3.5.37 \(\int \frac {-10 x-40 x^2+(-30-110 x+30 x^2+40 x^3) \log (\frac {1}{2} (x+2 x^2))+(30+60 x-30 x^2-60 x^3) \log ^2(\frac {1}{2} (x+2 x^2))+(-60 x-120 x^2+20 x^3+40 x^4) \log ^3(\frac {1}{2} (x+2 x^2))}{(1+2 x) \log ^3(\frac {1}{2} (x+2 x^2))} \, dx\)

Optimal. Leaf size=25 \[ 5 \left (3-x^2+\frac {x}{\log \left (\frac {x}{2}+x^2\right )}\right )^2 \]

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Rubi [F]  time = 1.00, 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 {-10 x-40 x^2+\left (-30-110 x+30 x^2+40 x^3\right ) \log \left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (30+60 x-30 x^2-60 x^3\right ) \log ^2\left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (-60 x-120 x^2+20 x^3+40 x^4\right ) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )}{(1+2 x) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10*x - 40*x^2 + (-30 - 110*x + 30*x^2 + 40*x^3)*Log[(x + 2*x^2)/2] + (30 + 60*x - 30*x^2 - 60*x^3)*Log[(
x + 2*x^2)/2]^2 + (-60*x - 120*x^2 + 20*x^3 + 40*x^4)*Log[(x + 2*x^2)/2]^3)/((1 + 2*x)*Log[(x + 2*x^2)/2]^3),x
]

[Out]

-30*x^2 + 5*x^4 + 10*Defer[Int][((-1 - 4*x)*x)/((1 + 2*x)*Log[x*(1/2 + x)]^3), x] + 10*Defer[Int][(-3 - 11*x +
3*x^2 + 4*x^3)/((1 + 2*x)*Log[x*(1/2 + x)]^2), x] + 30*Defer[Int][(1 - x^2)/Log[x*(1/2 + x)], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10 x-40 x^2+\left (-30-110 x+30 x^2+40 x^3\right ) \log \left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (30+60 x-30 x^2-60 x^3\right ) \log ^2\left (\frac {1}{2} \left (x+2 x^2\right )\right )+\left (-60 x-120 x^2+20 x^3+40 x^4\right ) \log ^3\left (\frac {1}{2} \left (x+2 x^2\right )\right )}{(1+2 x) \log ^3\left (\frac {1}{2} x (1+2 x)\right )} \, dx\\ &=\int \left (20 x \left (-3+x^2\right )+\frac {10 (-1-4 x) x}{(1+2 x) \log ^3\left (x \left (\frac {1}{2}+x\right )\right )}+\frac {10 \left (-3-11 x+3 x^2+4 x^3\right )}{(1+2 x) \log ^2\left (x \left (\frac {1}{2}+x\right )\right )}+\frac {30 \left (1-x^2\right )}{\log \left (x \left (\frac {1}{2}+x\right )\right )}\right ) \, dx\\ &=10 \int \frac {(-1-4 x) x}{(1+2 x) \log ^3\left (x \left (\frac {1}{2}+x\right )\right )} \, dx+10 \int \frac {-3-11 x+3 x^2+4 x^3}{(1+2 x) \log ^2\left (x \left (\frac {1}{2}+x\right )\right )} \, dx+20 \int x \left (-3+x^2\right ) \, dx+30 \int \frac {1-x^2}{\log \left (x \left (\frac {1}{2}+x\right )\right )} \, dx\\ &=10 \int \frac {(-1-4 x) x}{(1+2 x) \log ^3\left (x \left (\frac {1}{2}+x\right )\right )} \, dx+10 \int \frac {-3-11 x+3 x^2+4 x^3}{(1+2 x) \log ^2\left (x \left (\frac {1}{2}+x\right )\right )} \, dx+20 \int \left (-3 x+x^3\right ) \, dx+30 \int \frac {1-x^2}{\log \left (x \left (\frac {1}{2}+x\right )\right )} \, dx\\ &=-30 x^2+5 x^4+10 \int \frac {(-1-4 x) x}{(1+2 x) \log ^3\left (x \left (\frac {1}{2}+x\right )\right )} \, dx+10 \int \frac {-3-11 x+3 x^2+4 x^3}{(1+2 x) \log ^2\left (x \left (\frac {1}{2}+x\right )\right )} \, dx+30 \int \frac {1-x^2}{\log \left (x \left (\frac {1}{2}+x\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 43, normalized size = 1.72 \begin {gather*} 5 x \left (-6 x+x^3+\frac {x}{\log ^2\left (\frac {x}{2}+x^2\right )}-\frac {2 \left (-3+x^2\right )}{\log \left (\frac {x}{2}+x^2\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x - 40*x^2 + (-30 - 110*x + 30*x^2 + 40*x^3)*Log[(x + 2*x^2)/2] + (30 + 60*x - 30*x^2 - 60*x^3)
*Log[(x + 2*x^2)/2]^2 + (-60*x - 120*x^2 + 20*x^3 + 40*x^4)*Log[(x + 2*x^2)/2]^3)/((1 + 2*x)*Log[(x + 2*x^2)/2
]^3),x]

