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

Optimal. Leaf size=100 \[ \frac{3}{8} a^3 x \sqrt{a^2-b^2 x^2}+\frac{1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac{3 a^5 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{8 b} \]

[Out]

(3*a^3*x*Sqrt[a^2 - b^2*x^2])/8 + (a*x*(a^2 - b^2*x^2)^(3/2))/4 - (a^2 - b^2*x^2)^(5/2)/(5*b) + (3*a^5*ArcTan[
(b*x)/Sqrt[a^2 - b^2*x^2]])/(8*b)

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Rubi [A]  time = 0.0262414, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {641, 195, 217, 203} \[ \frac{3}{8} a^3 x \sqrt{a^2-b^2 x^2}+\frac{1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac{3 a^5 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^3*x*Sqrt[a^2 - b^2*x^2])/8 + (a*x*(a^2 - b^2*x^2)^(3/2))/4 - (a^2 - b^2*x^2)^(5/2)/(5*b) + (3*a^5*ArcTan[
(b*x)/Sqrt[a^2 - b^2*x^2]])/(8*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) \left (a^2-b^2 x^2\right )^{3/2} \, dx &=-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+a \int \left (a^2-b^2 x^2\right )^{3/2} \, dx\\ &=\frac{1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac{1}{4} \left (3 a^3\right ) \int \sqrt{a^2-b^2 x^2} \, dx\\ &=\frac{3}{8} a^3 x \sqrt{a^2-b^2 x^2}+\frac{1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac{1}{8} \left (3 a^5\right ) \int \frac{1}{\sqrt{a^2-b^2 x^2}} \, dx\\ &=\frac{3}{8} a^3 x \sqrt{a^2-b^2 x^2}+\frac{1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac{1}{8} \left (3 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{1+b^2 x^2} \, dx,x,\frac{x}{\sqrt{a^2-b^2 x^2}}\right )\\ &=\frac{3}{8} a^3 x \sqrt{a^2-b^2 x^2}+\frac{1}{4} a x \left (a^2-b^2 x^2\right )^{3/2}-\frac{\left (a^2-b^2 x^2\right )^{5/2}}{5 b}+\frac{3 a^5 \tan ^{-1}\left (\frac{b x}{\sqrt{a^2-b^2 x^2}}\right )}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.145797, size = 112, normalized size = 1.12 \[ \frac{\sqrt{a^2-b^2 x^2} \left (\sqrt{1-\frac{b^2 x^2}{a^2}} \left (16 a^2 b^2 x^2+25 a^3 b x-8 a^4-10 a b^3 x^3-8 b^4 x^4\right )+15 a^4 \sin ^{-1}\left (\frac{b x}{a}\right )\right )}{40 b \sqrt{1-\frac{b^2 x^2}{a^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 - b^2*x^2]*(Sqrt[1 - (b^2*x^2)/a^2]*(-8*a^4 + 25*a^3*b*x + 16*a^2*b^2*x^2 - 10*a*b^3*x^3 - 8*b^4*x^4
) + 15*a^4*ArcSin[(b*x)/a]))/(40*b*Sqrt[1 - (b^2*x^2)/a^2])

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Maple [A]  time = 0.047, size = 91, normalized size = 0.9 \begin{align*} -{\frac{1}{5\,b} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{ax}{4} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{3\,x{a}^{3}}{8}\sqrt{-{b}^{2}{x}^{2}+{a}^{2}}}+{\frac{3\,{a}^{5}}{8}\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)^(3/2),x)

[Out]

-1/5*(-b^2*x^2+a^2)^(5/2)/b+1/4*a*x*(-b^2*x^2+a^2)^(3/2)+3/8*a^3*x*(-b^2*x^2+a^2)^(1/2)+3/8*a^5/(b^2)^(1/2)*ar
ctan((b^2)^(1/2)*x/(-b^2*x^2+a^2)^(1/2))

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Maxima [A]  time = 1.68208, size = 112, normalized size = 1.12 \begin{align*} \frac{3 \, a^{5} \arcsin \left (\frac{b^{2} x}{\sqrt{a^{2} b^{2}}}\right )}{8 \, \sqrt{b^{2}}} + \frac{3}{8} \, \sqrt{-b^{2} x^{2} + a^{2}} a^{3} x + \frac{1}{4} \,{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac{3}{2}} a x - \frac{{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac{5}{2}}}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(-b^2*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

3/8*a^5*arcsin(b^2*x/sqrt(a^2*b^2))/sqrt(b^2) + 3/8*sqrt(-b^2*x^2 + a^2)*a^3*x + 1/4*(-b^2*x^2 + a^2)^(3/2)*a*
x - 1/5*(-b^2*x^2 + a^2)^(5/2)/b

