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

Optimal. Leaf size=56 \[ \frac{\sqrt{a+b x^2} (2 A+B x)}{2 b}-\frac{a B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0230157, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {780, 217, 206} \[ \frac{\sqrt{a+b x^2} (2 A+B x)}{2 b}-\frac{a B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x))/Sqrt[a + b*x^2],x]

[Out]

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

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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{x (A+B x)}{\sqrt{a+b x^2}} \, dx &=\frac{(2 A+B x) \sqrt{a+b x^2}}{2 b}-\frac{(a B) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{2 b}\\ &=\frac{(2 A+B x) \sqrt{a+b x^2}}{2 b}-\frac{(a B) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{2 b}\\ &=\frac{(2 A+B x) \sqrt{a+b x^2}}{2 b}-\frac{a B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0321117, size = 57, normalized size = 1.02 \[ \frac{\sqrt{b} \sqrt{a+b x^2} (2 A+B x)-a B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 55, normalized size = 1. \begin{align*}{\frac{Bx}{2\,b}\sqrt{b{x}^{2}+a}}-{\frac{Ba}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}}+{\frac{A}{b}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.79326, size = 275, normalized size = 4.91 \begin{align*} \left [\frac{B a \sqrt{b} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (B b x + 2 \, A b\right )} \sqrt{b x^{2} + a}}{4 \, b^{2}}, \frac{B a \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (B b x + 2 \, A b\right )} \sqrt{b x^{2} + a}}{2 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.2929, size = 70, normalized size = 1.25 \begin{align*} A \left (\begin{cases} \frac{x^{2}}{2 \sqrt{a}} & \text{for}\: b = 0 \\\frac{\sqrt{a + b x^{2}}}{b} & \text{otherwise} \end{cases}\right ) + \frac{B \sqrt{a} x \sqrt{1 + \frac{b x^{2}}{a}}}{2 b} - \frac{B a \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2 b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*Piecewise((x**2/(2*sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**2)/b, True)) + B*sqrt(a)*x*sqrt(1 + b*x**2/a)/(2*b) -
B*a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(3/2))

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Giac [A]  time = 1.18342, size = 68, normalized size = 1.21 \begin{align*} \frac{1}{2} \, \sqrt{b x^{2} + a}{\left (\frac{B x}{b} + \frac{2 \, A}{b}\right )} + \frac{B a \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{2 \, b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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