3.84.29 \(\int \frac {-6+84 x+12 x^2+(588+84 x) \log (x)+(x^2+2 x^3+(-1+12 x+28 x^2) \log (x)+(49+98 x) \log ^2(x)) \log ^2(-x^2+(1-14 x) \log (x)-49 \log ^2(x))}{(x^3+(-x+14 x^2) \log (x)+49 x \log ^2(x)) \log ^2(-x^2+(1-14 x) \log (x)-49 \log ^2(x))} \, dx\)

Optimal. Leaf size=25 \[ 2+2 x+\log (x)-\frac {6}{\log \left (\log (x)-(x+7 \log (x))^2\right )} \]

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Rubi [A]  time = 0.48, antiderivative size = 30, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 3, integrand size = 123, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6742, 43, 6686} \begin {gather*} -\frac {6}{\log \left (-x^2-49 \log ^2(x)-14 x \log (x)+\log (x)\right )}+2 x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6 + 84*x + 12*x^2 + (588 + 84*x)*Log[x] + (x^2 + 2*x^3 + (-1 + 12*x + 28*x^2)*Log[x] + (49 + 98*x)*Log[x
]^2)*Log[-x^2 + (1 - 14*x)*Log[x] - 49*Log[x]^2]^2)/((x^3 + (-x + 14*x^2)*Log[x] + 49*x*Log[x]^2)*Log[-x^2 + (
1 - 14*x)*Log[x] - 49*Log[x]^2]^2),x]

[Out]

2*x + Log[x] - 6/Log[-x^2 + Log[x] - 14*x*Log[x] - 49*Log[x]^2]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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 (\frac {1+2 x}{x}+\frac {6 \left (-1+14 x+2 x^2+98 \log (x)+14 x \log (x)\right )}{x \left (x^2-\log (x)+14 x \log (x)+49 \log ^2(x)\right ) \log ^2\left (-x^2+\log (x)-14 x \log (x)-49 \log ^2(x)\right )}\right ) \, dx\\ &=6 \int \frac {-1+14 x+2 x^2+98 \log (x)+14 x \log (x)}{x \left (x^2-\log (x)+14 x \log (x)+49 \log ^2(x)\right ) \log ^2\left (-x^2+\log (x)-14 x \log (x)-49 \log ^2(x)\right )} \, dx+\int \frac {1+2 x}{x} \, dx\\ &=-\frac {6}{\log \left (-x^2+\log (x)-14 x \log (x)-49 \log ^2(x)\right )}+\int \left (2+\frac {1}{x}\right ) \, dx\\ &=2 x+\log (x)-\frac {6}{\log \left (-x^2+\log (x)-14 x \log (x)-49 \log ^2(x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 30, normalized size = 1.20 \begin {gather*} 2 x+\log (x)-\frac {6}{\log \left (-x^2+\log (x)-14 x \log (x)-49 \log ^2(x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6 + 84*x + 12*x^2 + (588 + 84*x)*Log[x] + (x^2 + 2*x^3 + (-1 + 12*x + 28*x^2)*Log[x] + (49 + 98*x)
*Log[x]^2)*Log[-x^2 + (1 - 14*x)*Log[x] - 49*Log[x]^2]^2)/((x^3 + (-x + 14*x^2)*Log[x] + 49*x*Log[x]^2)*Log[-x
^2 + (1 - 14*x)*Log[x] - 49*Log[x]^2]^2),x]

[Out]

2*x + Log[x] - 6/Log[-x^2 + Log[x] - 14*x*Log[x] - 49*Log[x]^2]

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fricas [B]  time = 1.04, size = 56, normalized size = 2.24 \begin {gather*} \frac {{\left (2 \, x + \log \relax (x)\right )} \log \left (-x^{2} - {\left (14 \, x - 1\right )} \log \relax (x) - 49 \, \log \relax (x)^{2}\right ) - 6}{\log \left (-x^{2} - {\left (14 \, x - 1\right )} \log \relax (x) - 49 \, \log \relax (x)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((98*x+49)*log(x)^2+(28*x^2+12*x-1)*log(x)+2*x^3+x^2)*log(-49*log(x)^2+(-14*x+1)*log(x)-x^2)^2+(84*
x+588)*log(x)+12*x^2+84*x-6)/(49*x*log(x)^2+(14*x^2-x)*log(x)+x^3)/log(-49*log(x)^2+(-14*x+1)*log(x)-x^2)^2,x,
algorithm="fricas")

[Out]

((2*x + log(x))*log(-x^2 - (14*x - 1)*log(x) - 49*log(x)^2) - 6)/log(-x^2 - (14*x - 1)*log(x) - 49*log(x)^2)

