3.6.8 \(\int \frac {(a+b x)^2}{\sqrt {1-x^2}} \, dx\)

Optimal. Leaf size=54 \[ \frac {1}{2} \left (2 a^2+b^2\right ) \sin ^{-1}(x)-\frac {3}{2} a b \sqrt {1-x^2}-\frac {1}{2} b \sqrt {1-x^2} (a+b x) \]

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Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {743, 641, 216} \begin {gather*} \frac {1}{2} \left (2 a^2+b^2\right ) \sin ^{-1}(x)-\frac {3}{2} a b \sqrt {1-x^2}-\frac {1}{2} b \sqrt {1-x^2} (a+b x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/Sqrt[1 - x^2],x]

[Out]

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

Rule 216

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

Rule 641

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

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 38, normalized size = 0.70 \begin {gather*} \frac {1}{2} \left (\left (2 a^2+b^2\right ) \sin ^{-1}(x)-b \sqrt {1-x^2} (4 a+b x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/Sqrt[1 - x^2],x]

[Out]

(-(b*(4*a + b*x)*Sqrt[1 - x^2]) + (2*a^2 + b^2)*ArcSin[x])/2

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IntegrateAlgebraic [A]  time = 0.35, size = 55, normalized size = 1.02 \begin {gather*} \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {x}{\sqrt {1-x^2}-1}\right )+\frac {1}{2} \sqrt {1-x^2} \left (b^2 (-x)-4 a b\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^2/Sqrt[1 - x^2],x]

[Out]

((-4*a*b - b^2*x)*Sqrt[1 - x^2])/2 + (2*a^2 + b^2)*ArcTan[x/(-1 + Sqrt[1 - x^2])]

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fricas [A]  time = 0.41, size = 49, normalized size = 0.91 \begin {gather*} -{\left (2 \, a^{2} + b^{2}\right )} \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) - \frac {1}{2} \, {\left (b^{2} x + 4 \, a b\right )} \sqrt {-x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^2 + b^2)*arctan((sqrt(-x^2 + 1) - 1)/x) - 1/2*(b^2*x + 4*a*b)*sqrt(-x^2 + 1)

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giac [A]  time = 0.21, size = 35, normalized size = 0.65 \begin {gather*} \frac {1}{2} \, {\left (2 \, a^{2} + b^{2}\right )} \arcsin \relax (x) - \frac {1}{2} \, {\left (b^{2} x + 4 \, a b\right )} \sqrt {-x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 42, normalized size = 0.78 \begin {gather*} a^{2} \arcsin \relax (x )-2 \sqrt {-x^{2}+1}\, a b +\left (-\frac {\sqrt {-x^{2}+1}\, x}{2}+\frac {\arcsin \relax (x )}{2}\right ) b^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.94, size = 42, normalized size = 0.78 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} + 1} b^{2} x + a^{2} \arcsin \relax (x) + \frac {1}{2} \, b^{2} \arcsin \relax (x) - 2 \, \sqrt {-x^{2} + 1} a b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 35, normalized size = 0.65 \begin {gather*} \mathrm {asin}\relax (x)\,\left (a^2+\frac {b^2}{2}\right )-\left (\frac {x\,b^2}{2}+2\,a\,b\right )\,\sqrt {1-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/(1 - x^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 0.25, size = 42, normalized size = 0.78 \begin {gather*} a^{2} \operatorname {asin}{\relax (x )} - 2 a b \sqrt {1 - x^{2}} - \frac {b^{2} x \sqrt {1 - x^{2}}}{2} + \frac {b^{2} \operatorname {asin}{\relax (x )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*asin(x) - 2*a*b*sqrt(1 - x**2) - b**2*x*sqrt(1 - x**2)/2 + b**2*asin(x)/2

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