3.2.9 \(\int \sqrt {x} \log (-1+4 x+4 \sqrt {(-1+x) x}) \, dx\) [109]

Optimal. Leaf size=158 \[ \frac {\sqrt {x}}{12}-\frac {2 x^{3/2}}{9}-\frac {11 \sqrt {-x+x^2}}{12 \sqrt {x}}-\frac {2 \left (-x+x^2\right )^{3/2}}{9 x^{3/2}}+\frac {\sqrt {-x+x^2} \tan ^{-1}\left (\frac {2}{3} \sqrt {2} \sqrt {-1+x}\right )}{24 \sqrt {2} \sqrt {-1+x} \sqrt {x}}-\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{24 \sqrt {2}}+\frac {2}{3} x^{3/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \]

[Out]

-2/9*x^(3/2)-2/9*(x^2-x)^(3/2)/x^(3/2)+2/3*x^(3/2)*ln(-1+4*x+4*(x^2-x)^(1/2))-1/48*arctan(2*2^(1/2)*x^(1/2))*2
^(1/2)+1/12*x^(1/2)-11/12/x^(1/2)*(x^2-x)^(1/2)+1/48*arctan(2/3*2^(1/2)*(-1+x)^(1/2))*(x^2-x)^(1/2)*2^(1/2)/(-
1+x)^(1/2)/x^(1/2)

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Rubi [A]
time = 0.28, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {2617, 2615, 6865, 6874, 209, 1602, 2025, 1160, 455, 52, 65, 210} \begin {gather*} \frac {\sqrt {x^2-x} \text {ArcTan}\left (\frac {2}{3} \sqrt {2} \sqrt {x-1}\right )}{24 \sqrt {2} \sqrt {x-1} \sqrt {x}}-\frac {\text {ArcTan}\left (2 \sqrt {2} \sqrt {x}\right )}{24 \sqrt {2}}-\frac {2 x^{3/2}}{9}-\frac {11 \sqrt {x^2-x}}{12 \sqrt {x}}-\frac {2 \left (x^2-x\right )^{3/2}}{9 x^{3/2}}+\frac {2}{3} x^{3/2} \log \left (4 \sqrt {x^2-x}+4 x-1\right )+\frac {\sqrt {x}}{12} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[x]/12 - (2*x^(3/2))/9 - (11*Sqrt[-x + x^2])/(12*Sqrt[x]) - (2*(-x + x^2)^(3/2))/(9*x^(3/2)) + (Sqrt[-x +
x^2]*ArcTan[(2*Sqrt[2]*Sqrt[-1 + x])/3])/(24*Sqrt[2]*Sqrt[-1 + x]*Sqrt[x]) - ArcTan[2*Sqrt[2]*Sqrt[x]]/(24*Sqr
t[2]) + (2*x^(3/2)*Log[-1 + 4*x + 4*Sqrt[-x + x^2]])/3

Rule 52

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 65

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]

Rule 209

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

Rule 210

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

Rule 455

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 1160

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 1602

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 2025

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]

