3.8 \(\int \frac{(e+f x)^3 (A+B x+C x^2)}{\sqrt{1-d x} \sqrt{1+d x}} \, dx\)

Optimal. Leaf size=340 \[ \frac{\sqrt{1-d^2 x^2} \left (d^2 f x \left (-100 A d^2 e f^2-30 B d^2 e^2 f-45 B f^3+6 C d^2 e^3-71 C e f^2\right )+4 \left (C \left (-52 d^2 e^2 f^2+3 d^4 e^4-16 f^4\right )-5 d^2 f \left (4 A f \left (4 d^2 e^2+f^2\right )+3 B \left (d^2 e^3+4 e f^2\right )\right )\right )\right )}{120 d^6 f}+\frac{\sin ^{-1}(d x) \left (8 A d^4 e^3+12 A d^2 e f^2+12 B d^2 e^2 f+3 B f^3+4 C d^2 e^3+9 C e f^2\right )}{8 d^5}-\frac{\sqrt{1-d^2 x^2} (e+f x)^2 \left (4 f^2 \left (5 A d^2+4 C\right )-3 d^2 e (C e-5 B f)\right )}{60 d^4 f}+\frac{\sqrt{1-d^2 x^2} (e+f x)^3 (C e-5 B f)}{20 d^2 f}-\frac{C \sqrt{1-d^2 x^2} (e+f x)^4}{5 d^2 f} \]

[Out]

-((4*(4*C + 5*A*d^2)*f^2 - 3*d^2*e*(C*e - 5*B*f))*(e + f*x)^2*Sqrt[1 - d^2*x^2])/(60*d^4*f) + ((C*e - 5*B*f)*(
e + f*x)^3*Sqrt[1 - d^2*x^2])/(20*d^2*f) - (C*(e + f*x)^4*Sqrt[1 - d^2*x^2])/(5*d^2*f) + ((4*(C*(3*d^4*e^4 - 5
2*d^2*e^2*f^2 - 16*f^4) - 5*d^2*f*(4*A*f*(4*d^2*e^2 + f^2) + 3*B*(d^2*e^3 + 4*e*f^2))) + d^2*f*(6*C*d^2*e^3 -
30*B*d^2*e^2*f - 71*C*e*f^2 - 100*A*d^2*e*f^2 - 45*B*f^3)*x)*Sqrt[1 - d^2*x^2])/(120*d^6*f) + ((4*C*d^2*e^3 +
8*A*d^4*e^3 + 12*B*d^2*e^2*f + 9*C*e*f^2 + 12*A*d^2*e*f^2 + 3*B*f^3)*ArcSin[d*x])/(8*d^5)

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Rubi [A]  time = 0.632967, antiderivative size = 340, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.135, Rules used = {1609, 1654, 833, 780, 216} \[ -\frac{\sqrt{1-d^2 x^2} (e+f x)^2 \left (5 f (4 A f+3 B e)-C \left (3 e^2-\frac{16 f^2}{d^2}\right )\right )}{60 d^2 f}+\frac{\sqrt{1-d^2 x^2} \left (d^2 f x \left (-100 A d^2 e f^2-30 B d^2 e^2 f-45 B f^3+6 C d^2 e^3-71 C e f^2\right )+4 \left (C \left (-52 d^2 e^2 f^2+3 d^4 e^4-16 f^4\right )-5 d^2 f \left (4 A f \left (4 d^2 e^2+f^2\right )+3 B \left (d^2 e^3+4 e f^2\right )\right )\right )\right )}{120 d^6 f}+\frac{\sin ^{-1}(d x) \left (8 A d^4 e^3+12 A d^2 e f^2+12 B d^2 e^2 f+3 B f^3+4 C d^2 e^3+9 C e f^2\right )}{8 d^5}+\frac{\sqrt{1-d^2 x^2} (e+f x)^3 (C e-5 B f)}{20 d^2 f}-\frac{C \sqrt{1-d^2 x^2} (e+f x)^4}{5 d^2 f} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*(A + B*x + C*x^2))/(Sqrt[1 - d*x]*Sqrt[1 + d*x]),x]

[Out]

