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

Optimal. Leaf size=68 \[ -\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{3/2}}-\frac{2 \sqrt{x} (b B-A c)}{b c \sqrt{b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.050674, antiderivative size = 68, 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 = {788, 660, 207} \[ -\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{3/2}}-\frac{2 \sqrt{x} (b B-A c)}{b c \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 788

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((g*(c*d - b*e) + c*e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)*(2*c*d - b*e)), x] - Dist[(e*(m*(g
*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c*f - b*g)))/(c*(p + 1)*(2*c*d - b*e)), Int[(d + e*x)^(m - 1)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2,
0] && LtQ[p, -1] && GtQ[m, 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{\sqrt{x} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 (b B-A c) \sqrt{x}}{b c \sqrt{b x+c x^2}}+\frac{A \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{b}\\ &=-\frac{2 (b B-A c) \sqrt{x}}{b c \sqrt{b x+c x^2}}+\frac{(2 A) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )}{b}\\ &=-\frac{2 (b B-A c) \sqrt{x}}{b c \sqrt{b x+c x^2}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0341092, size = 69, normalized size = 1.01 \[ -\frac{2 \sqrt{x} \left (\sqrt{b} (b B-A c)+A c \sqrt{b+c x} \tanh ^{-1}\left (\frac{\sqrt{b+c x}}{\sqrt{b}}\right )\right )}{b^{3/2} c \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 63, normalized size = 0.9 \begin{align*} -2\,{\frac{\sqrt{x \left ( cx+b \right ) }}{{b}^{3/2}\sqrt{x} \left ( cx+b \right ) c} \left ( A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) c\sqrt{cx+b}-Ac\sqrt{b}+B{b}^{3/2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )} \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((B*x+A)*x^(1/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.95775, size = 429, normalized size = 6.31 \begin{align*} \left [\frac{{\left (A c^{2} x^{2} + A b c 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 \,{\left (B b^{2} - A b c\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{b^{2} c^{2} x^{2} + b^{3} c x}, \frac{2 \,{\left ({\left (A c^{2} x^{2} + A b c x\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) -{\left (B b^{2} - A b c\right )} \sqrt{c x^{2} + b x} \sqrt{x}\right )}}{b^{2} c^{2} x^{2} + b^{3} c x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x} \left (A + B x\right )}{\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((B*x+A)*x**(1/2)/(c*x**2+b*x)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.14161, size = 130, normalized size = 1.91 \begin{align*} \frac{2 \, A \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} - \frac{2 \,{\left (B b - A c\right )}}{\sqrt{c x + b} b c} - \frac{2 \,{\left (A \sqrt{b} c \arctan \left (\frac{\sqrt{b}}{\sqrt{-b}}\right ) - B \sqrt{-b} b + A \sqrt{-b} c\right )}}{\sqrt{-b} b^{\frac{3}{2}} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b) - 2*(B*b - A*c)/(sqrt(c*x + b)*b*c) - 2*(A*sqrt(b)*c*arctan(sq
rt(b)/sqrt(-b)) - B*sqrt(-b)*b + A*sqrt(-b)*c)/(sqrt(-b)*b^(3/2)*c)