3.102 \(\int x^2 \log (-1+4 x+4 \sqrt {(-1+x) x}) \, dx\)

Optimal. Leaf size=149 \[ -\frac {x^3}{18}+\frac {x^2}{96}-\frac {1}{18} \left (x^2-x\right )^{3/2}+\frac {5}{64} (1-2 x) \sqrt {x^2-x}-\frac {85 \sqrt {x^2-x}}{384}-\frac {\tanh ^{-1}\left (\frac {1-10 x}{6 \sqrt {x^2-x}}\right )}{3072}-\frac {223 \tanh ^{-1}\left (\frac {x}{\sqrt {x^2-x}}\right )}{1536}+\frac {1}{3} x^3 \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {x}{384}+\frac {\log (8 x+1)}{3072} \]

[Out]

-1/384*x+1/96*x^2-1/18*x^3-1/18*(x^2-x)^(3/2)-1/3072*arctanh(1/6*(1-10*x)/(x^2-x)^(1/2))-223/1536*arctanh(x/(x
^2-x)^(1/2))+1/3072*ln(1+8*x)+1/3*x^3*ln(-1+4*x+4*(x^2-x)^(1/2))-85/384*(x^2-x)^(1/2)+5/64*(1-2*x)*(x^2-x)^(1/
2)

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Rubi [A]  time = 0.29, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2537, 2535, 6742, 640, 620, 206, 612, 734, 843, 724} \[ -\frac {x^3}{18}+\frac {x^2}{96}-\frac {1}{18} \left (x^2-x\right )^{3/2}+\frac {5}{64} (1-2 x) \sqrt {x^2-x}-\frac {85 \sqrt {x^2-x}}{384}+\frac {1}{3} x^3 \log \left (4 \sqrt {x^2-x}+4 x-1\right )-\frac {\tanh ^{-1}\left (\frac {1-10 x}{6 \sqrt {x^2-x}}\right )}{3072}-\frac {223 \tanh ^{-1}\left (\frac {x}{\sqrt {x^2-x}}\right )}{1536}-\frac {x}{384}+\frac {\log (8 x+1)}{3072} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]],x]

[Out]

-x/384 + x^2/96 - x^3/18 - (85*Sqrt[-x + x^2])/384 + (5*(1 - 2*x)*Sqrt[-x + x^2])/64 - (-x + x^2)^(3/2)/18 - A
rcTanh[(1 - 10*x)/(6*Sqrt[-x + x^2])]/3072 - (223*ArcTanh[x/Sqrt[-x + x^2]])/1536 + Log[1 + 8*x]/3072 + (x^3*L
og[-1 + 4*x + 4*Sqrt[-x + x^2]])/3

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 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 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 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]

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 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 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 6742

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

Rubi steps

\begin {align*} \int x^2 \log \left (-1+4 x+4 \sqrt {(-1+x) x}\right ) \, dx &=\int x^2 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right ) \, dx\\ &=\frac {1}{3} x^3 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {8}{3} \int \frac {x^3}{-4 (1+2 x) \sqrt {-x+x^2}+8 \left (-x+x^2\right )} \, dx\\ &=\frac {1}{3} x^3 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {8}{3} \int \left (-\frac {1}{1024}+\frac {x}{128}-\frac {x^2}{16}+\frac {1}{1024 (1+8 x)}-\frac {x}{12 \sqrt {-x+x^2}}-\frac {11}{128} \sqrt {-x+x^2}-\frac {1}{16} x \sqrt {-x+x^2}+\frac {\sqrt {-x+x^2}}{384 (1+8 x)}\right ) \, dx\\ &=-\frac {x}{384}+\frac {x^2}{96}-\frac {x^3}{18}+\frac {\log (1+8 x)}{3072}+\frac {1}{3} x^3 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )+\frac {1}{144} \int \frac {\sqrt {-x+x^2}}{1+8 x} \, dx-\frac {1}{6} \int x \sqrt {-x+x^2} \, dx-\frac {2}{9} \int \frac {x}{\sqrt {-x+x^2}} \, dx-\frac {11}{48} \int \sqrt {-x+x^2} \, dx\\ &=-\frac {x}{384}+\frac {x^2}{96}-\frac {x^3}{18}-\frac {85}{384} \sqrt {-x+x^2}+\frac {11}{192} (1-2 x) \sqrt {-x+x^2}-\frac {1}{18} \left (-x+x^2\right )^{3/2}+\frac {\log (1+8 x)}{3072}+\frac {1}{3} x^3 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {\int \frac {-1+10 x}{(1+8 x) \sqrt {-x+x^2}} \, dx}{2304}+\frac {11}{384} \int \frac {1}{\sqrt {-x+x^2}} \, dx-\frac {1}{12} \int \sqrt {-x+x^2} \, dx-\frac {1}{9} \int \frac {1}{\sqrt {-x+x^2}} \, dx\\ &=-\frac {x}{384}+\frac {x^2}{96}-\frac {x^3}{18}-\frac {85}{384} \sqrt {-x+x^2}+\frac {5}{64} (1-2 x) \sqrt {-x+x^2}-\frac {1}{18} \left (-x+x^2\right )^{3/2}+\frac {\log (1+8 x)}{3072}+\frac {1}{3} x^3 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {5 \int \frac {1}{\sqrt {-x+x^2}} \, dx}{9216}+\frac {\int \frac {1}{(1+8 x) \sqrt {-x+x^2}} \, dx}{1024}+\frac {1}{96} \int \frac {1}{\sqrt {-x+x^2}} \, dx+\frac {11}{192} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )-\frac {2}{9} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )\\ &=-\frac {x}{384}+\frac {x^2}{96}-\frac {x^3}{18}-\frac {85}{384} \sqrt {-x+x^2}+\frac {5}{64} (1-2 x) \sqrt {-x+x^2}-\frac {1}{18} \left (-x+x^2\right )^{3/2}-\frac {95}{576} \tanh ^{-1}\left (\frac {x}{\sqrt {-x+x^2}}\right )+\frac {\log (1+8 x)}{3072}+\frac {1}{3} x^3 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )-\frac {5 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )}{4608}-\frac {1}{512} \operatorname {Subst}\left (\int \frac {1}{36-x^2} \, dx,x,\frac {1-10 x}{\sqrt {-x+x^2}}\right )+\frac {1}{48} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-x+x^2}}\right )\\ &=-\frac {x}{384}+\frac {x^2}{96}-\frac {x^3}{18}-\frac {85}{384} \sqrt {-x+x^2}+\frac {5}{64} (1-2 x) \sqrt {-x+x^2}-\frac {1}{18} \left (-x+x^2\right )^{3/2}-\frac {\tanh ^{-1}\left (\frac {1-10 x}{6 \sqrt {-x+x^2}}\right )}{3072}-\frac {223 \tanh ^{-1}\left (\frac {x}{\sqrt {-x+x^2}}\right )}{1536}+\frac {\log (1+8 x)}{3072}+\frac {1}{3} x^3 \log \left (-1+4 x+4 \sqrt {-x+x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 107, normalized size = 0.72 \[ \frac {-512 x^3+3072 x^3 \log \left (4 x+4 \sqrt {(x-1) x}-1\right )+96 x^2-8 \sqrt {(x-1) x} \left (64 x^2+116 x+165\right )-24 x+6 \log (8 x+1)-669 \log \left (-2 x-2 \sqrt {(x-1) x}+1\right )-3 \log \left (-10 x+6 \sqrt {(x-1) x}+1\right )}{9216} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]],x]