-((5*f*(3*B*e + 4*A*f) - C*(3*e^2 - (16*f^2)/d^2))*(e + f*x)^2*Sqrt[1 - d^2*x^2])/(60*d^2*f) + ((C*e - 5*B*f)*
(e + f*x)^3*Sqrt[1 - d^2*x^2])/(20*d^2*f) - (C*(e + f*x)^4*Sqrt[1 - d^2*x^2])/(5*d^2*f) + ((4*(C*(3*d^4*e^4 -
52*d^2*e^2*f^2 - 16*f^4) - 5*d^2*f*(4*A*f*(4*d^2*e^2 + f^2) + 3*B*(d^2*e^3 + 4*e*f^2))) + d^2*f*(6*C*d^2*e^3 -
30*B*d^2*e^2*f - 71*C*e*f^2 - 100*A*d^2*e*f^2 - 45*B*f^3)*x)*Sqrt[1 - d^2*x^2])/(120*d^6*f) + ((4*C*d^2*e^3 +
8*A*d^4*e^3 + 12*B*d^2*e^2*f + 9*C*e*f^2 + 12*A*d^2*e*f^2 + 3*B*f^3)*ArcSin[d*x])/(8*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 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 \frac{(e+f x)^3 \left (A+B x+C x^2\right )}{\sqrt{1-d x} \sqrt{1+d x}} \, dx &=\int \frac{(e+f x)^3 \left (A+B x+C x^2\right )}{\sqrt{1-d^2 x^2}} \, dx\\ &=-\frac{C (e+f x)^4 \sqrt{1-d^2 x^2}}{5 d^2 f}-\frac{\int \frac{(e+f x)^3 \left (-\left (4 C+5 A d^2\right ) f^2+d^2 f (C e-5 B f) x\right )}{\sqrt{1-d^2 x^2}} \, dx}{5 d^2 f^2}\\ &=\frac{(C e-5 B f) (e+f x)^3 \sqrt{1-d^2 x^2}}{20 d^2 f}-\frac{C (e+f x)^4 \sqrt{1-d^2 x^2}}{5 d^2 f}+\frac{\int \frac{(e+f x)^2 \left (d^2 f^2 \left (13 C e+20 A d^2 e+15 B f\right )+d^2 f \left (4 \left (4 C+5 A d^2\right ) f^2-3 d^2 e (C e-5 B f)\right ) x\right )}{\sqrt{1-d^2 x^2}} \, dx}{20 d^4 f^2}\\ &=-\frac{\left (4 \left (4 C+5 A d^2\right ) f^2-3 d^2 e (C e-5 B f)\right ) (e+f x)^2 \sqrt{1-d^2 x^2}}{60 d^4 f}+\frac{(C e-5 B f) (e+f x)^3 \sqrt{1-d^2 x^2}}{20 d^2 f}-\frac{C (e+f x)^4 \sqrt{1-d^2 x^2}}{5 d^2 f}-\frac{\int \frac{(e+f x) \left (-d^2 f^2 \left (33 C d^2 e^2+60 A d^4 e^2+75 B d^2 e f+32 C f^2+40 A d^2 f^2\right )+d^4 f \left (6 C d^2 e^3-30 B d^2 e^2 f-71 C e f^2-100 A d^2 e f^2-45 B f^3\right ) x\right )}{\sqrt{1-d^2 x^2}} \, dx}{60 d^6 f^2}\\ &=-\frac{\left (4 \left (4 C+5 A d^2\right ) f^2-3 d^2 e (C e-5 B f)\right ) (e+f x)^2 \sqrt{1-d^2 x^2}}{60 d^4 f}+\frac{(C e-5 B f) (e+f x)^3 \sqrt{1-d^2 x^2}}{20 d^2 f}-\frac{C (e+f x)^4 \sqrt{1-d^2 x^2}}{5 d^2 f}+\frac{\left (4 \left (C \left (3 d^4 e^4-52 d^2 e^2 f^2-16 f^4\right )-5 d^2 f \left (4 A f \left (4 d^2 e^2+f^2\right )+3 B \left (d^2 e^3+4 e f^2\right )\right )\right )+d^2 f \left (6 C d^2 e^3-30 B d^2 e^2 f-71 C e f^2-100 A d^2 e f^2-45 B f^3\right ) x\right ) \sqrt{1-d^2 x^2}}{120 d^6 f}+\frac{\left (4 C d^2 e^3+8 A d^4 e^3+12 B d^2 e^2 f+9 C e f^2+12 A d^2 e f^2+3 B f^3\right ) \int \frac{1}{\sqrt{1-d^2 x^2}} \, dx}{8 d^4}\\ &=-\frac{\left (4 \left (4 C+5 A d^2\right ) f^2-3 d^2 e (C e-5 B f)\right ) (e+f x)^2 \sqrt{1-d^2 x^2}}{60 d^4 f}+\frac{(C e-5 B f) (e+f x)^3 \sqrt{1-d^2 x^2}}{20 d^2 f}-\frac{C (e+f x)^4 \sqrt{1-d^2 x^2}}{5 d^2 f}+\frac{\left (4 \left (C \left (3 d^4 e^4-52 d^2 e^2 f^2-16 f^4\right )-5 d^2 f \left (4 A f \left (4 d^2 e^2+f^2\right )+3 B \left (d^2 e^3+4 e f^2\right )\right )\right )+d^2 f \left (6 C d^2 e^3-30 B d^2 e^2 f-71 C e f^2-100 A d^2 e f^2-45 B f^3\right ) x\right ) \sqrt{1-d^2 x^2}}{120 d^6 f}+\frac{\left (4 C d^2 e^3+8 A d^4 e^3+12 B d^2 e^2 f+9 C e f^2+12 A d^2 e f^2+3 B f^3\right ) \sin ^{-1}(d x)}{8 d^5}\\ \end{align*}

