3.31.11 \(\int \frac {72 x^2+8 x^3-144 x^4+(-30-10 x+180 x^2) \log (4)+(24 x^2-30 \log (4)) \log (\frac {-4 x^2+5 \log (4)}{\log (4)})}{-4 x^2+5 \log (4)} \, dx\)

Optimal. Leaf size=29 \[ x \left (-x+6 \left (-1+2 x^2-\log \left (5-\frac {4 x^2}{\log (4)}\right )\right )\right ) \]

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Rubi [B]  time = 0.25, antiderivative size = 71, normalized size of antiderivative = 2.45, number of steps used = 8, number of rules used = 6, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {6725, 1810, 207, 2448, 321, 206} \begin {gather*} 12 x^3-x^2-6 x \log \left (5-\frac {4 x^2}{\log (4)}\right )-6 x-6 \sqrt {\log (1024)} \tanh ^{-1}\left (\frac {2 x}{\sqrt {5 \log (4)}}\right )+6 \sqrt {5 \log (4)} \tanh ^{-1}\left (\frac {2 x}{\sqrt {5 \log (4)}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(72*x^2 + 8*x^3 - 144*x^4 + (-30 - 10*x + 180*x^2)*Log[4] + (24*x^2 - 30*Log[4])*Log[(-4*x^2 + 5*Log[4])/L
og[4]])/(-4*x^2 + 5*Log[4]),x]

[Out]

-6*x - x^2 + 12*x^3 + 6*ArcTanh[(2*x)/Sqrt[5*Log[4]]]*Sqrt[5*Log[4]] - 6*ArcTanh[(2*x)/Sqrt[5*Log[4]]]*Sqrt[Lo
g[1024]] - 6*x*Log[5 - (4*x^2)/Log[4]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (-4 x^3+72 x^4+15 \log (4)+5 x \log (4)-18 x^2 (2+5 \log (4))\right )}{4 x^2-5 \log (4)}-6 \log \left (5-\frac {4 x^2}{\log (4)}\right )\right ) \, dx\\ &=2 \int \frac {-4 x^3+72 x^4+15 \log (4)+5 x \log (4)-18 x^2 (2+5 \log (4))}{4 x^2-5 \log (4)} \, dx-6 \int \log \left (5-\frac {4 x^2}{\log (4)}\right ) \, dx\\ &=-6 x \log \left (5-\frac {4 x^2}{\log (4)}\right )+2 \int \left (-9-x+18 x^2-\frac {30 \log (4)}{4 x^2-5 \log (4)}\right ) \, dx-\frac {48 \int \frac {x^2}{5-\frac {4 x^2}{\log (4)}} \, dx}{\log (4)}\\ &=-6 x-x^2+12 x^3-6 x \log \left (5-\frac {4 x^2}{\log (4)}\right )-60 \int \frac {1}{5-\frac {4 x^2}{\log (4)}} \, dx-(60 \log (4)) \int \frac {1}{4 x^2-5 \log (4)} \, dx\\ &=-6 x-x^2+12 x^3+6 \tanh ^{-1}\left (\frac {2 x}{\sqrt {5 \log (4)}}\right ) \sqrt {5 \log (4)}-6 \tanh ^{-1}\left (\frac {2 x}{\sqrt {5 \log (4)}}\right ) \sqrt {\log (1024)}-6 x \log \left (5-\frac {4 x^2}{\log (4)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 33, normalized size = 1.14 \begin {gather*} 2 \left (-3 x-\frac {x^2}{2}+6 x^3-3 x \log \left (5-\frac {4 x^2}{\log (4)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(72*x^2 + 8*x^3 - 144*x^4 + (-30 - 10*x + 180*x^2)*Log[4] + (24*x^2 - 30*Log[4])*Log[(-4*x^2 + 5*Log
[4])/Log[4]])/(-4*x^2 + 5*Log[4]),x]

[Out]

2*(-3*x - x^2/2 + 6*x^3 - 3*x*Log[5 - (4*x^2)/Log[4]])

