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

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 100, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.14, size = 112, normalized size = 1.12 \[ \frac {\sqrt {a^2-b^2 x^2} \left (15 a^4 \sin ^{-1}\left (\frac {b x}{a}\right )+\sqrt {1-\frac {b^2 x^2}{a^2}} \left (-8 a^4+25 a^3 b x+16 a^2 b^2 x^2-10 a b^3 x^3-8 b^4 x^4\right )\right )}{40 b \sqrt {1-\frac {b^2 x^2}{a^2}}} \]

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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fricas [A]  time = 0.73, size = 94, normalized size = 0.94 \[ -\frac {30 \, a^{5} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) + {\left (8 \, b^{4} x^{4} + 10 \, a b^{3} x^{3} - 16 \, a^{2} b^{2} x^{2} - 25 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{40 \, b} \]

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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giac [A]  time = 0.21, size = 81, normalized size = 0.81 \[ \frac {3 \, a^{5} \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (b)}{8 \, {\left | b \right |}} - \frac {1}{40} \, \sqrt {-b^{2} x^{2} + a^{2}} {\left (\frac {8 \, a^{4}}{b} - {\left (25 \, a^{3} + 2 \, {\left (8 \, a^{2} b - {\left (4 \, b^{3} x + 5 \, a b^{2}\right )} x\right )} x\right )} x\right )} \]

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)

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maple [A]  time = 0.06, size = 91, normalized size = 0.91 \[ \frac {3 a^{5} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{8 \sqrt {b^{2}}}+\frac {3 \sqrt {-b^{2} x^{2}+a^{2}}\, a^{3} x}{8}+\frac {\left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}} a x}{4}-\frac {\left (-b^{2} x^{2}+a^{2}\right )^{\frac {5}{2}}}{5 b} \]

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

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maxima [A]  time = 2.98, size = 73, normalized size = 0.73 \[ \frac {3 \, a^{5} \arcsin \left (\frac {b x}{a}\right )}{8 \, b} + \frac {3}{8} \, \sqrt {-b^{2} x^{2} + a^{2}} a^{3} x + \frac {1}{4} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a x - \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {5}{2}}}{5 \, b} \]

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*x/a)/b + 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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mupad [B]  time = 0.73, size = 67, normalized size = 0.67 \[ \frac {a\,x\,{\left (a^2-b^2\,x^2\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ \frac {b^2\,x^2}{a^2}\right )}{{\left (1-\frac {b^2\,x^2}{a^2}\right )}^{3/2}}-\frac {{\left (a^2-b^2\,x^2\right )}^{5/2}}{5\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 - b^2*x^2)^(3/2)*(a + b*x),x)

[Out]

(a*x*(a^2 - b^2*x^2)^(3/2)*hypergeom([-3/2, 1/2], 3/2, (b^2*x^2)/a^2))/(1 - (b^2*x^2)/a^2)^(3/2) - (a^2 - b^2*
x^2)^(5/2)/(5*b)

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sympy [C]  time = 9.17, size = 435, normalized size = 4.35 \[ a^{3} \left (\begin {cases} - \frac {i a^{2} \operatorname {acosh}{\left (\frac {b x}{a} \right )}}{2 b} - \frac {i a x}{2 \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} + \frac {i b^{2} x^{3}}{2 a \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} & \text {for}\: \left |{\frac {b^{2} x^{2}}{a^{2}}}\right | > 1 \\\frac {a^{2} \operatorname {asin}{\left (\frac {b x}{a} \right )}}{2 b} + \frac {a x \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}}{2} & \text {otherwise} \end {cases}\right ) + a^{2} b \left (\begin {cases} \frac {x^{2} \sqrt {a^{2}}}{2} & \text {for}\: b^{2} = 0 \\- \frac {\left (a^{2} - b^{2} x^{2}\right )^{\frac {3}{2}}}{3 b^{2}} & \text {otherwise} \end {cases}\right ) - a b^{2} \left (\begin {cases} - \frac {i a^{4} \operatorname {acosh}{\left (\frac {b x}{a} \right )}}{8 b^{3}} + \frac {i a^{3} x}{8 b^{2} \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} - \frac {3 i a x^{3}}{8 \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} + \frac {i b^{2} x^{5}}{4 a \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} & \text {for}\: \left |{\frac {b^{2} x^{2}}{a^{2}}}\right | > 1 \\\frac {a^{4} \operatorname {asin}{\left (\frac {b x}{a} \right )}}{8 b^{3}} - \frac {a^{3} x}{8 b^{2} \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} + \frac {3 a x^{3}}{8 \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} - \frac {b^{2} x^{5}}{4 a \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} & \text {otherwise} \end {cases}\right ) - b^{3} \left (\begin {cases} - \frac {2 a^{4} \sqrt {a^{2} - b^{2} x^{2}}}{15 b^{4}} - \frac {a^{2} x^{2} \sqrt {a^{2} - b^{2} x^{2}}}{15 b^{2}} + \frac {x^{4} \sqrt {a^{2} - b^{2} x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {x^{4} \sqrt {a^{2}}}{4} & \text {otherwise} \end {cases}\right ) \]

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/a**2) > 1), (a**2*asin(b*x/a)/(2*b) + a*x*sqrt(1 - b**2*x**2/a**2)/2, True)) +
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*Piec
ewise((-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/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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