3.5.48 \(\int \frac {-3 x^2+4 x^3-2 x^4+(6 x-4 x^2) \log (2)-2 \log ^2(2)+(-x^2+2 x^3-x^4+(2 x-2 x^2) \log (2)-\log ^2(2)) \log (\frac {5}{x^2})}{x^2-2 x^3+x^4+(-2 x+2 x^2) \log (2)+\log ^2(2)} \, dx\)

Optimal. Leaf size=31 \[ 3+e^3+\frac {x^2}{-x+x^2+\log (2)}-x \left (4+\log \left (\frac {5}{x^2}\right )\right ) \]

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Rubi [B]  time = 0.64, antiderivative size = 410, normalized size of antiderivative = 13.23, number of steps used = 26, number of rules used = 13, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.123, Rules used = {6688, 6742, 738, 773, 634, 618, 204, 628, 800, 614, 722, 638, 2295} \begin {gather*} \frac {2 (1-2 x) \log ^2(2)}{(1-\log (16)) \left (-x^2+x-\log (2)\right )}+\frac {4 x^2 (x-\log (4))}{(1-\log (16)) \left (-x^2+x-\log (2)\right )}-\frac {2 x^2}{1-\log (16)}-x \log \left (\frac {5}{x^2}\right )-\frac {x (3+\log (16)) (x-\log (4))}{(1-\log (16)) \left (-x^2+x-\log (2)\right )}-\frac {(2-\log (256)) \log \left (-x^2+x-\log (2)\right )}{1-\log (16)}+2 \log \left (-x^2+x-\log (2)\right )+\frac {\log (64) (x-\log (4))}{(1-\log (16)) \left (-x^2+x-\log (2)\right )}-\frac {2 x^3 (x-\log (4))}{(1-\log (16)) \left (-x^2+x-\log (2)\right )}-2 x-\frac {2 x (2-\log (64))}{1-\log (16)}+\frac {4 x}{1-\log (16)}+\frac {8 \log ^2(2) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {\log (16)-1}}\right )}{(\log (16)-1)^{3/2}}-\frac {2 (2+\log (4) \log (64)-\log (4096)) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {\log (16)-1}}\right )}{(\log (16)-1)^{3/2}}-\frac {2 \log (64) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {\log (16)-1}}\right )}{(\log (16)-1)^{3/2}}+\frac {4 (1-\log (64)) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {\log (16)-1}}\right )}{(\log (16)-1)^{3/2}}+\frac {4 \log (2) (3+\log (16)) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {\log (16)-1}}\right )}{(\log (16)-1)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*x + (4*x)/(1 - Log[16]) - (2*x^2)/(1 - Log[16]) + (2*(1 - 2*x)*Log[2]^2)/((x - x^2 - Log[2])*(1 - Log[16]))
+ (4*x^2*(x - Log[4]))/((x - x^2 - Log[2])*(1 - Log[16])) - (2*x^3*(x - Log[4]))/((x - x^2 - Log[2])*(1 - Log
[16])) + (8*ArcTan[(1 - 2*x)/Sqrt[-1 + Log[16]]]*Log[2]^2)/(-1 + Log[16])^(3/2) - (x*(x - Log[4])*(3 + Log[16]
))/((x - x^2 - Log[2])*(1 - Log[16])) + (4*ArcTan[(1 - 2*x)/Sqrt[-1 + Log[16]]]*Log[2]*(3 + Log[16]))/(-1 + Lo
g[16])^(3/2) + (4*ArcTan[(1 - 2*x)/Sqrt[-1 + Log[16]]]*(1 - Log[64]))/(-1 + Log[16])^(3/2) - (2*x*(2 - Log[64]
))/(1 - Log[16]) + ((x - Log[4])*Log[64])/((x - x^2 - Log[2])*(1 - Log[16])) - (2*ArcTan[(1 - 2*x)/Sqrt[-1 + L
og[16]]]*Log[64])/(-1 + Log[16])^(3/2) - (2*ArcTan[(1 - 2*x)/Sqrt[-1 + Log[16]]]*(2 + Log[4]*Log[64] - Log[409
6]))/(-1 + Log[16])^(3/2) - x*Log[5/x^2] + 2*Log[x - x^2 - Log[2]] - ((2 - Log[256])*Log[x - x^2 - Log[2]])/(1
- Log[16])

