3.1.76 \(\int \frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}} \, dx\)

Optimal. Leaf size=87 \[ \frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{2 a^{5/2}}-\frac {3 b \sqrt {a x+b \sqrt {x}}}{2 a^2}+\frac {\sqrt {x} \sqrt {a x+b \sqrt {x}}}{a} \]

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Rubi [A]  time = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2018, 670, 640, 620, 206} \begin {gather*} \frac {3 b^2 \tanh ^{-1}\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} \end {gather*}

Antiderivative was successfully verified.

[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 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 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 670

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 2018

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*} \int \frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}} \, dx &=2 \operatorname {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) \operatorname {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 ) \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.12, size = 102, normalized size = 1.17 \begin {gather*} \frac {\sqrt {a} \sqrt {x} \left (2 a^2 x-a b \sqrt {x}-3 b^2\right )+3 b^{5/2} \sqrt [4]{x} \sqrt {\frac {a \sqrt {x}}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {a} \sqrt [4]{x}}{\sqrt {b}}\right )}{2 a^{5/2} \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*Sqrt[x]*(-3*b^2 - a*b*Sqrt[x] + 2*a^2*x) + 3*b^(5/2)*Sqrt[1 + (a*Sqrt[x])/b]*x^(1/4)*ArcSinh[(Sqrt[a]
*x^(1/4))/Sqrt[b]])/(2*a^(5/2)*Sqrt[b*Sqrt[x] + a*x])

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IntegrateAlgebraic [A]  time = 0.27, size = 86, normalized size = 0.99 \begin {gather*} \frac {\left (2 a \sqrt {x}-3 b\right ) \sqrt {a x+b \sqrt {x}}}{2 a^2}-\frac {3 b^2 \log \left (-2 a^{5/2} \sqrt {a x+b \sqrt {x}}+2 a^3 \sqrt {x}+a^2 b\right )}{4 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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*Log[a^2*b + 2*a^3*Sqrt[x] - 2*a^(5/2)*Sqrt[b*Sqr
t[x] + a*x]])/(4*a^(5/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.27, size = 69, normalized size = 0.79 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.05, size = 160, normalized size = 1.84 \begin {gather*} \frac {\sqrt {a x +b \sqrt {x}}\, \left (4 a \,b^{2} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-a \,b^{2} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+4 \sqrt {a x +b \sqrt {x}}\, a^{\frac {5}{2}} \sqrt {x}-8 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {3}{2}} b +2 \sqrt {a x +b \sqrt {x}}\, a^{\frac {3}{2}} b \right )}{4 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*x+b*x^(1/2))^(1/2)*(4*x^(1/2)*(a*x+b*x^(1/2))^(1/2)*a^(5/2)-8*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(3/2)*b+2
*(a*x+b*x^(1/2))^(1/2)*a^(3/2)*b+4*a*ln(1/2*(2*a*x^(1/2)+b+2*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*b
^2-b^2*ln(1/2*(2*a*x^(1/2)+b+2*(a*x+b*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*a)/((a*x^(1/2)+b)*x^(1/2))^(1/2)/a^(7/2
)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x}}{\sqrt {a x + b \sqrt {x}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x}}{\sqrt {a\,x+b\,\sqrt {x}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x}}{\sqrt {a x + b \sqrt {x}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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