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giac [A]  time = 0.53, size = 30, normalized size = 1.20 \begin {gather*} 2 \, x - \frac {6}{\log \left (-x^{2} - 14 \, x \log \relax (x) - 49 \, \log \relax (x)^{2} + \log \relax (x)\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((98*x+49)*log(x)^2+(28*x^2+12*x-1)*log(x)+2*x^3+x^2)*log(-49*log(x)^2+(-14*x+1)*log(x)-x^2)^2+(84*
x+588)*log(x)+12*x^2+84*x-6)/(49*x*log(x)^2+(14*x^2-x)*log(x)+x^3)/log(-49*log(x)^2+(-14*x+1)*log(x)-x^2)^2,x,
algorithm="giac")

[Out]

2*x - 6/log(-x^2 - 14*x*log(x) - 49*log(x)^2 + log(x)) + log(x)

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maple [A]  time = 0.04, size = 32, normalized size = 1.28

method	result	size
risch	\(2 x +\ln \relax (x )-\frac {6}{\ln \left (-49 \ln \relax (x )^{2}+\left (-14 x +1\right ) \ln \relax (x )-x^{2}\right )}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((98*x+49)*ln(x)^2+(28*x^2+12*x-1)*ln(x)+2*x^3+x^2)*ln(-49*ln(x)^2+(-14*x+1)*ln(x)-x^2)^2+(84*x+588)*ln(x
)+12*x^2+84*x-6)/(49*x*ln(x)^2+(14*x^2-x)*ln(x)+x^3)/ln(-49*ln(x)^2+(-14*x+1)*ln(x)-x^2)^2,x,method=_RETURNVER
BOSE)

[Out]

2*x+ln(x)-6/ln(-49*ln(x)^2+(-14*x+1)*ln(x)-x^2)

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maxima [B]  time = 0.41, size = 55, normalized size = 2.20 \begin {gather*} \frac {2 \, {\left (x \log \left (-x^{2} - {\left (14 \, x - 1\right )} \log \relax (x) - 49 \, \log \relax (x)^{2}\right ) - 3\right )}}{\log \left (-x^{2} - {\left (14 \, x - 1\right )} \log \relax (x) - 49 \, \log \relax (x)^{2}\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((98*x+49)*log(x)^2+(28*x^2+12*x-1)*log(x)+2*x^3+x^2)*log(-49*log(x)^2+(-14*x+1)*log(x)-x^2)^2+(84*
x+588)*log(x)+12*x^2+84*x-6)/(49*x*log(x)^2+(14*x^2-x)*log(x)+x^3)/log(-49*log(x)^2+(-14*x+1)*log(x)-x^2)^2,x,
algorithm="maxima")

[Out]

2*(x*log(-x^2 - (14*x - 1)*log(x) - 49*log(x)^2) - 3)/log(-x^2 - (14*x - 1)*log(x) - 49*log(x)^2) + log(x)

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mupad [B]  time = 5.27, size = 32, normalized size = 1.28 \begin {gather*} 2\,x+\ln \relax (x)-\frac {6}{\ln \left (-49\,{\ln \relax (x)}^2-\ln \relax (x)\,\left (14\,x-1\right )-x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((84*x + log(x)*(84*x + 588) + 12*x^2 + log(- 49*log(x)^2 - log(x)*(14*x - 1) - x^2)^2*(log(x)*(12*x + 28*x
^2 - 1) + x^2 + 2*x^3 + log(x)^2*(98*x + 49)) - 6)/(log(- 49*log(x)^2 - log(x)*(14*x - 1) - x^2)^2*(49*x*log(x
)^2 - log(x)*(x - 14*x^2) + x^3)),x)

[Out]

2*x + log(x) - 6/log(- 49*log(x)^2 - log(x)*(14*x - 1) - x^2)

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sympy [A]  time = 0.45, size = 27, normalized size = 1.08 \begin {gather*} 2 x + \log {\relax (x )} - \frac {6}{\log {\left (- x^{2} + \left (1 - 14 x\right ) \log {\relax (x )} - 49 \log {\relax (x )}^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((98*x+49)*ln(x)**2+(28*x**2+12*x-1)*ln(x)+2*x**3+x**2)*ln(-49*ln(x)**2+(-14*x+1)*ln(x)-x**2)**2+(8
4*x+588)*ln(x)+12*x**2+84*x-6)/(49*x*ln(x)**2+(14*x**2-x)*ln(x)+x**3)/ln(-49*ln(x)**2+(-14*x+1)*ln(x)-x**2)**2
,x)

[Out]

2*x + log(x) - 6/log(-x**2 + (1 - 14*x)*log(x) - 49*log(x)**2)

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