Mathematica [A]  time = 0.37035, size = 241, normalized size = 0.71 \[ \frac{15 d \sin ^{-1}(d x) \left (8 A d^4 e^3+12 A d^2 e f^2+12 B d^2 e^2 f+3 B f^3+4 C d^2 e^3+9 C e f^2\right )-\sqrt{1-d^2 x^2} \left (20 A d^2 f \left (d^2 \left (18 e^2+9 e f x+2 f^2 x^2\right )+4 f^2\right )+15 B \left (2 d^4 \left (6 e^2 f x+4 e^3+4 e f^2 x^2+f^3 x^3\right )+d^2 f^2 (16 e+3 f x)\right )+C \left (6 d^4 x \left (20 e^2 f x+10 e^3+15 e f^2 x^2+4 f^3 x^3\right )+d^2 f \left (240 e^2+135 e f x+32 f^2 x^2\right )+64 f^3\right )\right )}{120 d^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*(A + B*x + C*x^2))/(Sqrt[1 - d*x]*Sqrt[1 + d*x]),x]

[Out]

(-(Sqrt[1 - d^2*x^2]*(20*A*d^2*f*(4*f^2 + d^2*(18*e^2 + 9*e*f*x + 2*f^2*x^2)) + 15*B*(d^2*f^2*(16*e + 3*f*x) +
2*d^4*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3)) + C*(64*f^3 + d^2*f*(240*e^2 + 135*e*f*x + 32*f^2*x^2) + 6
*d^4*x*(10*e^3 + 20*e^2*f*x + 15*e*f^2*x^2 + 4*f^3*x^3)))) + 15*d*(4*C*d^2*e^3 + 8*A*d^4*e^3 + 12*B*d^2*e^2*f
+ 9*C*e*f^2 + 12*A*d^2*e*f^2 + 3*B*f^3)*ArcSin[d*x])/(120*d^6)

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Maple [C]  time = 0.025, size = 643, normalized size = 1.9 \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)^3*(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)*(-d^2*x^2+1)^(1/2)*x^4*d^4*f^3+30*B*csgn(d)*(-d^2*x^2+1)^(1/
2)*x^3*d^4*f^3+90*C*csgn(d)*(-d^2*x^2+1)^(1/2)*x^3*d^4*e*f^2+40*A*csgn(d)*(-d^2*x^2+1)^(1/2)*x^2*d^4*f^3+120*B
*csgn(d)*(-d^2*x^2+1)^(1/2)*x^2*d^4*e*f^2+120*C*csgn(d)*(-d^2*x^2+1)^(1/2)*x^2*d^4*e^2*f+180*A*csgn(d)*(-d^2*x
^2+1)^(1/2)*x*d^4*e*f^2+180*B*csgn(d)*(-d^2*x^2+1)^(1/2)*x*d^4*e^2*f+60*C*csgn(d)*(-d^2*x^2+1)^(1/2)*x*d^4*e^3
+360*A*csgn(d)*(-d^2*x^2+1)^(1/2)*d^4*e^2*f-120*A*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d^5*e^3+120*B*csgn(d)
*(-d^2*x^2+1)^(1/2)*d^4*e^3+32*C*csgn(d)*(-d^2*x^2+1)^(1/2)*x^2*d^2*f^3+45*B*csgn(d)*(-d^2*x^2+1)^(1/2)*x*d^2*
f^3+135*C*csgn(d)*(-d^2*x^2+1)^(1/2)*x*d^2*e*f^2+80*A*csgn(d)*(-d^2*x^2+1)^(1/2)*d^2*f^3-180*A*arctan(csgn(d)*
d*x/(-d^2*x^2+1)^(1/2))*d^3*e*f^2+240*B*csgn(d)*(-d^2*x^2+1)^(1/2)*d^2*e*f^2-180*B*arctan(csgn(d)*d*x/(-d^2*x^
2+1)^(1/2))*d^3*e^2*f+240*C*csgn(d)*(-d^2*x^2+1)^(1/2)*d^2*e^2*f-60*C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d
^3*e^3-45*B*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d*f^3+64*C*csgn(d)*(-d^2*x^2+1)^(1/2)*f^3-135*C*arctan(csgn
(d)*d*x/(-d^2*x^2+1)^(1/2))*d*e*f^2)*csgn(d)/d^6/(-d^2*x^2+1)^(1/2)

