3.7.23 \(\int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^2} \, dx\)

Optimal. Leaf size=54 \[ -\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}-\frac {\tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b} \]

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Rubi [A]  time = 0.01, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {663, 217, 203} \begin {gather*} -\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}-\frac {\tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 663

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1
, 0] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 51, normalized size = 0.94 \begin {gather*} -\frac {\frac {2 \sqrt {a^2-b^2 x^2}}{a+b x}+\tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.35, size = 73, normalized size = 1.35 \begin {gather*} -\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}-\frac {\sqrt {-b^2} \log \left (\sqrt {a^2-b^2 x^2}-\sqrt {-b^2} x\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 66, normalized size = 1.22 \begin {gather*} -\frac {2 \, {\left (b x - {\left (b x + a\right )} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) + a + \sqrt {-b^{2} x^{2} + a^{2}}\right )}}{b^{2} x + a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: abs(b)*(-(2*atan(i)-2*i)/b^2*sign((b*x+a
)^-1)*sign(b)-2*a*(sqrt(2*a*b*(b*x+a)^-1/b-1)*sign((b*x+a)^-1)*sign(b)-sign((b*x+a)^-1)*sign(b)*atan(sqrt(2*a*
b*(b*x+a)^-1/b-1)))/b^2/a)

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maple [B]  time = 0.05, size = 126, normalized size = 2.33 \begin {gather*} -\frac {\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 )}{\sqrt {b^{2}}}-\frac {\sqrt {2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}}}{a b}-\frac {\left (2 \left (x +\frac {a}{b}\right ) a b -\left (x +\frac {a}{b}\right )^{2} b^{2}\right )^{\frac {3}{2}}}{\left (x +\frac {a}{b}\right )^{2} a \,b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.95, size = 40, normalized size = 0.74 \begin {gather*} -\frac {\arcsin \left (\frac {b x}{a}\right )}{b} - \frac {2 \, \sqrt {-b^{2} x^{2} + a^{2}}}{b^{2} x + a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2+a^2)^(1/2)/(b*x+a)^2,x, algorithm="maxima")

[Out]

-arcsin(b*x/a)/b - 2*sqrt(-b^2*x^2 + a^2)/(b^2*x + a*b)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {a^2-b^2\,x^2}}{{\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- \left (- a + b x\right ) \left (a + b x\right )}}{\left (a + b x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-(-a + b*x)*(a + b*x))/(a + b*x)**2, x)

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