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

Optimal. Leaf size=168 \[ -\frac{\left (1-d^2 x^2\right )^{3/2} \left (4 \left (5 d^2 f (A f+B e)-C \left (3 d^2 e^2-2 f^2\right )\right )-3 d^2 f x (3 C e-5 B f)\right )}{60 d^4 f}+\frac{x \sqrt{1-d^2 x^2} \left (4 A d^2 e+B f+C e\right )}{8 d^2}+\frac{\sin ^{-1}(d x) \left (4 A d^2 e+B f+C e\right )}{8 d^3}-\frac{C \left (1-d^2 x^2\right )^{3/2} (e+f x)^2}{5 d^2 f} \]

[Out]

((C*e + 4*A*d^2*e + B*f)*x*Sqrt[1 - d^2*x^2])/(8*d^2) - (C*(e + f*x)^2*(1 - d^2*x^2)^(3/2))/(5*d^2*f) - ((4*(5
*d^2*f*(B*e + A*f) - C*(3*d^2*e^2 - 2*f^2)) - 3*d^2*f*(3*C*e - 5*B*f)*x)*(1 - d^2*x^2)^(3/2))/(60*d^4*f) + ((C
*e + 4*A*d^2*e + B*f)*ArcSin[d*x])/(8*d^3)

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Rubi [A]  time = 0.250389, antiderivative size = 170, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {1609, 1654, 780, 195, 216} \[ -\frac{\left (1-d^2 x^2\right )^{3/2} \left (4 \left (5 d^2 f (A f+B e)-\frac{1}{4} C \left (12 d^2 e^2-8 f^2\right )\right )-3 d^2 f x (3 C e-5 B f)\right )}{60 d^4 f}+\frac{x \sqrt{1-d^2 x^2} \left (4 A d^2 e+B f+C e\right )}{8 d^2}+\frac{\sin ^{-1}(d x) \left (4 A d^2 e+B f+C e\right )}{8 d^3}-\frac{C \left (1-d^2 x^2\right )^{3/2} (e+f x)^2}{5 d^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((C*e + 4*A*d^2*e + B*f)*x*Sqrt[1 - d^2*x^2])/(8*d^2) - (C*(e + f*x)^2*(1 - d^2*x^2)^(3/2))/(5*d^2*f) - ((4*(5
*d^2*f*(B*e + A*f) - (C*(12*d^2*e^2 - 8*f^2))/4) - 3*d^2*f*(3*C*e - 5*B*f)*x)*(1 - d^2*x^2)^(3/2))/(60*d^4*f)
+ ((C*e + 4*A*d^2*e + B*f)*ArcSin[d*x])/(8*d^3)

Rule 1609

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

Rule 1654

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[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} (e+f x) \left (A+B x+C x^2\right ) \, dx &=\int (e+f x) \left (A+B x+C x^2\right ) \sqrt{1-d^2 x^2} \, dx\\ &=-\frac{C (e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{5 d^2 f}-\frac{\int (e+f x) \left (-\left (2 C+5 A d^2\right ) f^2+d^2 f (3 C e-5 B f) x\right ) \sqrt{1-d^2 x^2} \, dx}{5 d^2 f^2}\\ &=-\frac{C (e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{5 d^2 f}-\frac{\left (4 \left (5 d^2 f (B e+A f)-\frac{1}{4} C \left (12 d^2 e^2-8 f^2\right )\right )-3 d^2 f (3 C e-5 B f) x\right ) \left (1-d^2 x^2\right )^{3/2}}{60 d^4 f}+\frac{\left (C e+4 A d^2 e+B f\right ) \int \sqrt{1-d^2 x^2} \, dx}{4 d^2}\\ &=\frac{\left (C e+4 A d^2 e+B f\right ) x \sqrt{1-d^2 x^2}}{8 d^2}-\frac{C (e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{5 d^2 f}-\frac{\left (4 \left (5 d^2 f (B e+A f)-\frac{1}{4} C \left (12 d^2 e^2-8 f^2\right )\right )-3 d^2 f (3 C e-5 B f) x\right ) \left (1-d^2 x^2\right )^{3/2}}{60 d^4 f}+\frac{\left (C e+4 A d^2 e+B f\right ) \int \frac{1}{\sqrt{1-d^2 x^2}} \, dx}{8 d^2}\\ &=\frac{\left (C e+4 A d^2 e+B f\right ) x \sqrt{1-d^2 x^2}}{8 d^2}-\frac{C (e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{5 d^2 f}-\frac{\left (4 \left (5 d^2 f (B e+A f)-\frac{1}{4} C \left (12 d^2 e^2-8 f^2\right )\right )-3 d^2 f (3 C e-5 B f) x\right ) \left (1-d^2 x^2\right )^{3/2}}{60 d^4 f}+\frac{\left (C e+4 A d^2 e+B f\right ) \sin ^{-1}(d x)}{8 d^3}\\ \end{align*}

