3.27.55 \(\int \frac {-2-8 x-2 x^2-10 x^3-8 x^4+(-4 x^2+22 x^3+24 x^4) \log (x)+(-2 x^2+11 x^3+12 x^4) \log ^2(x)+(2 x+8 x^2+(-2 x-8 x^2) \log (x)+(-x-4 x^2) \log ^2(x)) \log (1+4 x)}{(x+4 x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ \left (1+\frac {2}{\log (x)}\right ) \left (1+x \left (x+x^2-\log (1+4 x)\right )\right ) \]

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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 {-2-8 x-2 x^2-10 x^3-8 x^4+\left (-4 x^2+22 x^3+24 x^4\right ) \log (x)+\left (-2 x^2+11 x^3+12 x^4\right ) \log ^2(x)+\left (2 x+8 x^2+\left (-2 x-8 x^2\right ) \log (x)+\left (-x-4 x^2\right ) \log ^2(x)\right ) \log (1+4 x)}{\left (x+4 x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2 - 8*x - 2*x^2 - 10*x^3 - 8*x^4 + (-4*x^2 + 22*x^3 + 24*x^4)*Log[x] + (-2*x^2 + 11*x^3 + 12*x^4)*Log[x]
^2 + (2*x + 8*x^2 + (-2*x - 8*x^2)*Log[x] + (-x - 4*x^2)*Log[x]^2)*Log[1 + 4*x])/((x + 4*x^2)*Log[x]^2),x]

[Out]

