3.126 \(\int \frac{\sqrt{x}}{(b \sqrt{x}+a x)^{3/2}} \, dx\)

Optimal. Leaf size=60 \[ \frac{4 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b \sqrt{x}}}\right )}{a^{3/2}}-\frac{4 \sqrt{x}}{a \sqrt{a x+b \sqrt{x}}} \]

[Out]

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

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Rubi [A]  time = 0.0671288, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2018, 652, 620, 206} \[ \frac{4 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b \sqrt{x}}}\right )}{a^{3/2}}-\frac{4 \sqrt{x}}{a \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(b*Sqrt[x] + a*x)^(3/2),x]

[Out]

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

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]

Rule 652

Int[((d_.) + (e_.)*(x_))^2*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + b*x +
c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fr
eeQ[{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] &&  !IntegerQ[p] && LtQ[p,
-1]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.105056, size = 79, normalized size = 1.32 \[ \frac{4 \sqrt [4]{x} \left (\sqrt{b} \sqrt{\frac{a \sqrt{x}}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{a} \sqrt [4]{x}}{\sqrt{b}}\right )-\sqrt{a} \sqrt [4]{x}\right )}{a^{3/2} \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(b*Sqrt[x] + a*x)^(3/2),x]

[Out]

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

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Maple [B]  time = 0.006, size = 238, normalized size = 4. \begin{align*} 2\,{\frac{\sqrt{b\sqrt{x}+ax}}{{a}^{3/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }b \left ( b+a\sqrt{x} \right ) ^{2}} \left ( \ln \left ( 1/2\,{\frac{2\,\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }\sqrt{a}+2\,a\sqrt{x}+b}{\sqrt{a}}} \right ) x{a}^{2}b-2\,{a}^{5/2}x\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }+2\,\ln \left ( 1/2\,{\frac{2\,\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }\sqrt{a}+2\,a\sqrt{x}+b}{\sqrt{a}}} \right ) \sqrt{x}a{b}^{2}-4\,{a}^{3/2}\sqrt{x}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }b+2\,{a}^{3/2} \left ( \sqrt{x} \left ( b+a\sqrt{x} \right ) \right ) ^{3/2}+\ln \left ( 1/2\,{\frac{2\,\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }\sqrt{a}+2\,a\sqrt{x}+b}{\sqrt{a}}} \right ){b}^{3}-2\,\sqrt{a}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{b}^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(b*x^(1/2)+a*x)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{{\left (a x + b \sqrt{x}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(b*x^(1/2)+a*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x)/(a*x + b*sqrt(x))^(3/2), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(b*x^(1/2)+a*x)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x}}{\left (a x + b \sqrt{x}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)/(b*x**(1/2)+a*x)**(3/2),x)

[Out]

Integral(sqrt(x)/(a*x + b*sqrt(x))**(3/2), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(b*x^(1/2)+a*x)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError