### 3.780 $$\int (a+b x) \sqrt{a^2-b^2 x^2} \, dx$$

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 = 0.0201091, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {641, 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 + b*x)*Sqrt[a^2 - b^2*x^2],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 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 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 (a+b x) \sqrt{a^2-b^2 x^2} \, 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 = 0.0765633, 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 + b*x)*Sqrt[a^2 - b^2*x^2],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*Sq
rt[1 - (b^2*x^2)/a^2])

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Maple [A]  time = 0.048, size = 71, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,b} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{ax}{2}\sqrt{-{b}^{2}{x}^{2}+{a}^{2}}}+{\frac{{a}^{3}}{2}\arctan \left ({x\sqrt{{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}+{a}^{2}}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.7378, size = 85, normalized size = 1.12 \begin{align*} \frac{a^{3} \arcsin \left (\frac{b^{2} x}{\sqrt{a^{2} b^{2}}}\right )}{2 \, \sqrt{b^{2}}} + \frac{1}{2} \, \sqrt{-b^{2} x^{2} + a^{2}} a x - \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*x+a)*(-b^2*x^2+a^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.81042, 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*x+a)*(-b^2*x^2+a^2)^(1/2),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 = 3.34282, 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*x+a)*(-b**2*x**2+a**2)**(1/2),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 = 1.20121, 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*x+a)*(-b^2*x^2+a^2)^(1/2),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)