3.47.37 \(\int \frac {-45-8 x+5 x^2+e^x (12 x+3 x^2)+(-15-2 x+3 e^x x+x^2) \log (\frac {1}{3} (-60-8 x+12 e^x x+4 x^2))}{-15-2 x+3 e^x x+x^2} \, dx\)

Optimal. Leaf size=24 \[ x \left (3+\log \left (4 \left (-5+e^x x-\frac {1}{3} (2-x) x\right )\right )\right ) \]

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Rubi [A]  time = 1.08, antiderivative size = 27, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 2, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6742, 2548} \begin {gather*} x \log \left (-\frac {4}{3} \left (-x^2+\left (2-3 e^x\right ) x+15\right )\right )+3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-45 - 8*x + 5*x^2 + E^x*(12*x + 3*x^2) + (-15 - 2*x + 3*E^x*x + x^2)*Log[(-60 - 8*x + 12*E^x*x + 4*x^2)/3
])/(-15 - 2*x + 3*E^x*x + x^2),x]

[Out]

3*x + x*Log[(-4*(15 + (2 - 3*E^x)*x - x^2))/3]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[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 \left (4+x-\frac {-15-15 x-3 x^2+x^3}{-15-2 x+3 e^x x+x^2}+\log \left (\frac {4}{3} \left (-15+\left (-2+3 e^x\right ) x+x^2\right )\right )\right ) \, dx\\ &=4 x+\frac {x^2}{2}-\int \frac {-15-15 x-3 x^2+x^3}{-15-2 x+3 e^x x+x^2} \, dx+\int \log \left (\frac {4}{3} \left (-15+\left (-2+3 e^x\right ) x+x^2\right )\right ) \, dx\\ &=4 x+\frac {x^2}{2}+x \log \left (-\frac {4}{3} \left (15+\left (2-3 e^x\right ) x-x^2\right )\right )-\int \frac {x \left (-2 (-1+x)-3 e^x (1+x)\right )}{15-\left (-2+3 e^x\right ) x-x^2} \, dx-\int \left (-\frac {15}{-15-2 x+3 e^x x+x^2}-\frac {15 x}{-15-2 x+3 e^x x+x^2}-\frac {3 x^2}{-15-2 x+3 e^x x+x^2}+\frac {x^3}{-15-2 x+3 e^x x+x^2}\right ) \, dx\\ &=4 x+\frac {x^2}{2}+x \log \left (-\frac {4}{3} \left (15+\left (2-3 e^x\right ) x-x^2\right )\right )+3 \int \frac {x^2}{-15-2 x+3 e^x x+x^2} \, dx+15 \int \frac {1}{-15-2 x+3 e^x x+x^2} \, dx+15 \int \frac {x}{-15-2 x+3 e^x x+x^2} \, dx-\int \frac {x^3}{-15-2 x+3 e^x x+x^2} \, dx-\int \left (1+x-\frac {-15-15 x-3 x^2+x^3}{-15-2 x+3 e^x x+x^2}\right ) \, dx\\ &=3 x+x \log \left (-\frac {4}{3} \left (15+\left (2-3 e^x\right ) x-x^2\right )\right )+3 \int \frac {x^2}{-15-2 x+3 e^x x+x^2} \, dx+15 \int \frac {1}{-15-2 x+3 e^x x+x^2} \, dx+15 \int \frac {x}{-15-2 x+3 e^x x+x^2} \, dx-\int \frac {x^3}{-15-2 x+3 e^x x+x^2} \, dx+\int \frac {-15-15 x-3 x^2+x^3}{-15-2 x+3 e^x x+x^2} \, dx\\ &=3 x+x \log \left (-\frac {4}{3} \left (15+\left (2-3 e^x\right ) x-x^2\right )\right )+3 \int \frac {x^2}{-15-2 x+3 e^x x+x^2} \, dx+15 \int \frac {1}{-15-2 x+3 e^x x+x^2} \, dx+15 \int \frac {x}{-15-2 x+3 e^x x+x^2} \, dx-\int \frac {x^3}{-15-2 x+3 e^x x+x^2} \, dx+\int \left (-\frac {15}{-15-2 x+3 e^x x+x^2}-\frac {15 x}{-15-2 x+3 e^x x+x^2}-\frac {3 x^2}{-15-2 x+3 e^x x+x^2}+\frac {x^3}{-15-2 x+3 e^x x+x^2}\right ) \, dx\\ &=3 x+x \log \left (-\frac {4}{3} \left (15+\left (2-3 e^x\right ) x-x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 25, normalized size = 1.04 \begin {gather*} 3 x+x \log \left (\frac {4}{3} \left (-15+\left (-2+3 e^x\right ) x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45 - 8*x + 5*x^2 + E^x*(12*x + 3*x^2) + (-15 - 2*x + 3*E^x*x + x^2)*Log[(-60 - 8*x + 12*E^x*x + 4*
x^2)/3])/(-15 - 2*x + 3*E^x*x + x^2),x]