Mathematica [A]  time = 0.170314, size = 141, normalized size = 0.84 \[ \frac{\sqrt{1-d^2 x^2} \left (60 A d^4 e x+40 A d^2 f \left (d^2 x^2-1\right )+5 B d^2 \left (8 d^2 e x^2+6 d^2 f x^3-8 e-3 f x\right )+15 C d^2 e x \left (2 d^2 x^2-1\right )+8 C f \left (3 d^4 x^4-d^2 x^2-2\right )\right )+15 d \sin ^{-1}(d x) \left (4 A d^2 e+B f+C e\right )}{120 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - d^2*x^2]*(60*A*d^4*e*x + 40*A*d^2*f*(-1 + d^2*x^2) + 15*C*d^2*e*x*(-1 + 2*d^2*x^2) + 5*B*d^2*(-8*e -
3*f*x + 8*d^2*e*x^2 + 6*d^2*f*x^3) + 8*C*f*(-2 - d^2*x^2 + 3*d^4*x^4)) + 15*d*(C*e + 4*A*d^2*e + B*f)*ArcSin[
d*x])/(120*d^4)

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Maple [C]  time = 0.01, size = 377, normalized size = 2.2 \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{120\,{d}^{4}}\sqrt{-dx+1}\sqrt{dx+1} \left ( 24\,C{\it csgn} \left ( d \right ){x}^{4}{d}^{4}f\sqrt{-{d}^{2}{x}^{2}+1}+30\,B{\it csgn} \left ( d \right ){x}^{3}{d}^{4}f\sqrt{-{d}^{2}{x}^{2}+1}+30\,C{\it csgn} \left ( d \right ){x}^{3}{d}^{4}e\sqrt{-{d}^{2}{x}^{2}+1}+40\,A{\it csgn} \left ( d \right ){x}^{2}{d}^{4}f\sqrt{-{d}^{2}{x}^{2}+1}+40\,B{\it csgn} \left ( d \right ){x}^{2}{d}^{4}e\sqrt{-{d}^{2}{x}^{2}+1}+60\,A{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}x{d}^{4}e-8\,C{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}{x}^{2}{d}^{2}f-15\,B{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}x{d}^{2}f-15\,C{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}x{d}^{2}e-40\,A{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}{d}^{2}f+60\,A\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ){d}^{3}e-40\,B{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}{d}^{2}e+15\,B\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ) df-16\,C{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}f+15\,C\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ) de \right ){\frac{1}{\sqrt{-{d}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(-d*x+1)^(1/2)*(d*x+1)^(1/2)*(24*C*csgn(d)*x^4*d^4*f*(-d^2*x^2+1)^(1/2)+30*B*csgn(d)*x^3*d^4*f*(-d^2*x^2
+1)^(1/2)+30*C*csgn(d)*x^3*d^4*e*(-d^2*x^2+1)^(1/2)+40*A*csgn(d)*x^2*d^4*f*(-d^2*x^2+1)^(1/2)+40*B*csgn(d)*x^2
*d^4*e*(-d^2*x^2+1)^(1/2)+60*A*csgn(d)*(-d^2*x^2+1)^(1/2)*x*d^4*e-8*C*csgn(d)*(-d^2*x^2+1)^(1/2)*x^2*d^2*f-15*
B*csgn(d)*(-d^2*x^2+1)^(1/2)*x*d^2*f-15*C*csgn(d)*(-d^2*x^2+1)^(1/2)*x*d^2*e-40*A*csgn(d)*(-d^2*x^2+1)^(1/2)*d
^2*f+60*A*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d^3*e-40*B*csgn(d)*(-d^2*x^2+1)^(1/2)*d^2*e+15*B*arctan(csgn(
d)*d*x/(-d^2*x^2+1)^(1/2))*d*f-16*C*csgn(d)*(-d^2*x^2+1)^(1/2)*f+15*C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d
*e)*csgn(d)/d^4/(-d^2*x^2+1)^(1/2)