Rule 204

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

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

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

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 722

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*(2*p + 3)*(c*d
^2 - b*d*e + a*e^2))/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ
[{a, b, c, d, e}, 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] && EqQ[m +
2*p + 2, 0] && LtQ[p, -1]

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*g)*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 {4 x^3-2 x^4-2 \log ^2(2)-x^2 (3+\log (16))+x \log (64)-\left (-x+x^2+\log (2)\right )^2 \log \left (\frac {5}{x^2}\right )}{\left (x-x^2-\log (2)\right )^2} \, dx\\ &=\int \left (\frac {4 x^3}{\left (-x+x^2+\log (2)\right )^2}-\frac {2 x^4}{\left (-x+x^2+\log (2)\right )^2}-\frac {2 \log ^2(2)}{\left (-x+x^2+\log (2)\right )^2}-\frac {x^2 (3+\log (16))}{\left (-x+x^2+\log (2)\right )^2}+\frac {x \log (64)}{\left (-x+x^2+\log (2)\right )^2}-\log \left (\frac {5}{x^2}\right )\right ) \, dx\\ &=-\left (2 \int \frac {x^4}{\left (-x+x^2+\log (2)\right )^2} \, dx\right )+4 \int \frac {x^3}{\left (-x+x^2+\log (2)\right )^2} \, dx-\left (2 \log ^2(2)\right ) \int \frac {1}{\left (-x+x^2+\log (2)\right )^2} \, dx+(-3-\log (16)) \int \frac {x^2}{\left (-x+x^2+\log (2)\right )^2} \, dx+\log (64) \int \frac {x}{\left (-x+x^2+\log (2)\right )^2} \, dx-\int \log \left (\frac {5}{x^2}\right ) \, dx\\ &=-2 x+\frac {2 (1-2 x) \log ^2(2)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {4 x^2 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {2 x^3 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {x (x-\log (4)) (3+\log (16))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {(x-\log (4)) \log (64)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-x \log \left (\frac {5}{x^2}\right )+\frac {2 \int \frac {x^2 (-2 x+\log (64))}{-x+x^2+\log (2)} \, dx}{1-\log (16)}-\frac {4 \int \frac {x (-x+\log (16))}{-x+x^2+\log (2)} \, dx}{1-\log (16)}+\frac {\left (4 \log ^2(2)\right ) \int \frac {1}{-x+x^2+\log (2)} \, dx}{1-\log (16)}+\frac {(2 \log (2) (3+\log (16))) \int \frac {1}{-x+x^2+\log (2)} \, dx}{1-\log (16)}-\frac {\log (64) \int \frac {1}{-x+x^2+\log (2)} \, dx}{1-\log (16)}\\ &=-2 x+\frac {4 x}{1-\log (16)}+\frac {2 (1-2 x) \log ^2(2)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {4 x^2 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {2 x^3 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {x (x-\log (4)) (3+\log (16))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {(x-\log (4)) \log (64)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-x \log \left (\frac {5}{x^2}\right )+\frac {2 \int \left (-2-2 x+\log (64)-\frac {-\log (2) (2-\log (64))+x (2-\log (256))}{-x+x^2+\log (2)}\right ) \, dx}{1-\log (16)}-\frac {4 \int \frac {\log (2)+x (-1+\log (16))}{-x+x^2+\log (2)} \, dx}{1-\log (16)}-\frac {\left (8 \log ^2(2)\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2-4 \log (2)} \, dx,x,-1+2 x\right )}{1-\log (16)}-\frac {(4 \log (2) (3+\log (16))) \operatorname {Subst}\left (\int \frac {1}{1-x^2-4 \log (2)} \, dx,x,-1+2 x\right )}{1-\log (16)}+\frac {(2 \log (64)) \operatorname {Subst}\left (\int \frac {1}{1-x^2-4 \log (2)} \, dx,x,-1+2 x\right )}{1-\log (16)}\\ &=-2 x+\frac {4 x}{1-\log (16)}-\frac {2 x^2}{1-\log (16)}+\frac {2 (1-2 x) \log ^2(2)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {4 x^2 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {2 x^3 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {8 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log ^2(2)}{(-1+\log (16))^{3/2}}-\frac {x (x-\log (4)) (3+\log (16))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {4 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log (2) (3+\log (16))}{(-1+\log (16))^{3/2}}-\frac {2 x (2-\log (64))}{1-\log (16)}+\frac {(x-\log (4)) \log (64)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {2 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log (64)}{(-1+\log (16))^{3/2}}-x \log \left (\frac {5}{x^2}\right )+2 \int \frac {-1+2 x}{-x+x^2+\log (2)} \, dx-\frac {2 \int \frac {-\log (2) (2-\log (64))+x (2-\log (256))}{-x+x^2+\log (2)} \, dx}{1-\log (16)}+\frac {(2 (1-\log (64))) \int \frac {1}{-x+x^2+\log (2)} \, dx}{1-\log (16)}\\ &=-2 x+\frac {4 x}{1-\log (16)}-\frac {2 x^2}{1-\log (16)}+\frac {2 (1-2 x) \log ^2(2)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {4 x^2 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {2 x^3 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {8 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log ^2(2)}{(-1+\log (16))^{3/2}}-\frac {x (x-\log (4)) (3+\log (16))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {4 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log (2) (3+\log (16))}{(-1+\log (16))^{3/2}}-\frac {2 x (2-\log (64))}{1-\log (16)}+\frac {(x-\log (4)) \log (64)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {2 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log (64)}{(-1+\log (16))^{3/2}}-x \log \left (\frac {5}{x^2}\right )+2 \log \left (x-x^2-\log (2)\right )-\frac {(4 (1-\log (64))) \operatorname {Subst}\left (\int \frac {1}{1-x^2-4 \log (2)} \, dx,x,-1+2 x\right )}{1-\log (16)}-\frac {(2-\log (256)) \int \frac {-1+2 x}{-x+x^2+\log (2)} \, dx}{1-\log (16)}-\frac {(2+\log (4) \log (64)-\log (4096)) \int \frac {1}{-x+x^2+\log (2)} \, dx}{1-\log (16)}\\ &=-2 x+\frac {4 x}{1-\log (16)}-\frac {2 x^2}{1-\log (16)}+\frac {2 (1-2 x) \log ^2(2)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {4 x^2 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {2 x^3 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {8 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log ^2(2)}{(-1+\log (16))^{3/2}}-\frac {x (x-\log (4)) (3+\log (16))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {4 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log (2) (3+\log (16))}{(-1+\log (16))^{3/2}}+\frac {4 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) (1-\log (64))}{(-1+\log (16))^{3/2}}-\frac {2 x (2-\log (64))}{1-\log (16)}+\frac {(x-\log (4)) \log (64)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {2 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log (64)}{(-1+\log (16))^{3/2}}-x \log \left (\frac {5}{x^2}\right )+2 \log \left (x-x^2-\log (2)\right )-\frac {(2-\log (256)) \log \left (x-x^2-\log (2)\right )}{1-\log (16)}+\frac {(2 (2+\log (4) \log (64)-\log (4096))) \operatorname {Subst}\left (\int \frac {1}{1-x^2-4 \log (2)} \, dx,x,-1+2 x\right )}{1-\log (16)}\\ &=-2 x+\frac {4 x}{1-\log (16)}-\frac {2 x^2}{1-\log (16)}+\frac {2 (1-2 x) \log ^2(2)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {4 x^2 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {2 x^3 (x-\log (4))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {8 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log ^2(2)}{(-1+\log (16))^{3/2}}-\frac {x (x-\log (4)) (3+\log (16))}{\left (x-x^2-\log (2)\right ) (1-\log (16))}+\frac {4 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log (2) (3+\log (16))}{(-1+\log (16))^{3/2}}+\frac {4 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) (1-\log (64))}{(-1+\log (16))^{3/2}}-\frac {2 x (2-\log (64))}{1-\log (16)}+\frac {(x-\log (4)) \log (64)}{\left (x-x^2-\log (2)\right ) (1-\log (16))}-\frac {2 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) \log (64)}{(-1+\log (16))^{3/2}}-\frac {2 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {-1+\log (16)}}\right ) (2+\log (4) \log (64)-\log (4096))}{(-1+\log (16))^{3/2}}-x \log \left (\frac {5}{x^2}\right )+2 \log \left (x-x^2-\log (2)\right )-\frac {(2-\log (256)) \log \left (x-x^2-\log (2)\right )}{1-\log (16)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.21, size = 69, normalized size = 2.23 \begin {gather*} -4 x-\frac {x-4 \log ^2(2)+8 x \log ^2(2)-x (1+\log (4)) \log (16)-\log (2) (1+\log (16))+\log (4) \log (64)}{\left (-x+x^2+\log (2)\right ) (-1+\log (16))}-x \log \left (\frac {5}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*x - (x - 4*Log[2]^2 + 8*x*Log[2]^2 - x*(1 + Log[4])*Log[16] - Log[2]*(1 + Log[16]) + Log[4]*Log[64])/((-x +
x^2 + Log[2])*(-1 + Log[16])) - x*Log[5/x^2]

