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

Optimal. Leaf size=107 \[ \frac {5}{8} a^2 x \sqrt {a^2-b^2 x^2}-\frac {5 a \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{4 b}+\frac {5 a^4 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{8 b} \]

[Out]

-5/12*a*(-b^2*x^2+a^2)^(3/2)/b-1/4*(b*x+a)*(-b^2*x^2+a^2)^(3/2)/b+5/8*a^4*arctan(b*x/(-b^2*x^2+a^2)^(1/2))/b+5
/8*a^2*x*(-b^2*x^2+a^2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {671, 641, 195, 217, 203} \[ \frac {5}{8} a^2 x \sqrt {a^2-b^2 x^2}-\frac {5 a \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{4 b}+\frac {5 a^4 \tan ^{-1}\left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^2*x*Sqrt[a^2 - b^2*x^2])/8 - (5*a*(a^2 - b^2*x^2)^(3/2))/(12*b) - ((a + b*x)*(a^2 - b^2*x^2)^(3/2))/(4*b)
+ (5*a^4*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]

Rule 671

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[(2*c*d*(m + p))/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rubi steps

\begin {align*} \int (a+b x)^2 \sqrt {a^2-b^2 x^2} \, dx &=-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{4 b}+\frac {1}{4} (5 a) \int (a+b x) \sqrt {a^2-b^2 x^2} \, dx\\ &=-\frac {5 a \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{4 b}+\frac {1}{4} \left (5 a^2\right ) \int \sqrt {a^2-b^2 x^2} \, dx\\ &=\frac {5}{8} a^2 x \sqrt {a^2-b^2 x^2}-\frac {5 a \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{4 b}+\frac {1}{8} \left (5 a^4\right ) \int \frac {1}{\sqrt {a^2-b^2 x^2}} \, dx\\ &=\frac {5}{8} a^2 x \sqrt {a^2-b^2 x^2}-\frac {5 a \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{4 b}+\frac {1}{8} \left (5 a^4\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 {5}{8} a^2 x \sqrt {a^2-b^2 x^2}-\frac {5 a \left (a^2-b^2 x^2\right )^{3/2}}{12 b}-\frac {(a+b x) \left (a^2-b^2 x^2\right )^{3/2}}{4 b}+\frac {5 a^4 \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 = 0.18, size = 101, normalized size = 0.94 \[ \frac {\sqrt {a^2-b^2 x^2} \left (15 a^3 \sin ^{-1}\left (\frac {b x}{a}\right )+\sqrt {1-\frac {b^2 x^2}{a^2}} \left (-16 a^3+9 a^2 b x+16 a b^2 x^2+6 b^3 x^3\right )\right )}{24 b \sqrt {1-\frac {b^2 x^2}{a^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 - b^2*x^2]*(Sqrt[1 - (b^2*x^2)/a^2]*(-16*a^3 + 9*a^2*b*x + 16*a*b^2*x^2 + 6*b^3*x^3) + 15*a^3*ArcSin
[(b*x)/a]))/(24*b*Sqrt[1 - (b^2*x^2)/a^2])

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fricas [A]  time = 0.93, size = 84, normalized size = 0.79 \[ -\frac {30 \, a^{4} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) - {\left (6 \, b^{3} x^{3} + 16 \, a b^{2} x^{2} + 9 \, a^{2} b x - 16 \, a^{3}\right )} \sqrt {-b^{2} x^{2} + a^{2}}}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(30*a^4*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) - (6*b^3*x^3 + 16*a*b^2*x^2 + 9*a^2*b*x - 16*a^3)*sqrt
(-b^2*x^2 + a^2))/b

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giac [A]  time = 0.20, size = 69, normalized size = 0.64 \[ \frac {5 \, a^{4} \arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (b)}{8 \, {\left | b \right |}} - \frac {1}{24} \, \sqrt {-b^{2} x^{2} + a^{2}} {\left (\frac {16 \, a^{3}}{b} - {\left (9 \, a^{2} + 2 \, {\left (3 \, b^{2} x + 8 \, a b\right )} x\right )} x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*a^4*arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) - 1/24*sqrt(-b^2*x^2 + a^2)*(16*a^3/b - (9*a^2 + 2*(3*b^2*x + 8*a*b
)*x)*x)

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maple [A]  time = 0.05, size = 91, normalized size = 0.85 \[ \frac {5 a^{4} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right )}{8 \sqrt {b^{2}}}+\frac {5 \sqrt {-b^{2} x^{2}+a^{2}}\, a^{2} x}{8}-\frac {\left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}} x}{4}-\frac {2 \left (-b^{2} x^{2}+a^{2}\right )^{\frac {3}{2}} a}{3 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.80, size = 73, normalized size = 0.68 \[ \frac {5 \, a^{4} \arcsin \left (\frac {b x}{a}\right )}{8 \, b} + \frac {5}{8} \, \sqrt {-b^{2} x^{2} + a^{2}} a^{2} x - \frac {1}{4} \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} x - \frac {2 \, {\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} a}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {a^2-b^2\,x^2}\,{\left (a+b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 7.73, size = 350, normalized size = 3.27 \[ a^{2} \left (\begin {cases} - \frac {i a^{2} \operatorname {acosh}{\left (\frac {b x}{a} \right )}}{2 b} - \frac {i a x}{2 \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} + \frac {i b^{2} x^{3}}{2 a \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} & \text {for}\: \left |{\frac {b^{2} x^{2}}{a^{2}}}\right | > 1 \\\frac {a^{2} \operatorname {asin}{\left (\frac {b x}{a} \right )}}{2 b} + \frac {a x \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}}{2} & \text {otherwise} \end {cases}\right ) + 2 a b \left (\begin {cases} \frac {x^{2} \sqrt {a^{2}}}{2} & \text {for}\: b^{2} = 0 \\- \frac {\left (a^{2} - b^{2} x^{2}\right )^{\frac {3}{2}}}{3 b^{2}} & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} - \frac {i a^{4} \operatorname {acosh}{\left (\frac {b x}{a} \right )}}{8 b^{3}} + \frac {i a^{3} x}{8 b^{2} \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} - \frac {3 i a x^{3}}{8 \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} + \frac {i b^{2} x^{5}}{4 a \sqrt {-1 + \frac {b^{2} x^{2}}{a^{2}}}} & \text {for}\: \left |{\frac {b^{2} x^{2}}{a^{2}}}\right | > 1 \\\frac {a^{4} \operatorname {asin}{\left (\frac {b x}{a} \right )}}{8 b^{3}} - \frac {a^{3} x}{8 b^{2} \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} + \frac {3 a x^{3}}{8 \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} - \frac {b^{2} x^{5}}{4 a \sqrt {1 - \frac {b^{2} x^{2}}{a^{2}}}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*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)) +
2*a*b*Piecewise((x**2*sqrt(a**2)/2, Eq(b**2, 0)), (-(a**2 - b**2*x**2)**(3/2)/(3*b**2), True)) + b**2*Piecewi
se((-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))

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