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Maxima [A]  time = 3.19919, size = 524, normalized size = 1.54 \begin{align*} -\frac{\sqrt{-d^{2} x^{2} + 1} C f^{3} x^{4}}{5 \, d^{2}} + \frac{A e^{3} \arcsin \left (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{\sqrt{d^{2}}} - \frac{\sqrt{-d^{2} x^{2} + 1} B e^{3}}{d^{2}} - \frac{3 \, \sqrt{-d^{2} x^{2} + 1} A e^{2} f}{d^{2}} - \frac{4 \, \sqrt{-d^{2} x^{2} + 1} C f^{3} x^{2}}{15 \, d^{4}} - \frac{{\left (3 \, C e f^{2} + B f^{3}\right )} \sqrt{-d^{2} x^{2} + 1} x^{3}}{4 \, d^{2}} - \frac{{\left (3 \, C e^{2} f + 3 \, B e f^{2} + A f^{3}\right )} \sqrt{-d^{2} x^{2} + 1} x^{2}}{3 \, d^{2}} - \frac{{\left (C e^{3} + 3 \, B e^{2} f + 3 \, A e f^{2}\right )} \sqrt{-d^{2} x^{2} + 1} x}{2 \, d^{2}} + \frac{{\left (C e^{3} + 3 \, B e^{2} f + 3 \, A e f^{2}\right )} \arcsin \left (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{2 \, \sqrt{d^{2}} d^{2}} - \frac{8 \, \sqrt{-d^{2} x^{2} + 1} C f^{3}}{15 \, d^{6}} - \frac{3 \,{\left (3 \, C e f^{2} + B f^{3}\right )} \sqrt{-d^{2} x^{2} + 1} x}{8 \, d^{4}} - \frac{2 \,{\left (3 \, C e^{2} f + 3 \, B e f^{2} + A f^{3}\right )} \sqrt{-d^{2} x^{2} + 1}}{3 \, d^{4}} + \frac{3 \,{\left (3 \, C e f^{2} + B f^{3}\right )} \arcsin \left (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{8 \, \sqrt{d^{2}} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*(C*x^2+B*x+A)/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.14323, size = 644, normalized size = 1.89 \begin{align*} -\frac{{\left (24 \, C d^{4} f^{3} x^{4} + 120 \, B d^{4} e^{3} + 240 \, B d^{2} e f^{2} + 120 \,{\left (3 \, A d^{4} + 2 \, C d^{2}\right )} e^{2} f + 16 \,{\left (5 \, A d^{2} + 4 \, C\right )} f^{3} + 30 \,{\left (3 \, C d^{4} e f^{2} + B d^{4} f^{3}\right )} x^{3} + 8 \,{\left (15 \, C d^{4} e^{2} f + 15 \, B d^{4} e f^{2} +{\left (5 \, A d^{4} + 4 \, C d^{2}\right )} f^{3}\right )} x^{2} + 15 \,{\left (4 \, C d^{4} e^{3} + 12 \, B d^{4} e^{2} f + 3 \, B d^{2} f^{3} + 3 \,{\left (4 \, A d^{4} + 3 \, C d^{2}\right )} e f^{2}\right )} x\right )} \sqrt{d x + 1} \sqrt{-d x + 1} + 30 \,{\left (12 \, B d^{3} e^{2} f + 3 \, B d f^{3} + 4 \,{\left (2 \, A d^{5} + C d^{3}\right )} e^{3} + 3 \,{\left (4 \, A d^{3} + 3 \, C d\right )} e f^{2}\right )} \arctan \left (\frac{\sqrt{d x + 1} \sqrt{-d x + 1} - 1}{d x}\right )}{120 \, d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*(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^3*x^4 + 120*B*d^4*e^3 + 240*B*d^2*e*f^2 + 120*(3*A*d^4 + 2*C*d^2)*e^2*f + 16*(5*A*d^2 + 4*
C)*f^3 + 30*(3*C*d^4*e*f^2 + B*d^4*f^3)*x^3 + 8*(15*C*d^4*e^2*f + 15*B*d^4*e*f^2 + (5*A*d^4 + 4*C*d^2)*f^3)*x^
2 + 15*(4*C*d^4*e^3 + 12*B*d^4*e^2*f + 3*B*d^2*f^3 + 3*(4*A*d^4 + 3*C*d^2)*e*f^2)*x)*sqrt(d*x + 1)*sqrt(-d*x +
1) + 30*(12*B*d^3*e^2*f + 3*B*d*f^3 + 4*(2*A*d^5 + C*d^3)*e^3 + 3*(4*A*d^3 + 3*C*d)*e*f^2)*arctan((sqrt(d*x +
1)*sqrt(-d*x + 1) - 1)/(d*x)))/d^6

