3.7.64 \(\int \frac {1}{(-5+2 x)^2 \sqrt [4]{4-4 x+x^2}} \, dx\)

Optimal. Leaf size=53 \[ \frac {\left ((x-2)^2\right )^{3/4} \left (\frac {\sqrt {x-2}}{5-2 x}+\frac {\tanh ^{-1}\left (\sqrt {2} \sqrt {x-2}\right )}{\sqrt {2}}\right )}{(x-2)^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {646, 51, 63, 207} \begin {gather*} \frac {\sqrt {x-2} \tanh ^{-1}\left (\sqrt {2} \sqrt {x-2}\right )}{\sqrt {2} \sqrt [4]{x^2-4 x+4}}-\frac {2-x}{(5-2 x) \sqrt [4]{x^2-4 x+4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((-5 + 2*x)^2*(4 - 4*x + x^2)^(1/4)),x]

[Out]

-((2 - x)/((5 - 2*x)*(4 - 4*x + x^2)^(1/4))) + (Sqrt[-2 + x]*ArcTanh[Sqrt[2]*Sqrt[-2 + x]])/(Sqrt[2]*(4 - 4*x
+ x^2)^(1/4))

Rule 51

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

Rule 207

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {1}{(-5+2 x)^2 \sqrt [4]{4-4 x+x^2}} \, dx &=\frac {\sqrt {-2+x} \int \frac {1}{\sqrt {-2+x} (-5+2 x)^2} \, dx}{\sqrt [4]{4-4 x+x^2}}\\ &=-\frac {2-x}{(5-2 x) \sqrt [4]{4-4 x+x^2}}-\frac {\sqrt {-2+x} \int \frac {1}{\sqrt {-2+x} (-5+2 x)} \, dx}{2 \sqrt [4]{4-4 x+x^2}}\\ &=-\frac {2-x}{(5-2 x) \sqrt [4]{4-4 x+x^2}}-\frac {\sqrt {-2+x} \operatorname {Subst}\left (\int \frac {1}{-1+2 x^2} \, dx,x,\sqrt {-2+x}\right )}{\sqrt [4]{4-4 x+x^2}}\\ &=-\frac {2-x}{(5-2 x) \sqrt [4]{4-4 x+x^2}}+\frac {\sqrt {-2+x} \tanh ^{-1}\left (\sqrt {2} \sqrt {-2+x}\right )}{\sqrt {2} \sqrt [4]{4-4 x+x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 59, normalized size = 1.11 \begin {gather*} \frac {x-2}{(1-2 (x-2)) \sqrt [4]{(x-2)^2}}+\frac {\sqrt {x-2} \tanh ^{-1}\left (\sqrt {2} \sqrt {x-2}\right )}{\sqrt {2} \sqrt [4]{(x-2)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((-5 + 2*x)^2*(4 - 4*x + x^2)^(1/4)),x]

[Out]

(-2 + x)/((1 - 2*(-2 + x))*((-2 + x)^2)^(1/4)) + (Sqrt[-2 + x]*ArcTanh[Sqrt[2]*Sqrt[-2 + x]])/(Sqrt[2]*((-2 +
x)^2)^(1/4))

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IntegrateAlgebraic [A]  time = 4.71, size = 56, normalized size = 1.06 \begin {gather*} \frac {\left ((-2+x)^2\right )^{3/4} \left (-\frac {\sqrt {-2+x}}{-1+2 (-2+x)}+\frac {\tanh ^{-1}\left (\sqrt {2} \sqrt {-2+x}\right )}{\sqrt {2}}\right )}{(-2+x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((-5 + 2*x)^2*(4 - 4*x + x^2)^(1/4)),x]

[Out]

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

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fricas [A]  time = 0.61, size = 60, normalized size = 1.13 \begin {gather*} \frac {\sqrt {2} {\left (2 \, x - 5\right )} \log \left (\frac {2 \, x + 2 \, \sqrt {2} {\left (x^{2} - 4 \, x + 4\right )}^{\frac {1}{4}} - 3}{2 \, x - 5}\right ) - 4 \, {\left (x^{2} - 4 \, x + 4\right )}^{\frac {1}{4}}}{4 \, {\left (2 \, x - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+2*x)^2/(x^2-4*x+4)^(1/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} - 4 \, x + 4\right )}^{\frac {1}{4}} {\left (2 \, x - 5\right )}^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+2*x)^2/(x^2-4*x+4)^(1/4),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 48, normalized size = 0.91

method	result	size
risch	\(-\frac {-2+x}{\left (-5+2 x \right ) \left (\left (-2+x \right )^{2}\right )^{\frac {1}{4}}}+\frac {\arctanh \left (\sqrt {2}\, \sqrt {-2+x}\right ) \sqrt {2}\, \sqrt {-2+x}}{2 \left (\left (-2+x \right )^{2}\right )^{\frac {1}{4}}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-5+2*x)^2/(x^2-4*x+4)^(1/4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} - 4 \, x + 4\right )}^{\frac {1}{4}} {\left (2 \, x - 5\right )}^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+2*x)^2/(x^2-4*x+4)^(1/4),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (2\,x-5\right )}^2\,{\left (x^2-4\,x+4\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (2 x - 5\right )^{2} \sqrt [4]{\left (x - 2\right )^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+2*x)**2/(x**2-4*x+4)**(1/4),x)

[Out]

Integral(1/((2*x - 5)**2*((x - 2)**2)**(1/4)), x)

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