3.110 \(\int \frac{\log (-1+4 x+4 \sqrt{(-1+x) x})}{\sqrt{x}} \, dx\)

Optimal. Leaf size=118 \[ -\frac{2 \sqrt{x^2-x}}{\sqrt{x}}+2 \sqrt{x} \log \left (4 \sqrt{x^2-x}+4 x-1\right )-\frac{\sqrt{x^2-x} \tan ^{-1}\left (\frac{2}{3} \sqrt{2} \sqrt{x-1}\right )}{\sqrt{2} \sqrt{x-1} \sqrt{x}}-2 \sqrt{x}+\frac{\tan ^{-1}\left (2 \sqrt{2} \sqrt{x}\right )}{\sqrt{2}} \]

[Out]

-2*Sqrt[x] - (2*Sqrt[-x + x^2])/Sqrt[x] - (Sqrt[-x + x^2]*ArcTan[(2*Sqrt[2]*Sqrt[-1 + x])/3])/(Sqrt[2]*Sqrt[-1
+ x]*Sqrt[x]) + ArcTan[2*Sqrt[2]*Sqrt[x]]/Sqrt[2] + 2*Sqrt[x]*Log[-1 + 4*x + 4*Sqrt[-x + x^2]]

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Rubi [A]  time = 0.385141, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {2537, 2535, 6733, 6742, 203, 1588, 1146, 444, 50, 63} \[ -\frac{2 \sqrt{x^2-x}}{\sqrt{x}}+2 \sqrt{x} \log \left (4 \sqrt{x^2-x}+4 x-1\right )-\frac{\sqrt{x^2-x} \tan ^{-1}\left (\frac{2}{3} \sqrt{2} \sqrt{x-1}\right )}{\sqrt{2} \sqrt{x-1} \sqrt{x}}-2 \sqrt{x}+\frac{\tan ^{-1}\left (2 \sqrt{2} \sqrt{x}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]]/Sqrt[x],x]

[Out]

-2*Sqrt[x] - (2*Sqrt[-x + x^2])/Sqrt[x] - (Sqrt[-x + x^2]*ArcTan[(2*Sqrt[2]*Sqrt[-1 + x])/3])/(Sqrt[2]*Sqrt[-1
+ x]*Sqrt[x]) + ArcTan[2*Sqrt[2]*Sqrt[x]]/Sqrt[2] + 2*Sqrt[x]*Log[-1 + 4*x + 4*Sqrt[-x + x^2]]

Rule 2537

Int[Log[(d_.) + (f_.)*Sqrt[u_] + (e_.)*(x_)]*(v_.), x_Symbol] :> Int[v*Log[d + e*x + f*Sqrt[ExpandToSum[u, x]]
], x] /; FreeQ[{d, e, f}, x] && QuadraticQ[u, x] &&  !QuadraticMatchQ[u, x] && (EqQ[v, 1] || MatchQ[v, ((g_.)*
x)^(m_.) /; FreeQ[{g, m}, x]])

Rule 2535

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

Rule 6733

Int[(u_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k, Subst[Int[x^(k*(m + 1) - 1)*(u /. x -> x^k
), x], x, x^(1/k)], x]] /; FractionQ[m]

Rule 6742

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

Rule 203

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

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 1146

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(b*x^2 + c*x^4)^FracPart
[p]/(x^(2*FracPart[p])*(b + c*x^2)^FracPart[p]), Int[x^(2*p)*(d + e*x^2)^q*(b + c*x^2)^p, x], x] /; FreeQ[{b,
c, d, e, p, q}, x] &&  !IntegerQ[p]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rubi steps

