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

Optimal. Leaf size=286 \[ \frac{\left (1-d^2 x^2\right )^{3/2} \left (8 \left (C \left (d^2 e^3-4 e f^2\right )-2 f \left (5 A d^2 e f+B \left (d^2 e^2+f^2\right )\right )\right )-3 f x \left (5 f^2 \left (2 A d^2+C\right )-2 d^2 e (C e-2 B f)\right )\right )}{120 d^4 f}+\frac{x \sqrt{1-d^2 x^2} \left (2 d^2 \left (A \left (4 d^2 e^2+f^2\right )+2 B e f\right )+C \left (2 d^2 e^2+f^2\right )\right )}{16 d^4}+\frac{\sin ^{-1}(d x) \left (2 d^2 \left (A \left (4 d^2 e^2+f^2\right )+2 B e f\right )+C \left (2 d^2 e^2+f^2\right )\right )}{16 d^5}+\frac{\left (1-d^2 x^2\right )^{3/2} (e+f x)^2 (C e-2 B f)}{10 d^2 f}-\frac{C \left (1-d^2 x^2\right )^{3/2} (e+f x)^3}{6 d^2 f} \]

[Out]

((C*(2*d^2*e^2 + f^2) + 2*d^2*(2*B*e*f + A*(4*d^2*e^2 + f^2)))*x*Sqrt[1 - d^2*x^2])/(16*d^4) + ((C*e - 2*B*f)*
(e + f*x)^2*(1 - d^2*x^2)^(3/2))/(10*d^2*f) - (C*(e + f*x)^3*(1 - d^2*x^2)^(3/2))/(6*d^2*f) + ((8*(C*(d^2*e^3
- 4*e*f^2) - 2*f*(5*A*d^2*e*f + B*(d^2*e^2 + f^2))) - 3*f*(5*(C + 2*A*d^2)*f^2 - 2*d^2*e*(C*e - 2*B*f))*x)*(1
- d^2*x^2)^(3/2))/(120*d^4*f) + ((C*(2*d^2*e^2 + f^2) + 2*d^2*(2*B*e*f + A*(4*d^2*e^2 + f^2)))*ArcSin[d*x])/(1
6*d^5)

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Rubi [A]  time = 0.563279, antiderivative size = 286, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {1609, 1654, 833, 780, 195, 216} \[ \frac{\left (1-d^2 x^2\right )^{3/2} \left (8 \left (C \left (d^2 e^3-4 e f^2\right )-2 f \left (5 A d^2 e f+B \left (d^2 e^2+f^2\right )\right )\right )-3 f x \left (5 f^2 \left (2 A d^2+C\right )-2 d^2 e (C e-2 B f)\right )\right )}{120 d^4 f}+\frac{x \sqrt{1-d^2 x^2} \left (2 d^2 \left (A \left (4 d^2 e^2+f^2\right )+2 B e f\right )+C \left (2 d^2 e^2+f^2\right )\right )}{16 d^4}+\frac{\sin ^{-1}(d x) \left (2 d^2 \left (A \left (4 d^2 e^2+f^2\right )+2 B e f\right )+C \left (2 d^2 e^2+f^2\right )\right )}{16 d^5}+\frac{\left (1-d^2 x^2\right )^{3/2} (e+f x)^2 (C e-2 B f)}{10 d^2 f}-\frac{C \left (1-d^2 x^2\right )^{3/2} (e+f x)^3}{6 d^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((C*(2*d^2*e^2 + f^2) + 2*d^2*(2*B*e*f + A*(4*d^2*e^2 + f^2)))*x*Sqrt[1 - d^2*x^2])/(16*d^4) + ((C*e - 2*B*f)*
(e + f*x)^2*(1 - d^2*x^2)^(3/2))/(10*d^2*f) - (C*(e + f*x)^3*(1 - d^2*x^2)^(3/2))/(6*d^2*f) + ((8*(C*(d^2*e^3
- 4*e*f^2) - 2*f*(5*A*d^2*e*f + B*(d^2*e^2 + f^2))) - 3*f*(5*(C + 2*A*d^2)*f^2 - 2*d^2*e*(C*e - 2*B*f))*x)*(1
- d^2*x^2)^(3/2))/(120*d^4*f) + ((C*(2*d^2*e^2 + f^2) + 2*d^2*(2*B*e*f + A*(4*d^2*e^2 + f^2)))*ArcSin[d*x])/(1
6*d^5)

