3.298 \(\int \frac{(-1+2 \sqrt{x})^{5/4}}{x^2} \, dx\)

Optimal. Leaf size=193 \[ -\frac{\left (2 \sqrt{x}-1\right )^{5/4}}{x}-\frac{5 \sqrt [4]{2 \sqrt{x}-1}}{2 \sqrt{x}}-\frac{5 \log \left (\sqrt{2 \sqrt{x}-1}-\sqrt{2} \sqrt [4]{2 \sqrt{x}-1}+1\right )}{4 \sqrt{2}}+\frac{5 \log \left (\sqrt{2 \sqrt{x}-1}+\sqrt{2} \sqrt [4]{2 \sqrt{x}-1}+1\right )}{4 \sqrt{2}}-\frac{5 \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{2 \sqrt{x}-1}\right )}{2 \sqrt{2}}+\frac{5 \tan ^{-1}\left (\sqrt{2} \sqrt [4]{2 \sqrt{x}-1}+1\right )}{2 \sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.113249, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529, Rules used = {266, 47, 63, 211, 1165, 628, 1162, 617, 204} \[ -\frac{\left (2 \sqrt{x}-1\right )^{5/4}}{x}-\frac{5 \sqrt [4]{2 \sqrt{x}-1}}{2 \sqrt{x}}-\frac{5 \log \left (\sqrt{2 \sqrt{x}-1}-\sqrt{2} \sqrt [4]{2 \sqrt{x}-1}+1\right )}{4 \sqrt{2}}+\frac{5 \log \left (\sqrt{2 \sqrt{x}-1}+\sqrt{2} \sqrt [4]{2 \sqrt{x}-1}+1\right )}{4 \sqrt{2}}-\frac{5 \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{2 \sqrt{x}-1}\right )}{2 \sqrt{2}}+\frac{5 \tan ^{-1}\left (\sqrt{2} \sqrt [4]{2 \sqrt{x}-1}+1\right )}{2 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[(-1 + 2*Sqrt[x])^(5/4)/x^2,x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 47

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

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

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

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{\left (-1+2 \sqrt{x}\right )^{5/4}}{x^2} \, dx &=2 \operatorname{Subst}\left (\int \frac{(-1+2 x)^{5/4}}{x^3} \, dx,x,\sqrt{x}\right )\\ &=-\frac{\left (-1+2 \sqrt{x}\right )^{5/4}}{x}+\frac{5}{2} \operatorname{Subst}\left (\int \frac{\sqrt [4]{-1+2 x}}{x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{\left (-1+2 \sqrt{x}\right )^{5/4}}{x}-\frac{5 \sqrt [4]{-1+2 \sqrt{x}}}{2 \sqrt{x}}+\frac{5}{4} \operatorname{Subst}\left (\int \frac{1}{x (-1+2 x)^{3/4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{\left (-1+2 \sqrt{x}\right )^{5/4}}{x}-\frac{5 \sqrt [4]{-1+2 \sqrt{x}}}{2 \sqrt{x}}+\frac{5}{2} \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2}+\frac{x^4}{2}} \, dx,x,\sqrt [4]{-1+2 \sqrt{x}}\right )\\ &=-\frac{\left (-1+2 \sqrt{x}\right )^{5/4}}{x}-\frac{5 \sqrt [4]{-1+2 \sqrt{x}}}{2 \sqrt{x}}+\frac{5}{4} \operatorname{Subst}\left (\int \frac{1-x^2}{\frac{1}{2}+\frac{x^4}{2}} \, dx,x,\sqrt [4]{-1+2 \sqrt{x}}\right )+\frac{5}{4} \operatorname{Subst}\left (\int \frac{1+x^2}{\frac{1}{2}+\frac{x^4}{2}} \, dx,x,\sqrt [4]{-1+2 \sqrt{x}}\right )\\ &=-\frac{\left (-1+2 \sqrt{x}\right )^{5/4}}{x}-\frac{5 \sqrt [4]{-1+2 \sqrt{x}}}{2 \sqrt{x}}+\frac{5}{4} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt [4]{-1+2 \sqrt{x}}\right )+\frac{5}{4} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt [4]{-1+2 \sqrt{x}}\right )-\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt [4]{-1+2 \sqrt{x}}\right )}{4 \sqrt{2}}-\frac{5 \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt [4]{-1+2 \sqrt{x}}\right )}{4 \sqrt{2}}\\ &=-\frac{\left (-1+2 \sqrt{x}\right )^{5/4}}{x}-\frac{5 \sqrt [4]{-1+2 \sqrt{x}}}{2 \sqrt{x}}-\frac{5 \log \left (1-\sqrt{2} \sqrt [4]{-1+2 \sqrt{x}}+\sqrt{-1+2 \sqrt{x}}\right )}{4 \sqrt{2}}+\frac{5 \log \left (1+\sqrt{2} \sqrt [4]{-1+2 \sqrt{x}}+\sqrt{-1+2 \sqrt{x}}\right )}{4 \sqrt{2}}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt [4]{-1+2 \sqrt{x}}\right )}{2 \sqrt{2}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt [4]{-1+2 \sqrt{x}}\right )}{2 \sqrt{2}}\\ &=-\frac{\left (-1+2 \sqrt{x}\right )^{5/4}}{x}-\frac{5 \sqrt [4]{-1+2 \sqrt{x}}}{2 \sqrt{x}}-\frac{5 \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{-1+2 \sqrt{x}}\right )}{2 \sqrt{2}}+\frac{5 \tan ^{-1}\left (1+\sqrt{2} \sqrt [4]{-1+2 \sqrt{x}}\right )}{2 \sqrt{2}}-\frac{5 \log \left (1-\sqrt{2} \sqrt [4]{-1+2 \sqrt{x}}+\sqrt{-1+2 \sqrt{x}}\right )}{4 \sqrt{2}}+\frac{5 \log \left (1+\sqrt{2} \sqrt [4]{-1+2 \sqrt{x}}+\sqrt{-1+2 \sqrt{x}}\right )}{4 \sqrt{2}}\\ \end{align*}

