3.10.23 \(\int \frac {25-20 x-20 x^2+4 x^4+(-25+10 x+20 x^2-4 x^3-4 x^4) \log (\frac {-10 x+4 x^2+4 x^3}{-5+2 x^2})}{1125 x^2-450 x^3-900 x^4+180 x^5+180 x^6+(-750 x+300 x^2+600 x^3-120 x^4-120 x^5) \log (\frac {-10 x+4 x^2+4 x^3}{-5+2 x^2})+(125-50 x-100 x^2+20 x^3+20 x^4) \log ^2(\frac {-10 x+4 x^2+4 x^3}{-5+2 x^2})} \, dx\)

Optimal. Leaf size=32 \[ \frac {x}{5 \left (3 x-\log \left (2 x+\frac {2 x^2}{-\frac {5}{2}+x^2}\right )\right )} \]

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Rubi [A]  time = 0.54, antiderivative size = 36, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, integrand size = 188, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6688, 12, 6711, 32} \begin {gather*} -\frac {1}{15 \left (1-\frac {3 x}{\log \left (\frac {2 x \left (-2 x^2-2 x+5\right )}{5-2 x^2}\right )}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 - 20*x - 20*x^2 + 4*x^4 + (-25 + 10*x + 20*x^2 - 4*x^3 - 4*x^4)*Log[(-10*x + 4*x^2 + 4*x^3)/(-5 + 2*x^
2)])/(1125*x^2 - 450*x^3 - 900*x^4 + 180*x^5 + 180*x^6 + (-750*x + 300*x^2 + 600*x^3 - 120*x^4 - 120*x^5)*Log[
(-10*x + 4*x^2 + 4*x^3)/(-5 + 2*x^2)] + (125 - 50*x - 100*x^2 + 20*x^3 + 20*x^4)*Log[(-10*x + 4*x^2 + 4*x^3)/(
-5 + 2*x^2)]^2),x]

[Out]

-1/15*1/(1 - (3*x)/Log[(2*x*(5 - 2*x - 2*x^2))/(5 - 2*x^2)])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25-20 x-20 x^2+4 x^4+\left (-25+10 x+20 x^2-4 x^3-4 x^4\right ) \log \left (\frac {2 x \left (-5+2 x+2 x^2\right )}{-5+2 x^2}\right )}{5 \left (25-10 x-20 x^2+4 x^3+4 x^4\right ) \left (3 x-\log \left (\frac {2 x \left (-5+2 x+2 x^2\right )}{-5+2 x^2}\right )\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {25-20 x-20 x^2+4 x^4+\left (-25+10 x+20 x^2-4 x^3-4 x^4\right ) \log \left (\frac {2 x \left (-5+2 x+2 x^2\right )}{-5+2 x^2}\right )}{\left (25-10 x-20 x^2+4 x^3+4 x^4\right ) \left (3 x-\log \left (\frac {2 x \left (-5+2 x+2 x^2\right )}{-5+2 x^2}\right )\right )^2} \, dx\\ &=-\left (\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{(-1+3 x)^2} \, dx,x,\frac {x}{\log \left (\frac {2 x \left (-5+2 x+2 x^2\right )}{-5+2 x^2}\right )}\right )\right )\\ &=-\frac {1}{15 \left (1-\frac {3 x}{\log \left (\frac {2 x \left (5-2 x-2 x^2\right )}{5-2 x^2}\right )}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 34, normalized size = 1.06 \begin {gather*} -\frac {x}{5 \left (-3 x+\log \left (\frac {2 x \left (-5+2 x+2 x^2\right )}{-5+2 x^2}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 - 20*x - 20*x^2 + 4*x^4 + (-25 + 10*x + 20*x^2 - 4*x^3 - 4*x^4)*Log[(-10*x + 4*x^2 + 4*x^3)/(-5
+ 2*x^2)])/(1125*x^2 - 450*x^3 - 900*x^4 + 180*x^5 + 180*x^6 + (-750*x + 300*x^2 + 600*x^3 - 120*x^4 - 120*x^5
)*Log[(-10*x + 4*x^2 + 4*x^3)/(-5 + 2*x^2)] + (125 - 50*x - 100*x^2 + 20*x^3 + 20*x^4)*Log[(-10*x + 4*x^2 + 4*
x^3)/(-5 + 2*x^2)]^2),x]

[Out]

-1/5*x/(-3*x + Log[(2*x*(-5 + 2*x + 2*x^2))/(-5 + 2*x^2)])

