3.96.58 \(\int \frac {(-160 x-80 x^2) \log ^2(4)+(100 x^2+25 x^3) \log ^4(4)+((-160-80 x) \log ^2(4)+100 x \log ^4(4)) \log (x^2)-25 x \log ^4(4) \log ^2(x^2)}{128 x-160 x^2 \log ^2(4)+50 x^3 \log ^4(4)} \, dx\)

Optimal. Leaf size=22 \[ \frac {5 \left (x+\log \left (x^2\right )\right )^2}{10 x-\frac {16}{\log ^2(4)}} \]

________________________________________________________________________________________

Rubi [B]  time = 0.85, antiderivative size = 87, normalized size of antiderivative = 3.95, number of steps used = 19, number of rules used = 13, integrand size = 92, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.141, Rules used = {1594, 27, 12, 6688, 6742, 698, 2357, 2301, 2314, 31, 2317, 2391, 2318} \begin {gather*} -\frac {5}{16} \log ^2(4) \log ^2\left (x^2\right )-\frac {5 x \log ^2(4) \log \left (x^2\right )}{8-5 x \log ^2(4)}-\frac {25 x \log ^4(4) \log ^2\left (x^2\right )}{16 \left (8-5 x \log ^2(4)\right )}+\frac {x}{2}-\frac {32}{5 \log ^2(4) \left (8-5 x \log ^2(4)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-160*x - 80*x^2)*Log[4]^2 + (100*x^2 + 25*x^3)*Log[4]^4 + ((-160 - 80*x)*Log[4]^2 + 100*x*Log[4]^4)*Log[
x^2] - 25*x*Log[4]^4*Log[x^2]^2)/(128*x - 160*x^2*Log[4]^2 + 50*x^3*Log[4]^4),x]

[Out]

x/2 - 32/(5*Log[4]^2*(8 - 5*x*Log[4]^2)) - (5*x*Log[4]^2*Log[x^2])/(8 - 5*x*Log[4]^2) - (5*Log[4]^2*Log[x^2]^2
)/16 - (25*x*Log[4]^4*Log[x^2]^2)/(16*(8 - 5*x*Log[4]^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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2318

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_))^2, x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])
^p)/(d*(d + e*x)), x] - Dist[(b*n*p)/d, Int[(a + b*Log[c*x^n])^(p - 1)/(d + e*x), x], x] /; FreeQ[{a, b, c, d,
e, n, p}, x] && GtQ[p, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 6688

