3.4 \(\int \sqrt{1-d x} \sqrt{1+d x} (A+B x+C x^2) \, dx\)

Optimal. Leaf size=95 \[ \frac{x \sqrt{1-d^2 x^2} \left (4 A d^2+C\right )}{8 d^2}+\frac{\left (4 A d^2+C\right ) \sin ^{-1}(d x)}{8 d^3}-\frac{B \left (1-d^2 x^2\right )^{3/2}}{3 d^2}-\frac{C x \left (1-d^2 x^2\right )^{3/2}}{4 d^2} \]

[Out]

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

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Rubi [A]  time = 0.0729961, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {899, 1815, 641, 195, 216} \[ \frac{x \sqrt{1-d^2 x^2} \left (4 A d^2+C\right )}{8 d^2}+\frac{\left (4 A d^2+C\right ) \sin ^{-1}(d x)}{8 d^3}-\frac{B \left (1-d^2 x^2\right )^{3/2}}{3 d^2}-\frac{C x \left (1-d^2 x^2\right )^{3/2}}{4 d^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - d*x]*Sqrt[1 + d*x]*(A + B*x + C*x^2),x]

[Out]

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

Rule 899

Int[((d_) + (e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :>
Int[(d*f + e*g*x^2)^m*(a + b*x + c*x^2)^p, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[m - n, 0] &&
EqQ[e*f + d*g, 0] && (IntegerQ[m] || (GtQ[d, 0] && GtQ[f, 0]))

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]

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 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0622698, size = 71, normalized size = 0.75 \[ \frac{d \sqrt{1-d^2 x^2} \left (12 A d^2 x+8 B d^2 x^2-8 B+6 C d^2 x^3-3 C x\right )+3 \left (4 A d^2+C\right ) \sin ^{-1}(d x)}{24 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - d*x]*Sqrt[1 + d*x]*(A + B*x + C*x^2),x]

[Out]

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

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Maple [C]  time = 0.011, size = 185, normalized size = 2. \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{24\,{d}^{3}}\sqrt{-dx+1}\sqrt{dx+1} \left ( 6\,C{\it csgn} \left ( d \right ){x}^{3}{d}^{3}\sqrt{-{d}^{2}{x}^{2}+1}+8\,B{\it csgn} \left ( d \right ){x}^{2}{d}^{3}\sqrt{-{d}^{2}{x}^{2}+1}+12\,A{\it csgn} \left ( d \right ){d}^{3}\sqrt{-{d}^{2}{x}^{2}+1}x-3\,C{\it csgn} \left ( d \right ) d\sqrt{-{d}^{2}{x}^{2}+1}x+12\,A\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ){d}^{2}-8\,B\sqrt{-{d}^{2}{x}^{2}+1}{\it csgn} \left ( d \right ) d+3\,C\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ) \right ){\frac{1}{\sqrt{-{d}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(-d*x+1)^(1/2)*(d*x+1)^(1/2)*(6*C*csgn(d)*x^3*d^3*(-d^2*x^2+1)^(1/2)+8*B*csgn(d)*x^2*d^3*(-d^2*x^2+1)^(1/
2)+12*A*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*x-3*C*csgn(d)*d*(-d^2*x^2+1)^(1/2)*x+12*A*arctan(csgn(d)*d*x/(-d^2*x^2+
1)^(1/2))*d^2-8*B*(-d^2*x^2+1)^(1/2)*csgn(d)*d+3*C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2)))*csgn(d)/(-d^2*x^2+1
)^(1/2)/d^3

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Maxima [A]  time = 4.04605, size = 154, normalized size = 1.62 \begin{align*} \frac{1}{2} \, \sqrt{-d^{2} x^{2} + 1} A x - \frac{{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}} C x}{4 \, d^{2}} + \frac{A \arcsin \left (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{2 \, \sqrt{d^{2}}} - \frac{{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}} B}{3 \, d^{2}} + \frac{\sqrt{-d^{2} x^{2} + 1} C x}{8 \, d^{2}} + \frac{C \arcsin \left (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{8 \, \sqrt{d^{2}} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.04244, size = 224, normalized size = 2.36 \begin{align*} \frac{{\left (6 \, C d^{3} x^{3} + 8 \, B d^{3} x^{2} - 8 \, B d + 3 \,{\left (4 \, A d^{3} - C d\right )} x\right )} \sqrt{d x + 1} \sqrt{-d x + 1} - 6 \,{\left (4 \, A d^{2} + C\right )} \arctan \left (\frac{\sqrt{d x + 1} \sqrt{-d x + 1} - 1}{d x}\right )}{24 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.89606, size = 198, normalized size = 2.08 \begin{align*} \frac{\frac{8 \,{\left (d x + 1\right )}^{\frac{3}{2}}{\left (d x - 1\right )} \sqrt{-d x + 1} B}{d} + 12 \,{\left (\sqrt{d x + 1} \sqrt{-d x + 1} d x + 2 \, \arcsin \left (\frac{1}{2} \, \sqrt{2} \sqrt{d x + 1}\right )\right )} A + 3 \,{\left ({\left ({\left (d x + 1\right )}{\left (2 \,{\left (d x + 1\right )}{\left (\frac{d x + 1}{d^{2}} - \frac{3}{d^{2}}\right )} + \frac{5}{d^{2}}\right )} - \frac{1}{d^{2}}\right )} \sqrt{d x + 1} \sqrt{-d x + 1} + \frac{2 \, \arcsin \left (\frac{1}{2} \, \sqrt{2} \sqrt{d x + 1}\right )}{d^{2}}\right )} C}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*(d*x + 1)^(3/2)*(d*x - 1)*sqrt(-d*x + 1)*B/d + 12*(sqrt(d*x + 1)*sqrt(-d*x + 1)*d*x + 2*arcsin(1/2*sqr
t(2)*sqrt(d*x + 1)))*A + 3*(((d*x + 1)*(2*(d*x + 1)*((d*x + 1)/d^2 - 3/d^2) + 5/d^2) - 1/d^2)*sqrt(d*x + 1)*sq
rt(-d*x + 1) + 2*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^2)*C)/d