Mathematica [C]  time = 0.0077126, size = 34, normalized size = 0.18 \[ \frac{32}{9} \left (2 \sqrt{x}-1\right )^{9/4} \, _2F_1\left (\frac{9}{4},3;\frac{13}{4};1-2 \sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*(-1 + 2*Sqrt[x])^(9/4)*Hypergeometric2F1[9/4, 3, 13/4, 1 - 2*Sqrt[x]])/9

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Maple [A]  time = 0.013, size = 130, normalized size = 0.7 \begin{align*} 8\,{\frac{1}{x} \left ( -{\frac{9\, \left ( -1+2\,\sqrt{x} \right ) ^{5/4}}{32}}-{\frac{5\,\sqrt [4]{-1+2\,\sqrt{x}}}{32}} \right ) }+{\frac{5\,\sqrt{2}}{4}\arctan \left ( 1+\sqrt{2}\sqrt [4]{-1+2\,\sqrt{x}} \right ) }+{\frac{5\,\sqrt{2}}{4}\arctan \left ( -1+\sqrt{2}\sqrt [4]{-1+2\,\sqrt{x}} \right ) }+{\frac{5\,\sqrt{2}}{8}\ln \left ({ \left ( 1+\sqrt{2}\sqrt [4]{-1+2\,\sqrt{x}}+\sqrt{-1+2\,\sqrt{x}} \right ) \left ( 1-\sqrt{2}\sqrt [4]{-1+2\,\sqrt{x}}+\sqrt{-1+2\,\sqrt{x}} \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*(-9/32*(-1+2*x^(1/2))^(5/4)-5/32*(-1+2*x^(1/2))^(1/4))/x+5/4*arctan(1+2^(1/2)*(-1+2*x^(1/2))^(1/4))*2^(1/2)+
5/4*arctan(-1+2^(1/2)*(-1+2*x^(1/2))^(1/4))*2^(1/2)+5/8*2^(1/2)*ln((1+2^(1/2)*(-1+2*x^(1/2))^(1/4)+(-1+2*x^(1/
2))^(1/2))/(1-2^(1/2)*(-1+2*x^(1/2))^(1/4)+(-1+2*x^(1/2))^(1/2)))

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Maxima [A]  time = 1.46602, size = 212, normalized size = 1.1 \begin{align*} \frac{5}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \,{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}}\right )}\right ) + \frac{5}{4} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \,{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}}\right )}\right ) + \frac{5}{8} \, \sqrt{2} \log \left (\sqrt{2}{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}} + \sqrt{2 \, \sqrt{x} - 1} + 1\right ) - \frac{5}{8} \, \sqrt{2} \log \left (-\sqrt{2}{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}} + \sqrt{2 \, \sqrt{x} - 1} + 1\right ) - \frac{9 \,{\left (2 \, \sqrt{x} - 1\right )}^{\frac{5}{4}} + 5 \,{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}}}{{\left (2 \, \sqrt{x} - 1\right )}^{2} + 4 \, \sqrt{x} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(2*sqrt(x) - 1)^(1/4))) + 5/4*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)
- 2*(2*sqrt(x) - 1)^(1/4))) + 5/8*sqrt(2)*log(sqrt(2)*(2*sqrt(x) - 1)^(1/4) + sqrt(2*sqrt(x) - 1) + 1) - 5/8*
sqrt(2)*log(-sqrt(2)*(2*sqrt(x) - 1)^(1/4) + sqrt(2*sqrt(x) - 1) + 1) - (9*(2*sqrt(x) - 1)^(5/4) + 5*(2*sqrt(x
) - 1)^(1/4))/((2*sqrt(x) - 1)^2 + 4*sqrt(x) - 1)

