\(\int \frac {(-1+2 \sqrt {x})^{5/4}}{x^2} \, dx\) [298]

Optimal result

Integrand size = 17, antiderivative size = 193 \[ \int \frac {\left (-1+2 \sqrt {x}\right )^{5/4}}{x^2} \, dx=-\frac {\left (-1+2 \sqrt {x}\right )^{5/4}}{x}-\frac {5 \sqrt [4]{-1+2 \sqrt {x}}}{2 \sqrt {x}}-\frac {5 \arctan \left (1-\sqrt {2} \sqrt [4]{-1+2 \sqrt {x}}\right )}{2 \sqrt {2}}+\frac {5 \arctan \left (1+\sqrt {2} \sqrt [4]{-1+2 \sqrt {x}}\right )}{2 \sqrt {2}}-\frac {5 \log \left (1-\sqrt {2} \sqrt [4]{-1+2 \sqrt {x}}+\sqrt {-1+2 \sqrt {x}}\right )}{4 \sqrt {2}}+\frac {5 \log \left (1+\sqrt {2} \sqrt [4]{-1+2 \sqrt {x}}+\sqrt {-1+2 \sqrt {x}}\right )}{4 \sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {272, 43, 65, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {\left (-1+2 \sqrt {x}\right )^{5/4}}{x^2} \, dx=-\frac {5 \arctan \left (1-\sqrt {2} \sqrt [4]{2 \sqrt {x}-1}\right )}{2 \sqrt {2}}+\frac {5 \arctan \left (\sqrt {2} \sqrt [4]{2 \sqrt {x}-1}+1\right )}{2 \sqrt {2}}-\frac {\left (2 \sqrt {x}-1\right )^{5/4}}{x}-\frac {5 \sqrt [4]{2 \sqrt {x}-1}}{2 \sqrt {x}}-\frac {5 \log \left (\sqrt {2 \sqrt {x}-1}-\sqrt {2} \sqrt [4]{2 \sqrt {x}-1}+1\right )}{4 \sqrt {2}}+\frac {5 \log \left (\sqrt {2 \sqrt {x}-1}+\sqrt {2} \sqrt [4]{2 \sqrt {x}-1}+1\right )}{4 \sqrt {2}} \]

[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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 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 217

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 272

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 631

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 642

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 1176

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 1179

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]

Rubi steps \begin{align*} \text {integral}& = 2 \text {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} \text {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} \text {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} \text {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} \text {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} \text {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} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+2 \sqrt {x}}\right )+\frac {5}{4} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt [4]{-1+2 \sqrt {x}}\right )-\frac {5 \text {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 \text {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 \text {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 \text {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 \arctan \left (1-\sqrt {2} \sqrt [4]{-1+2 \sqrt {x}}\right )}{2 \sqrt {2}}+\frac {5 \arctan \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 [A] (verified)

Time = 0.36 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.62 \[ \int \frac {\left (-1+2 \sqrt {x}\right )^{5/4}}{x^2} \, dx=\frac {2 \left (2-9 \sqrt {x}\right ) \sqrt [4]{-1+2 \sqrt {x}}+5 \sqrt {2} x \arctan \left (\frac {-1+\sqrt {-1+2 \sqrt {x}}}{\sqrt {2} \sqrt [4]{-1+2 \sqrt {x}}}\right )+5 \sqrt {2} x \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+2 \sqrt {x}}}{1+\sqrt {-1+2 \sqrt {x}}}\right )}{4 x} \]

[In]

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

[Out]

(2*(2 - 9*Sqrt[x])*(-1 + 2*Sqrt[x])^(1/4) + 5*Sqrt[2]*x*ArcTan[(-1 + Sqrt[-1 + 2*Sqrt[x]])/(Sqrt[2]*(-1 + 2*Sq
rt[x])^(1/4))] + 5*Sqrt[2]*x*ArcTanh[(Sqrt[2]*(-1 + 2*Sqrt[x])^(1/4))/(1 + Sqrt[-1 + 2*Sqrt[x]])])/(4*x)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.33 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.44