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fricas [A]  time = 0.97, size = 55, normalized size = 1.77 \begin {gather*} -\frac {4 \, x^{3} - 4 \, x^{2} + {\left (4 \, x + 1\right )} \log \relax (2) + {\left (x^{3} - x^{2} + x \log \relax (2)\right )} \log \left (\frac {5}{x^{2}}\right ) - x}{x^{2} - x + \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(2)^2+(-2*x^2+2*x)*log(2)-x^4+2*x^3-x^2)*log(5/x^2)-2*log(2)^2+(-4*x^2+6*x)*log(2)-2*x^4+4*x^3
-3*x^2)/(log(2)^2+(2*x^2-2*x)*log(2)+x^4-2*x^3+x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.32, size = 32, normalized size = 1.03 \begin {gather*} -x {\left (\log \relax (5) + 4\right )} + x \log \left (x^{2}\right ) + \frac {x - \log \relax (2)}{x^{2} - x + \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(2)^2+(-2*x^2+2*x)*log(2)-x^4+2*x^3-x^2)*log(5/x^2)-2*log(2)^2+(-4*x^2+6*x)*log(2)-2*x^4+4*x^3
-3*x^2)/(log(2)^2+(2*x^2-2*x)*log(2)+x^4-2*x^3+x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.27, size = 32, normalized size = 1.03

