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

Optimal. Leaf size=48 \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{b^{3/2}}-\frac{2 \sqrt{x}}{b \sqrt{a+b x}} \]

[Out]

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

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Rubi [A]  time = 0.0161718, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {47, 63, 217, 206} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{b^{3/2}}-\frac{2 \sqrt{x}}{b \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(a + b*x)^(3/2),x]

[Out]

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

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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

Mathematica [A]  time = 0.0684489, size = 64, normalized size = 1.33 \[ \frac{2 \left (\sqrt{a} \sqrt{\frac{b x}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )-\sqrt{b} \sqrt{x}\right )}{b^{3/2} \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(a + b*x)^(3/2),x]

[Out]

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

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{x} \left ( bx+a \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.87175, size = 308, normalized size = 6.42 \begin{align*} \left [\frac{{\left (b x + a\right )} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \, \sqrt{b x + a} b \sqrt{x}}{b^{3} x + a b^{2}}, -\frac{2 \,{\left ({\left (b x + a\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) + \sqrt{b x + a} b \sqrt{x}\right )}}{b^{3} x + a b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((b*x + a)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*sqrt(b*x + a)*b*sqrt(x))/(b^3*x + a*b
^2), -2*((b*x + a)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + sqrt(b*x + a)*b*sqrt(x))/(b^3*x + a*b
^2)]

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Sympy [A]  time = 2.10617, size = 46, normalized size = 0.96 \begin{align*} \frac{2 \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{b^{\frac{3}{2}}} - \frac{2 \sqrt{x}}{\sqrt{a} b \sqrt{1 + \frac{b x}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*asinh(sqrt(b)*sqrt(x)/sqrt(a))/b**(3/2) - 2*sqrt(x)/(sqrt(a)*b*sqrt(1 + b*x/a))

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Giac [B]  time = 59.6491, size = 115, normalized size = 2.4 \begin{align*} -\frac{{\left (\frac{4 \, a \sqrt{b}}{{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} + a b} + \frac{\log \left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{\sqrt{b}}\right )}{\left | b \right |}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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