3.2468 \(\int \frac{A+B x}{\sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=67 \[ \frac{B \sqrt{a+b x+c x^2}}{c}-\frac{(b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 c^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0235932, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {640, 621, 206} \[ \frac{B \sqrt{a+b x+c x^2}}{c}-\frac{(b B-2 A c) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 640

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

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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

Mathematica [A]  time = 0.0963428, size = 66, normalized size = 0.99 \[ \frac{(2 A c-b B) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{2 c^{3/2}}+\frac{B \sqrt{a+x (b+c x)}}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0., size = 81, normalized size = 1.2 \begin{align*}{\frac{B}{c}\sqrt{c{x}^{2}+bx+a}}-{\frac{bB}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{A\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \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+a)^(1/2),x)

[Out]

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

________________________________________________________________________________________

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+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.38225, size = 400, normalized size = 5.97 \begin{align*} \left [\frac{4 \, \sqrt{c x^{2} + b x + a} B c -{\left (B b - 2 \, A c\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right )}{4 \, c^{2}}, \frac{2 \, \sqrt{c x^{2} + b x + a} B c +{\left (B b - 2 \, A c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right )}{2 \, 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+a)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(4*sqrt(c*x^2 + b*x + a)*B*c - (B*b - 2*A*c)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x
+ a)*(2*c*x + b)*sqrt(c) - 4*a*c))/c^2, 1/2*(2*sqrt(c*x^2 + b*x + a)*B*c + (B*b - 2*A*c)*sqrt(-c)*arctan(1/2*s
qrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)))/c^2]

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.16049, size = 84, normalized size = 1.25 \begin{align*} \frac{\sqrt{c x^{2} + b x + a} B}{c} + \frac{{\left (B b - 2 \, A c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{2 \, 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+a)^(1/2),x, algorithm="giac")

[Out]

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