method	result	size
derivativedivides	\(-x \ln \relax (5)-x \ln \left (\frac {1}{x^{2}}\right )-4 x +\frac {1}{\frac {\ln \relax (2)}{x^{2}}-\frac {1}{x}+1}\)	\(32\)
default	\(-x \ln \relax (5)-x \ln \left (\frac {1}{x^{2}}\right )-4 x +\frac {1}{\frac {\ln \relax (2)}{x^{2}}-\frac {1}{x}+1}\)	\(32\)
risch	\(-x \ln \left (\frac {5}{x^{2}}\right )-\frac {4 x^{3}+4 x \ln \relax (2)-4 x^{2}+\ln \relax (2)-x}{\ln \relax (2)+x^{2}-x}\)	\(45\)
norman	\(\frac {x^{2} \ln \left (\frac {5}{x^{2}}\right )+\left (5-4 \ln \relax (2)\right ) x -4 x^{3}-\ln \left (\frac {5}{x^{2}}\right ) x^{3}-\ln \relax (2) \ln \left (\frac {5}{x^{2}}\right ) x -5 \ln \relax (2)}{\ln \relax (2)+x^{2}-x}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-ln(2)^2+(-2*x^2+2*x)*ln(2)-x^4+2*x^3-x^2)*ln(5/x^2)-2*ln(2)^2+(-4*x^2+6*x)*ln(2)-2*x^4+4*x^3-3*x^2)/(ln
(2)^2+(2*x^2-2*x)*ln(2)+x^4-2*x^3+x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 1.06, size = 243, normalized size = 7.84 \begin {gather*} -2 \, {\left (\frac {2 \, x - 1}{x^{2} {\left (4 \, \log \relax (2) - 1\right )} - x {\left (4 \, \log \relax (2) - 1\right )} + 4 \, \log \relax (2)^{2} - \log \relax (2)} + \frac {4 \, \arctan \left (\frac {2 \, x - 1}{\sqrt {4 \, \log \relax (2) - 1}}\right )}{{\left (4 \, \log \relax (2) - 1\right )}^{\frac {3}{2}}}\right )} \log \relax (2)^{2} + \frac {8 \, \arctan \left (\frac {2 \, x - 1}{\sqrt {4 \, \log \relax (2) - 1}}\right ) \log \relax (2)^{2}}{{\left (4 \, \log \relax (2) - 1\right )}^{\frac {3}{2}}} - \frac {{\left (4 \, {\left (\log \relax (5) + 4\right )} \log \relax (2) - \log \relax (5) - 4\right )} x^{3} - {\left (4 \, {\left (\log \relax (5) + 4\right )} \log \relax (2) - \log \relax (5) - 4\right )} x^{2} + {\left (4 \, {\left (\log \relax (5) + 3\right )} \log \relax (2)^{2} - {\left (\log \relax (5) + 8\right )} \log \relax (2) + 1\right )} x + 6 \, \log \relax (2)^{2} - 2 \, {\left (x^{3} {\left (4 \, \log \relax (2) - 1\right )} - x^{2} {\left (4 \, \log \relax (2) - 1\right )} + {\left (4 \, \log \relax (2)^{2} - \log \relax (2)\right )} x\right )} \log \relax (x) - \log \relax (2)}{x^{2} {\left (4 \, \log \relax (2) - 1\right )} - x {\left (4 \, \log \relax (2) - 1\right )} + 4 \, \log \relax (2)^{2} - \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(2)^2+(-2*x^2+2*x)*log(2)-x^4+2*x^3-x^2)*log(5/x^2)-2*log(2)^2+(-4*x^2+6*x)*log(2)-2*x^4+4*x^3
-3*x^2)/(log(2)^2+(2*x^2-2*x)*log(2)+x^4-2*x^3+x^2),x, algorithm="maxima")

[Out]

-2*((2*x - 1)/(x^2*(4*log(2) - 1) - x*(4*log(2) - 1) + 4*log(2)^2 - log(2)) + 4*arctan((2*x - 1)/sqrt(4*log(2)
- 1))/(4*log(2) - 1)^(3/2))*log(2)^2 + 8*arctan((2*x - 1)/sqrt(4*log(2) - 1))*log(2)^2/(4*log(2) - 1)^(3/2) -
((4*(log(5) + 4)*log(2) - log(5) - 4)*x^3 - (4*(log(5) + 4)*log(2) - log(5) - 4)*x^2 + (4*(log(5) + 3)*log(2)
^2 - (log(5) + 8)*log(2) + 1)*x + 6*log(2)^2 - 2*(x^3*(4*log(2) - 1) - x^2*(4*log(2) - 1) + (4*log(2)^2 - log(
2))*x)*log(x) - log(2))/(x^2*(4*log(2) - 1) - x*(4*log(2) - 1) + 4*log(2)^2 - log(2))

