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

[Out]

-1/5*C*(f*x+e)^2*(-d^2*x^2+1)^(3/2)/d^2/f-1/60*(20*d^2*f*(A*f+B*e)-4*C*(3*d^2*e^2-2*f^2)-3*d^2*f*(-5*B*f+3*C*e
)*x)*(-d^2*x^2+1)^(3/2)/d^4/f+1/8*(4*A*d^2*e+B*f+C*e)*arcsin(d*x)/d^3+1/8*(4*A*d^2*e+B*f+C*e)*x*(-d^2*x^2+1)^(
1/2)/d^2

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Rubi [A]  time = 0.25, 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 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]

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 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]))

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*}

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Mathematica [A]  time = 0.21, size = 141, normalized size = 0.84 \[ \frac {15 d \sin ^{-1}(d x) \left (4 A d^2 e+B f+C e\right )+\sqrt {1-d^2 x^2} \left (60 A d^4 e x+40 A d^2 f \left (d^2 x^2-1\right )+5 B d^2 \left (8 d^2 e x^2+6 d^2 f x^3-8 e-3 f x\right )+15 C d^2 e x \left (2 d^2 x^2-1\right )+8 C f \left (3 d^4 x^4-d^2 x^2-2\right )\right )}{120 d^4} \]

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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fricas [A]  time = 0.91, size = 170, normalized size = 1.01 \[ \frac {{\left (24 \, C d^{4} f x^{4} - 40 \, B d^{2} e + 30 \, {\left (C d^{4} e + B d^{4} f\right )} x^{3} + 8 \, {\left (5 \, B d^{4} e + {\left (5 \, A d^{4} - C d^{2}\right )} f\right )} x^{2} - 8 \, {\left (5 \, A d^{2} + 2 \, C\right )} f - 15 \, {\left (B d^{2} f - {\left (4 \, A d^{4} - C d^{2}\right )} e\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - 30 \, {\left (B d f + {\left (4 \, A d^{3} + C d\right )} e\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{120 \, d^{4}} \]

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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giac [B]  time = 2.00, size = 782, normalized size = 4.65 \[ \text {result too large to display} \]

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*(20*(sqrt(d*x + 1)*sqrt(-d*x + 1)*((d*x + 1)*(2*(d*x + 1)/d^2 - 7/d^2) + 9/d^2) + 6*arcsin(1/2*sqrt(2)*s
qrt(d*x + 1))/d^2)*A*d*f + 5*(((d*x + 1)*(2*(d*x + 1)*(3*(d*x + 1)/d^3 - 13/d^3) + 43/d^3) - 39/d^3)*sqrt(d*x
+ 1)*sqrt(-d*x + 1) - 18*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^3)*B*d*f + (((2*(d*x + 1)*(3*(d*x + 1)*(4*(d*x +
1)/d^4 - 21/d^4) + 133/d^4) - 295/d^4)*(d*x + 1) + 195/d^4)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 90*arcsin(1/2*sqrt(
2)*sqrt(d*x + 1))/d^4)*C*d*f + 20*(sqrt(d*x + 1)*sqrt(-d*x + 1)*((d*x + 1)*(2*(d*x + 1)/d^2 - 7/d^2) + 9/d^2)
+ 6*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^2)*B*d*e + 5*(((d*x + 1)*(2*(d*x + 1)*(3*(d*x + 1)/d^3 - 13/d^3) + 43/
d^3) - 39/d^3)*sqrt(d*x + 1)*sqrt(-d*x + 1) - 18*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^3)*C*d*e + 20*(sqrt(d*x +
1)*sqrt(-d*x + 1)*((d*x + 1)*(2*(d*x + 1)/d^2 - 7/d^2) + 9/d^2) + 6*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^2)*B*
f + 5*(((d*x + 1)*(2*(d*x + 1)*(3*(d*x + 1)/d^3 - 13/d^3) + 43/d^3) - 39/d^3)*sqrt(d*x + 1)*sqrt(-d*x + 1) - 1
8*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^3)*C*f + 60*(sqrt(d*x + 1)*(d*x - 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(
2)*sqrt(d*x + 1)))*A*e + 120*(sqrt(d*x + 1)*sqrt(-d*x + 1) + 2*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*A*e + 20*(sq
rt(d*x + 1)*sqrt(-d*x + 1)*((d*x + 1)*(2*(d*x + 1)/d^2 - 7/d^2) + 9/d^2) + 6*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))
/d^2)*C*e + 60*(sqrt(d*x + 1)*(d*x - 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*A*f/d + 60*(sqrt
(d*x + 1)*(d*x - 2)*sqrt(-d*x + 1) - 2*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))*B*e/d)/d