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 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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)^2 \left (A+B x+C x^2\right ) \, dx &=\int (e+f x)^2 \left (A+B x+C x^2\right ) \sqrt{1-d^2 x^2} \, dx\\ &=-\frac{C (e+f x)^3 \left (1-d^2 x^2\right )^{3/2}}{6 d^2 f}-\frac{\int (e+f x)^2 \left (-3 \left (C+2 A d^2\right ) f^2+3 d^2 f (C e-2 B f) x\right ) \sqrt{1-d^2 x^2} \, dx}{6 d^2 f^2}\\ &=\frac{(C e-2 B f) (e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{10 d^2 f}-\frac{C (e+f x)^3 \left (1-d^2 x^2\right )^{3/2}}{6 d^2 f}+\frac{\int (e+f x) \left (3 d^2 f^2 \left (3 C e+10 A d^2 e+4 B f\right )+3 d^2 f \left (5 \left (C+2 A d^2\right ) f^2-2 d^2 e (C e-2 B f)\right ) x\right ) \sqrt{1-d^2 x^2} \, dx}{30 d^4 f^2}\\ &=\frac{(C e-2 B f) (e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{10 d^2 f}-\frac{C (e+f x)^3 \left (1-d^2 x^2\right )^{3/2}}{6 d^2 f}+\frac{\left (8 \left (C \left (d^2 e^3-4 e f^2\right )-2 f \left (5 A d^2 e f+B \left (d^2 e^2+f^2\right )\right )\right )-3 f \left (5 \left (C+2 A d^2\right ) f^2-2 d^2 e (C e-2 B f)\right ) x\right ) \left (1-d^2 x^2\right )^{3/2}}{120 d^4 f}+\frac{\left (C \left (2 d^2 e^2+f^2\right )+2 d^2 \left (2 B e f+A \left (4 d^2 e^2+f^2\right )\right )\right ) \int \sqrt{1-d^2 x^2} \, dx}{8 d^4}\\ &=\frac{\left (C \left (2 d^2 e^2+f^2\right )+2 d^2 \left (2 B e f+A \left (4 d^2 e^2+f^2\right )\right )\right ) x \sqrt{1-d^2 x^2}}{16 d^4}+\frac{(C e-2 B f) (e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{10 d^2 f}-\frac{C (e+f x)^3 \left (1-d^2 x^2\right )^{3/2}}{6 d^2 f}+\frac{\left (8 \left (C \left (d^2 e^3-4 e f^2\right )-2 f \left (5 A d^2 e f+B \left (d^2 e^2+f^2\right )\right )\right )-3 f \left (5 \left (C+2 A d^2\right ) f^2-2 d^2 e (C e-2 B f)\right ) x\right ) \left (1-d^2 x^2\right )^{3/2}}{120 d^4 f}+\frac{\left (C \left (2 d^2 e^2+f^2\right )+2 d^2 \left (2 B e f+A \left (4 d^2 e^2+f^2\right )\right )\right ) \int \frac{1}{\sqrt{1-d^2 x^2}} \, dx}{16 d^4}\\ &=\frac{\left (C \left (2 d^2 e^2+f^2\right )+2 d^2 \left (2 B e f+A \left (4 d^2 e^2+f^2\right )\right )\right ) x \sqrt{1-d^2 x^2}}{16 d^4}+\frac{(C e-2 B f) (e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{10 d^2 f}-\frac{C (e+f x)^3 \left (1-d^2 x^2\right )^{3/2}}{6 d^2 f}+\frac{\left (8 \left (C \left (d^2 e^3-4 e f^2\right )-2 f \left (5 A d^2 e f+B \left (d^2 e^2+f^2\right )\right )\right )-3 f \left (5 \left (C+2 A d^2\right ) f^2-2 d^2 e (C e-2 B f)\right ) x\right ) \left (1-d^2 x^2\right )^{3/2}}{120 d^4 f}+\frac{\left (C \left (2 d^2 e^2+f^2\right )+2 d^2 \left (2 B e f+A \left (4 d^2 e^2+f^2\right )\right )\right ) \sin ^{-1}(d x)}{16 d^5}\\ \end{align*}

Mathematica [A]  time = 0.325953, size = 244, normalized size = 0.85 \[ \frac{d \sqrt{1-d^2 x^2} \left (10 A d^2 \left (12 d^2 e^2 x+16 e f \left (d^2 x^2-1\right )+3 f^2 x \left (2 d^2 x^2-1\right )\right )+4 B \left (2 d^4 x^2 \left (10 e^2+15 e f x+6 f^2 x^2\right )-d^2 \left (20 e^2+15 e f x+4 f^2 x^2\right )-8 f^2\right )+C \left (30 d^2 e^2 x \left (2 d^2 x^2-1\right )+32 e f \left (3 d^4 x^4-d^2 x^2-2\right )+5 f^2 x \left (8 d^4 x^4-2 d^2 x^2-3\right )\right )\right )+15 \sin ^{-1}(d x) \left (2 d^2 \left (A \left (4 d^2 e^2+f^2\right )+2 B e f\right )+C \left (2 d^2 e^2+f^2\right )\right )}{240 d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[1 - d^2*x^2]*(10*A*d^2*(12*d^2*e^2*x + 16*e*f*(-1 + d^2*x^2) + 3*f^2*x*(-1 + 2*d^2*x^2)) + 4*B*(-8*f^2
- d^2*(20*e^2 + 15*e*f*x + 4*f^2*x^2) + 2*d^4*x^2*(10*e^2 + 15*e*f*x + 6*f^2*x^2)) + C*(30*d^2*e^2*x*(-1 + 2*
d^2*x^2) + 32*e*f*(-2 - d^2*x^2 + 3*d^4*x^4) + 5*f^2*x*(-3 - 2*d^2*x^2 + 8*d^4*x^4))) + 15*(C*(2*d^2*e^2 + f^2
) + 2*d^2*(2*B*e*f + A*(4*d^2*e^2 + f^2)))*ArcSin[d*x])/(240*d^5)