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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)**3*(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 = 2.27081, size = 551, normalized size = 1.62 \begin{align*} -\frac{{\left (360 \, A d^{29} f e^{2} - 180 \, A d^{28} f^{2} e + 120 \, A d^{27} f^{3} + 120 \, B d^{29} e^{3} - 180 \, B d^{28} f e^{2} + 360 \, B d^{27} f^{2} e - 75 \, B d^{26} f^{3} - 60 \, C d^{28} e^{3} + 360 \, C d^{27} f e^{2} - 225 \, C d^{26} f^{2} e + 120 \, C d^{25} f^{3} +{\left (180 \, A d^{28} f^{2} e - 80 \, A d^{27} f^{3} + 180 \, B d^{28} f e^{2} - 240 \, B d^{27} f^{2} e + 135 \, B d^{26} f^{3} + 60 \, C d^{28} e^{3} - 240 \, C d^{27} f e^{2} + 405 \, C d^{26} f^{2} e - 160 \, C d^{25} f^{3} + 2 \,{\left (20 \, A d^{27} f^{3} + 60 \, B d^{27} f^{2} e - 45 \, B d^{26} f^{3} + 60 \, C d^{27} f e^{2} - 135 \, C d^{26} f^{2} e + 88 \, C d^{25} f^{3} + 3 \,{\left (4 \,{\left (d x + 1\right )} C d^{25} f^{3} + 5 \, B d^{26} f^{3} + 15 \, C d^{26} f^{2} e - 16 \, C d^{25} f^{3}\right )}{\left (d x + 1\right )}\right )}{\left (d x + 1\right )}\right )}{\left (d x + 1\right )}\right )} \sqrt{d x + 1} \sqrt{-d x + 1} - 30 \,{\left (8 \, A d^{30} e^{3} + 12 \, A d^{28} f^{2} e + 12 \, B d^{28} f e^{2} + 3 \, B d^{26} f^{3} + 4 \, C d^{28} e^{3} + 9 \, C d^{26} f^{2} e\right )} \arcsin \left (\frac{1}{2} \, \sqrt{2} \sqrt{d x + 1}\right )}{2211840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*(C*x^2+B*x+A)/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="giac")

[Out]

-1/2211840*((360*A*d^29*f*e^2 - 180*A*d^28*f^2*e + 120*A*d^27*f^3 + 120*B*d^29*e^3 - 180*B*d^28*f*e^2 + 360*B*
d^27*f^2*e - 75*B*d^26*f^3 - 60*C*d^28*e^3 + 360*C*d^27*f*e^2 - 225*C*d^26*f^2*e + 120*C*d^25*f^3 + (180*A*d^2
8*f^2*e - 80*A*d^27*f^3 + 180*B*d^28*f*e^2 - 240*B*d^27*f^2*e + 135*B*d^26*f^3 + 60*C*d^28*e^3 - 240*C*d^27*f*
e^2 + 405*C*d^26*f^2*e - 160*C*d^25*f^3 + 2*(20*A*d^27*f^3 + 60*B*d^27*f^2*e - 45*B*d^26*f^3 + 60*C*d^27*f*e^2
- 135*C*d^26*f^2*e + 88*C*d^25*f^3 + 3*(4*(d*x + 1)*C*d^25*f^3 + 5*B*d^26*f^3 + 15*C*d^26*f^2*e - 16*C*d^25*f
^3)*(d*x + 1))*(d*x + 1))*(d*x + 1))*sqrt(d*x + 1)*sqrt(-d*x + 1) - 30*(8*A*d^30*e^3 + 12*A*d^28*f^2*e + 12*B*
d^28*f*e^2 + 3*B*d^26*f^3 + 4*C*d^28*e^3 + 9*C*d^26*f^2*e)*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))/d