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maple [C]  time = 0.01, size = 377, normalized size = 2.24 \[ \frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \left (24 \sqrt {-d^{2} x^{2}+1}\, C \,d^{4} f \,x^{4} \mathrm {csgn}\relax (d )+30 \sqrt {-d^{2} x^{2}+1}\, B \,d^{4} f \,x^{3} \mathrm {csgn}\relax (d )+30 \sqrt {-d^{2} x^{2}+1}\, C \,d^{4} e \,x^{3} \mathrm {csgn}\relax (d )+40 \sqrt {-d^{2} x^{2}+1}\, A \,d^{4} f \,x^{2} \mathrm {csgn}\relax (d )+40 \sqrt {-d^{2} x^{2}+1}\, B \,d^{4} e \,x^{2} \mathrm {csgn}\relax (d )+60 \sqrt {-d^{2} x^{2}+1}\, A \,d^{4} e x \,\mathrm {csgn}\relax (d )-8 \sqrt {-d^{2} x^{2}+1}\, C \,d^{2} f \,x^{2} \mathrm {csgn}\relax (d )+60 A \,d^{3} e \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )-15 \sqrt {-d^{2} x^{2}+1}\, B \,d^{2} f x \,\mathrm {csgn}\relax (d )-15 \sqrt {-d^{2} x^{2}+1}\, C \,d^{2} e x \,\mathrm {csgn}\relax (d )-40 \sqrt {-d^{2} x^{2}+1}\, A \,d^{2} f \,\mathrm {csgn}\relax (d )-40 \sqrt {-d^{2} x^{2}+1}\, B \,d^{2} e \,\mathrm {csgn}\relax (d )+15 B d f \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )+15 C d e \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )-16 \sqrt {-d^{2} x^{2}+1}\, C f \,\mathrm {csgn}\relax (d )\right ) \mathrm {csgn}\relax (d )}{120 \sqrt {-d^{2} x^{2}+1}\, d^{4}} \]

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(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d))*d^3*e-40*B*csgn(d)*(-d^2*x^2+1)^(1/2)*d^2*e+15*B*arctan(1/(
-d^2*x^2+1)^(1/2)*d*x*csgn(d))*d*f-16*C*csgn(d)*(-d^2*x^2+1)^(1/2)*f+15*C*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn
(d))*d*e)*csgn(d)/d^4/(-d^2*x^2+1)^(1/2)

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maxima [A]  time = 1.07, size = 174, normalized size = 1.04 \[ \frac {1}{2} \, \sqrt {-d^{2} x^{2} + 1} A e x - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} C f x^{2}}{5 \, d^{2}} + \frac {A e \arcsin \left (d x\right )}{2 \, d} - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} B e}{3 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} A f}{3 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (C e + B f\right )} x}{4 \, d^{2}} + \frac {\sqrt {-d^{2} x^{2} + 1} {\left (C e + B f\right )} x}{8 \, d^{2}} - \frac {2 \, {\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} C f}{15 \, d^{4}} + \frac {{\left (C e + B f\right )} \arcsin \left (d x\right )}{8 \, d^{3}} \]

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*x)/d - 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*x)/d^3

