3.90.39 \(\int \frac {54-104 x-24 x^2+e^x (-2 x-56 x^2-8 x^3)+(56+16 x+e^x (56 x+8 x^2)) \log (x)}{-x-28 x^2-4 x^3+(28 x+4 x^2) \log (x)} \, dx\)

Optimal. Leaf size=32 \[ \log \left (\frac {1}{25} e^{2 e^x} x^2 \left (\frac {1}{4}-(7+x) (-x+\log (x))\right )^2\right ) \]

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Rubi [F]  time = 1.59, 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 {54-104 x-24 x^2+e^x \left (-2 x-56 x^2-8 x^3\right )+\left (56+16 x+e^x \left (56 x+8 x^2\right )\right ) \log (x)}{-x-28 x^2-4 x^3+\left (28 x+4 x^2\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(54 - 104*x - 24*x^2 + E^x*(-2*x - 56*x^2 - 8*x^3) + (56 + 16*x + E^x*(56*x + 8*x^2))*Log[x])/(-x - 28*x^2
- 4*x^3 + (28*x + 4*x^2)*Log[x]),x]

[Out]

2*E^x + 2*Log[x] + 2*Log[7 + x] + 48*Defer[Int][(1 + 28*x + 4*x^2 - 28*Log[x] - 4*x*Log[x])^(-1), x] - 56*Defe
r[Int][1/(x*(1 + 28*x + 4*x^2 - 28*Log[x] - 4*x*Log[x])), x] + 8*Defer[Int][x/(1 + 28*x + 4*x^2 - 28*Log[x] -
4*x*Log[x]), x] - 2*Defer[Int][1/((7 + x)*(1 + 28*x + 4*x^2 - 28*Log[x] - 4*x*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 e^x+\frac {104}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)}-\frac {54}{x \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )}+\frac {24 x}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)}-\frac {56 \log (x)}{x \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )}+\frac {16 \log (x)}{-1-28 x-4 x^2+28 \log (x)+4 x \log (x)}\right ) \, dx\\ &=2 \int e^x \, dx+16 \int \frac {\log (x)}{-1-28 x-4 x^2+28 \log (x)+4 x \log (x)} \, dx+24 \int \frac {x}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx-54 \int \frac {1}{x \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )} \, dx-56 \int \frac {\log (x)}{x \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )} \, dx+104 \int \frac {1}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx\\ &=2 e^x+16 \int \left (\frac {1}{4 (7+x)}+\frac {-1-28 x-4 x^2}{4 (7+x) \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )}\right ) \, dx+24 \int \frac {x}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx-54 \int \frac {1}{x \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )} \, dx-56 \int \left (-\frac {1}{4 x (7+x)}+\frac {1+28 x+4 x^2}{4 x (7+x) \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )}\right ) \, dx+104 \int \frac {1}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx\\ &=2 e^x+4 \log (7+x)+4 \int \frac {-1-28 x-4 x^2}{(7+x) \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )} \, dx+14 \int \frac {1}{x (7+x)} \, dx-14 \int \frac {1+28 x+4 x^2}{x (7+x) \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )} \, dx+24 \int \frac {x}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx-54 \int \frac {1}{x \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )} \, dx+104 \int \frac {1}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx\\ &=2 e^x+4 \log (7+x)+2 \int \frac {1}{x} \, dx-2 \int \frac {1}{7+x} \, dx+4 \int \left (-\frac {4 x}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)}-\frac {1}{(7+x) \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )}\right ) \, dx-14 \int \left (\frac {4}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)}+\frac {1}{7 x \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )}-\frac {1}{7 (7+x) \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )}\right ) \, dx+24 \int \frac {x}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx-54 \int \frac {1}{x \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )} \, dx+104 \int \frac {1}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx\\ &=2 e^x+2 \log (x)+2 \log (7+x)-2 \int \frac {1}{x \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )} \, dx+2 \int \frac {1}{(7+x) \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )} \, dx-4 \int \frac {1}{(7+x) \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )} \, dx-16 \int \frac {x}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx+24 \int \frac {x}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx-54 \int \frac {1}{x \left (1+28 x+4 x^2-28 \log (x)-4 x \log (x)\right )} \, dx-56 \int \frac {1}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx+104 \int \frac {1}{1+28 x+4 x^2-28 \log (x)-4 x \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.54, size = 26, normalized size = 0.81 \begin {gather*} 2 \left (e^x+\log (x)+\log \left (1+28 x+4 x^2-4 (7+x) \log (x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(54 - 104*x - 24*x^2 + E^x*(-2*x - 56*x^2 - 8*x^3) + (56 + 16*x + E^x*(56*x + 8*x^2))*Log[x])/(-x -
28*x^2 - 4*x^3 + (28*x + 4*x^2)*Log[x]),x]

[Out]

2*(E^x + Log[x] + Log[1 + 28*x + 4*x^2 - 4*(7 + x)*Log[x]])

