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

Optimal. Leaf size=73 \[ -\frac{2 A \left (b x+c x^2\right )^{3/2}}{3 b x^3}-\frac{2 B \sqrt{b x+c x^2}}{x}+2 B \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0745639, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {792, 662, 620, 206} \[ -\frac{2 A \left (b x+c x^2\right )^{3/2}}{3 b x^3}-\frac{2 B \sqrt{b x+c x^2}}{x}+2 B \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*B*Sqrt[b*x + c*x^2])/x - (2*A*(b*x + c*x^2)^(3/2))/(3*b*x^3) + 2*B*Sqrt[c]*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x +
c*x^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 662

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

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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

Mathematica [C]  time = 0.0681139, size = 86, normalized size = 1.18 \[ \frac{2 \sqrt{x (b+c x)} \left ((b+c x) \sqrt{\frac{c x}{b}+1} (b B-A c)-b^2 B \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};-\frac{c x}{b}\right )\right )}{3 b c x^2 \sqrt{\frac{c x}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x*(b + c*x)]*((b*B - A*c)*(b + c*x)*Sqrt[1 + (c*x)/b] - b^2*B*Hypergeometric2F1[-3/2, -3/2, -1/2, -((c
*x)/b)]))/(3*b*c*x^2*Sqrt[1 + (c*x)/b])

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Maple [A]  time = 0.009, size = 89, normalized size = 1.2 \begin{align*} -{\frac{2\,A}{3\,b{x}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-2\,{\frac{B \left ( c{x}^{2}+bx \right ) ^{3/2}}{b{x}^{2}}}+2\,{\frac{Bc\sqrt{c{x}^{2}+bx}}{b}}+B\sqrt{c}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.03001, size = 335, normalized size = 4.59 \begin{align*} \left [\frac{3 \, B b \sqrt{c} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \, \sqrt{c x^{2} + b x}{\left (A b +{\left (3 \, B b + A c\right )} x\right )}}{3 \, b x^{2}}, -\frac{2 \,{\left (3 \, B b \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) + \sqrt{c x^{2} + b x}{\left (A b +{\left (3 \, B b + A c\right )} x\right )}\right )}}{3 \, b x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*B*b*sqrt(c)*x^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*sqrt(c*x^2 + b*x)*(A*b + (3*B*b + A*c
)*x))/(b*x^2), -2/3*(3*B*b*sqrt(-c)*x^2*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + sqrt(c*x^2 + b*x)*(A*b + (3
*B*b + A*c)*x))/(b*x^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.15488, size = 204, normalized size = 2.79 \begin{align*} -B \sqrt{c} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right ) + \frac{2 \,{\left (3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b \sqrt{c} + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A c^{\frac{3}{2}} + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b c + A b^{2} \sqrt{c}\right )}}{3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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