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

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 \frac {\left (-160 x-80 x^2\right ) \log ^2(4)+\left (100 x^2+25 x^3\right ) \log ^4(4)+\left ((-160-80 x) \log ^2(4)+100 x \log ^4(4)\right ) \log \left (x^2\right )-25 x \log ^4(4) \log ^2\left (x^2\right )}{x \left (128-160 x \log ^2(4)+50 x^2 \log ^4(4)\right )} \, dx\\ &=\int \frac {\left (-160 x-80 x^2\right ) \log ^2(4)+\left (100 x^2+25 x^3\right ) \log ^4(4)+\left ((-160-80 x) \log ^2(4)+100 x \log ^4(4)\right ) \log \left (x^2\right )-25 x \log ^4(4) \log ^2\left (x^2\right )}{2 x \left (-8+5 x \log ^2(4)\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {\left (-160 x-80 x^2\right ) \log ^2(4)+\left (100 x^2+25 x^3\right ) \log ^4(4)+\left ((-160-80 x) \log ^2(4)+100 x \log ^4(4)\right ) \log \left (x^2\right )-25 x \log ^4(4) \log ^2\left (x^2\right )}{x \left (-8+5 x \log ^2(4)\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {5 \log ^2(4) \left (x+\log \left (x^2\right )\right ) \left (-32+5 x^2 \log ^2(4)+4 x \left (-4+5 \log ^2(4)\right )-5 x \log ^2(4) \log \left (x^2\right )\right )}{x \left (8-5 x \log ^2(4)\right )^2} \, dx\\ &=\frac {1}{2} \left (5 \log ^2(4)\right ) \int \frac {\left (x+\log \left (x^2\right )\right ) \left (-32+5 x^2 \log ^2(4)+4 x \left (-4+5 \log ^2(4)\right )-5 x \log ^2(4) \log \left (x^2\right )\right )}{x \left (8-5 x \log ^2(4)\right )^2} \, dx\\ &=\frac {1}{2} \left (5 \log ^2(4)\right ) \int \left (\frac {-32+5 x^2 \log ^2(4)-4 x \left (4-5 \log ^2(4)\right )}{\left (8-5 x \log ^2(4)\right )^2}+\frac {4 \left (-8-x \left (4-5 \log ^2(4)\right )\right ) \log \left (x^2\right )}{x \left (8-5 x \log ^2(4)\right )^2}-\frac {5 \log ^2(4) \log ^2\left (x^2\right )}{\left (-8+5 x \log ^2(4)\right )^2}\right ) \, dx\\ &=\frac {1}{2} \left (5 \log ^2(4)\right ) \int \frac {-32+5 x^2 \log ^2(4)-4 x \left (4-5 \log ^2(4)\right )}{\left (8-5 x \log ^2(4)\right )^2} \, dx+\left (10 \log ^2(4)\right ) \int \frac {\left (-8-x \left (4-5 \log ^2(4)\right )\right ) \log \left (x^2\right )}{x \left (8-5 x \log ^2(4)\right )^2} \, dx-\frac {1}{2} \left (25 \log ^4(4)\right ) \int \frac {\log ^2\left (x^2\right )}{\left (-8+5 x \log ^2(4)\right )^2} \, dx\\ &=-\frac {25 x \log ^4(4) \log ^2\left (x^2\right )}{16 \left (8-5 x \log ^2(4)\right )}+\frac {1}{2} \left (5 \log ^2(4)\right ) \int \left (\frac {1}{5 \log ^2(4)}-\frac {64}{5 \log ^2(4) \left (-8+5 x \log ^2(4)\right )^2}+\frac {4}{-8+5 x \log ^2(4)}\right ) \, dx+\left (10 \log ^2(4)\right ) \int \left (-\frac {\log \left (x^2\right )}{8 x}-\frac {4 \log \left (x^2\right )}{\left (-8+5 x \log ^2(4)\right )^2}+\frac {5 \log ^2(4) \log \left (x^2\right )}{8 \left (-8+5 x \log ^2(4)\right )}\right ) \, dx-\frac {1}{4} \left (25 \log ^4(4)\right ) \int \frac {\log \left (x^2\right )}{-8+5 x \log ^2(4)} \, dx\\ &=\frac {x}{2}-\frac {32}{5 \log ^2(4) \left (8-5 x \log ^2(4)\right )}-\frac {25 x \log ^4(4) \log ^2\left (x^2\right )}{16 \left (8-5 x \log ^2(4)\right )}+2 \log \left (8-5 x \log ^2(4)\right )-\frac {5}{4} \log ^2(4) \log \left (x^2\right ) \log \left (1-\frac {5}{8} x \log ^2(4)\right )-\frac {1}{4} \left (5 \log ^2(4)\right ) \int \frac {\log \left (x^2\right )}{x} \, dx+\frac {1}{2} \left (5 \log ^2(4)\right ) \int \frac {\log \left (1-\frac {5}{8} x \log ^2(4)\right )}{x} \, dx-\left (40 \log ^2(4)\right ) \int \frac {\log \left (x^2\right )}{\left (-8+5 x \log ^2(4)\right )^2} \, dx+\frac {1}{4} \left (25 \log ^4(4)\right ) \int \frac {\log \left (x^2\right )}{-8+5 x \log ^2(4)} \, dx\\ &=\frac {x}{2}-\frac {32}{5 \log ^2(4) \left (8-5 x \log ^2(4)\right )}-\frac {5 x \log ^2(4) \log \left (x^2\right )}{8-5 x \log ^2(4)}-\frac {5}{16} \log ^2(4) \log ^2\left (x^2\right )-\frac {25 x \log ^4(4) \log ^2\left (x^2\right )}{16 \left (8-5 x \log ^2(4)\right )}+2 \log \left (8-5 x \log ^2(4)\right )-\frac {5}{2} \log ^2(4) \text {Li}_2\left (\frac {5}{8} x \log ^2(4)\right )-\frac {1}{2} \left (5 \log ^2(4)\right ) \int \frac {\log \left (1-\frac {5}{8} x \log ^2(4)\right )}{x} \, dx-\left (10 \log ^2(4)\right ) \int \frac {1}{-8+5 x \log ^2(4)} \, dx\\ &=\frac {x}{2}-\frac {32}{5 \log ^2(4) \left (8-5 x \log ^2(4)\right )}-\frac {5 x \log ^2(4) \log \left (x^2\right )}{8-5 x \log ^2(4)}-\frac {5}{16} \log ^2(4) \log ^2\left (x^2\right )-\frac {25 x \log ^4(4) \log ^2\left (x^2\right )}{16 \left (8-5 x \log ^2(4)\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.13, size = 60, normalized size = 2.73 \begin {gather*} \frac {64-40 x \log ^2(4)+25 x^2 \log ^4(4)+50 x \log ^4(4) \log \left (x^2\right )+25 \log ^4(4) \log ^2\left (x^2\right )}{10 \log ^2(4) \left (-8+5 x \log ^2(4)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-160*x - 80*x^2)*Log[4]^2 + (100*x^2 + 25*x^3)*Log[4]^4 + ((-160 - 80*x)*Log[4]^2 + 100*x*Log[4]^4
)*Log[x^2] - 25*x*Log[4]^4*Log[x^2]^2)/(128*x - 160*x^2*Log[4]^2 + 50*x^3*Log[4]^4),x]