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Maxima [A]  time = 3.29509, size = 263, normalized size = 1.57 \begin{align*} \frac{1}{2} \, \sqrt{-d^{2} x^{2} + 1} A e x - \frac{{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}} C f x^{2}}{5 \, d^{2}} + \frac{A e \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 e}{3 \, d^{2}} - \frac{{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}} A f}{3 \, d^{2}} - \frac{{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}}{\left (C e + B f\right )} x}{4 \, d^{2}} + \frac{\sqrt{-d^{2} x^{2} + 1}{\left (C e + B f\right )} x}{8 \, d^{2}} - \frac{2 \,{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}} C f}{15 \, d^{4}} + \frac{{\left (C e + B f\right )} \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((f*x+e)*(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*e*x - 1/5*(-d^2*x^2 + 1)^(3/2)*C*f*x^2/d^2 + 1/2*A*e*arcsin(d^2*x/sqrt(d^2))/sqrt(d^2
) - 1/3*(-d^2*x^2 + 1)^(3/2)*B*e/d^2 - 1/3*(-d^2*x^2 + 1)^(3/2)*A*f/d^2 - 1/4*(-d^2*x^2 + 1)^(3/2)*(C*e + B*f)
*x/d^2 + 1/8*sqrt(-d^2*x^2 + 1)*(C*e + B*f)*x/d^2 - 2/15*(-d^2*x^2 + 1)^(3/2)*C*f/d^4 + 1/8*(C*e + B*f)*arcsin
(d^2*x/sqrt(d^2))/(sqrt(d^2)*d^2)

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Fricas [A]  time = 1.08591, size = 386, normalized size = 2.3 \begin{align*} \frac{{\left (24 \, C d^{4} f x^{4} - 40 \, B d^{2} e + 30 \,{\left (C d^{4} e + B d^{4} f\right )} x^{3} + 8 \,{\left (5 \, B d^{4} e +{\left (5 \, A d^{4} - C d^{2}\right )} f\right )} x^{2} - 8 \,{\left (5 \, A d^{2} + 2 \, C\right )} f - 15 \,{\left (B d^{2} f -{\left (4 \, A d^{4} - C d^{2}\right )} e\right )} x\right )} \sqrt{d x + 1} \sqrt{-d x + 1} - 30 \,{\left (B d f +{\left (4 \, A d^{3} + C d\right )} e\right )} \arctan \left (\frac{\sqrt{d x + 1} \sqrt{-d x + 1} - 1}{d x}\right )}{120 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*((24*C*d^4*f*x^4 - 40*B*d^2*e + 30*(C*d^4*e + B*d^4*f)*x^3 + 8*(5*B*d^4*e + (5*A*d^4 - C*d^2)*f)*x^2 - 8
*(5*A*d^2 + 2*C)*f - 15*(B*d^2*f - (4*A*d^4 - C*d^2)*e)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) - 30*(B*d*f + (4*A*d^3
+ C*d)*e)*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/d^4

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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((f*x+e)*(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 [B]  time = 3.04403, size = 429, normalized size = 2.55 \begin{align*} \frac{8 \,{\left ({\left (d x + 1\right )}{\left (3 \,{\left (d x + 1\right )}{\left (\frac{d x + 1}{d^{3}} - \frac{4}{d^{3}}\right )} + \frac{17}{d^{3}}\right )} - \frac{10}{d^{3}}\right )}{\left (d x + 1\right )}^{\frac{3}{2}} \sqrt{-d x + 1} C f + \frac{40 \,{\left (d x + 1\right )}^{\frac{3}{2}}{\left (d x - 1\right )} \sqrt{-d x + 1} A f}{d} + \frac{40 \,{\left (d x + 1\right )}^{\frac{3}{2}}{\left (d x - 1\right )} \sqrt{-d x + 1} B e}{d} + 15 \,{\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 )} B f + 60 \,{\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 e + 15 \,{\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 e}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(8*((d*x + 1)*(3*(d*x + 1)*((d*x + 1)/d^3 - 4/d^3) + 17/d^3) - 10/d^3)*(d*x + 1)^(3/2)*sqrt(-d*x + 1)*C*
f + 40*(d*x + 1)^(3/2)*(d*x - 1)*sqrt(-d*x + 1)*A*f/d + 40*(d*x + 1)^(3/2)*(d*x - 1)*sqrt(-d*x + 1)*B*e/d + 15
*(((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)*sqrt(-d*x + 1) + 2*arcsin(1/
2*sqrt(2)*sqrt(d*x + 1))/d^2)*B*f + 60*(sqrt(d*x + 1)*sqrt(-d*x + 1)*d*x + 2*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))
)*A*e + 15*(((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)*sqrt(-d*x + 1) + 2
*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^2)*C*e)/d