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

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

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Rubi [A]  time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 {gather*}

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 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.08, size = 90, normalized size = 1.18 \begin {gather*} \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}}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.26, size = 92, normalized size = 1.21 \begin {gather*} \frac {\left (2 a^2+3 a b x-2 b^2 x^2\right ) \sqrt {a^2-b^2 x^2}}{6 b}+\frac {a^3 \sqrt {-b^2} \log \left (\sqrt {a^2-b^2 x^2}-\sqrt {-b^2} x\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2 + 3*a*b*x - 2*b^2*x^2)*Sqrt[a^2 - b^2*x^2])/(6*b) + (a^3*Sqrt[-b^2]*Log[-(Sqrt[-b^2]*x) + Sqrt[a^2 - b
^2*x^2]])/(2*b^2)

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fricas [A]  time = 0.41, size = 72, normalized size = 0.95 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.21, size = 56, normalized size = 0.74 \begin {gather*} \frac {a^{3} \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (b)}{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 {gather*}

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)

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maple [A]  time = 0.05, size = 113, normalized size = 1.49 \begin {gather*} \frac {a^{3} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}}}\right )}{2 \sqrt {b^{2}}}+\frac {\sqrt {2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}}\, a x}{2}+\frac {\left (2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}\right )^{\frac {3}{2}}}{3 b} \end {gather*}

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

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maxima [C]  time = 3.00, size = 88, normalized size = 1.16 \begin {gather*} -\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 {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a^2-b^2\,x^2\right )}^{3/2}}{a+b\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 4.41, size = 144, normalized size = 1.89 \begin {gather*} 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}\: \left |{\frac {b^{2} x^{2}}{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 {gather*}

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/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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