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

Optimal. Leaf size=66 \[ -\frac{(2 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{3/2}}-\frac{A \sqrt{b x+c x^2}}{b x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0511594, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {792, 660, 207} \[ -\frac{(2 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{3/2}}-\frac{A \sqrt{b x+c x^2}}{b x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(3/2)*Sqrt[b*x + c*x^2]),x]

[Out]

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

Rule 792

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

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

Mathematica [A]  time = 0.0368084, size = 73, normalized size = 1.11 \[ \frac{-x \sqrt{b+c x} (2 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b+c x}}{\sqrt{b}}\right )-A \sqrt{b} (b+c x)}{b^{3/2} \sqrt{x} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.015, size = 71, normalized size = 1.1 \begin{align*}{\sqrt{x \left ( cx+b \right ) } \left ( A{\it Artanh} \left ({\sqrt{cx+b}{\frac{1}{\sqrt{b}}}} \right ) xc-2\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) xb-A\sqrt{cx+b}\sqrt{b} \right ){b}^{-{\frac{3}{2}}}{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{cx+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.89751, size = 352, normalized size = 5.33 \begin{align*} \left [-\frac{{\left (2 \, B b - A c\right )} \sqrt{b} x^{2} \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} A b \sqrt{x}}{2 \, b^{2} x^{2}}, \frac{{\left (2 \, B b - A c\right )} \sqrt{-b} x^{2} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) - \sqrt{c x^{2} + b x} A b \sqrt{x}}{b^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((2*B*b - A*c)*sqrt(b)*x^2*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)*A*b*sqrt(x))/(b^2*x^2), ((2*B*b - A*c)*sqrt(-b)*x^2*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) - sqrt
(c*x^2 + b*x)*A*b*sqrt(x))/(b^2*x^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20931, size = 78, normalized size = 1.18 \begin{align*} -\frac{\frac{\sqrt{c x + b} A c}{b x} - \frac{{\left (2 \, B b c - A c^{2}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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