Rule 2615

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 2617

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 6865

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 6874

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

Rubi steps

\begin {align*} \int \sqrt {x} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx &=\int \sqrt {x} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \, dx\\ &=\frac {2}{3} x^{3/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {16}{3} \int \frac {x^{3/2}}{-4 (1+2 x) \sqrt {-x+x^2}+8 \left (-x+x^2\right )} \, dx\\ &=\frac {2}{3} x^{3/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {32}{3} \text {Subst}\left (\int \frac {x^4}{-4 \left (1+2 x^2\right ) \sqrt {-x^2+x^4}+8 \left (-x^2+x^4\right )} \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{3} x^{3/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {32}{3} \text {Subst}\left (\int \left (\frac {1}{128}-\frac {x^2}{16}-\frac {1}{128 \left (1+8 x^2\right )}-\frac {x^2}{12 \sqrt {-x^2+x^4}}-\frac {1}{16} \sqrt {-x^2+x^4}+\frac {\sqrt {-x^2+x^4}}{48 \left (-1-8 x^2\right )}\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {\sqrt {x}}{12}-\frac {2 x^{3/2}}{9}+\frac {2}{3} x^{3/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {1}{12} \text {Subst}\left (\int \frac {1}{1+8 x^2} \, dx,x,\sqrt {x}\right )+\frac {2}{9} \text {Subst}\left (\int \frac {\sqrt {-x^2+x^4}}{-1-8 x^2} \, dx,x,\sqrt {x}\right )-\frac {2}{3} \text {Subst}\left (\int \sqrt {-x^2+x^4} \, dx,x,\sqrt {x}\right )-\frac {8}{9} \text {Subst}\left (\int \frac {x^2}{\sqrt {-x^2+x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {\sqrt {x}}{12}-\frac {2 x^{3/2}}{9}-\frac {8 \sqrt {-x+x^2}}{9 \sqrt {x}}-\frac {2 \left (-x+x^2\right )^{3/2}}{9 x^{3/2}}-\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{24 \sqrt {2}}+\frac {2}{3} x^{3/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {\left (2 \sqrt {-x+x^2}\right ) \text {Subst}\left (\int \frac {x \sqrt {-1+x^2}}{-1-8 x^2} \, dx,x,\sqrt {x}\right )}{9 \sqrt {-1+x} \sqrt {x}}\\ &=\frac {\sqrt {x}}{12}-\frac {2 x^{3/2}}{9}-\frac {8 \sqrt {-x+x^2}}{9 \sqrt {x}}-\frac {2 \left (-x+x^2\right )^{3/2}}{9 x^{3/2}}-\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{24 \sqrt {2}}+\frac {2}{3} x^{3/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {\sqrt {-x+x^2} \text {Subst}\left (\int \frac {\sqrt {-1+x}}{-1-8 x} \, dx,x,x\right )}{9 \sqrt {-1+x} \sqrt {x}}\\ &=\frac {\sqrt {x}}{12}-\frac {2 x^{3/2}}{9}-\frac {11 \sqrt {-x+x^2}}{12 \sqrt {x}}-\frac {2 \left (-x+x^2\right )^{3/2}}{9 x^{3/2}}-\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{24 \sqrt {2}}+\frac {2}{3} x^{3/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {\sqrt {-x+x^2} \text {Subst}\left (\int \frac {1}{(-1-8 x) \sqrt {-1+x}} \, dx,x,x\right )}{8 \sqrt {-1+x} \sqrt {x}}\\ &=\frac {\sqrt {x}}{12}-\frac {2 x^{3/2}}{9}-\frac {11 \sqrt {-x+x^2}}{12 \sqrt {x}}-\frac {2 \left (-x+x^2\right )^{3/2}}{9 x^{3/2}}-\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{24 \sqrt {2}}+\frac {2}{3} x^{3/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {\sqrt {-x+x^2} \text {Subst}\left (\int \frac {1}{-9-8 x^2} \, dx,x,\sqrt {-1+x}\right )}{4 \sqrt {-1+x} \sqrt {x}}\\ &=\frac {\sqrt {x}}{12}-\frac {2 x^{3/2}}{9}-\frac {11 \sqrt {-x+x^2}}{12 \sqrt {x}}-\frac {2 \left (-x+x^2\right )^{3/2}}{9 x^{3/2}}+\frac {\sqrt {-x+x^2} \tan ^{-1}\left (\frac {2}{3} \sqrt {2} \sqrt {-1+x}\right )}{24 \sqrt {2} \sqrt {-1+x} \sqrt {x}}-\frac {\tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )}{24 \sqrt {2}}+\frac {2}{3} x^{3/2} \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.16, size = 209, normalized size = 1.32 \begin {gather*} \frac {1}{576} \left (48 \sqrt {x}-128 x^{3/2}-\frac {400 \sqrt {(-1+x) x}}{\sqrt {x}}-128 \sqrt {x} \sqrt {(-1+x) x}-12 \sqrt {2} \tan ^{-1}\left (2 \sqrt {2} \sqrt {x}\right )+12 \sqrt {2} \tan ^{-1}\left (\frac {2 \sqrt {2} \sqrt {(-1+x) x}}{3 \sqrt {x}}\right )+6 i \sqrt {2} \log \left (4 (1+8 x)^2\right )-3 i \sqrt {2} \log \left ((1+8 x) \left (1-10 x-6 \sqrt {(-1+x) x}\right )\right )+384 x^{3/2} \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right )-3 i \sqrt {2} \log \left ((1+8 x) \left (1-10 x+6 \sqrt {(-1+x) x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(48*Sqrt[x] - 128*x^(3/2) - (400*Sqrt[(-1 + x)*x])/Sqrt[x] - 128*Sqrt[x]*Sqrt[(-1 + x)*x] - 12*Sqrt[2]*ArcTan[
2*Sqrt[2]*Sqrt[x]] + 12*Sqrt[2]*ArcTan[(2*Sqrt[2]*Sqrt[(-1 + x)*x])/(3*Sqrt[x])] + (6*I)*Sqrt[2]*Log[4*(1 + 8*
x)^2] - (3*I)*Sqrt[2]*Log[(1 + 8*x)*(1 - 10*x - 6*Sqrt[(-1 + x)*x])] + 384*x^(3/2)*Log[-1 + 4*x + 4*Sqrt[(-1 +
x)*x]] - (3*I)*Sqrt[2]*Log[(1 + 8*x)*(1 - 10*x + 6*Sqrt[(-1 + x)*x])])/576