[Out]

5*x*(-6*x + x^3 + x/Log[x/2 + x^2]^2 - (2*(-3 + x^2))/Log[x/2 + x^2])

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fricas [B]  time = 0.99, size = 53, normalized size = 2.12 \begin {gather*} \frac {5 \, {\left ({\left (x^{4} - 6 \, x^{2}\right )} \log \left (x^{2} + \frac {1}{2} \, x\right )^{2} + x^{2} - 2 \, {\left (x^{3} - 3 \, x\right )} \log \left (x^{2} + \frac {1}{2} \, x\right )\right )}}{\log \left (x^{2} + \frac {1}{2} \, x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^4+20*x^3-120*x^2-60*x)*log(x^2+1/2*x)^3+(-60*x^3-30*x^2+60*x+30)*log(x^2+1/2*x)^2+(40*x^3+30*
x^2-110*x-30)*log(x^2+1/2*x)-40*x^2-10*x)/(2*x+1)/log(x^2+1/2*x)^3,x, algorithm="fricas")

[Out]

5*((x^4 - 6*x^2)*log(x^2 + 1/2*x)^2 + x^2 - 2*(x^3 - 3*x)*log(x^2 + 1/2*x))/log(x^2 + 1/2*x)^2

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giac [B]  time = 0.81, size = 53, normalized size = 2.12 \begin {gather*} 5 \, x^{4} - 30 \, x^{2} - \frac {5 \, {\left (2 \, x^{3} \log \left (x^{2} + \frac {1}{2} \, x\right ) - x^{2} - 6 \, x \log \left (x^{2} + \frac {1}{2} \, x\right )\right )}}{\log \left (x^{2} + \frac {1}{2} \, x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^4+20*x^3-120*x^2-60*x)*log(x^2+1/2*x)^3+(-60*x^3-30*x^2+60*x+30)*log(x^2+1/2*x)^2+(40*x^3+30*
x^2-110*x-30)*log(x^2+1/2*x)-40*x^2-10*x)/(2*x+1)/log(x^2+1/2*x)^3,x, algorithm="giac")

[Out]

5*x^4 - 30*x^2 - 5*(2*x^3*log(x^2 + 1/2*x) - x^2 - 6*x*log(x^2 + 1/2*x))/log(x^2 + 1/2*x)^2

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maple [B]  time = 0.26, size = 52, normalized size = 2.08

method	result	size
risch	\(5 x^{4}-30 x^{2}-\frac {5 \left (2 \ln \left (x^{2}+\frac {1}{2} x \right ) x^{2}-x -6 \ln \left (x^{2}+\frac {1}{2} x \right )\right ) x}{\ln \left (x^{2}+\frac {1}{2} x \right )^{2}}\)	\(52\)
norman	\(\frac {5 x^{2}+30 \ln \left (x^{2}+\frac {1}{2} x \right ) x -10 \ln \left (x^{2}+\frac {1}{2} x \right ) x^{3}-30 \ln \left (x^{2}+\frac {1}{2} x \right )^{2} x^{2}+5 \ln \left (x^{2}+\frac {1}{2} x \right )^{2} x^{4}}{\ln \left (x^{2}+\frac {1}{2} x \right )^{2}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((40*x^4+20*x^3-120*x^2-60*x)*ln(x^2+1/2*x)^3+(-60*x^3-30*x^2+60*x+30)*ln(x^2+1/2*x)^2+(40*x^3+30*x^2-110*
x-30)*ln(x^2+1/2*x)-40*x^2-10*x)/(2*x+1)/ln(x^2+1/2*x)^3,x,method=_RETURNVERBOSE)