method	result	size
meijerg	\(\frac {5 \operatorname {signum}\left (-1+2 \sqrt {x}\right )^{\frac {5}{4}} \left (-\frac {2 \Gamma \left (\frac {3}{4}\right )}{5 x}+\frac {2 \Gamma \left (\frac {3}{4}\right )}{\sqrt {x}}+\frac {\left (-2 \ln \left (2\right )+\frac {\pi }{2}-\frac {3}{2}+\frac {\ln \left (x \right )}{2}+i \pi \right ) \Gamma \left (\frac {3}{4}\right )}{2}+\frac {\Gamma \left (\frac {3}{4}\right ) \sqrt {x}\, {}_{3}^{}{\moversetsp {}{\mundersetsp {}{F_{2}^{}}}}\left (1,1,\frac {7}{4};2,4;2 \sqrt {x}\right )}{4}\right )}{2 \Gamma \left (\frac {3}{4}\right ) \left (-\operatorname {signum}\left (-1+2 \sqrt {x}\right )\right )^{\frac {5}{4}}}\)	\(85\)
derivativedivides	\(\frac {-\frac {9 \left (-1+2 \sqrt {x}\right )^{\frac {5}{4}}}{4}-\frac {5 \left (-1+2 \sqrt {x}\right )^{\frac {1}{4}}}{4}}{x}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (-1+2 \sqrt {x}\right )^{\frac {1}{4}}+\sqrt {-1+2 \sqrt {x}}}{1-\sqrt {2}\, \left (-1+2 \sqrt {x}\right )^{\frac {1}{4}}+\sqrt {-1+2 \sqrt {x}}}\right )+2 \arctan \left (1+\sqrt {2}\, \left (-1+2 \sqrt {x}\right )^{\frac {1}{4}}\right )+2 \arctan \left (-1+\sqrt {2}\, \left (-1+2 \sqrt {x}\right )^{\frac {1}{4}}\right )\right )}{8}\)	\(125\)
default	\(\frac {-\frac {9 \left (-1+2 \sqrt {x}\right )^{\frac {5}{4}}}{4}-\frac {5 \left (-1+2 \sqrt {x}\right )^{\frac {1}{4}}}{4}}{x}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (-1+2 \sqrt {x}\right )^{\frac {1}{4}}+\sqrt {-1+2 \sqrt {x}}}{1-\sqrt {2}\, \left (-1+2 \sqrt {x}\right )^{\frac {1}{4}}+\sqrt {-1+2 \sqrt {x}}}\right )+2 \arctan \left (1+\sqrt {2}\, \left (-1+2 \sqrt {x}\right )^{\frac {1}{4}}\right )+2 \arctan \left (-1+\sqrt {2}\, \left (-1+2 \sqrt {x}\right )^{\frac {1}{4}}\right )\right )}{8}\)	\(125\)

[In]

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

[Out]

5/2/GAMMA(3/4)*signum(-1+2*x^(1/2))^(5/4)/(-signum(-1+2*x^(1/2)))^(5/4)*(-2/5*GAMMA(3/4)/x+2*GAMMA(3/4)/x^(1/2
)+1/2*(-2*ln(2)+1/2*Pi-3/2+1/2*ln(x)+I*Pi)*GAMMA(3/4)+1/4*GAMMA(3/4)*x^(1/2)*hypergeom([1,1,7/4],[2,4],2*x^(1/
2)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.62 \[ \int \frac {\left (-1+2 \sqrt {x}\right )^{5/4}}{x^2} \, dx=\frac {\left (5 i + 5\right ) \, \sqrt {2} x \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}}\right ) - \left (5 i - 5\right ) \, \sqrt {2} x \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}}\right ) + \left (5 i - 5\right ) \, \sqrt {2} x \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}}\right ) - \left (5 i + 5\right ) \, \sqrt {2} x \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}}\right ) - 4 \, {\left (9 \, \sqrt {x} - 2\right )} {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}}}{8 \, x} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.75 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.23 \[ \int \frac {\left (-1+2 \sqrt {x}\right )^{5/4}}{x^2} \, dx=- \frac {4 \cdot \sqrt [4]{2} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e^{2 i \pi }}{2 \sqrt {x}}} \right )}}{x^{\frac {3}{8}} \Gamma \left (\frac {7}{4}\right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.81 \[ \int \frac {\left (-1+2 \sqrt {x}\right )^{5/4}}{x^2} \, dx=\frac {5}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {5}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {5}{8} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}} + \sqrt {2 \, \sqrt {x} - 1} + 1\right ) - \frac {5}{8} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}} + \sqrt {2 \, \sqrt {x} - 1} + 1\right ) - \frac {9 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {5}{4}} + 5 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}}}{{\left (2 \, \sqrt {x} - 1\right )}^{2} + 4 \, \sqrt {x} - 1} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-1+2 \sqrt {x}\right )^{5/4}}{x^2} \, dx=\frac {5}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {5}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {5}{8} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}} + \sqrt {2 \, \sqrt {x} - 1} + 1\right ) - \frac {5}{8} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}} + \sqrt {2 \, \sqrt {x} - 1} + 1\right ) - \frac {9 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {5}{4}} + 5 \, {\left (2 \, \sqrt {x} - 1\right )}^{\frac {1}{4}}}{4 \, x} \]

[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

Mupad [B] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.40 \[ \int \frac {\left (-1+2 \sqrt {x}\right )^{5/4}}{x^2} \, dx=-\frac {5\,{\left (2\,\sqrt {x}-1\right )}^{1/4}}{4\,x}-\frac {9\,{\left (2\,\sqrt {x}-1\right )}^{5/4}}{4\,x}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (2\,\sqrt {x}-1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {5}{4}+\frac {5}{4}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (2\,\sqrt {x}-1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {5}{4}-\frac {5}{4}{}\mathrm {i}\right ) \]

[In]

int((2*x^(1/2) - 1)^(5/4)/x^2,x)

[Out]

2^(1/2)*atan(2^(1/2)*(2*x^(1/2) - 1)^(1/4)*(1/2 - 1i/2))*(5/4 + 5i/4) - (9*(2*x^(1/2) - 1)^(5/4))/(4*x) - (5*(
2*x^(1/2) - 1)^(1/4))/(4*x) + 2^(1/2)*atan(2^(1/2)*(2*x^(1/2) - 1)^(1/4)*(1/2 + 1i/2))*(5/4 - 5i/4)