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

Optimal. Leaf size=92 \[ -\frac{b (b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{3/2}}-\frac{\sqrt{b x+c x^2} (b B-4 A c)}{4 c}+\frac{B \left (b x+c x^2\right )^{3/2}}{2 c x} \]

[Out]

-((b*B - 4*A*c)*Sqrt[b*x + c*x^2])/(4*c) + (B*(b*x + c*x^2)^(3/2))/(2*c*x) - (b*(b*B - 4*A*c)*ArcTanh[(Sqrt[c]
*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2))

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Rubi [A]  time = 0.0724537, antiderivative size = 92, 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 = {794, 664, 620, 206} \[ -\frac{b (b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{3/2}}-\frac{\sqrt{b x+c x^2} (b B-4 A c)}{4 c}+\frac{B \left (b x+c x^2\right )^{3/2}}{2 c x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*B - 4*A*c)*Sqrt[b*x + c*x^2])/(4*c) + (B*(b*x + c*x^2)^(3/2))/(2*c*x) - (b*(b*B - 4*A*c)*ArcTanh[(Sqrt[c]
*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2))

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 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} \, dx &=\frac{B \left (b x+c x^2\right )^{3/2}}{2 c x}+\frac{\left (b B-A c+\frac{3}{2} (-b B+2 A c)\right ) \int \frac{\sqrt{b x+c x^2}}{x} \, dx}{2 c}\\ &=-\frac{(b B-4 A c) \sqrt{b x+c x^2}}{4 c}+\frac{B \left (b x+c x^2\right )^{3/2}}{2 c x}-\frac{(b (b B-4 A c)) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{8 c}\\ &=-\frac{(b B-4 A c) \sqrt{b x+c x^2}}{4 c}+\frac{B \left (b x+c x^2\right )^{3/2}}{2 c x}-\frac{(b (b B-4 A c)) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{4 c}\\ &=-\frac{(b B-4 A c) \sqrt{b x+c x^2}}{4 c}+\frac{B \left (b x+c x^2\right )^{3/2}}{2 c x}-\frac{b (b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.150339, size = 89, normalized size = 0.97 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} (4 A c+b B+2 B c x)-\frac{\sqrt{b} (b B-4 A c) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{4 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(b*B + 4*A*c + 2*B*c*x) - (Sqrt[b]*(b*B - 4*A*c)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]
])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(4*c^(3/2))

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Maple [A]  time = 0.009, size = 112, normalized size = 1.2 \begin{align*}{\frac{Bx}{2}\sqrt{c{x}^{2}+bx}}+{\frac{bB}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{{b}^{2}B}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{3}{2}}}}+A\sqrt{c{x}^{2}+bx}+{\frac{Ab}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){\frac{1}{\sqrt{c}}}} \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,x)

[Out]

1/2*B*x*(c*x^2+b*x)^(1/2)+1/4*B/c*(c*x^2+b*x)^(1/2)*b-1/8*B*b^2/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/
2))+A*(c*x^2+b*x)^(1/2)+1/2*A*b*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))/c^(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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.0023, size = 362, normalized size = 3.93 \begin{align*} \left [-\frac{{\left (B b^{2} - 4 \, A b c\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (2 \, B c^{2} x + B b c + 4 \, A c^{2}\right )} \sqrt{c x^{2} + b x}}{8 \, c^{2}}, \frac{{\left (B b^{2} - 4 \, A b c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (2 \, B c^{2} x + B b c + 4 \, A c^{2}\right )} \sqrt{c x^{2} + b x}}{4 \, c^{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,x, algorithm="fricas")

[Out]

[-1/8*((B*b^2 - 4*A*b*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(2*B*c^2*x + B*b*c + 4*A*c^2
)*sqrt(c*x^2 + b*x))/c^2, 1/4*((B*b^2 - 4*A*b*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (2*B*c^2*
x + B*b*c + 4*A*c^2)*sqrt(c*x^2 + b*x))/c^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}\, 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,x)

[Out]

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

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Giac [A]  time = 1.18574, size = 104, normalized size = 1.13 \begin{align*} \frac{1}{4} \, \sqrt{c x^{2} + b x}{\left (2 \, B x + \frac{B b + 4 \, A c}{c}\right )} + \frac{{\left (B b^{2} - 4 \, A b c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{8 \, c^{\frac{3}{2}}} \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,x, algorithm="giac")

[Out]

1/4*sqrt(c*x^2 + b*x)*(2*B*x + (B*b + 4*A*c)/c) + 1/8*(B*b^2 - 4*A*b*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b
*x))*sqrt(c) - b))/c^(3/2)