3.8.81 \(\int \frac {\sqrt {a^2-b^2 x^2}}{a+b x} \, dx\) [781]

Optimal. Leaf size=46 \[ \frac {\sqrt {a^2-b^2 x^2}}{b}+\frac {a \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 46, 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 = {679, 223, 209} \begin {gather*} \frac {a \text {ArcTan}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b}+\frac {\sqrt {a^2-b^2 x^2}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 - b^2*x^2]/(a + b*x),x]

[Out]

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

Rule 209

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

Rule 223

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 679

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

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Mathematica [A]
time = 0.11, size = 63, normalized size = 1.37 \begin {gather*} \frac {\sqrt {a^2-b^2 x^2}}{b}-\frac {a \log \left (-\sqrt {-b^2} x+\sqrt {a^2-b^2 x^2}\right )}{\sqrt {-b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.47, size = 78, normalized size = 1.70

method	result	size
risch	\(\frac {\sqrt {-b^{2} x^{2}+a^{2}}}{b}+\frac {a \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{\sqrt {b^{2}}}\)	\(49\)
default	\(\frac {\sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}+\frac {a b \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}\right )}{\sqrt {b^{2}}}}{b}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 31, normalized size = 0.67 \begin {gather*} \frac {a \arcsin \left (\frac {b x}{a}\right )}{b} + \frac {\sqrt {-b^{2} x^{2} + a^{2}}}{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),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.60, size = 52, normalized size = 1.13 \begin {gather*} -\frac {2 \, a \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) - \sqrt {-b^{2} x^{2} + a^{2}}}{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),x, algorithm="fricas")

[Out]

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

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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 )}}{a + b x}\, 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),x)

[Out]

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

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Giac [A]
time = 1.38, size = 36, normalized size = 0.78 \begin {gather*} \frac {a \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (b\right )}{{\left | b \right |}} + \frac {\sqrt {-b^{2} x^{2} + a^{2}}}{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),x, algorithm="giac")

[Out]

a*arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) + sqrt(-b^2*x^2 + a^2)/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}}{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)^(1/2)/(a + b*x),x)

[Out]

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

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