### 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*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)

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Rubi [A]  time = 0.0339475, antiderivative size = 107, normalized size of antiderivative = 1., 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 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
]

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)^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*}

Mathematica [A]  time = 0.179711, size = 101, normalized size = 0.94 $\frac{\sqrt{a^2-b^2 x^2} \left (\sqrt{1-\frac{b^2 x^2}{a^2}} \left (9 a^2 b x-16 a^3+16 a b^2 x^2+6 b^3 x^3\right )+15 a^3 \sin ^{-1}\left (\frac{b x}{a}\right )\right )}{24 b \sqrt{1-\frac{b^2 x^2}{a^2}}}$

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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Maple [A]  time = 0.048, size = 91, normalized size = 0.9 \begin{align*} -{\frac{x}{4} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{a}^{2}x}{8}\sqrt{-{b}^{2}{x}^{2}+{a}^{2}}}+{\frac{5\,{a}^{4}}{8}\arctan \left ({x\sqrt{{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}+{a}^{2}}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{2\,a}{3\,b} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

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

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Maxima [A]  time = 1.67389, size = 112, normalized size = 1.05 \begin{align*} \frac{5 \, a^{4} \arcsin \left (\frac{b^{2} x}{\sqrt{a^{2} b^{2}}}\right )}{8 \, \sqrt{b^{2}}} + \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} \end{align*}

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^2*x/sqrt(a^2*b^2))/sqrt(b^2) + 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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Fricas [A]  time = 1.82214, size = 177, normalized size = 1.65 \begin{align*} -\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} \end{align*}

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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Sympy [C]  time = 6.28458, size = 354, normalized size = 3.31 \begin{align*} 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}\: \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 ) + 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}\: \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 ) \end{align*}

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)/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)) + 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*Pi
ecewise((-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))

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Giac [A]  time = 1.24354, size = 93, normalized size = 0.87 \begin{align*} \frac{5 \, a^{4} \arcsin \left (\frac{b x}{a}\right ) \mathrm{sgn}\left (a\right ) \mathrm{sgn}\left (b\right )}{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 )} \end{align*}

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)