3.791 \(\int \frac{(a^2-b^2 x^2)^{3/2}}{a+b x} \, dx\)

Optimal. Leaf size=76 \[ \frac{1}{2} a x \sqrt{a^2-b^2 x^2}+\frac{\left (a^2-b^2 x^2\right )^{3/2}}{3 b}+\frac{a^3 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{2 b} \]

[Out]

(a*x*Sqrt[a^2 - b^2*x^2])/2 + (a^2 - b^2*x^2)^(3/2)/(3*b) + (a^3*ArcTan[(b*x)/Sqrt[a^2 - b^2*x^2]])/(2*b)

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Rubi [A]  time = 0.0221436, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {665, 195, 217, 203} \[ \frac{1}{2} a x \sqrt{a^2-b^2 x^2}+\frac{\left (a^2-b^2 x^2\right )^{3/2}}{3 b}+\frac{a^3 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 - b^2*x^2)^(3/2)/(a + b*x),x]

[Out]

(a*x*Sqrt[a^2 - b^2*x^2])/2 + (a^2 - b^2*x^2)^(3/2)/(3*b) + (a^3*ArcTan[(b*x)/Sqrt[a^2 - b^2*x^2]])/(2*b)

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] - Dist[(2*c*d*p)/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + 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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0784287, size = 90, normalized size = 1.18 \[ \frac{\sqrt{a^2-b^2 x^2} \left (\left (2 a^2+3 a b x-2 b^2 x^2\right ) \sqrt{1-\frac{b^2 x^2}{a^2}}+3 a^2 \sin ^{-1}\left (\frac{b x}{a}\right )\right )}{6 b \sqrt{1-\frac{b^2 x^2}{a^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 - b^2*x^2)^(3/2)/(a + b*x),x]

[Out]

(Sqrt[a^2 - b^2*x^2]*((2*a^2 + 3*a*b*x - 2*b^2*x^2)*Sqrt[1 - (b^2*x^2)/a^2] + 3*a^2*ArcSin[(b*x)/a]))/(6*b*Sqr
t[1 - (b^2*x^2)/a^2])

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Maple [A]  time = 0.047, size = 113, normalized size = 1.5 \begin{align*}{\frac{1}{3\,b} \left ( - \left ( x+{\frac{a}{b}} \right ) ^{2}{b}^{2}+2\, \left ( x+{\frac{a}{b}} \right ) ab \right ) ^{{\frac{3}{2}}}}+{\frac{ax}{2}\sqrt{- \left ( x+{\frac{a}{b}} \right ) ^{2}{b}^{2}+2\, \left ( x+{\frac{a}{b}} \right ) ab}}+{\frac{{a}^{3}}{2}\arctan \left ({x\sqrt{{b}^{2}}{\frac{1}{\sqrt{- \left ( x+{\frac{a}{b}} \right ) ^{2}{b}^{2}+2\, \left ( x+{\frac{a}{b}} \right ) ab}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b^2*x^2+a^2)^(3/2)/(b*x+a),x)

[Out]

1/3/b*(-(x+1/b*a)^2*b^2+2*(x+1/b*a)*a*b)^(3/2)+1/2*a*(-(x+1/b*a)^2*b^2+2*(x+1/b*a)*a*b)^(1/2)*x+1/2*a^3/(b^2)^
(1/2)*arctan((b^2)^(1/2)*x/(-(x+1/b*a)^2*b^2+2*(x+1/b*a)*a*b)^(1/2))

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Maxima [C]  time = 1.73481, size = 119, normalized size = 1.57 \begin{align*} -\frac{i \, a^{3} \arcsin \left (\frac{b x}{a} + 2\right )}{2 \, b} + \frac{1}{2} \, \sqrt{b^{2} x^{2} + 4 \, a b x + 3 \, a^{2}} a x + \frac{\sqrt{b^{2} x^{2} + 4 \, a b x + 3 \, a^{2}} a^{2}}{b} + \frac{{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac{3}{2}}}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(3/2)/(b*x+a),x, algorithm="maxima")

[Out]

-1/2*I*a^3*arcsin(b*x/a + 2)/b + 1/2*sqrt(b^2*x^2 + 4*a*b*x + 3*a^2)*a*x + sqrt(b^2*x^2 + 4*a*b*x + 3*a^2)*a^2
/b + 1/3*(-b^2*x^2 + a^2)^(3/2)/b

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Fricas [A]  time = 2.01181, size = 150, normalized size = 1.97 \begin{align*} -\frac{6 \, a^{3} \arctan \left (-\frac{a - \sqrt{-b^{2} x^{2} + a^{2}}}{b x}\right ) +{\left (2 \, b^{2} x^{2} - 3 \, a b x - 2 \, a^{2}\right )} \sqrt{-b^{2} x^{2} + a^{2}}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(3/2)/(b*x+a),x, algorithm="fricas")

[Out]

-1/6*(6*a^3*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) + (2*b^2*x^2 - 3*a*b*x - 2*a^2)*sqrt(-b^2*x^2 + a^2))/b

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Sympy [C]  time = 4.00543, size = 146, normalized size = 1.92 \begin{align*} a \left (\begin{cases} - \frac{i a^{2} \operatorname{acosh}{\left (\frac{b x}{a} \right )}}{2 b} - \frac{i a x}{2 \sqrt{-1 + \frac{b^{2} x^{2}}{a^{2}}}} + \frac{i b^{2} x^{3}}{2 a \sqrt{-1 + \frac{b^{2} x^{2}}{a^{2}}}} & \text{for}\: \frac{\left |{b^{2} x^{2}}\right |}{\left |{a^{2}}\right |} > 1 \\\frac{a^{2} \operatorname{asin}{\left (\frac{b x}{a} \right )}}{2 b} + \frac{a x \sqrt{1 - \frac{b^{2} x^{2}}{a^{2}}}}{2} & \text{otherwise} \end{cases}\right ) - b \left (\begin{cases} \frac{x^{2} \sqrt{a^{2}}}{2} & \text{for}\: b^{2} = 0 \\- \frac{\left (a^{2} - b^{2} x^{2}\right )^{\frac{3}{2}}}{3 b^{2}} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b**2*x**2+a**2)**(3/2)/(b*x+a),x)

[Out]

a*Piecewise((-I*a**2*acosh(b*x/a)/(2*b) - I*a*x/(2*sqrt(-1 + b**2*x**2/a**2)) + I*b**2*x**3/(2*a*sqrt(-1 + b**
2*x**2/a**2)), Abs(b**2*x**2)/Abs(a**2) > 1), (a**2*asin(b*x/a)/(2*b) + a*x*sqrt(1 - b**2*x**2/a**2)/2, True))
- b*Piecewise((x**2*sqrt(a**2)/2, Eq(b**2, 0)), (-(a**2 - b**2*x**2)**(3/2)/(3*b**2), True))

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Giac [A]  time = 1.23443, size = 76, normalized size = 1. \begin{align*} \frac{a^{3} \arcsin \left (\frac{b x}{a}\right ) \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right )}{2 \,{\left | b \right |}} - \frac{1}{6} \, \sqrt{-b^{2} x^{2} + a^{2}}{\left ({\left (2 \, b x - 3 \, a\right )} x - \frac{2 \, a^{2}}{b}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(3/2)/(b*x+a),x, algorithm="giac")

[Out]

1/2*a^3*arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) - 1/6*sqrt(-b^2*x^2 + a^2)*((2*b*x - 3*a)*x - 2*a^2/b)