[Out]

(64 - 40*x*Log[4]^2 + 25*x^2*Log[4]^4 + 50*x*Log[4]^4*Log[x^2] + 25*Log[4]^4*Log[x^2]^2)/(10*Log[4]^2*(-8 + 5*
x*Log[4]^2))

________________________________________________________________________________________

fricas [B]  time = 0.64, size = 59, normalized size = 2.68 \begin {gather*} \frac {25 \, x^{2} \log \relax (2)^{4} + 50 \, x \log \relax (2)^{4} \log \left (x^{2}\right ) + 25 \, \log \relax (2)^{4} \log \left (x^{2}\right )^{2} - 10 \, x \log \relax (2)^{2} + 4}{10 \, {\left (5 \, x \log \relax (2)^{4} - 2 \, \log \relax (2)^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-400*x*log(2)^4*log(x^2)^2+(1600*x*log(2)^4+4*(-80*x-160)*log(2)^2)*log(x^2)+16*(25*x^3+100*x^2)*lo
g(2)^4+4*(-80*x^2-160*x)*log(2)^2)/(800*x^3*log(2)^4-640*x^2*log(2)^2+128*x),x, algorithm="fricas")

[Out]

1/10*(25*x^2*log(2)^4 + 50*x*log(2)^4*log(x^2) + 25*log(2)^4*log(x^2)^2 - 10*x*log(2)^2 + 4)/(5*x*log(2)^4 - 2
*log(2)^2)

________________________________________________________________________________________

giac [B]  time = 0.21, size = 66, normalized size = 3.00 \begin {gather*} \frac {5 \, \log \relax (2)^{2} \log \left (x^{2}\right )^{2}}{2 \, {\left (5 \, x \log \relax (2)^{2} - 2\right )}} + \frac {1}{2} \, x + \frac {2 \, \log \left (x^{2}\right )}{5 \, x \log \relax (2)^{2} - 2} + \frac {2}{5 \, {\left (5 \, x \log \relax (2)^{4} - 2 \, \log \relax (2)^{2}\right )}} + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-400*x*log(2)^4*log(x^2)^2+(1600*x*log(2)^4+4*(-80*x-160)*log(2)^2)*log(x^2)+16*(25*x^3+100*x^2)*lo
g(2)^4+4*(-80*x^2-160*x)*log(2)^2)/(800*x^3*log(2)^4-640*x^2*log(2)^2+128*x),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.15, size = 46, normalized size = 2.09

method	result	size
norman	\(\frac {5 \ln \relax (2)^{2} \ln \left (x^{2}\right ) x +\frac {5 x^{2} \ln \relax (2)^{2}}{2}+\frac {5 \ln \relax (2)^{2} \ln \left (x^{2}\right )^{2}}{2}}{5 x \ln \relax (2)^{2}-2}\)	\(46\)
risch	\(\frac {5 \ln \relax (2)^{2} \ln \left (x^{2}\right )^{2}}{2 \left (5 x \ln \relax (2)^{2}-2\right )}+\frac {2 \ln \left (x^{2}\right )}{5 x \ln \relax (2)^{2}-2}+\frac {100 \ln \relax (2)^{4} \ln \left (-x \right ) x +25 x^{2} \ln \relax (2)^{4}-40 \ln \left (-x \right ) \ln \relax (2)^{2}-10 x \ln \relax (2)^{2}+4}{10 \ln \relax (2)^{2} \left (5 x \ln \relax (2)^{2}-2\right )}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-400*x*ln(2)^4*ln(x^2)^2+(1600*x*ln(2)^4+4*(-80*x-160)*ln(2)^2)*ln(x^2)+16*(25*x^3+100*x^2)*ln(2)^4+4*(-8
0*x^2-160*x)*ln(2)^2)/(800*x^3*ln(2)^4-640*x^2*ln(2)^2+128*x),x,method=_RETURNVERBOSE)

[Out]

(5*ln(2)^2*ln(x^2)*x+5/2*x^2*ln(2)^2+5/2*ln(2)^2*ln(x^2)^2)/(5*x*ln(2)^2-2)

