{(3-x) (5 x-\log (4 (-3+x)))} \, dx\\ &=\frac {1}{5} \int \left (\frac {2 \left (-16 x+5 x^2-15 x \log (5 x-\log (4 (-3+x)))+5 x^2 \log (5 x-\log (4 (-3+x)))+3 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))-x \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))\right )}{(-3+x) (5 x-\log (4 (-3+x)))}+\frac {e^x \left (-16 x+5 x^2-15 x \log (5 x-\log (4 (-3+x)))-10 x^2 \log (5 x-\log (4 (-3+x)))+5 x^3 \log (5 x-\log (4 (-3+x)))+3 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))+2 x \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))-x^2 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))\right )}{(-3+x) (5 x-\log (4 (-3+x)))}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^x \left (-16 x+5 x^2-15 x \log (5 x-\log (4 (-3+x)))-10 x^2 \log (5 x-\log (4 (-3+x)))+5 x^3 \log (5 x-\log (4 (-3+x)))+3 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))+2 x \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))-x^2 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))\right )}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx+\frac {2}{5} \int \frac {-16 x+5 x^2-15 x \log (5 x-\log (4 (-3+x)))+5 x^2 \log (5 x-\log (4 (-3+x)))+3 \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))-x \log (4 (-3+x)) \log (5 x-\log (4 (-3+x)))}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx\\ &=\frac {1}{5} \int \frac {e^x \left (-\frac {x (-16+5 x)}{5 x-\log (4 (-3+x))}-\left (-3-2 x+x^2\right ) \log (5 x-\log (4 (-3+x)))\right )}{3-x} \, dx+\frac {2}{5} \int \left (\frac {x (-16+5 x)}{(-3+x) (5 x-\log (4 (-3+x)))}+\log (5 x-\log (4 (-3+x)))\right ) \, dx\\ &=\frac {1}{5} \int \left (\frac {e^x x (-16+5 x)}{(-3+x) (5 x-\log (4 (-3+x)))}+e^x (1+x) \log (5 x-\log (4 (-3+x)))\right ) \, dx+\frac {2}{5} \int \frac {x (-16+5 x)}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx+\frac {2}{5} \int \log (5 x-\log (4 (-3+x))) \, dx\\ &=\frac {1}{5} \int \frac {e^x x (-16+5 x)}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx+\frac {1}{5} \int e^x (1+x) \log (5 x-\log (4 (-3+x))) \, dx+\frac {2}{5} \int \left (-\frac {3}{(-3+x) (5 x-\log (4 (-3+x)))}+\frac {5 x}{5 x-\log (4 (-3+x))}+\frac {1}{-5 x+\log (4 (-3+x))}\right ) \, dx+\frac {2}{5} \int \log (5 x-\log (4 (-3+x))) \, dx\\ &=\frac {1}{5} \int \left (-\frac {3 e^x}{(-3+x) (5 x-\log (4 (-3+x)))}+\frac {5 e^x x}{5 x-\log (4 (-3+x))}+\frac {e^x}{-5 x+\log (4 (-3+x))}\right ) \, dx+\frac {1}{5} \int \left (e^x \log (5 x-\log (4 (-3+x)))+e^x x \log (5 x-\log (4 (-3+x)))\right ) \, dx+\frac {2}{5} \int \frac {1}{-5 x+\log (4 (-3+x))} \, dx+\frac {2}{5} \int \log (5 x-\log (4 (-3+x))) \, dx-\frac {6}{5} \int \frac {1}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx+2 \int \frac {x}{5 x-\log (4 (-3+x))} \, dx\\ &=\frac {1}{5} \int \frac {e^x}{-5 x+\log (4 (-3+x))} \, dx+\frac {1}{5} \int e^x \log (5 x-\log (4 (-3+x))) \, dx+\frac {1}{5} \int e^x x \log (5 x-\log (4 (-3+x))) \, dx+\frac {2}{5} \int \frac {1}{-5 x+\log (4 (-3+x))} \, dx+\frac {2}{5} \int \log (5 x-\log (4 (-3+x))) \, dx-\frac {3}{5} \int \frac {e^x}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx-\frac {6}{5} \int \frac {1}{(-3+x) (5 x-\log (4 (-3+x)))} \, dx+2 \int \frac {x}{5 x-\log (4 (-3+x))} \, dx+\int \frac {e^x x}{5 x-\log (4 (-3+x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.69, size = 23, normalized size = 1.00 \begin {gather*} \frac {1}{5} \left (2+e^x\right ) x \log (5 x-\log (4 (-3+x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32*x - 10*x^2 + E^x*(16*x - 5*x^2) + (30*x - 10*x^2 + E^x*(15*x + 10*x^2 - 5*x^3) + (-6 + 2*x + E^x
*(-3 - 2*x + x^2))*Log[-12 + 4*x])*Log[5*x - Log[-12 + 4*x]])/(75*x - 25*x^2 + (-15 + 5*x)*Log[-12 + 4*x]),x]