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fricas [A]  time = 0.66, size = 37, normalized size = 1.16 \begin {gather*} \frac {x}{5 \, {\left (3 \, x - \log \left (\frac {2 \, {\left (2 \, x^{3} + 2 \, x^{2} - 5 \, x\right )}}{2 \, x^{2} - 5}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^4-4*x^3+20*x^2+10*x-25)*log((4*x^3+4*x^2-10*x)/(2*x^2-5))+4*x^4-20*x^2-20*x+25)/((20*x^4+20*x
^3-100*x^2-50*x+125)*log((4*x^3+4*x^2-10*x)/(2*x^2-5))^2+(-120*x^5-120*x^4+600*x^3+300*x^2-750*x)*log((4*x^3+4
*x^2-10*x)/(2*x^2-5))+180*x^6+180*x^5-900*x^4-450*x^3+1125*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.89, size = 37, normalized size = 1.16 \begin {gather*} \frac {x}{5 \, {\left (3 \, x - \log \left (\frac {2 \, {\left (2 \, x^{3} + 2 \, x^{2} - 5 \, x\right )}}{2 \, x^{2} - 5}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^4-4*x^3+20*x^2+10*x-25)*log((4*x^3+4*x^2-10*x)/(2*x^2-5))+4*x^4-20*x^2-20*x+25)/((20*x^4+20*x
^3-100*x^2-50*x+125)*log((4*x^3+4*x^2-10*x)/(2*x^2-5))^2+(-120*x^5-120*x^4+600*x^3+300*x^2-750*x)*log((4*x^3+4
*x^2-10*x)/(2*x^2-5))+180*x^6+180*x^5-900*x^4-450*x^3+1125*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 37, normalized size = 1.16

method	result	size
norman	\(\frac {x}{15 x -5 \ln \left (\frac {4 x^{3}+4 x^{2}-10 x}{2 x^{2}-5}\right )}\)	\(37\)
risch	\(\frac {x}{15 x -5 \ln \left (\frac {4 x^{3}+4 x^{2}-10 x}{2 x^{2}-5}\right )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^4-4*x^3+20*x^2+10*x-25)*ln((4*x^3+4*x^2-10*x)/(2*x^2-5))+4*x^4-20*x^2-20*x+25)/((20*x^4+20*x^3-100*
x^2-50*x+125)*ln((4*x^3+4*x^2-10*x)/(2*x^2-5))^2+(-120*x^5-120*x^4+600*x^3+300*x^2-750*x)*ln((4*x^3+4*x^2-10*x
)/(2*x^2-5))+180*x^6+180*x^5-900*x^4-450*x^3+1125*x^2),x,method=_RETURNVERBOSE)

[Out]

1/5*x/(3*x-ln((4*x^3+4*x^2-10*x)/(2*x^2-5)))

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maxima [A]  time = 0.89, size = 38, normalized size = 1.19 \begin {gather*} \frac {x}{5 \, {\left (3 \, x - \log \relax (2) - \log \left (2 \, x^{2} + 2 \, x - 5\right ) + \log \left (2 \, x^{2} - 5\right ) - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^4-4*x^3+20*x^2+10*x-25)*log((4*x^3+4*x^2-10*x)/(2*x^2-5))+4*x^4-20*x^2-20*x+25)/((20*x^4+20*x
^3-100*x^2-50*x+125)*log((4*x^3+4*x^2-10*x)/(2*x^2-5))^2+(-120*x^5-120*x^4+600*x^3+300*x^2-750*x)*log((4*x^3+4
*x^2-10*x)/(2*x^2-5))+180*x^6+180*x^5-900*x^4-450*x^3+1125*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.33, size = 35, normalized size = 1.09 \begin {gather*} \frac {x}{5\,\left (3\,x-\ln \left (\frac {4\,x^3+4\,x^2-10\,x}{2\,x^2-5}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x + log((4*x^2 - 10*x + 4*x^3)/(2*x^2 - 5))*(4*x^3 - 20*x^2 - 10*x + 4*x^4 + 25) + 20*x^2 - 4*x^4 - 2
5)/(log((4*x^2 - 10*x + 4*x^3)/(2*x^2 - 5))^2*(20*x^3 - 100*x^2 - 50*x + 20*x^4 + 125) + 1125*x^2 - 450*x^3 -
900*x^4 + 180*x^5 + 180*x^6 - log((4*x^2 - 10*x + 4*x^3)/(2*x^2 - 5))*(750*x - 300*x^2 - 600*x^3 + 120*x^4 + 1
20*x^5)),x)

[Out]

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

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sympy [A]  time = 0.29, size = 29, normalized size = 0.91 \begin {gather*} - \frac {x}{- 15 x + 5 \log {\left (\frac {4 x^{3} + 4 x^{2} - 10 x}{2 x^{2} - 5} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**4-4*x**3+20*x**2+10*x-25)*ln((4*x**3+4*x**2-10*x)/(2*x**2-5))+4*x**4-20*x**2-20*x+25)/((20*x
**4+20*x**3-100*x**2-50*x+125)*ln((4*x**3+4*x**2-10*x)/(2*x**2-5))**2+(-120*x**5-120*x**4+600*x**3+300*x**2-75
0*x)*ln((4*x**3+4*x**2-10*x)/(2*x**2-5))+180*x**6+180*x**5-900*x**4-450*x**3+1125*x**2),x)

[Out]

-x/(-15*x + 5*log((4*x**3 + 4*x**2 - 10*x)/(2*x**2 - 5)))

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