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Fricas [A]  time = 1.89518, size = 643, normalized size = 3.33 \begin{align*} -\frac{20 \, \sqrt{2} x \arctan \left (\sqrt{2} \sqrt{\sqrt{2}{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}} + \sqrt{2 \, \sqrt{x} - 1} + 1} - \sqrt{2}{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}} - 1\right ) + 20 \, \sqrt{2} x \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{-4 \, \sqrt{2}{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}} + 4 \, \sqrt{2 \, \sqrt{x} - 1} + 4} - \sqrt{2}{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}} + 1\right ) - 5 \, \sqrt{2} x \log \left (4 \, \sqrt{2}{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}} + 4 \, \sqrt{2 \, \sqrt{x} - 1} + 4\right ) + 5 \, \sqrt{2} x \log \left (-4 \, \sqrt{2}{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}} + 4 \, \sqrt{2 \, \sqrt{x} - 1} + 4\right ) + 4 \,{\left (9 \, \sqrt{x} - 2\right )}{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}}}{8 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(20*sqrt(2)*x*arctan(sqrt(2)*sqrt(sqrt(2)*(2*sqrt(x) - 1)^(1/4) + sqrt(2*sqrt(x) - 1) + 1) - sqrt(2)*(2*s
qrt(x) - 1)^(1/4) - 1) + 20*sqrt(2)*x*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*(2*sqrt(x) - 1)^(1/4) + 4*sqrt(2*sqrt
(x) - 1) + 4) - sqrt(2)*(2*sqrt(x) - 1)^(1/4) + 1) - 5*sqrt(2)*x*log(4*sqrt(2)*(2*sqrt(x) - 1)^(1/4) + 4*sqrt(
2*sqrt(x) - 1) + 4) + 5*sqrt(2)*x*log(-4*sqrt(2)*(2*sqrt(x) - 1)^(1/4) + 4*sqrt(2*sqrt(x) - 1) + 4) + 4*(9*sqr
t(x) - 2)*(2*sqrt(x) - 1)^(1/4))/x

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Sympy [C]  time = 19.9278, size = 44, normalized size = 0.23 \begin{align*} - \frac{4 \sqrt [4]{2} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{e^{2 i \pi }}{2 \sqrt{x}}} \right )}}{x^{\frac{3}{8}} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*2**(1/4)*gamma(3/4)*hyper((-5/4, 3/4), (7/4,), exp_polar(2*I*pi)/(2*sqrt(x)))/(x**(3/8)*gamma(7/4))

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Giac [A]  time = 1.07839, size = 192, normalized size = 0.99 \begin{align*} \frac{5}{4} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \,{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}}\right )}\right ) + \frac{5}{4} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \,{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}}\right )}\right ) + \frac{5}{8} \, \sqrt{2} \log \left (\sqrt{2}{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}} + \sqrt{2 \, \sqrt{x} - 1} + 1\right ) - \frac{5}{8} \, \sqrt{2} \log \left (-\sqrt{2}{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}} + \sqrt{2 \, \sqrt{x} - 1} + 1\right ) - \frac{9 \,{\left (2 \, \sqrt{x} - 1\right )}^{\frac{5}{4}} + 5 \,{\left (2 \, \sqrt{x} - 1\right )}^{\frac{1}{4}}}{4 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(2*sqrt(x) - 1)^(1/4))) + 5/4*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)
- 2*(2*sqrt(x) - 1)^(1/4))) + 5/8*sqrt(2)*log(sqrt(2)*(2*sqrt(x) - 1)^(1/4) + sqrt(2*sqrt(x) - 1) + 1) - 5/8*
sqrt(2)*log(-sqrt(2)*(2*sqrt(x) - 1)^(1/4) + sqrt(2*sqrt(x) - 1) + 1) - 1/4*(9*(2*sqrt(x) - 1)^(5/4) + 5*(2*sq
rt(x) - 1)^(1/4))/x