3.103 $$\int x \log (-1+4 x+4 \sqrt{(-1+x) x}) \, dx$$

Optimal. Leaf size=127 $-\frac{x^2}{8}+\frac{1}{16} (1-2 x) \sqrt{x^2-x}-\frac{11 \sqrt{x^2-x}}{32}+\frac{1}{2} x^2 \log \left (4 \sqrt{x^2-x}+4 x-1\right )+\frac{1}{256} \tanh ^{-1}\left (\frac{1-10 x}{6 \sqrt{x^2-x}}\right )-\frac{33}{128} \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-x}}\right )+\frac{x}{32}-\frac{1}{256} \log (8 x+1)$

[Out]

x/32 - x^2/8 - (11*Sqrt[-x + x^2])/32 + ((1 - 2*x)*Sqrt[-x + x^2])/16 + ArcTanh[(1 - 10*x)/(6*Sqrt[-x + x^2])]
/256 - (33*ArcTanh[x/Sqrt[-x + x^2]])/128 - Log[1 + 8*x]/256 + (x^2*Log[-1 + 4*x + 4*Sqrt[-x + x^2]])/2

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Rubi [A]  time = 0.245347, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 10, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.526, Rules used = {2537, 2535, 6742, 640, 620, 206, 612, 734, 843, 724} $-\frac{x^2}{8}+\frac{1}{16} (1-2 x) \sqrt{x^2-x}-\frac{11 \sqrt{x^2-x}}{32}+\frac{1}{2} x^2 \log \left (4 \sqrt{x^2-x}+4 x-1\right )+\frac{1}{256} \tanh ^{-1}\left (\frac{1-10 x}{6 \sqrt{x^2-x}}\right )-\frac{33}{128} \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-x}}\right )+\frac{x}{32}-\frac{1}{256} \log (8 x+1)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x/32 - x^2/8 - (11*Sqrt[-x + x^2])/32 + ((1 - 2*x)*Sqrt[-x + x^2])/16 + ArcTanh[(1 - 10*x)/(6*Sqrt[-x + x^2])]
/256 - (33*ArcTanh[x/Sqrt[-x + x^2]])/128 - Log[1 + 8*x]/256 + (x^2*Log[-1 + 4*x + 4*Sqrt[-x + x^2]])/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 6742

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

Rule 640

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

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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 612

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

Rule 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), 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] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 843

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

Rule 724

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

Rubi steps

\begin{align*} \int x \log \left (-1+4 x+4 \sqrt{(-1+x) x}\right ) \, dx &=\int x \log \left (-1+4 x+4 \sqrt{-x+x^2}\right ) \, dx\\ &=\frac{1}{2} x^2 \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+4 \int \frac{x^2}{-4 (1+2 x) \sqrt{-x+x^2}+8 \left (-x+x^2\right )} \, dx\\ &=\frac{1}{2} x^2 \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+4 \int \left (\frac{1}{128}-\frac{x}{16}-\frac{1}{128 (1+8 x)}-\frac{x}{12 \sqrt{-x+x^2}}-\frac{1}{16} \sqrt{-x+x^2}+\frac{\sqrt{-x+x^2}}{48 (-1-8 x)}\right ) \, dx\\ &=\frac{x}{32}-\frac{x^2}{8}-\frac{1}{256} \log (1+8 x)+\frac{1}{2} x^2 \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+\frac{1}{12} \int \frac{\sqrt{-x+x^2}}{-1-8 x} \, dx-\frac{1}{4} \int \sqrt{-x+x^2} \, dx-\frac{1}{3} \int \frac{x}{\sqrt{-x+x^2}} \, dx\\ &=\frac{x}{32}-\frac{x^2}{8}-\frac{11}{32} \sqrt{-x+x^2}+\frac{1}{16} (1-2 x) \sqrt{-x+x^2}-\frac{1}{256} \log (1+8 x)+\frac{1}{2} x^2 \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+\frac{1}{192} \int \frac{1-10 x}{(-1-8 x) \sqrt{-x+x^2}} \, dx+\frac{1}{32} \int \frac{1}{\sqrt{-x+x^2}} \, dx-\frac{1}{6} \int \frac{1}{\sqrt{-x+x^2}} \, dx\\ &=\frac{x}{32}-\frac{x^2}{8}-\frac{11}{32} \sqrt{-x+x^2}+\frac{1}{16} (1-2 x) \sqrt{-x+x^2}-\frac{1}{256} \log (1+8 x)+\frac{1}{2} x^2 \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+\frac{5}{768} \int \frac{1}{\sqrt{-x+x^2}} \, dx+\frac{3}{256} \int \frac{1}{(-1-8 x) \sqrt{-x+x^2}} \, dx+\frac{1}{16} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{x}{\sqrt{-x+x^2}}\right )-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{x}{\sqrt{-x+x^2}}\right )\\ &=\frac{x}{32}-\frac{x^2}{8}-\frac{11}{32} \sqrt{-x+x^2}+\frac{1}{16} (1-2 x) \sqrt{-x+x^2}-\frac{13}{48} \tanh ^{-1}\left (\frac{x}{\sqrt{-x+x^2}}\right )-\frac{1}{256} \log (1+8 x)+\frac{1}{2} x^2 \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )+\frac{5}{384} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{x}{\sqrt{-x+x^2}}\right )-\frac{3}{128} \operatorname{Subst}\left (\int \frac{1}{36-x^2} \, dx,x,\frac{-1+10 x}{\sqrt{-x+x^2}}\right )\\ &=\frac{x}{32}-\frac{x^2}{8}-\frac{11}{32} \sqrt{-x+x^2}+\frac{1}{16} (1-2 x) \sqrt{-x+x^2}+\frac{1}{256} \tanh ^{-1}\left (\frac{1-10 x}{6 \sqrt{-x+x^2}}\right )-\frac{33}{128} \tanh ^{-1}\left (\frac{x}{\sqrt{-x+x^2}}\right )-\frac{1}{256} \log (1+8 x)+\frac{1}{2} x^2 \log \left (-1+4 x+4 \sqrt{-x+x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.269203, size = 102, normalized size = 0.8 $\frac{1}{256} \left (-32 x^2+128 x^2 \log \left (4 x+4 \sqrt{(x-1) x}-1\right )-32 \sqrt{(x-1) x} x+8 x-72 \sqrt{(x-1) x}-2 \log (8 x+1)-33 \log \left (-2 x-2 \sqrt{(x-1) x}+1\right )+\log \left (-10 x+6 \sqrt{(x-1) x}+1\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(8*x - 32*x^2 - 72*Sqrt[(-1 + x)*x] - 32*x*Sqrt[(-1 + x)*x] - 2*Log[1 + 8*x] - 33*Log[1 - 2*x - 2*Sqrt[(-1 + x
)*x]] + 128*x^2*Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]] + Log[1 - 10*x + 6*Sqrt[(-1 + x)*x]])/256