\begin{align*} \int \frac{\log \left (-1+4 x+4 \sqrt{(-1+x) x}\right )}{\sqrt{x}} \, dx &=\int \frac{\log \left (-1+4 x+4 \sqrt{-x+x^2}\right )}{\sqrt{x}} \, dx\\ &=2 \sqrt{x} \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+16 \int \frac{\sqrt{x}}{-4 (1+2 x) \sqrt{-x+x^2}+8 \left (-x+x^2\right )} \, dx\\ &=2 \sqrt{x} \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+32 \operatorname{Subst}\left (\int \frac{x^2}{-4 \left (1+2 x^2\right ) \sqrt{-x^2+x^4}+8 \left (-x^2+x^4\right )} \, dx,x,\sqrt{x}\right )\\ &=2 \sqrt{x} \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+32 \operatorname{Subst}\left (\int \left (-\frac{1}{16}+\frac{1}{16 \left (1+8 x^2\right )}-\frac{x^2}{12 \sqrt{-x^2+x^4}}+\frac{\sqrt{-x^2+x^4}}{6 \left (1+8 x^2\right )}\right ) \, dx,x,\sqrt{x}\right )\\ &=-2 \sqrt{x}+2 \sqrt{x} \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+2 \operatorname{Subst}\left (\int \frac{1}{1+8 x^2} \, dx,x,\sqrt{x}\right )-\frac{8}{3} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{-x^2+x^4}} \, dx,x,\sqrt{x}\right )+\frac{16}{3} \operatorname{Subst}\left (\int \frac{\sqrt{-x^2+x^4}}{1+8 x^2} \, dx,x,\sqrt{x}\right )\\ &=-2 \sqrt{x}-\frac{8 \sqrt{-x+x^2}}{3 \sqrt{x}}+\frac{\tan ^{-1}\left (2 \sqrt{2} \sqrt{x}\right )}{\sqrt{2}}+2 \sqrt{x} \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+\frac{\left (16 \sqrt{-x+x^2}\right ) \operatorname{Subst}\left (\int \frac{x \sqrt{-1+x^2}}{1+8 x^2} \, dx,x,\sqrt{x}\right )}{3 \sqrt{-1+x} \sqrt{x}}\\ &=-2 \sqrt{x}-\frac{8 \sqrt{-x+x^2}}{3 \sqrt{x}}+\frac{\tan ^{-1}\left (2 \sqrt{2} \sqrt{x}\right )}{\sqrt{2}}+2 \sqrt{x} \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+\frac{\left (8 \sqrt{-x+x^2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{-1+x}}{1+8 x} \, dx,x,x\right )}{3 \sqrt{-1+x} \sqrt{x}}\\ &=-2 \sqrt{x}-\frac{2 \sqrt{-x+x^2}}{\sqrt{x}}+\frac{\tan ^{-1}\left (2 \sqrt{2} \sqrt{x}\right )}{\sqrt{2}}+2 \sqrt{x} \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )-\frac{\left (3 \sqrt{-x+x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} (1+8 x)} \, dx,x,x\right )}{\sqrt{-1+x} \sqrt{x}}\\ &=-2 \sqrt{x}-\frac{2 \sqrt{-x+x^2}}{\sqrt{x}}+\frac{\tan ^{-1}\left (2 \sqrt{2} \sqrt{x}\right )}{\sqrt{2}}+2 \sqrt{x} \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )-\frac{\left (6 \sqrt{-x+x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{9+8 x^2} \, dx,x,\sqrt{-1+x}\right )}{\sqrt{-1+x} \sqrt{x}}\\ &=-2 \sqrt{x}-\frac{2 \sqrt{-x+x^2}}{\sqrt{x}}-\frac{\sqrt{-x+x^2} \tan ^{-1}\left (\frac{2}{3} \sqrt{2} \sqrt{-1+x}\right )}{\sqrt{2} \sqrt{-1+x} \sqrt{x}}+\frac{\tan ^{-1}\left (2 \sqrt{2} \sqrt{x}\right )}{\sqrt{2}}+2 \sqrt{x} \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )\\ \end{align*}