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fricas [A]  time = 1.13, size = 34, normalized size = 1.17 \begin {gather*} 12 \, x^{3} - x^{2} - 6 \, x \log \left (-\frac {2 \, x^{2} - 5 \, \log \relax (2)}{\log \relax (2)}\right ) - 6 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-60*log(2)+24*x^2)*log(1/2*(10*log(2)-4*x^2)/log(2))+2*(180*x^2-10*x-30)*log(2)-144*x^4+8*x^3+72*x
^2)/(10*log(2)-4*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.40, size = 33, normalized size = 1.14 \begin {gather*} 12 \, x^{3} - x^{2} + 6 \, x {\left (\log \left (\log \relax (2)\right ) - 1\right )} - 6 \, x \log \left (-2 \, x^{2} + 5 \, \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-60*log(2)+24*x^2)*log(1/2*(10*log(2)-4*x^2)/log(2))+2*(180*x^2-10*x-30)*log(2)-144*x^4+8*x^3+72*x
^2)/(10*log(2)-4*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.47, size = 35, normalized size = 1.21

method	result	size
default	\(12 x^{3}-x^{2}-6 x -6 x \ln \left (-2 x^{2}+5 \ln \relax (2)\right )+6 x \ln \left (\ln \relax (2)\right )\)	\(35\)
norman	\(-6 x -x^{2}+12 x^{3}-6 x \ln \left (\frac {10 \ln \relax (2)-4 x^{2}}{2 \ln \relax (2)}\right )\)	\(35\)
risch	\(-6 x -x^{2}+12 x^{3}-6 x \ln \left (\frac {10 \ln \relax (2)-4 x^{2}}{2 \ln \relax (2)}\right )\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-60*ln(2)+24*x^2)*ln(1/2*(10*ln(2)-4*x^2)/ln(2))+2*(180*x^2-10*x-30)*ln(2)-144*x^4+8*x^3+72*x^2)/(10*ln(
2)-4*x^2),x,method=_RETURNVERBOSE)

[Out]

12*x^3-x^2-6*x-6*x*ln(-2*x^2+5*ln(2))+6*x*ln(ln(2))

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maxima [B]  time = 0.98, size = 124, normalized size = 4.28 \begin {gather*} 12 \, x^{3} + \frac {45}{2} \, \sqrt {10} \log \relax (2)^{\frac {3}{2}} \log \left (\frac {2 \, x - \sqrt {10} \sqrt {\log \relax (2)}}{2 \, x + \sqrt {10} \sqrt {\log \relax (2)}}\right ) - x^{2} + 6 \, x {\left (\log \left (\log \relax (2)\right ) + 2\right )} - \frac {45}{2} \, {\left (\sqrt {10} \sqrt {\log \relax (2)} \log \left (\frac {2 \, x - \sqrt {10} \sqrt {\log \relax (2)}}{2 \, x + \sqrt {10} \sqrt {\log \relax (2)}}\right ) + 4 \, x\right )} \log \relax (2) + 90 \, x \log \relax (2) - 6 \, x \log \left (-2 \, x^{2} + 5 \, \log \relax (2)\right ) - 18 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-60*log(2)+24*x^2)*log(1/2*(10*log(2)-4*x^2)/log(2))+2*(180*x^2-10*x-30)*log(2)-144*x^4+8*x^3+72*x
^2)/(10*log(2)-4*x^2),x, algorithm="maxima")

[Out]

12*x^3 + 45/2*sqrt(10)*log(2)^(3/2)*log((2*x - sqrt(10)*sqrt(log(2)))/(2*x + sqrt(10)*sqrt(log(2)))) - x^2 + 6
*x*(log(log(2)) + 2) - 45/2*(sqrt(10)*sqrt(log(2))*log((2*x - sqrt(10)*sqrt(log(2)))/(2*x + sqrt(10)*sqrt(log(
2)))) + 4*x)*log(2) + 90*x*log(2) - 6*x*log(-2*x^2 + 5*log(2)) - 18*x

