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

Optimal. Leaf size=83 \[ \frac{\sqrt{b x+c x^2} (2 A c+b B)}{b}+\frac{(2 A c+b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{\sqrt{c}}-\frac{2 A \left (b x+c x^2\right )^{3/2}}{b x^2} \]

[Out]

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

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Rubi [A]  time = 0.0863942, antiderivative size = 83, 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, 664, 620, 206} \[ \frac{\sqrt{b x+c x^2} (2 A c+b B)}{b}+\frac{(2 A c+b B) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{\sqrt{c}}-\frac{2 A \left (b x+c x^2\right )^{3/2}}{b x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 664

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 + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), 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] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*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^2} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{3/2}}{b x^2}+\frac{\left (2 \left (-2 (-b B+A c)+\frac{3}{2} (-b B+2 A c)\right )\right ) \int \frac{\sqrt{b x+c x^2}}{x} \, dx}{b}\\ &=\frac{(b B+2 A c) \sqrt{b x+c x^2}}{b}-\frac{2 A \left (b x+c x^2\right )^{3/2}}{b x^2}+\frac{1}{2} (b B+2 A c) \int \frac{1}{\sqrt{b x+c x^2}} \, dx\\ &=\frac{(b B+2 A c) \sqrt{b x+c x^2}}{b}-\frac{2 A \left (b x+c x^2\right )^{3/2}}{b x^2}+(b B+2 A c) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )\\ &=\frac{(b B+2 A c) \sqrt{b x+c x^2}}{b}-\frac{2 A \left (b x+c x^2\right )^{3/2}}{b x^2}+\frac{(b B+2 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{\sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.145768, size = 75, normalized size = 0.9 \[ \frac{\sqrt{x (b+c x)} \left (\frac{\sqrt{x} (2 A c+b B) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{b} \sqrt{c} \sqrt{\frac{c x}{b}+1}}-2 A+B x\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(-2*A + B*x + ((b*B + 2*A*c)*Sqrt[x]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[b]*Sqrt[c]*S
qrt[1 + (c*x)/b])))/x

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Maple [A]  time = 0.01, size = 113, normalized size = 1.4 \begin{align*} B\sqrt{c{x}^{2}+bx}+{\frac{bB}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){\frac{1}{\sqrt{c}}}}-2\,{\frac{A \left ( c{x}^{2}+bx \right ) ^{3/2}}{b{x}^{2}}}+2\,{\frac{Ac\sqrt{c{x}^{2}+bx}}{b}}+A\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^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85168, size = 319, normalized size = 3.84 \begin{align*} \left [\frac{{\left (B b + 2 \, A c\right )} \sqrt{c} x \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (B c x - 2 \, A c\right )} \sqrt{c x^{2} + b x}}{2 \, c x}, -\frac{{\left (B b + 2 \, A c\right )} \sqrt{-c} x \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (B c x - 2 \, A c\right )} \sqrt{c x^{2} + b x}}{c x}\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^2,x, algorithm="fricas")

[Out]

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

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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^{2}}\, 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**2,x)

[Out]

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

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Giac [A]  time = 1.25631, size = 111, normalized size = 1.34 \begin{align*} \sqrt{c x^{2} + b x} B - \frac{{\left (B b + 2 \, A c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{2 \, \sqrt{c}} + \frac{2 \, A b}{\sqrt{c} x - \sqrt{c x^{2} + b x}} \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^2,x, algorithm="giac")

[Out]

sqrt(c*x^2 + b*x)*B - 1/2*(B*b + 2*A*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/sqrt(c) + 2*A
*b/(sqrt(c)*x - sqrt(c*x^2 + b*x))