[Out]

((2 + E^x)*x*Log[5*x - Log[4*(-3 + x)]])/5

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fricas [A]  time = 1.26, size = 23, normalized size = 1.00 \begin {gather*} \frac {1}{5} \, {\left (x e^{x} + 2 \, x\right )} \log \left (5 \, x - \log \left (4 \, x - 12\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^2-2*x-3)*exp(x)+2*x-6)*log(4*x-12)+(-5*x^3+10*x^2+15*x)*exp(x)-10*x^2+30*x)*log(-log(4*x-12)+5
*x)+(-5*x^2+16*x)*exp(x)-10*x^2+32*x)/((5*x-15)*log(4*x-12)-25*x^2+75*x),x, algorithm="fricas")

[Out]

1/5*(x*e^x + 2*x)*log(5*x - log(4*x - 12))

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giac [A]  time = 0.46, size = 35, normalized size = 1.52 \begin {gather*} \frac {1}{5} \, x e^{x} \log \left (5 \, x - \log \left (4 \, x - 12\right )\right ) + \frac {2}{5} \, x \log \left (5 \, x - \log \left (4 \, x - 12\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^2-2*x-3)*exp(x)+2*x-6)*log(4*x-12)+(-5*x^3+10*x^2+15*x)*exp(x)-10*x^2+30*x)*log(-log(4*x-12)+5
*x)+(-5*x^2+16*x)*exp(x)-10*x^2+32*x)/((5*x-15)*log(4*x-12)-25*x^2+75*x),x, algorithm="giac")

[Out]

1/5*x*e^x*log(5*x - log(4*x - 12)) + 2/5*x*log(5*x - log(4*x - 12))

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maple [A]  time = 0.42, size = 24, normalized size = 1.04

method	result	size
risch	\(\left (\frac {2 x}{5}+\frac {{\mathrm e}^{x} x}{5}\right ) \ln \left (-\ln \left (4 x -12\right )+5 x \right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((x^2-2*x-3)*exp(x)+2*x-6)*ln(4*x-12)+(-5*x^3+10*x^2+15*x)*exp(x)-10*x^2+30*x)*ln(-ln(4*x-12)+5*x)+(-5*x
^2+16*x)*exp(x)-10*x^2+32*x)/((5*x-15)*ln(4*x-12)-25*x^2+75*x),x,method=_RETURNVERBOSE)

[Out]

(2/5*x+1/5*exp(x)*x)*ln(-ln(4*x-12)+5*x)

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maxima [A]  time = 0.74, size = 25, normalized size = 1.09 \begin {gather*} \frac {1}{5} \, {\left (x e^{x} + 2 \, x\right )} \log \left (5 \, x - 2 \, \log \relax (2) - \log \left (x - 3\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^2-2*x-3)*exp(x)+2*x-6)*log(4*x-12)+(-5*x^3+10*x^2+15*x)*exp(x)-10*x^2+30*x)*log(-log(4*x-12)+5
*x)+(-5*x^2+16*x)*exp(x)-10*x^2+32*x)/((5*x-15)*log(4*x-12)-25*x^2+75*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.68, size = 20, normalized size = 0.87 \begin {gather*} \frac {x\,\ln \left (5\,x-\ln \left (4\,x-12\right )\right )\,\left ({\mathrm {e}}^x+2\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x + log(5*x - log(4*x - 12))*(30*x - 10*x^2 + exp(x)*(15*x + 10*x^2 - 5*x^3) - log(4*x - 12)*(exp(x)*(
2*x - x^2 + 3) - 2*x + 6)) + exp(x)*(16*x - 5*x^2) - 10*x^2)/(75*x + log(4*x - 12)*(5*x - 15) - 25*x^2),x)

[Out]

(x*log(5*x - log(4*x - 12))*(exp(x) + 2))/5

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sympy [B]  time = 22.83, size = 53, normalized size = 2.30 \begin {gather*} \frac {x e^{x} \log {\left (5 x - \log {\left (4 x - 12 \right )} \right )}}{5} + \left (\frac {2 x}{5} - \frac {3}{5}\right ) \log {\left (5 x - \log {\left (4 x - 12 \right )} \right )} + \frac {3 \log {\left (- 5 x + \log {\left (4 x - 12 \right )} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x**2-2*x-3)*exp(x)+2*x-6)*ln(4*x-12)+(-5*x**3+10*x**2+15*x)*exp(x)-10*x**2+30*x)*ln(-ln(4*x-12)+
5*x)+(-5*x**2+16*x)*exp(x)-10*x**2+32*x)/((5*x-15)*ln(4*x-12)-25*x**2+75*x),x)

[Out]

x*exp(x)*log(5*x - log(4*x - 12))/5 + (2*x/5 - 3/5)*log(5*x - log(4*x - 12)) + 3*log(-5*x + log(4*x - 12))/5

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