3.3 \(\int (A+B x+C x^2) \sqrt {d^2-e^2 x^2} \, dx\)

Optimal. Leaf size=125 \[ \frac {1}{8} x \sqrt {d^2-e^2 x^2} \left (4 A+\frac {C d^2}{e^2}\right )+\frac {d^2 \left (4 A e^2+C d^2\right ) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}-\frac {B \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2} \]

[Out]

-1/3*B*(-e^2*x^2+d^2)^(3/2)/e^2-1/4*C*x*(-e^2*x^2+d^2)^(3/2)/e^2+1/8*d^2*(4*A*e^2+C*d^2)*arctan(e*x/(-e^2*x^2+
d^2)^(1/2))/e^3+1/8*(4*A+C*d^2/e^2)*x*(-e^2*x^2+d^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1815, 641, 195, 217, 203} \[ \frac {1}{8} x \sqrt {d^2-e^2 x^2} \left (4 A+\frac {C d^2}{e^2}\right )+\frac {d^2 \left (4 A e^2+C d^2\right ) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}-\frac {B \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)*Sqrt[d^2 - e^2*x^2],x]

[Out]

((4*A + (C*d^2)/e^2)*x*Sqrt[d^2 - e^2*x^2])/8 - (B*(d^2 - e^2*x^2)^(3/2))/(3*e^2) - (C*x*(d^2 - e^2*x^2)^(3/2)
)/(4*e^2) + (d^2*(C*d^2 + 4*A*e^2)*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

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 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2} \, dx &=-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac {\int \left (-C d^2-4 A e^2-4 B e^2 x\right ) \sqrt {d^2-e^2 x^2} \, dx}{4 e^2}\\ &=-\frac {B \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac {\left (-C d^2-4 A e^2\right ) \int \sqrt {d^2-e^2 x^2} \, dx}{4 e^2}\\ &=\frac {\left (C d^2+4 A e^2\right ) x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {B \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac {\left (d^2 \left (-C d^2-4 A e^2\right )\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=\frac {\left (C d^2+4 A e^2\right ) x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {B \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac {\left (d^2 \left (-C d^2-4 A e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^2}\\ &=\frac {\left (C d^2+4 A e^2\right ) x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {B \left (d^2-e^2 x^2\right )^{3/2}}{3 e^2}-\frac {C x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}+\frac {d^2 \left (C d^2+4 A e^2\right ) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 121, normalized size = 0.97 \[ \frac {\sqrt {d^2-e^2 x^2} \left (e \sqrt {1-\frac {e^2 x^2}{d^2}} \left (12 A e^2 x-8 B d^2+8 B e^2 x^2-3 C d^2 x+6 C e^2 x^3\right )+3 \left (4 A d e^2+C d^3\right ) \sin ^{-1}\left (\frac {e x}{d}\right )\right )}{24 e^3 \sqrt {1-\frac {e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x + C*x^2)*Sqrt[d^2 - e^2*x^2],x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(e*Sqrt[1 - (e^2*x^2)/d^2]*(-8*B*d^2 - 3*C*d^2*x + 12*A*e^2*x + 8*B*e^2*x^2 + 6*C*e^2*x^3
) + 3*(C*d^3 + 4*A*d*e^2)*ArcSin[(e*x)/d]))/(24*e^3*Sqrt[1 - (e^2*x^2)/d^2])

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fricas [A]  time = 0.97, size = 108, normalized size = 0.86 \[ -\frac {6 \, {\left (C d^{4} + 4 \, A d^{2} e^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (6 \, C e^{3} x^{3} + 8 \, B e^{3} x^{2} - 8 \, B d^{2} e - 3 \, {\left (C d^{2} e - 4 \, A e^{3}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/24*(6*(C*d^4 + 4*A*d^2*e^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (6*C*e^3*x^3 + 8*B*e^3*x^2 - 8*B*d^
2*e - 3*(C*d^2*e - 4*A*e^3)*x)*sqrt(-e^2*x^2 + d^2))/e^3