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Maple [C]  time = 0.013, size = 652, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(-d*x+1)^(1/2)*(d*x+1)^(1/2)*(-15*C*csgn(d)*d*(-d^2*x^2+1)^(1/2)*x*f^2+40*C*csgn(d)*x^5*d^5*f^2*(-d^2*x^
2+1)^(1/2)+48*B*csgn(d)*x^4*d^5*f^2*(-d^2*x^2+1)^(1/2)+60*A*csgn(d)*x^3*d^5*f^2*(-d^2*x^2+1)^(1/2)+60*C*csgn(d
)*x^3*d^5*e^2*(-d^2*x^2+1)^(1/2)+80*B*csgn(d)*x^2*d^5*e^2*(-d^2*x^2+1)^(1/2)-10*C*csgn(d)*d^3*(-d^2*x^2+1)^(1/
2)*x^3*f^2-16*B*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*x^2*f^2-160*A*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*e*f-30*C*csgn(d)*d
^3*(-d^2*x^2+1)^(1/2)*x*e^2-30*A*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*x*f^2+120*A*csgn(d)*d^5*(-d^2*x^2+1)^(1/2)*x*e
^2-80*B*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*e^2+120*A*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d^4*e^2+30*A*arctan(cs
gn(d)*d*x/(-d^2*x^2+1)^(1/2))*d^2*f^2+30*C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d^2*e^2+15*C*arctan(csgn(d)*
d*x/(-d^2*x^2+1)^(1/2))*f^2+60*B*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d^2*e*f-32*B*csgn(d)*d*(-d^2*x^2+1)^(1
/2)*f^2-64*C*csgn(d)*d*(-d^2*x^2+1)^(1/2)*e*f-60*B*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*x*e*f+96*C*csgn(d)*x^4*d^5*e
*f*(-d^2*x^2+1)^(1/2)+120*B*csgn(d)*x^3*d^5*e*f*(-d^2*x^2+1)^(1/2)+160*A*csgn(d)*x^2*d^5*e*f*(-d^2*x^2+1)^(1/2
)-32*C*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*x^2*e*f)*csgn(d)/(-d^2*x^2+1)^(1/2)/d^5

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Maxima [A]  time = 3.75575, size = 459, normalized size = 1.6 \begin{align*} -\frac{{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}} C f^{2} x^{3}}{6 \, d^{2}} + \frac{1}{2} \, \sqrt{-d^{2} x^{2} + 1} A e^{2} x + \frac{A e^{2} \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^{2}}{3 \, d^{2}} - \frac{2 \,{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}} A e f}{3 \, d^{2}} - \frac{{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}}{\left (2 \, C e f + B f^{2}\right )} x^{2}}{5 \, d^{2}} - \frac{{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}}{\left (C e^{2} + 2 \, B e f + A f^{2}\right )} x}{4 \, d^{2}} - \frac{{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}} C f^{2} x}{8 \, d^{4}} + \frac{\sqrt{-d^{2} x^{2} + 1}{\left (C e^{2} + 2 \, B e f + A f^{2}\right )} x}{8 \, d^{2}} + \frac{\sqrt{-d^{2} x^{2} + 1} C f^{2} x}{16 \, d^{4}} + \frac{{\left (C e^{2} + 2 \, B e f + A f^{2}\right )} \arcsin \left (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{8 \, \sqrt{d^{2}} d^{2}} + \frac{C f^{2} \arcsin \left (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{16 \, \sqrt{d^{2}} d^{4}} - \frac{2 \,{\left (-d^{2} x^{2} + 1\right )}^{\frac{3}{2}}{\left (2 \, C e f + B f^{2}\right )}}{15 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(-d^2*x^2 + 1)^(3/2)*C*f^2*x^3/d^2 + 1/2*sqrt(-d^2*x^2 + 1)*A*e^2*x + 1/2*A*e^2*arcsin(d^2*x/sqrt(d^2))/s
qrt(d^2) - 1/3*(-d^2*x^2 + 1)^(3/2)*B*e^2/d^2 - 2/3*(-d^2*x^2 + 1)^(3/2)*A*e*f/d^2 - 1/5*(-d^2*x^2 + 1)^(3/2)*
(2*C*e*f + B*f^2)*x^2/d^2 - 1/4*(-d^2*x^2 + 1)^(3/2)*(C*e^2 + 2*B*e*f + A*f^2)*x/d^2 - 1/8*(-d^2*x^2 + 1)^(3/2
)*C*f^2*x/d^4 + 1/8*sqrt(-d^2*x^2 + 1)*(C*e^2 + 2*B*e*f + A*f^2)*x/d^2 + 1/16*sqrt(-d^2*x^2 + 1)*C*f^2*x/d^4 +
1/8*(C*e^2 + 2*B*e*f + A*f^2)*arcsin(d^2*x/sqrt(d^2))/(sqrt(d^2)*d^2) + 1/16*C*f^2*arcsin(d^2*x/sqrt(d^2))/(s
qrt(d^2)*d^4) - 2/15*(-d^2*x^2 + 1)^(3/2)*(2*C*e*f + B*f^2)/d^4