[Out]

3*x + x*Log[(4*(-15 + (-2 + 3*E^x)*x + x^2))/3]

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fricas [A]  time = 0.92, size = 22, normalized size = 0.92 \begin {gather*} x \log \left (\frac {4}{3} \, x^{2} + 4 \, x e^{x} - \frac {8}{3} \, x - 20\right ) + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x+x^2-2*x-15)*log(4*exp(x)*x+4/3*x^2-8/3*x-20)+(3*x^2+12*x)*exp(x)+5*x^2-8*x-45)/(3*exp(x
)*x+x^2-2*x-15),x, algorithm="fricas")

[Out]

x*log(4/3*x^2 + 4*x*e^x - 8/3*x - 20) + 3*x

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giac [A]  time = 0.21, size = 22, normalized size = 0.92 \begin {gather*} x \log \left (\frac {4}{3} \, x^{2} + 4 \, x e^{x} - \frac {8}{3} \, x - 20\right ) + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x+x^2-2*x-15)*log(4*exp(x)*x+4/3*x^2-8/3*x-20)+(3*x^2+12*x)*exp(x)+5*x^2-8*x-45)/(3*exp(x
)*x+x^2-2*x-15),x, algorithm="giac")

[Out]

x*log(4/3*x^2 + 4*x*e^x - 8/3*x - 20) + 3*x

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maple [A]  time = 0.05, size = 23, normalized size = 0.96

method	result	size
norman	\(\ln \left (4 \,{\mathrm e}^{x} x +\frac {4 x^{2}}{3}-\frac {8 x}{3}-20\right ) x +3 x\)	\(23\)
risch	\(\ln \left (4 \,{\mathrm e}^{x} x +\frac {4 x^{2}}{3}-\frac {8 x}{3}-20\right ) x +3 x\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*exp(x)*x+x^2-2*x-15)*ln(4*exp(x)*x+4/3*x^2-8/3*x-20)+(3*x^2+12*x)*exp(x)+5*x^2-8*x-45)/(3*exp(x)*x+x^2
-2*x-15),x,method=_RETURNVERBOSE)

[Out]

ln(4*exp(x)*x+4/3*x^2-8/3*x-20)*x+3*x

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maxima [A]  time = 0.50, size = 28, normalized size = 1.17 \begin {gather*} -x {\left (\log \relax (3) - 2 \, \log \relax (2) - 3\right )} + x \log \left (x^{2} + 3 \, x e^{x} - 2 \, x - 15\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x+x^2-2*x-15)*log(4*exp(x)*x+4/3*x^2-8/3*x-20)+(3*x^2+12*x)*exp(x)+5*x^2-8*x-45)/(3*exp(x
)*x+x^2-2*x-15),x, algorithm="maxima")

[Out]

-x*(log(3) - 2*log(2) - 3) + x*log(x^2 + 3*x*e^x - 2*x - 15)

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mupad [B]  time = 3.56, size = 20, normalized size = 0.83 \begin {gather*} x\,\left (\ln \left (4\,x\,{\mathrm {e}}^x-\frac {8\,x}{3}+\frac {4\,x^2}{3}-20\right )+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + log(4*x*exp(x) - (8*x)/3 + (4*x^2)/3 - 20)*(2*x - 3*x*exp(x) - x^2 + 15) - exp(x)*(12*x + 3*x^2) -
5*x^2 + 45)/(2*x - 3*x*exp(x) - x^2 + 15),x)

[Out]

x*(log(4*x*exp(x) - (8*x)/3 + (4*x^2)/3 - 20) + 3)

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sympy [A]  time = 0.39, size = 26, normalized size = 1.08 \begin {gather*} x \log {\left (\frac {4 x^{2}}{3} + 4 x e^{x} - \frac {8 x}{3} - 20 \right )} + 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*exp(x)*x+x**2-2*x-15)*ln(4*exp(x)*x+4/3*x**2-8/3*x-20)+(3*x**2+12*x)*exp(x)+5*x**2-8*x-45)/(3*ex
p(x)*x+x**2-2*x-15),x)

[Out]

x*log(4*x**2/3 + 4*x*exp(x) - 8*x/3 - 20) + 3*x

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