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

Optimal. Leaf size=56 \[ \frac{2 \sqrt{x}}{b \sqrt{b x+c x^2}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0216995, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {666, 660, 207} \[ \frac{2 \sqrt{x}}{b \sqrt{b x+c x^2}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(b*x + c*x^2)^(3/2),x]

[Out]

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

Rule 666

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

Rule 660

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

Rule 207

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

Rubi steps

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

Mathematica [C]  time = 0.0088132, size = 37, normalized size = 0.66 \[ \frac{2 \sqrt{x} \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c x}{b}+1\right )}{b \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (c*x)/b])/(b*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.181, size = 51, normalized size = 0.9 \begin{align*} -2\,{\frac{\sqrt{x \left ( cx+b \right ) }}{{b}^{3/2}\sqrt{x} \left ( cx+b \right ) } \left ({\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) \sqrt{cx+b}-\sqrt{b} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.0726, size = 359, normalized size = 6.41 \begin{align*} \left [\frac{{\left (c x^{2} + b x\right )} \sqrt{b} \log \left (-\frac{c x^{2} + 2 \, b x - 2 \, \sqrt{c x^{2} + b x} \sqrt{b} \sqrt{x}}{x^{2}}\right ) + 2 \, \sqrt{c x^{2} + b x} b \sqrt{x}}{b^{2} c x^{2} + b^{3} x}, \frac{2 \,{\left ({\left (c x^{2} + b x\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) + \sqrt{c x^{2} + b x} b \sqrt{x}\right )}}{b^{2} c x^{2} + b^{3} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23874, size = 90, normalized size = 1.61 \begin{align*} \frac{2 \, \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} - \frac{2 \,{\left (\sqrt{b} \arctan \left (\frac{\sqrt{b}}{\sqrt{-b}}\right ) + \sqrt{-b}\right )}}{\sqrt{-b} b^{\frac{3}{2}}} + \frac{2}{\sqrt{c x + b} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b) - 2*(sqrt(b)*arctan(sqrt(b)/sqrt(-b)) + sqrt(-b))/(sqrt(-b)*b^(3
/2)) + 2/(sqrt(c*x + b)*b)