x^2 + x^3 + Log[1 + 4*x]/4 - ((1 + 4*x)*Log[1 + 4*x])/4 - 8*Defer[Int][1/((1 + 4*x)*Log[x]^2), x] - 2*Defer[In
t][1/(x*(1 + 4*x)*Log[x]^2), x] - 2*Defer[Int][x/((1 + 4*x)*Log[x]^2), x] - 10*Defer[Int][x^2/((1 + 4*x)*Log[x
]^2), x] - 8*Defer[Int][x^3/((1 + 4*x)*Log[x]^2), x] + 2*Defer[Int][(x*(-2 + 11*x + 12*x^2))/((1 + 4*x)*Log[x]
), x] + 2*Defer[Int][Log[1 + 4*x]/Log[x]^2, x] - 2*Defer[Int][Log[1 + 4*x]/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2-8 x-2 x^2-10 x^3-8 x^4+\left (-4 x^2+22 x^3+24 x^4\right ) \log (x)+\left (-2 x^2+11 x^3+12 x^4\right ) \log ^2(x)+\left (2 x+8 x^2+\left (-2 x-8 x^2\right ) \log (x)+\left (-x-4 x^2\right ) \log ^2(x)\right ) \log (1+4 x)}{x (1+4 x) \log ^2(x)} \, dx\\ &=\int \left (\frac {x \left (-2+11 x+12 x^2\right )}{1+4 x}-\frac {8}{(1+4 x) \log ^2(x)}-\frac {2}{x (1+4 x) \log ^2(x)}-\frac {2 x}{(1+4 x) \log ^2(x)}-\frac {10 x^2}{(1+4 x) \log ^2(x)}-\frac {8 x^3}{(1+4 x) \log ^2(x)}+\frac {2 x \left (-2+11 x+12 x^2\right )}{(1+4 x) \log (x)}-\frac {\left (-2+2 \log (x)+\log ^2(x)\right ) \log (1+4 x)}{\log ^2(x)}\right ) \, dx\\ &=-\left (2 \int \frac {1}{x (1+4 x) \log ^2(x)} \, dx\right )-2 \int \frac {x}{(1+4 x) \log ^2(x)} \, dx+2 \int \frac {x \left (-2+11 x+12 x^2\right )}{(1+4 x) \log (x)} \, dx-8 \int \frac {1}{(1+4 x) \log ^2(x)} \, dx-8 \int \frac {x^3}{(1+4 x) \log ^2(x)} \, dx-10 \int \frac {x^2}{(1+4 x) \log ^2(x)} \, dx+\int \frac {x \left (-2+11 x+12 x^2\right )}{1+4 x} \, dx-\int \frac {\left (-2+2 \log (x)+\log ^2(x)\right ) \log (1+4 x)}{\log ^2(x)} \, dx\\ &=-\left (2 \int \frac {1}{x (1+4 x) \log ^2(x)} \, dx\right )-2 \int \frac {x}{(1+4 x) \log ^2(x)} \, dx+2 \int \frac {x \left (-2+11 x+12 x^2\right )}{(1+4 x) \log (x)} \, dx-8 \int \frac {1}{(1+4 x) \log ^2(x)} \, dx-8 \int \frac {x^3}{(1+4 x) \log ^2(x)} \, dx-10 \int \frac {x^2}{(1+4 x) \log ^2(x)} \, dx+\int \left (-1+2 x+3 x^2+\frac {1}{1+4 x}\right ) \, dx-\int \left (\log (1+4 x)-\frac {2 \log (1+4 x)}{\log ^2(x)}+\frac {2 \log (1+4 x)}{\log (x)}\right ) \, dx\\ &=-x+x^2+x^3+\frac {1}{4} \log (1+4 x)-2 \int \frac {1}{x (1+4 x) \log ^2(x)} \, dx-2 \int \frac {x}{(1+4 x) \log ^2(x)} \, dx+2 \int \frac {x \left (-2+11 x+12 x^2\right )}{(1+4 x) \log (x)} \, dx+2 \int \frac {\log (1+4 x)}{\log ^2(x)} \, dx-2 \int \frac {\log (1+4 x)}{\log (x)} \, dx-8 \int \frac {1}{(1+4 x) \log ^2(x)} \, dx-8 \int \frac {x^3}{(1+4 x) \log ^2(x)} \, dx-10 \int \frac {x^2}{(1+4 x) \log ^2(x)} \, dx-\int \log (1+4 x) \, dx\\ &=-x+x^2+x^3+\frac {1}{4} \log (1+4 x)-\frac {1}{4} \operatorname {Subst}(\int \log (x) \, dx,x,1+4 x)-2 \int \frac {1}{x (1+4 x) \log ^2(x)} \, dx-2 \int \frac {x}{(1+4 x) \log ^2(x)} \, dx+2 \int \frac {x \left (-2+11 x+12 x^2\right )}{(1+4 x) \log (x)} \, dx+2 \int \frac {\log (1+4 x)}{\log ^2(x)} \, dx-2 \int \frac {\log (1+4 x)}{\log (x)} \, dx-8 \int \frac {1}{(1+4 x) \log ^2(x)} \, dx-8 \int \frac {x^3}{(1+4 x) \log ^2(x)} \, dx-10 \int \frac {x^2}{(1+4 x) \log ^2(x)} \, dx\\ &=x^2+x^3+\frac {1}{4} \log (1+4 x)-\frac {1}{4} (1+4 x) \log (1+4 x)-2 \int \frac {1}{x (1+4 x) \log ^2(x)} \, dx-2 \int \frac {x}{(1+4 x) \log ^2(x)} \, dx+2 \int \frac {x \left (-2+11 x+12 x^2\right )}{(1+4 x) \log (x)} \, dx+2 \int \frac {\log (1+4 x)}{\log ^2(x)} \, dx-2 \int \frac {\log (1+4 x)}{\log (x)} \, dx-8 \int \frac {1}{(1+4 x) \log ^2(x)} \, dx-8 \int \frac {x^3}{(1+4 x) \log ^2(x)} \, dx-10 \int \frac {x^2}{(1+4 x) \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 38, normalized size = 1.46 \begin {gather*} x^2+x^3+\frac {2 \left (1+x^2+x^3\right )}{\log (x)}-\frac {x (2+\log (x)) \log (1+4 x)}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 8*x - 2*x^2 - 10*x^3 - 8*x^4 + (-4*x^2 + 22*x^3 + 24*x^4)*Log[x] + (-2*x^2 + 11*x^3 + 12*x^4)*
Log[x]^2 + (2*x + 8*x^2 + (-2*x - 8*x^2)*Log[x] + (-x - 4*x^2)*Log[x]^2)*Log[1 + 4*x])/((x + 4*x^2)*Log[x]^2),
x]

[Out]

x^2 + x^3 + (2*(1 + x^2 + x^3))/Log[x] - (x*(2 + Log[x])*Log[1 + 4*x])/Log[x]

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fricas [A]  time = 0.59, size = 43, normalized size = 1.65 \begin {gather*} \frac {2 \, x^{3} + 2 \, x^{2} - {\left (x \log \relax (x) + 2 \, x\right )} \log \left (4 \, x + 1\right ) + {\left (x^{3} + x^{2}\right )} \log \relax (x) + 2}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2-x)*log(x)^2+(-8*x^2-2*x)*log(x)+8*x^2+2*x)*log(4*x+1)+(12*x^4+11*x^3-2*x^2)*log(x)^2+(24*x
^4+22*x^3-4*x^2)*log(x)-8*x^4-10*x^3-2*x^2-8*x-2)/(4*x^2+x)/log(x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 38, normalized size = 1.46 \begin {gather*} x^{3} + x^{2} - {\left (x + \frac {2 \, x}{\log \relax (x)}\right )} \log \left (4 \, x + 1\right ) + \frac {2 \, {\left (x^{3} + x^{2} + 1\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2-x)*log(x)^2+(-8*x^2-2*x)*log(x)+8*x^2+2*x)*log(4*x+1)+(12*x^4+11*x^3-2*x^2)*log(x)^2+(24*x
^4+22*x^3-4*x^2)*log(x)-8*x^4-10*x^3-2*x^2-8*x-2)/(4*x^2+x)/log(x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.40, size = 48, normalized size = 1.85

method	result	size
risch	\(-\frac {x \left (\ln \relax (x )+2\right ) \ln \left (4 x +1\right )}{\ln \relax (x )}+\frac {x^{3} \ln \relax (x )+2 x^{3}+x^{2} \ln \relax (x )+2 x^{2}+2}{\ln \relax (x )}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^2-x)*ln(x)^2+(-8*x^2-2*x)*ln(x)+8*x^2+2*x)*ln(4*x+1)+(12*x^4+11*x^3-2*x^2)*ln(x)^2+(24*x^4+22*x^3-
4*x^2)*ln(x)-8*x^4-10*x^3-2*x^2-8*x-2)/(4*x^2+x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-x*(ln(x)+2)/ln(x)*ln(4*x+1)+(x^3*ln(x)+2*x^3+x^2*ln(x)+2*x^2+2)/ln(x)

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maxima [A]  time = 0.79, size = 43, normalized size = 1.65 \begin {gather*} \frac {2 \, x^{3} + 2 \, x^{2} - {\left (x \log \relax (x) + 2 \, x\right )} \log \left (4 \, x + 1\right ) + {\left (x^{3} + x^{2}\right )} \log \relax (x) + 2}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^2-x)*log(x)^2+(-8*x^2-2*x)*log(x)+8*x^2+2*x)*log(4*x+1)+(12*x^4+11*x^3-2*x^2)*log(x)^2+(24*x
^4+22*x^3-4*x^2)*log(x)-8*x^4-10*x^3-2*x^2-8*x-2)/(4*x^2+x)/log(x)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.65, size = 60, normalized size = 2.31 \begin {gather*} \frac {2\,x^2+2\,x^3-2\,x^2\,\ln \relax (x)\,\left (3\,x+2\right )+2}{\ln \relax (x)}+5\,x^2+7\,x^3-\frac {\ln \left (4\,x+1\right )\,\left (2\,x+x\,\ln \relax (x)\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x - log(x)*(22*x^3 - 4*x^2 + 24*x^4) - log(x)^2*(11*x^3 - 2*x^2 + 12*x^4) - log(4*x + 1)*(2*x - log(x)
*(2*x + 8*x^2) - log(x)^2*(x + 4*x^2) + 8*x^2) + 2*x^2 + 10*x^3 + 8*x^4 + 2)/(log(x)^2*(x + 4*x^2)),x)

[Out]

(2*x^2 + 2*x^3 - 2*x^2*log(x)*(3*x + 2) + 2)/log(x) + 5*x^2 + 7*x^3 - (log(4*x + 1)*(2*x + x*log(x)))/log(x)

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sympy [B]  time = 0.43, size = 41, normalized size = 1.58 \begin {gather*} x^{3} + x^{2} + \frac {\left (- x \log {\relax (x )} - 2 x\right ) \log {\left (4 x + 1 \right )}}{\log {\relax (x )}} + \frac {2 x^{3} + 2 x^{2} + 2}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x**2-x)*ln(x)**2+(-8*x**2-2*x)*ln(x)+8*x**2+2*x)*ln(4*x+1)+(12*x**4+11*x**3-2*x**2)*ln(x)**2+(
24*x**4+22*x**3-4*x**2)*ln(x)-8*x**4-10*x**3-2*x**2-8*x-2)/(4*x**2+x)/ln(x)**2,x)

[Out]

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

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