[Out]

(-24*x + 96*x^2 - 512*x^3 - 8*Sqrt[(-1 + x)*x]*(165 + 116*x + 64*x^2) + 6*Log[1 + 8*x] - 669*Log[1 - 2*x - 2*S
qrt[(-1 + x)*x]] + 3072*x^3*Log[-1 + 4*x + 4*Sqrt[(-1 + x)*x]] - 3*Log[1 - 10*x + 6*Sqrt[(-1 + x)*x]])/9216

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fricas [A]  time = 0.99, size = 124, normalized size = 0.83 \[ -\frac {1}{18} \, x^{3} + \frac {1}{96} \, x^{2} + \frac {1}{3} \, {\left (x^{3} + 1\right )} \log \left (4 \, x + 4 \, \sqrt {x^{2} - x} - 1\right ) - \frac {1}{1152} \, {\left (64 \, x^{2} + 116 \, x + 165\right )} \sqrt {x^{2} - x} - \frac {1}{384} \, x - \frac {511}{3072} \, \log \left (8 \, x + 1\right ) + \frac {245}{1024} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} + 1\right ) - \frac {511}{3072} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x} - 1\right ) + \frac {511}{3072} \, \log \left (-4 \, x + 4 \, \sqrt {x^{2} - x} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*x^3 + 1/96*x^2 + 1/3*(x^3 + 1)*log(4*x + 4*sqrt(x^2 - x) - 1) - 1/1152*(64*x^2 + 116*x + 165)*sqrt(x^2 -
x) - 1/384*x - 511/3072*log(8*x + 1) + 245/1024*log(-2*x + 2*sqrt(x^2 - x) + 1) - 511/3072*log(-2*x + 2*sqrt(
x^2 - x) - 1) + 511/3072*log(-4*x + 4*sqrt(x^2 - x) + 1)

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giac [A]  time = 0.31, size = 124, normalized size = 0.83 \[ \frac {1}{3} \, x^{3} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right ) - \frac {1}{18} \, x^{3} + \frac {1}{96} \, x^{2} - \frac {1}{1152} \, {\left (4 \, {\left (16 \, x + 29\right )} x + 165\right )} \sqrt {x^{2} - x} - \frac {1}{384} \, x + \frac {1}{3072} \, \log \left ({\left | 8 \, x + 1 \right |}\right ) + \frac {223}{3072} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} + 1 \right |}\right ) + \frac {1}{3072} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x} - 1 \right |}\right ) - \frac {1}{3072} \, \log \left ({\left | -4 \, x + 4 \, \sqrt {x^{2} - x} + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log(4*x + 4*sqrt((x - 1)*x) - 1) - 1/18*x^3 + 1/96*x^2 - 1/1152*(4*(16*x + 29)*x + 165)*sqrt(x^2 - x)
- 1/384*x + 1/3072*log(abs(8*x + 1)) + 223/3072*log(abs(-2*x + 2*sqrt(x^2 - x) + 1)) + 1/3072*log(abs(-2*x + 2
*sqrt(x^2 - x) - 1)) - 1/3072*log(abs(-4*x + 4*sqrt(x^2 - x) + 1))

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int x^{2} \ln \left (4 x -1+4 \sqrt {\left (x -1\right ) x}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \log \left (4 \, x + 4 \, \sqrt {{\left (x - 1\right )} x} - 1\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*log(4*x + 4*sqrt((x - 1)*x) - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\ln \left (4\,x+4\,\sqrt {x\,\left (x-1\right )}-1\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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