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Fricas [A]  time = 2.11073, size = 200, normalized size = 2. \begin{align*} -\frac{30 \, a^{5} \arctan \left (-\frac{a - \sqrt{-b^{2} x^{2} + a^{2}}}{b x}\right ) +{\left (8 \, b^{4} x^{4} + 10 \, a b^{3} x^{3} - 16 \, a^{2} b^{2} x^{2} - 25 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt{-b^{2} x^{2} + a^{2}}}{40 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(-b^2*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

-1/40*(30*a^5*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) + (8*b^4*x^4 + 10*a*b^3*x^3 - 16*a^2*b^2*x^2 - 25*a^3*
b*x + 8*a^4)*sqrt(-b^2*x^2 + a^2))/b

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Sympy [C]  time = 7.83493, size = 439, normalized size = 4.39 \begin{align*} a^{3} \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 ) + a^{2} 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 ) - a b^{2} \left (\begin{cases} - \frac{i a^{4} \operatorname{acosh}{\left (\frac{b x}{a} \right )}}{8 b^{3}} + \frac{i a^{3} x}{8 b^{2} \sqrt{-1 + \frac{b^{2} x^{2}}{a^{2}}}} - \frac{3 i a x^{3}}{8 \sqrt{-1 + \frac{b^{2} x^{2}}{a^{2}}}} + \frac{i b^{2} x^{5}}{4 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^{4} \operatorname{asin}{\left (\frac{b x}{a} \right )}}{8 b^{3}} - \frac{a^{3} x}{8 b^{2} \sqrt{1 - \frac{b^{2} x^{2}}{a^{2}}}} + \frac{3 a x^{3}}{8 \sqrt{1 - \frac{b^{2} x^{2}}{a^{2}}}} - \frac{b^{2} x^{5}}{4 a \sqrt{1 - \frac{b^{2} x^{2}}{a^{2}}}} & \text{otherwise} \end{cases}\right ) - b^{3} \left (\begin{cases} - \frac{2 a^{4} \sqrt{a^{2} - b^{2} x^{2}}}{15 b^{4}} - \frac{a^{2} x^{2} \sqrt{a^{2} - b^{2} x^{2}}}{15 b^{2}} + \frac{x^{4} \sqrt{a^{2} - b^{2} x^{2}}}{5} & \text{for}\: b \neq 0 \\\frac{x^{4} \sqrt{a^{2}}}{4} & \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)**(3/2),x)

[Out]

a**3*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, Tru
e)) + a**2*b*Piecewise((x**2*sqrt(a**2)/2, Eq(b**2, 0)), (-(a**2 - b**2*x**2)**(3/2)/(3*b**2), True)) - a*b**2
*Piecewise((-I*a**4*acosh(b*x/a)/(8*b**3) + I*a**3*x/(8*b**2*sqrt(-1 + b**2*x**2/a**2)) - 3*I*a*x**3/(8*sqrt(-
1 + b**2*x**2/a**2)) + I*b**2*x**5/(4*a*sqrt(-1 + b**2*x**2/a**2)), Abs(b**2*x**2)/Abs(a**2) > 1), (a**4*asin(
b*x/a)/(8*b**3) - a**3*x/(8*b**2*sqrt(1 - b**2*x**2/a**2)) + 3*a*x**3/(8*sqrt(1 - b**2*x**2/a**2)) - b**2*x**5
/(4*a*sqrt(1 - b**2*x**2/a**2)), True)) - b**3*Piecewise((-2*a**4*sqrt(a**2 - b**2*x**2)/(15*b**4) - a**2*x**2
*sqrt(a**2 - b**2*x**2)/(15*b**2) + x**4*sqrt(a**2 - b**2*x**2)/5, Ne(b, 0)), (x**4*sqrt(a**2)/4, True))

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Giac [A]  time = 1.2365, size = 109, normalized size = 1.09 \begin{align*} \frac{3 \, a^{5} \arcsin \left (\frac{b x}{a}\right ) \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right )}{8 \,{\left | b \right |}} - \frac{1}{40} \, \sqrt{-b^{2} x^{2} + a^{2}}{\left (\frac{8 \, a^{4}}{b} -{\left (25 \, a^{3} + 2 \,{\left (8 \, a^{2} b -{\left (4 \, b^{3} x + 5 \, a b^{2}\right )} x\right )} x\right )} x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(-b^2*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

3/8*a^5*arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) - 1/40*sqrt(-b^2*x^2 + a^2)*(8*a^4/b - (25*a^3 + 2*(8*a^2*b - (4*b^
3*x + 5*a*b^2)*x)*x)*x)