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Fricas [A]  time = 1.10435, size = 617, normalized size = 2.16 \begin{align*} \frac{{\left (40 \, C d^{5} f^{2} x^{5} - 80 \, B d^{3} e^{2} + 48 \,{\left (2 \, C d^{5} e f + B d^{5} f^{2}\right )} x^{4} - 32 \, B d f^{2} + 10 \,{\left (6 \, C d^{5} e^{2} + 12 \, B d^{5} e f +{\left (6 \, A d^{5} - C d^{3}\right )} f^{2}\right )} x^{3} - 32 \,{\left (5 \, A d^{3} + 2 \, C d\right )} e f + 16 \,{\left (5 \, B d^{5} e^{2} - B d^{3} f^{2} + 2 \,{\left (5 \, A d^{5} - C d^{3}\right )} e f\right )} x^{2} - 15 \,{\left (4 \, B d^{3} e f - 2 \,{\left (4 \, A d^{5} - C d^{3}\right )} e^{2} +{\left (2 \, A d^{3} + C d\right )} f^{2}\right )} x\right )} \sqrt{d x + 1} \sqrt{-d x + 1} - 30 \,{\left (4 \, B d^{2} e f + 2 \,{\left (4 \, A d^{4} + C d^{2}\right )} e^{2} +{\left (2 \, A d^{2} + C\right )} f^{2}\right )} \arctan \left (\frac{\sqrt{d x + 1} \sqrt{-d x + 1} - 1}{d x}\right )}{240 \, d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*((40*C*d^5*f^2*x^5 - 80*B*d^3*e^2 + 48*(2*C*d^5*e*f + B*d^5*f^2)*x^4 - 32*B*d*f^2 + 10*(6*C*d^5*e^2 + 12
*B*d^5*e*f + (6*A*d^5 - C*d^3)*f^2)*x^3 - 32*(5*A*d^3 + 2*C*d)*e*f + 16*(5*B*d^5*e^2 - B*d^3*f^2 + 2*(5*A*d^5
- C*d^3)*e*f)*x^2 - 15*(4*B*d^3*e*f - 2*(4*A*d^5 - C*d^3)*e^2 + (2*A*d^3 + C*d)*f^2)*x)*sqrt(d*x + 1)*sqrt(-d*
x + 1) - 30*(4*B*d^2*e*f + 2*(4*A*d^4 + C*d^2)*e^2 + (2*A*d^2 + C)*f^2)*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) -
1)/(d*x)))/d^5

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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)**2*(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 = 1.65372, size = 772, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(16*((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)*B
*f^2 + 32*((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*e + 160*(d*x + 1)^(3/2)*(d*x - 1)*sqrt(-d*x + 1)*A*f*e/d + 30*(((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)*A*f^2 + 5*(((2*(
(d*x + 1)*(4*(d*x + 1)*((d*x + 1)/d^4 - 5/d^4) + 39/d^4) - 37/d^4)*(d*x + 1) + 31/d^4)*(d*x + 1) - 3/d^4)*sqrt
(d*x + 1)*sqrt(-d*x + 1) + 6*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^4)*C*f^2 + 80*(d*x + 1)^(3/2)*(d*x - 1)*sqrt(
-d*x + 1)*B*e^2/d + 60*(((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*e + 120*(sqrt(d*x + 1)*sqrt(-d*x + 1)*d*x + 2*arcsin(1
/2*sqrt(2)*sqrt(d*x + 1)))*A*e^2 + 30*(((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^2)/d