________________________________________________________________________________________

maxima [B]  time = 0.51, size = 197, normalized size = 8.95 \begin {gather*} -\frac {1}{10} \, {\left (\frac {4}{5 \, x \log \relax (2)^{8} - 2 \, \log \relax (2)^{6}} - \frac {5 \, x}{\log \relax (2)^{4}} - \frac {4 \, \log \left (5 \, x \log \relax (2)^{2} - 2\right )}{\log \relax (2)^{6}}\right )} \log \relax (2)^{4} - 2 \, {\left (\frac {2}{5 \, x \log \relax (2)^{6} - 2 \, \log \relax (2)^{4}} - \frac {\log \left (5 \, x \log \relax (2)^{2} - 2\right )}{\log \relax (2)^{4}}\right )} \log \relax (2)^{4} + \frac {2}{5} \, {\left (\frac {2}{5 \, x \log \relax (2)^{6} - 2 \, \log \relax (2)^{4}} - \frac {\log \left (5 \, x \log \relax (2)^{2} - 2\right )}{\log \relax (2)^{4}}\right )} \log \relax (2)^{2} + \frac {4 \, \log \relax (2)^{2}}{5 \, x \log \relax (2)^{4} - 2 \, \log \relax (2)^{2}} + \frac {2 \, {\left (5 \, \log \relax (2)^{2} \log \relax (x)^{2} + 2 \, \log \relax (x)\right )}}{5 \, x \log \relax (2)^{2} - 2} - 2 \, \log \left (5 \, x \log \relax (2)^{2} - 2\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-400*x*log(2)^4*log(x^2)^2+(1600*x*log(2)^4+4*(-80*x-160)*log(2)^2)*log(x^2)+16*(25*x^3+100*x^2)*lo
g(2)^4+4*(-80*x^2-160*x)*log(2)^2)/(800*x^3*log(2)^4-640*x^2*log(2)^2+128*x),x, algorithm="maxima")

[Out]

-1/10*(4/(5*x*log(2)^8 - 2*log(2)^6) - 5*x/log(2)^4 - 4*log(5*x*log(2)^2 - 2)/log(2)^6)*log(2)^4 - 2*(2/(5*x*l
og(2)^6 - 2*log(2)^4) - log(5*x*log(2)^2 - 2)/log(2)^4)*log(2)^4 + 2/5*(2/(5*x*log(2)^6 - 2*log(2)^4) - log(5*
x*log(2)^2 - 2)/log(2)^4)*log(2)^2 + 4*log(2)^2/(5*x*log(2)^4 - 2*log(2)^2) + 2*(5*log(2)^2*log(x)^2 + 2*log(x
))/(5*x*log(2)^2 - 2) - 2*log(5*x*log(2)^2 - 2) + 2*log(x)

________________________________________________________________________________________

mupad [B]  time = 9.78, size = 65, normalized size = 2.95 \begin {gather*} \frac {x}{2}+2\,\ln \relax (x)+\frac {2\,\ln \left (x^2\right )}{5\,x\,{\ln \relax (2)}^2-2}+\frac {2}{5\,{\ln \relax (2)}^2\,\left (5\,x\,{\ln \relax (2)}^2-2\right )}+\frac {5\,{\ln \left (x^2\right )}^2\,{\ln \relax (2)}^2}{2\,\left (5\,x\,{\ln \relax (2)}^2-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^2)*(4*log(2)^2*(80*x + 160) - 1600*x*log(2)^4) + 4*log(2)^2*(160*x + 80*x^2) - 16*log(2)^4*(100*x^
2 + 25*x^3) + 400*x*log(x^2)^2*log(2)^4)/(128*x - 640*x^2*log(2)^2 + 800*x^3*log(2)^4),x)

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 0.42, size = 65, normalized size = 2.95 \begin {gather*} \frac {x}{2} + 2 \log {\relax (x )} + \frac {2}{25 x \log {\relax (2 )}^{4} - 10 \log {\relax (2 )}^{2}} + \frac {5 \log {\relax (2 )}^{2} \log {\left (x^{2} \right )}^{2}}{10 x \log {\relax (2 )}^{2} - 4} + \frac {2 \log {\left (x^{2} \right )}}{5 x \log {\relax (2 )}^{2} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-400*x*ln(2)**4*ln(x**2)**2+(1600*x*ln(2)**4+4*(-80*x-160)*ln(2)**2)*ln(x**2)+16*(25*x**3+100*x**2)
*ln(2)**4+4*(-80*x**2-160*x)*ln(2)**2)/(800*x**3*ln(2)**4-640*x**2*ln(2)**2+128*x),x)

[Out]

x/2 + 2*log(x) + 2/(25*x*log(2)**4 - 10*log(2)**2) + 5*log(2)**2*log(x**2)**2/(10*x*log(2)**2 - 4) + 2*log(x**
2)/(5*x*log(2)**2 - 2)

________________________________________________________________________________________