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Maple [F]  time = 0.004, size = 0, normalized size = 0. \begin{align*} \int x\ln \left ( -1+4\,x+4\,\sqrt{ \left ( -1+x \right ) x} \right ) \, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*log(4*x + 4*sqrt((x - 1)*x) - 1), x)

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Fricas [A]  time = 2.57492, size = 327, normalized size = 2.57 \begin{align*} -\frac{1}{8} \, x^{2} + \frac{1}{2} \,{\left (x^{2} - 1\right )} \log \left (4 \, x + 4 \, \sqrt{x^{2} - x} - 1\right ) - \frac{1}{32} \, \sqrt{x^{2} - x}{\left (4 \, x + 9\right )} + \frac{1}{32} \, x + \frac{63}{256} \, \log \left (8 \, x + 1\right ) - \frac{31}{256} \, \log \left (-2 \, x + 2 \, \sqrt{x^{2} - x} + 1\right ) + \frac{63}{256} \, \log \left (-2 \, x + 2 \, \sqrt{x^{2} - x} - 1\right ) - \frac{63}{256} \, \log \left (-4 \, x + 4 \, \sqrt{x^{2} - x} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*x^2 + 1/2*(x^2 - 1)*log(4*x + 4*sqrt(x^2 - x) - 1) - 1/32*sqrt(x^2 - x)*(4*x + 9) + 1/32*x + 63/256*log(8
*x + 1) - 31/256*log(-2*x + 2*sqrt(x^2 - x) + 1) + 63/256*log(-2*x + 2*sqrt(x^2 - x) - 1) - 63/256*log(-4*x +
4*sqrt(x^2 - x) + 1)

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

[Out]

Timed out

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Giac [A]  time = 1.29707, size = 154, normalized size = 1.21 \begin{align*} \frac{1}{2} \, x^{2} \log \left (4 \, x + 4 \, \sqrt{{\left (x - 1\right )} x} - 1\right ) - \frac{1}{8} \, x^{2} - \frac{1}{32} \, \sqrt{x^{2} - x}{\left (4 \, x + 9\right )} + \frac{1}{32} \, x - \frac{1}{256} \, \log \left ({\left | 8 \, x + 1 \right |}\right ) + \frac{33}{256} \, \log \left ({\left | -2 \, x + 2 \, \sqrt{x^{2} - x} + 1 \right |}\right ) - \frac{1}{256} \, \log \left ({\left | -2 \, x + 2 \, \sqrt{x^{2} - x} - 1 \right |}\right ) + \frac{1}{256} \, \log \left ({\left | -4 \, x + 4 \, \sqrt{x^{2} - x} + 1 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log(4*x + 4*sqrt((x - 1)*x) - 1) - 1/8*x^2 - 1/32*sqrt(x^2 - x)*(4*x + 9) + 1/32*x - 1/256*log(abs(8*x
+ 1)) + 33/256*log(abs(-2*x + 2*sqrt(x^2 - x) + 1)) - 1/256*log(abs(-2*x + 2*sqrt(x^2 - x) - 1)) + 1/256*log(
abs(-4*x + 4*sqrt(x^2 - x) + 1))