[Out]

5*x^4-30*x^2-5*(2*ln(x^2+1/2*x)*x^2-x-6*ln(x^2+1/2*x))*x/ln(x^2+1/2*x)^2

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maxima [B]  time = 0.55, size = 173, normalized size = 6.92 \begin {gather*} \frac {5 \, {\left (x^{4} \log \relax (2)^{2} + 2 \, x^{3} \log \relax (2) - {\left (6 \, \log \relax (2)^{2} - 1\right )} x^{2} + {\left (x^{4} - 6 \, x^{2}\right )} \log \left (2 \, x + 1\right )^{2} + {\left (x^{4} - 6 \, x^{2}\right )} \log \relax (x)^{2} - 6 \, x \log \relax (2) - 2 \, {\left (x^{4} \log \relax (2) + x^{3} - 6 \, x^{2} \log \relax (2) - {\left (x^{4} - 6 \, x^{2}\right )} \log \relax (x) - 3 \, x\right )} \log \left (2 \, x + 1\right ) - 2 \, {\left (x^{4} \log \relax (2) + x^{3} - 6 \, x^{2} \log \relax (2) - 3 \, x\right )} \log \relax (x)\right )}}{\log \relax (2)^{2} - 2 \, {\left (\log \relax (2) - \log \relax (x)\right )} \log \left (2 \, x + 1\right ) + \log \left (2 \, x + 1\right )^{2} - 2 \, \log \relax (2) \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^4+20*x^3-120*x^2-60*x)*log(x^2+1/2*x)^3+(-60*x^3-30*x^2+60*x+30)*log(x^2+1/2*x)^2+(40*x^3+30*
x^2-110*x-30)*log(x^2+1/2*x)-40*x^2-10*x)/(2*x+1)/log(x^2+1/2*x)^3,x, algorithm="maxima")

[Out]

5*(x^4*log(2)^2 + 2*x^3*log(2) - (6*log(2)^2 - 1)*x^2 + (x^4 - 6*x^2)*log(2*x + 1)^2 + (x^4 - 6*x^2)*log(x)^2
- 6*x*log(2) - 2*(x^4*log(2) + x^3 - 6*x^2*log(2) - (x^4 - 6*x^2)*log(x) - 3*x)*log(2*x + 1) - 2*(x^4*log(2) +
x^3 - 6*x^2*log(2) - 3*x)*log(x))/(log(2)^2 - 2*(log(2) - log(x))*log(2*x + 1) + log(2*x + 1)^2 - 2*log(2)*lo
g(x) + log(x)^2)

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mupad [B]  time = 0.67, size = 48, normalized size = 1.92 \begin {gather*} \frac {5\,x^2-5\,x\,\ln \left (x^2+\frac {x}{2}\right )\,\left (2\,x^2-6\right )}{{\ln \left (x^2+\frac {x}{2}\right )}^2}-5\,x\,\left (6\,x-x^3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x + log(x/2 + x^2)^3*(60*x + 120*x^2 - 20*x^3 - 40*x^4) + log(x/2 + x^2)*(110*x - 30*x^2 - 40*x^3 + 3
0) + 40*x^2 - log(x/2 + x^2)^2*(60*x - 30*x^2 - 60*x^3 + 30))/(log(x/2 + x^2)^3*(2*x + 1)),x)

[Out]

(5*x^2 - 5*x*log(x/2 + x^2)*(2*x^2 - 6))/log(x/2 + x^2)^2 - 5*x*(6*x - x^3)

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sympy [B]  time = 0.18, size = 41, normalized size = 1.64 \begin {gather*} 5 x^{4} - 30 x^{2} + \frac {5 x^{2} + \left (- 10 x^{3} + 30 x\right ) \log {\left (x^{2} + \frac {x}{2} \right )}}{\log {\left (x^{2} + \frac {x}{2} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x**4+20*x**3-120*x**2-60*x)*ln(x**2+1/2*x)**3+(-60*x**3-30*x**2+60*x+30)*ln(x**2+1/2*x)**2+(40*
x**3+30*x**2-110*x-30)*ln(x**2+1/2*x)-40*x**2-10*x)/(2*x+1)/ln(x**2+1/2*x)**3,x)

[Out]

5*x**4 - 30*x**2 + (5*x**2 + (-10*x**3 + 30*x)*log(x**2 + x/2))/log(x**2 + x/2)**2

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