\(\int \frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}} \, dx\) [119]

Optimal result

Integrand size = 21, antiderivative size = 87 \[ \int \frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}} \, dx=-\frac {3 b \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {\sqrt {x} \sqrt {b \sqrt {x}+a x}}{a}+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{2 a^{5/2}} \]

[Out]

3/2*b^2*arctanh(a^(1/2)*x^(1/2)/(b*x^(1/2)+a*x)^(1/2))/a^(5/2)-3/2*b*(b*x^(1/2)+a*x)^(1/2)/a^2+x^(1/2)*(b*x^(1
/2)+a*x)^(1/2)/a

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2043, 684, 654, 634, 212} \[ \int \frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}} \, dx=\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{2 a^{5/2}}-\frac {3 b \sqrt {a x+b \sqrt {x}}}{2 a^2}+\frac {\sqrt {x} \sqrt {a x+b \sqrt {x}}}{a} \]

[In]

Int[Sqrt[x]/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

(-3*b*Sqrt[b*Sqrt[x] + a*x])/(2*a^2) + (Sqrt[x]*Sqrt[b*Sqrt[x] + a*x])/a + (3*b^2*ArcTanh[(Sqrt[a]*Sqrt[x])/Sq
rt[b*Sqrt[x] + a*x]])/(2*a^(5/2))

Rule 212

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

Rule 634

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 654

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 684

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

Rule 2043

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {\sqrt {x} \sqrt {b \sqrt {x}+a x}}{a}-\frac {(3 b) \text {Subst}\left (\int \frac {x}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{2 a} \\ & = -\frac {3 b \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {\sqrt {x} \sqrt {b \sqrt {x}+a x}}{a}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{4 a^2} \\ & = -\frac {3 b \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {\sqrt {x} \sqrt {b \sqrt {x}+a x}}{a}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{2 a^2} \\ & = -\frac {3 b \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {\sqrt {x} \sqrt {b \sqrt {x}+a x}}{a}+\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{2 a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}} \, dx=\frac {\left (-3 b+2 a \sqrt {x}\right ) \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {b \sqrt {x}+a x}}{b+a \sqrt {x}}\right )}{2 a^{5/2}} \]

[In]

Integrate[Sqrt[x]/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

((-3*b + 2*a*Sqrt[x])*Sqrt[b*Sqrt[x] + a*x])/(2*a^2) + (3*b^2*ArcTanh[(Sqrt[a]*Sqrt[b*Sqrt[x] + a*x])/(b + a*S
qrt[x])])/(2*a^(5/2))

Maple [A] (verified)

Time = 2.41 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {\sqrt {x}\, \sqrt {b \sqrt {x}+a x}}{a}-\frac {3 b \left (\frac {\sqrt {b \sqrt {x}+a x}}{a}-\frac {b \ln \left (\frac {\frac {b}{2}+a \sqrt {x}}{\sqrt {a}}+\sqrt {b \sqrt {x}+a x}\right )}{2 a^{\frac {3}{2}}}\right )}{2 a}\)	\(74\)
default	\(\frac {\sqrt {b \sqrt {x}+a x}\, \left (4 \sqrt {x}\, \sqrt {b \sqrt {x}+a x}\, a^{\frac {5}{2}}+2 \sqrt {b \sqrt {x}+a x}\, a^{\frac {3}{2}} b +4 \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a \,b^{2}-8 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {3}{2}} b -b^{2} \ln \left (\frac {2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+2 a \sqrt {x}+b}{2 \sqrt {a}}\right ) a \right )}{4 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {7}{2}}}\)	\(160\)

[In]

int(x^(1/2)/(b*x^(1/2)+a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

x^(1/2)*(b*x^(1/2)+a*x)^(1/2)/a-3/2*b/a*((b*x^(1/2)+a*x)^(1/2)/a-1/2*b/a^(3/2)*ln((1/2*b+a*x^(1/2))/a^(1/2)+(b
*x^(1/2)+a*x)^(1/2)))

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}} \, dx=\text {Timed out} \]

[In]

integrate(x^(1/2)/(b*x^(1/2)+a*x)^(1/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.64 \[ \int \frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}} \, dx=2 \left (\begin {cases} \left (\frac {\sqrt {x}}{2 a} - \frac {3 b}{4 a^{2}}\right ) \sqrt {a x + b \sqrt {x}} + \frac {3 b^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {a} \sqrt {a x + b \sqrt {x}} + 2 a \sqrt {x} + b \right )}}{\sqrt {a}} & \text {for}\: \frac {b^{2}}{a} \neq 0 \\\frac {\left (\sqrt {x} + \frac {b}{2 a}\right ) \log {\left (\sqrt {x} + \frac {b}{2 a} \right )}}{\sqrt {a \left (\sqrt {x} + \frac {b}{2 a}\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 a^{2}} & \text {for}\: a \neq 0 \\\frac {2 \left (b \sqrt {x}\right )^{\frac {5}{2}}}{5 b^{3}} & \text {for}\: b \neq 0 \\\tilde {\infty } x^{\frac {3}{2}} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate(x**(1/2)/(b*x**(1/2)+a*x)**(1/2),x)

[Out]

2*Piecewise(((sqrt(x)/(2*a) - 3*b/(4*a**2))*sqrt(a*x + b*sqrt(x)) + 3*b**2*Piecewise((log(2*sqrt(a)*sqrt(a*x +
b*sqrt(x)) + 2*a*sqrt(x) + b)/sqrt(a), Ne(b**2/a, 0)), ((sqrt(x) + b/(2*a))*log(sqrt(x) + b/(2*a))/sqrt(a*(sq
rt(x) + b/(2*a))**2), True))/(8*a**2), Ne(a, 0)), (2*(b*sqrt(x))**(5/2)/(5*b**3), Ne(b, 0)), (zoo*x**(3/2), Tr
ue))

Maxima [F]

\[ \int \frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}} \, dx=\int { \frac {\sqrt {x}}{\sqrt {a x + b \sqrt {x}}} \,d x } \]

[In]

integrate(x^(1/2)/(b*x^(1/2)+a*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x)/sqrt(a*x + b*sqrt(x)), x)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}} \, dx=\frac {1}{2} \, \sqrt {a x + b \sqrt {x}} {\left (\frac {2 \, \sqrt {x}}{a} - \frac {3 \, b}{a^{2}}\right )} - \frac {3 \, b^{2} \log \left ({\left | 2 \, \sqrt {a} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} + b \right |}\right )}{4 \, a^{\frac {5}{2}}} \]

[In]

integrate(x^(1/2)/(b*x^(1/2)+a*x)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}} \, dx=\int \frac {\sqrt {x}}{\sqrt {a\,x+b\,\sqrt {x}}} \,d x \]

[In]

int(x^(1/2)/(a*x + b*x^(1/2))^(1/2),x)

[Out]

int(x^(1/2)/(a*x + b*x^(1/2))^(1/2), x)