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mupad [B]  time = 12.06, size = 736, normalized size = 4.38 \[ \frac {\frac {B\,f\,\left (\sqrt {1-d\,x}-1\right )}{2\,\left (\sqrt {d\,x+1}-1\right )}-\frac {35\,B\,f\,{\left (\sqrt {1-d\,x}-1\right )}^3}{2\,{\left (\sqrt {d\,x+1}-1\right )}^3}+\frac {273\,B\,f\,{\left (\sqrt {1-d\,x}-1\right )}^5}{2\,{\left (\sqrt {d\,x+1}-1\right )}^5}-\frac {715\,B\,f\,{\left (\sqrt {1-d\,x}-1\right )}^7}{2\,{\left (\sqrt {d\,x+1}-1\right )}^7}+\frac {715\,B\,f\,{\left (\sqrt {1-d\,x}-1\right )}^9}{2\,{\left (\sqrt {d\,x+1}-1\right )}^9}-\frac {273\,B\,f\,{\left (\sqrt {1-d\,x}-1\right )}^{11}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{11}}+\frac {35\,B\,f\,{\left (\sqrt {1-d\,x}-1\right )}^{13}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{13}}-\frac {B\,f\,{\left (\sqrt {1-d\,x}-1\right )}^{15}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{15}}}{d^3\,{\left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )}^8}-\sqrt {1-d\,x}\,\left (\frac {2\,C\,f\,\sqrt {d\,x+1}}{15\,d^4}-\frac {C\,f\,x^4\,\sqrt {d\,x+1}}{5}+\frac {C\,f\,x^2\,\sqrt {d\,x+1}}{15\,d^2}\right )+\frac {\frac {C\,e\,\left (\sqrt {1-d\,x}-1\right )}{2\,\left (\sqrt {d\,x+1}-1\right )}-\frac {35\,C\,e\,{\left (\sqrt {1-d\,x}-1\right )}^3}{2\,{\left (\sqrt {d\,x+1}-1\right )}^3}+\frac {273\,C\,e\,{\left (\sqrt {1-d\,x}-1\right )}^5}{2\,{\left (\sqrt {d\,x+1}-1\right )}^5}-\frac {715\,C\,e\,{\left (\sqrt {1-d\,x}-1\right )}^7}{2\,{\left (\sqrt {d\,x+1}-1\right )}^7}+\frac {715\,C\,e\,{\left (\sqrt {1-d\,x}-1\right )}^9}{2\,{\left (\sqrt {d\,x+1}-1\right )}^9}-\frac {273\,C\,e\,{\left (\sqrt {1-d\,x}-1\right )}^{11}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{11}}+\frac {35\,C\,e\,{\left (\sqrt {1-d\,x}-1\right )}^{13}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{13}}-\frac {C\,e\,{\left (\sqrt {1-d\,x}-1\right )}^{15}}{2\,{\left (\sqrt {d\,x+1}-1\right )}^{15}}}{d^3\,{\left (\frac {{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+1\right )}^8}-\frac {B\,f\,\mathrm {atan}\left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{2\,d^3}-\frac {C\,e\,\mathrm {atan}\left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )}{2\,d^3}+\frac {A\,e\,x\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}}{2}-\frac {A\,\sqrt {d}\,e\,\ln \left (\sqrt {-d}\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}-d^{3/2}\,x\right )}{2\,{\left (-d\right )}^{3/2}}+\frac {A\,f\,\left (d^2\,x^2-1\right )\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}}{3\,d^2}+\frac {B\,e\,\left (d^2\,x^2-1\right )\,\sqrt {1-d\,x}\,\sqrt {d\,x+1}}{3\,d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*f*((1 - d*x)^(1/2) - 1))/(2*((d*x + 1)^(1/2) - 1)) - (35*B*f*((1 - d*x)^(1/2) - 1)^3)/(2*((d*x + 1)^(1/2)
- 1)^3) + (273*B*f*((1 - d*x)^(1/2) - 1)^5)/(2*((d*x + 1)^(1/2) - 1)^5) - (715*B*f*((1 - d*x)^(1/2) - 1)^7)/(2
*((d*x + 1)^(1/2) - 1)^7) + (715*B*f*((1 - d*x)^(1/2) - 1)^9)/(2*((d*x + 1)^(1/2) - 1)^9) - (273*B*f*((1 - d*x
)^(1/2) - 1)^11)/(2*((d*x + 1)^(1/2) - 1)^11) + (35*B*f*((1 - d*x)^(1/2) - 1)^13)/(2*((d*x + 1)^(1/2) - 1)^13)
- (B*f*((1 - d*x)^(1/2) - 1)^15)/(2*((d*x + 1)^(1/2) - 1)^15))/(d^3*(((1 - d*x)^(1/2) - 1)^2/((d*x + 1)^(1/2)
- 1)^2 + 1)^8) - (1 - d*x)^(1/2)*((2*C*f*(d*x + 1)^(1/2))/(15*d^4) - (C*f*x^4*(d*x + 1)^(1/2))/5 + (C*f*x^2*(
d*x + 1)^(1/2))/(15*d^2)) + ((C*e*((1 - d*x)^(1/2) - 1))/(2*((d*x + 1)^(1/2) - 1)) - (35*C*e*((1 - d*x)^(1/2)
- 1)^3)/(2*((d*x + 1)^(1/2) - 1)^3) + (273*C*e*((1 - d*x)^(1/2) - 1)^5)/(2*((d*x + 1)^(1/2) - 1)^5) - (715*C*e
*((1 - d*x)^(1/2) - 1)^7)/(2*((d*x + 1)^(1/2) - 1)^7) + (715*C*e*((1 - d*x)^(1/2) - 1)^9)/(2*((d*x + 1)^(1/2)
- 1)^9) - (273*C*e*((1 - d*x)^(1/2) - 1)^11)/(2*((d*x + 1)^(1/2) - 1)^11) + (35*C*e*((1 - d*x)^(1/2) - 1)^13)/
(2*((d*x + 1)^(1/2) - 1)^13) - (C*e*((1 - d*x)^(1/2) - 1)^15)/(2*((d*x + 1)^(1/2) - 1)^15))/(d^3*(((1 - d*x)^(
1/2) - 1)^2/((d*x + 1)^(1/2) - 1)^2 + 1)^8) - (B*f*atan(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/(2*d^3)
- (C*e*atan(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/(2*d^3) + (A*e*x*(1 - d*x)^(1/2)*(d*x + 1)^(1/2))/2
- (A*d^(1/2)*e*log((-d)^(1/2)*(1 - d*x)^(1/2)*(d*x + 1)^(1/2) - d^(3/2)*x))/(2*(-d)^(3/2)) + (A*f*(d^2*x^2 - 1
)*(1 - d*x)^(1/2)*(d*x + 1)^(1/2))/(3*d^2) + (B*e*(d^2*x^2 - 1)*(1 - d*x)^(1/2)*(d*x + 1)^(1/2))/(3*d^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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