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fricas [A]  time = 0.61, size = 42, normalized size = 1.31 \begin {gather*} 2 \, e^{x} + 2 \, \log \left (x^{2} + 7 \, x\right ) + 2 \, \log \left (-\frac {4 \, x^{2} - 4 \, {\left (x + 7\right )} \log \relax (x) + 28 \, x + 1}{x + 7}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+56*x)*exp(x)+16*x+56)*log(x)+(-8*x^3-56*x^2-2*x)*exp(x)-24*x^2-104*x+54)/((4*x^2+28*x)*log(
x)-4*x^3-28*x^2-x),x, algorithm="fricas")

[Out]

2*e^x + 2*log(x^2 + 7*x) + 2*log(-(4*x^2 - 4*(x + 7)*log(x) + 28*x + 1)/(x + 7))

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giac [A]  time = 0.57, size = 31, normalized size = 0.97 \begin {gather*} 2 \, e^{x} + 2 \, \log \left (-4 \, x^{2} + 4 \, x \log \relax (x) - 28 \, x + 28 \, \log \relax (x) - 1\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+56*x)*exp(x)+16*x+56)*log(x)+(-8*x^3-56*x^2-2*x)*exp(x)-24*x^2-104*x+54)/((4*x^2+28*x)*log(
x)-4*x^3-28*x^2-x),x, algorithm="giac")

[Out]

2*e^x + 2*log(-4*x^2 + 4*x*log(x) - 28*x + 28*log(x) - 1) + 2*log(x)

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maple [A]  time = 0.08, size = 32, normalized size = 1.00

method	result	size
default	\(2 \ln \relax (x )+2 \ln \left (4 x^{2}-4 x \ln \relax (x )+28 x -28 \ln \relax (x )+1\right )+2 \,{\mathrm e}^{x}\)	\(32\)
norman	\(2 \ln \relax (x )+2 \ln \left (4 x^{2}-4 x \ln \relax (x )+28 x -28 \ln \relax (x )+1\right )+2 \,{\mathrm e}^{x}\)	\(32\)
risch	\(2 \ln \left (x^{2}+7 x \right )+2 \,{\mathrm e}^{x}+2 \ln \left (\ln \relax (x )-\frac {4 x^{2}+28 x +1}{4 \left (x +7\right )}\right )\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x^2+56*x)*exp(x)+16*x+56)*ln(x)+(-8*x^3-56*x^2-2*x)*exp(x)-24*x^2-104*x+54)/((4*x^2+28*x)*ln(x)-4*x^3
-28*x^2-x),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)+2*ln(4*x^2-4*x*ln(x)+28*x-28*ln(x)+1)+2*exp(x)

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maxima [A]  time = 0.40, size = 42, normalized size = 1.31 \begin {gather*} 2 \, e^{x} + 2 \, \log \left (x + 7\right ) + 2 \, \log \relax (x) + 2 \, \log \left (-\frac {4 \, x^{2} - 4 \, {\left (x + 7\right )} \log \relax (x) + 28 \, x + 1}{4 \, {\left (x + 7\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+56*x)*exp(x)+16*x+56)*log(x)+(-8*x^3-56*x^2-2*x)*exp(x)-24*x^2-104*x+54)/((4*x^2+28*x)*log(
x)-4*x^3-28*x^2-x),x, algorithm="maxima")

[Out]

2*e^x + 2*log(x + 7) + 2*log(x) + 2*log(-1/4*(4*x^2 - 4*(x + 7)*log(x) + 28*x + 1)/(x + 7))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {104\,x-\ln \relax (x)\,\left (16\,x+{\mathrm {e}}^x\,\left (8\,x^2+56\,x\right )+56\right )+24\,x^2+{\mathrm {e}}^x\,\left (8\,x^3+56\,x^2+2\,x\right )-54}{x-\ln \relax (x)\,\left (4\,x^2+28\,x\right )+28\,x^2+4\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((104*x - log(x)*(16*x + exp(x)*(56*x + 8*x^2) + 56) + 24*x^2 + exp(x)*(2*x + 56*x^2 + 8*x^3) - 54)/(x - lo
g(x)*(28*x + 4*x^2) + 28*x^2 + 4*x^3),x)

[Out]

int((104*x - log(x)*(16*x + exp(x)*(56*x + 8*x^2) + 56) + 24*x^2 + exp(x)*(2*x + 56*x^2 + 8*x^3) - 54)/(x - lo
g(x)*(28*x + 4*x^2) + 28*x^2 + 4*x^3), x)

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sympy [A]  time = 0.76, size = 37, normalized size = 1.16 \begin {gather*} 2 e^{x} + 2 \log {\left (x^{2} + 7 x \right )} + 2 \log {\left (\log {\relax (x )} + \frac {- 4 x^{2} - 28 x - 1}{4 x + 28} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x**2+56*x)*exp(x)+16*x+56)*ln(x)+(-8*x**3-56*x**2-2*x)*exp(x)-24*x**2-104*x+54)/((4*x**2+28*x)*
ln(x)-4*x**3-28*x**2-x),x)

[Out]

2*exp(x) + 2*log(x**2 + 7*x) + 2*log(log(x) + (-4*x**2 - 28*x - 1)/(4*x + 28))

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