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \sqrt {x}\, \ln \left (-1+4 x +4 \sqrt {\left (-1+x \right ) x}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x^(3/2)*log(4*sqrt(x - 1)*sqrt(x) + 4*x - 1) - 4/9*x^(3/2) - 2/3*sqrt(x) + integrate(1/3*(2*x^2 + x)/(4*x^
(5/2) + 4*(x^2 - x)*sqrt(x - 1) - 5*x^(3/2) + sqrt(x)), x) + 1/3*log(sqrt(x) + 1) - 1/3*log(sqrt(x) - 1)

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Fricas [A]
time = 0.39, size = 100, normalized size = 0.63 \begin {gather*} \frac {96 \, x^{\frac {5}{2}} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - 3 \, \sqrt {2} x \arctan \left (2 \, \sqrt {2} \sqrt {x}\right ) - 3 \, \sqrt {2} x \arctan \left (\frac {3 \, \sqrt {2} \sqrt {x}}{4 \, \sqrt {x^{2} - x}}\right ) - 4 \, \sqrt {x^{2} - x} {\left (8 \, x + 25\right )} \sqrt {x} - 4 \, {\left (8 \, x^{2} - 3 \, x\right )} \sqrt {x}}{144 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*(96*x^(5/2)*log(4*x + 4*sqrt(x^2 - x) - 1) - 3*sqrt(2)*x*arctan(2*sqrt(2)*sqrt(x)) - 3*sqrt(2)*x*arctan(
3/4*sqrt(2)*sqrt(x)/sqrt(x^2 - x)) - 4*sqrt(x^2 - x)*(8*x + 25)*sqrt(x) - 4*(8*x^2 - 3*x)*sqrt(x))/x

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [C] Result contains complex when optimal does not.
time = 3.56, size = 118, normalized size = 0.75 \begin {gather*} \frac {2}{3} \, x^{\frac {3}{2}} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) - \frac {1}{96} i \, \sqrt {2} \pi - \frac {1}{96} \, \sqrt {2} {\left (\pi - 2 \, \arctan \left (\frac {\sqrt {2} {\left ({\left (\sqrt {x - 1} - \sqrt {x}\right )}^{2} - 1\right )}}{3 \, {\left (\sqrt {x - 1} - \sqrt {x}\right )}}\right )\right )} - \frac {1}{36} \, {\left (8 \, x + 25\right )} \sqrt {x - 1} - \frac {2}{9} \, x^{\frac {3}{2}} - \frac {1}{48} \, \sqrt {2} \arctan \left (2 \, \sqrt {2} \sqrt {x}\right ) - \frac {1}{48} \, \sqrt {2} \arctan \left (\frac {2}{3} i \, \sqrt {2}\right ) + \frac {1}{12} \, \sqrt {x} + \frac {25}{36} i \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x^(3/2)*log(4*x + 4*sqrt((x - 1)*x) - 1) - 1/96*I*sqrt(2)*pi - 1/96*sqrt(2)*(pi - 2*arctan(1/3*sqrt(2)*((s
qrt(x - 1) - sqrt(x))^2 - 1)/(sqrt(x - 1) - sqrt(x)))) - 1/36*(8*x + 25)*sqrt(x - 1) - 2/9*x^(3/2) - 1/48*sqrt
(2)*arctan(2*sqrt(2)*sqrt(x)) - 1/48*sqrt(2)*arctan(2/3*I*sqrt(2)) + 1/12*sqrt(x) + 25/36*I

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {x}\,\ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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