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giac [A]  time = 0.20, size = 85, normalized size = 0.68 \[ \frac {1}{8} \, {\left (C d^{4} + 4 \, A d^{2} e^{2}\right )} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\relax (d) - \frac {1}{24} \, {\left (8 \, B d^{2} e^{\left (-2\right )} - {\left (2 \, {\left (3 \, C x + 4 \, B\right )} x - 3 \, {\left (C d^{2} e^{2} - 4 \, A e^{4}\right )} e^{\left (-4\right )}\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

1/8*(C*d^4 + 4*A*d^2*e^2)*arcsin(x*e/d)*e^(-3)*sgn(d) - 1/24*(8*B*d^2*e^(-2) - (2*(3*C*x + 4*B)*x - 3*(C*d^2*e
^2 - 4*A*e^4)*e^(-4))*x)*sqrt(-x^2*e^2 + d^2)

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maple [A]  time = 0.01, size = 154, normalized size = 1.23 \[ \frac {A \,d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}+\frac {C \,d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{2}}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, A x}{2}+\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, C \,d^{2} x}{8 e^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} C x}{4 e^{2}}-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} B}{3 e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)*(-e^2*x^2+d^2)^(1/2),x)

[Out]

-1/4*C*x*(-e^2*x^2+d^2)^(3/2)/e^2+1/8*C*d^2/e^2*x*(-e^2*x^2+d^2)^(1/2)+1/8*C*d^4/e^2/(e^2)^(1/2)*arctan((e^2)^
(1/2)/(-e^2*x^2+d^2)^(1/2)*x)-1/3*B*(-e^2*x^2+d^2)^(3/2)/e^2+1/2*A*x*(-e^2*x^2+d^2)^(1/2)+1/2*A*d^2/(e^2)^(1/2
)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)

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maxima [A]  time = 0.96, size = 116, normalized size = 0.93 \[ \frac {C d^{4} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{3}} + \frac {A d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e} + \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} A x + \frac {\sqrt {-e^{2} x^{2} + d^{2}} C d^{2} x}{8 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} C x}{4 \, e^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} B}{3 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

1/8*C*d^4*arcsin(e*x/d)/e^3 + 1/2*A*d^2*arcsin(e*x/d)/e + 1/2*sqrt(-e^2*x^2 + d^2)*A*x + 1/8*sqrt(-e^2*x^2 + d
^2)*C*d^2*x/e^2 - 1/4*(-e^2*x^2 + d^2)^(3/2)*C*x/e^2 - 1/3*(-e^2*x^2 + d^2)^(3/2)*B/e^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d^2 - e^2*x^2)^(1/2)*(A + B*x + C*x^2),x)

[Out]

int((d^2 - e^2*x^2)^(1/2)*(A + B*x + C*x^2), x)

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sympy [C]  time = 7.11, size = 343, normalized size = 2.74 \[ A \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e} - \frac {i d x}{2 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e} + \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} \frac {x^{2} \sqrt {d^{2}}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\left (d^{2} - e^{2} x^{2}\right )^{\frac {3}{2}}}{3 e^{2}} & \text {otherwise} \end {cases}\right ) + C \left (\begin {cases} - \frac {i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{3}} + \frac {i d^{3} x}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {3 i d x^{3}}{8 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{3}} - \frac {d^{3} x}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {3 d x^{3}}{8 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)*(-e**2*x**2+d**2)**(1/2),x)

[Out]

A*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d*x/(2*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**3/(2*d*sqrt(-1 + e**
2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**2*asin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, True)) + B*
Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + C*Piecewise((-I*d**
4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - 3*I*d*x**3/(8*sqrt(-1 + e**2*x**2/d**2
)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**4*asin(e*x/d)/(8*e**3) - d**3*
x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*d*x**3/(8*sqrt(1 - e**2*x**2/d**2)) - e**2*x**5/(4*d*sqrt(1 - e**2*x**
2/d**2)), True))

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