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

Optimal. Leaf size=61 \[ \frac{2 B \sqrt{x} \sqrt{b x+c x^2}}{3 c}-\frac{2 \sqrt{b x+c x^2} (2 b B-3 A c)}{3 c^2 \sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.0407592, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {794, 648} \[ \frac{2 B \sqrt{x} \sqrt{b x+c x^2}}{3 c}-\frac{2 \sqrt{b x+c x^2} (2 b B-3 A c)}{3 c^2 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rule 648

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

Rubi steps

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

Mathematica [A]  time = 0.0255501, size = 36, normalized size = 0.59 \[ \frac{2 \sqrt{x (b+c x)} (3 A c-2 b B+B c x)}{3 c^2 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 38, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,cx+2\,b \right ) \left ( Bcx+3\,Ac-2\,bB \right ) }{3\,{c}^{2}}\sqrt{x}{\frac{1}{\sqrt{c{x}^{2}+bx}}}} \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)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.08302, size = 61, normalized size = 1. \begin{align*} \frac{2 \, \sqrt{c x + b} A}{c} + \frac{2 \,{\left (c^{2} x^{2} - b c x - 2 \, b^{2}\right )} B}{3 \, \sqrt{c x + b} c^{2}} \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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.56172, size = 82, normalized size = 1.34 \begin{align*} \frac{2 \,{\left (B c x - 2 \, B b + 3 \, A c\right )} \sqrt{c x^{2} + b x}}{3 \, c^{2} \sqrt{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)^(1/2),x, algorithm="fricas")

[Out]

2/3*(B*c*x - 2*B*b + 3*A*c)*sqrt(c*x^2 + b*x)/(c^2*sqrt(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 )}{\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**(1/2)/(c*x**2+b*x)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.14612, size = 77, normalized size = 1.26 \begin{align*} \frac{2 \,{\left ({\left (c x + b\right )}^{\frac{3}{2}} B - 3 \, \sqrt{c x + b} B b + 3 \, \sqrt{c x + b} A c\right )}}{3 \, c^{2}} + \frac{2 \,{\left (2 \, B b^{\frac{3}{2}} - 3 \, A \sqrt{b} c\right )}}{3 \, c^{2}} \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)^(1/2),x, algorithm="giac")

[Out]

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