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mupad [B]  time = 6.41, size = 251, normalized size = 8.66 \begin {gather*} 2\,\mathrm {atan}\left (\frac {4\,x\,\sqrt {-\frac {45\,\ln \relax (2)}{2}}}{30\,\ln \relax (2)-450\,{\ln \relax (2)}^2}-\frac {60\,x\,\ln \relax (2)\,\sqrt {-\frac {45\,\ln \relax (2)}{2}}}{30\,\ln \relax (2)-450\,{\ln \relax (2)}^2}\right )\,\sqrt {-\frac {45\,\ln \relax (2)}{2}}-6\,x+\frac {5\,\ln \relax (2)\,\ln \left (x^2-\frac {5\,\ln \relax (2)}{2}\right )}{2}-\frac {\ln \left (32\right )\,\ln \left (x^2-\frac {5\,\ln \relax (2)}{2}\right )}{2}-90\,x\,\ln \relax (2)+18\,x\,\ln \left (32\right )-6\,x\,\ln \left (\ln \left (32\right )-2\,x^2\right )+6\,x\,\ln \left (\ln \relax (2)\right )-x^2+12\,x^3-30\,\ln \relax (2)\,\mathrm {atan}\left (\frac {4\,x\,\sqrt {-\frac {45\,\ln \relax (2)}{2}}}{30\,\ln \relax (2)-450\,{\ln \relax (2)}^2}-\frac {60\,x\,\ln \relax (2)\,\sqrt {-\frac {45\,\ln \relax (2)}{2}}}{30\,\ln \relax (2)-450\,{\ln \relax (2)}^2}\right )\,\sqrt {-\frac {45\,\ln \relax (2)}{2}}-6\,\sqrt {10}\,\mathrm {atanh}\left (\frac {\sqrt {10}\,x}{5\,\sqrt {\ln \relax (2)}}\right )\,\sqrt {\ln \relax (2)}+9\,\sqrt {2}\,\sqrt {5}\,\sqrt {\ln \relax (2)}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {5}\,x}{5\,\sqrt {\ln \relax (2)}}\right )-\frac {225\,\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,x}{\sqrt {\ln \left (32\right )}}\right )\,{\ln \relax (2)}^2}{\sqrt {\ln \left (32\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(2)*(10*x - 180*x^2 + 30) + log((5*log(2) - 2*x^2)/log(2))*(60*log(2) - 24*x^2) - 72*x^2 - 8*x^3 +
144*x^4)/(10*log(2) - 4*x^2),x)

[Out]

2*atan((4*x*(-(45*log(2))/2)^(1/2))/(30*log(2) - 450*log(2)^2) - (60*x*log(2)*(-(45*log(2))/2)^(1/2))/(30*log(
2) - 450*log(2)^2))*(-(45*log(2))/2)^(1/2) - 6*x + (5*log(2)*log(x^2 - (5*log(2))/2))/2 - (log(32)*log(x^2 - (
5*log(2))/2))/2 - 90*x*log(2) + 18*x*log(32) - 6*x*log(log(32) - 2*x^2) + 6*x*log(log(2)) - x^2 + 12*x^3 - 30*
log(2)*atan((4*x*(-(45*log(2))/2)^(1/2))/(30*log(2) - 450*log(2)^2) - (60*x*log(2)*(-(45*log(2))/2)^(1/2))/(30
*log(2) - 450*log(2)^2))*(-(45*log(2))/2)^(1/2) - 6*10^(1/2)*atanh((10^(1/2)*x)/(5*log(2)^(1/2)))*log(2)^(1/2)
+ 9*2^(1/2)*5^(1/2)*log(2)^(1/2)*atanh((2^(1/2)*5^(1/2)*x)/(5*log(2)^(1/2))) - (225*2^(1/2)*atanh((2^(1/2)*x)
/log(32)^(1/2))*log(2)^2)/log(32)^(1/2)

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sympy [A]  time = 0.17, size = 29, normalized size = 1.00 \begin {gather*} 12 x^{3} - x^{2} - 6 x \log {\left (\frac {- 2 x^{2} + 5 \log {\relax (2 )}}{\log {\relax (2 )}} \right )} - 6 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-60*ln(2)+24*x**2)*ln(1/2*(10*ln(2)-4*x**2)/ln(2))+2*(180*x**2-10*x-30)*ln(2)-144*x**4+8*x**3+72*x
**2)/(10*ln(2)-4*x**2),x)

[Out]

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

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