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mupad [B]  time = 0.90, size = 239, normalized size = 7.71 \begin {gather*} \left (\sum _{k=1}^4\ln \left (\mathrm {root}\left ({\ln \relax (2)}^2+16\,{\ln \relax (2)}^4-8\,{\ln \relax (2)}^3,z,k\right )\,\ln \relax (2)\,2-\mathrm {root}\left ({\ln \relax (2)}^2+16\,{\ln \relax (2)}^4-8\,{\ln \relax (2)}^3,z,k\right )\,x\,2-\mathrm {root}\left ({\ln \relax (2)}^2+16\,{\ln \relax (2)}^4-8\,{\ln \relax (2)}^3,z,k\right )\,{\ln \relax (2)}^2\,14+\mathrm {root}\left ({\ln \relax (2)}^2+16\,{\ln \relax (2)}^4-8\,{\ln \relax (2)}^3,z,k\right )\,{\ln \relax (2)}^3\,24-4\,x\,{\ln \relax (2)}^2+8\,x\,{\ln \relax (2)}^3+{\ln \relax (2)}^2+\mathrm {root}\left ({\ln \relax (2)}^2+16\,{\ln \relax (2)}^4-8\,{\ln \relax (2)}^3,z,k\right )\,x\,\ln \relax (2)\,16-\mathrm {root}\left ({\ln \relax (2)}^2+16\,{\ln \relax (2)}^4-8\,{\ln \relax (2)}^3,z,k\right )\,x\,{\ln \relax (2)}^2\,36+\mathrm {root}\left ({\ln \relax (2)}^2+16\,{\ln \relax (2)}^4-8\,{\ln \relax (2)}^3,z,k\right )\,x\,{\ln \relax (2)}^3\,16\right )\,\mathrm {root}\left ({\ln \relax (2)}^2+16\,{\ln \relax (2)}^4-8\,{\ln \relax (2)}^3,z,k\right )\right )-x\,\ln \left (\frac {1}{x^2}\right )-4\,x-x\,\ln \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5/x^2)*(log(2)^2 - log(2)*(2*x - 2*x^2) + x^2 - 2*x^3 + x^4) - log(2)*(6*x - 4*x^2) + 2*log(2)^2 + 3
*x^2 - 4*x^3 + 2*x^4)/(log(2)^2 - log(2)*(2*x - 2*x^2) + x^2 - 2*x^3 + x^4),x)

[Out]

symsum(log(2*root(log(2)^2 + 16*log(2)^4 - 8*log(2)^3, z, k)*log(2) - 2*root(log(2)^2 + 16*log(2)^4 - 8*log(2)
^3, z, k)*x - 14*root(log(2)^2 + 16*log(2)^4 - 8*log(2)^3, z, k)*log(2)^2 + 24*root(log(2)^2 + 16*log(2)^4 - 8
*log(2)^3, z, k)*log(2)^3 - 4*x*log(2)^2 + 8*x*log(2)^3 + log(2)^2 + 16*root(log(2)^2 + 16*log(2)^4 - 8*log(2)
^3, z, k)*x*log(2) - 36*root(log(2)^2 + 16*log(2)^4 - 8*log(2)^3, z, k)*x*log(2)^2 + 16*root(log(2)^2 + 16*log
(2)^4 - 8*log(2)^3, z, k)*x*log(2)^3)*root(log(2)^2 + 16*log(2)^4 - 8*log(2)^3, z, k), k, 1, 4) - x*log(1/x^2)
- 4*x - x*log(5)

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sympy [A]  time = 0.37, size = 26, normalized size = 0.84 \begin {gather*} - x \log {\left (\frac {5}{x^{2}} \right )} - 4 x - \frac {- x + \log {\relax (2 )}}{x^{2} - x + \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-ln(2)**2+(-2*x**2+2*x)*ln(2)-x**4+2*x**3-x**2)*ln(5/x**2)-2*ln(2)**2+(-4*x**2+6*x)*ln(2)-2*x**4+4
*x**3-3*x**2)/(ln(2)**2+(2*x**2-2*x)*ln(2)+x**4-2*x**3+x**2),x)

[Out]

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

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