Mathematica [C]  time = 0.544953, size = 186, normalized size = 1.58 \[ \frac{1}{8} \left (-16 \sqrt{x}-\frac{16 \sqrt{(x-1) x}}{\sqrt{x}}-2 i \sqrt{2} \log \left (4 (8 x+1)^2\right )+i \sqrt{2} \log \left ((8 x+1) \left (-10 x-6 \sqrt{(x-1) x}+1\right )\right )+16 \sqrt{x} \log \left (4 x+4 \sqrt{(x-1) x}-1\right )+i \sqrt{2} \log \left ((8 x+1) \left (-10 x+6 \sqrt{(x-1) x}+1\right )\right )+4 \sqrt{2} \tan ^{-1}\left (2 \sqrt{2} \sqrt{x}\right )-4 \sqrt{2} \tan ^{-1}\left (\frac{2 \sqrt{2} \sqrt{(x-1) x}}{3 \sqrt{x}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]]/Sqrt[x],x]

[Out]

(-16*Sqrt[x] - (16*Sqrt[(-1 + x)*x])/Sqrt[x] + 4*Sqrt[2]*ArcTan[2*Sqrt[2]*Sqrt[x]] - 4*Sqrt[2]*ArcTan[(2*Sqrt[
2]*Sqrt[(-1 + x)*x])/(3*Sqrt[x])] - (2*I)*Sqrt[2]*Log[4*(1 + 8*x)^2] + I*Sqrt[2]*Log[(1 + 8*x)*(1 - 10*x - 6*S
qrt[(-1 + x)*x])] + 16*Sqrt[x]*Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]] + I*Sqrt[2]*Log[(1 + 8*x)*(1 - 10*x + 6*Sqrt
[(-1 + x)*x])])/8

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Maple [F]  time = 0.009, size = 0, normalized size = 0. \begin{align*} \int{\ln \left ( -1+4\,x+4\,\sqrt{ \left ( -1+x \right ) x} \right ){\frac{1}{\sqrt{x}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \, \sqrt{x} \log \left (4 \, \sqrt{x - 1} \sqrt{x} + 4 \, x - 1\right ) - 4 \, \sqrt{x} + \int \frac{2 \, x^{2} + x}{4 \, x^{\frac{7}{2}} - 5 \, x^{\frac{5}{2}} + 4 \,{\left (x^{3} - x^{2}\right )} \sqrt{x - 1} + x^{\frac{3}{2}}}\,{d x} + \log \left (\sqrt{x} + 1\right ) - \log \left (\sqrt{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-1+4*x+4*((-1+x)*x)^(1/2))/x^(1/2),x, algorithm="maxima")

[Out]

2*sqrt(x)*log(4*sqrt(x - 1)*sqrt(x) + 4*x - 1) - 4*sqrt(x) + integrate((2*x^2 + x)/(4*x^(7/2) - 5*x^(5/2) + 4*
(x^3 - x^2)*sqrt(x - 1) + x^(3/2)), x) + log(sqrt(x) + 1) - log(sqrt(x) - 1)

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Fricas [A]  time = 2.18175, size = 243, normalized size = 2.06 \begin{align*} \frac{\sqrt{2} x \arctan \left (2 \, \sqrt{2} \sqrt{x}\right ) + \sqrt{2} x \arctan \left (\frac{3 \, \sqrt{2} \sqrt{x}}{4 \, \sqrt{x^{2} - x}}\right ) + 4 \, x^{\frac{3}{2}} \log \left (4 \, x + 4 \, \sqrt{x^{2} - x} - 1\right ) - 4 \, x^{\frac{3}{2}} - 4 \, \sqrt{x^{2} - x} \sqrt{x}}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-1+4*x+4*((-1+x)*x)^(1/2))/x^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(-1+4*x+4*((-1+x)*x)**(1/2))/x**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(-1+4*x+4*((-1+x)*x)^(1/2))/x^(1/2),x, algorithm="giac")

[Out]

undef