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

Optimal. Leaf size=228 \[ \frac {\sin ^{-1}(d x) \left (4 d^2 \left (A \left (2 d^2 e^2+f^2\right )+2 B e f\right )+C \left (4 d^2 e^2+3 f^2\right )\right )}{8 d^5}+\frac {\sqrt {1-d^2 x^2} \left (4 \left (C \left (d^2 e^3-8 e f^2\right )-4 f \left (3 A d^2 e f+B \left (d^2 e^2+f^2\right )\right )\right )-f x \left (3 f^2 \left (4 A d^2+3 C\right )-2 d^2 e (C e-4 B f)\right )\right )}{24 d^4 f}+\frac {\sqrt {1-d^2 x^2} (e+f x)^2 (C e-4 B f)}{12 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^3}{4 d^2 f} \]

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Rubi [A]  time = 0.49, antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \begin {gather*} \frac {\sqrt {1-d^2 x^2} \left (4 \left (C \left (d^2 e^3-8 e f^2\right )-4 f \left (3 A d^2 e f+B \left (d^2 e^2+f^2\right )\right )\right )-f x \left (3 f^2 \left (4 A d^2+3 C\right )-2 d^2 e (C e-4 B f)\right )\right )}{24 d^4 f}+\frac {\sin ^{-1}(d x) \left (4 d^2 \left (A \left (2 d^2 e^2+f^2\right )+2 B e f\right )+C \left (4 d^2 e^2+3 f^2\right )\right )}{8 d^5}+\frac {\sqrt {1-d^2 x^2} (e+f x)^2 (C e-4 B f)}{12 d^2 f}-\frac {C \sqrt {1-d^2 x^2} (e+f x)^3}{4 d^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((C*e - 4*B*f)*(e + f*x)^2*Sqrt[1 - d^2*x^2])/(12*d^2*f) - (C*(e + f*x)^3*Sqrt[1 - d^2*x^2])/(4*d^2*f) + ((4*(
C*(d^2*e^3 - 8*e*f^2) - 4*f*(3*A*d^2*e*f + B*(d^2*e^2 + f^2))) - f*(3*(3*C + 4*A*d^2)*f^2 - 2*d^2*e*(C*e - 4*B
*f))*x)*Sqrt[1 - d^2*x^2])/(24*d^4*f) + ((C*(4*d^2*e^2 + 3*f^2) + 4*d^2*(2*B*e*f + A*(2*d^2*e^2 + f^2)))*ArcSi
n[d*x])/(8*d^5)

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 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 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 \frac {(e+f x)^2 \left (A+B x+C x^2\right )}{\sqrt {1-d x} \sqrt {1+d x}} \, dx &=\int \frac {(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 \sqrt {1-d^2 x^2}}{4 d^2 f}-\frac {\int \frac {(e+f x)^2 \left (-\left (\left (3 C+4 A d^2\right ) f^2\right )+d^2 f (C e-4 B f) x\right )}{\sqrt {1-d^2 x^2}} \, dx}{4 d^2 f^2}\\ &=\frac {(C e-4 B f) (e+f x)^2 \sqrt {1-d^2 x^2}}{12 d^2 f}-\frac {C (e+f x)^3 \sqrt {1-d^2 x^2}}{4 d^2 f}+\frac {\int \frac {(e+f x) \left (d^2 f^2 \left (7 C e+12 A d^2 e+8 B f\right )+d^2 f \left (3 \left (3 C+4 A d^2\right ) f^2-2 d^2 e (C e-4 B f)\right ) x\right )}{\sqrt {1-d^2 x^2}} \, dx}{12 d^4 f^2}\\ &=\frac {(C e-4 B f) (e+f x)^2 \sqrt {1-d^2 x^2}}{12 d^2 f}-\frac {C (e+f x)^3 \sqrt {1-d^2 x^2}}{4 d^2 f}+\frac {\left (4 \left (C \left (d^2 e^3-8 e f^2\right )-4 f \left (3 A d^2 e f+B \left (d^2 e^2+f^2\right )\right )\right )-f \left (3 \left (3 C+4 A d^2\right ) f^2-2 d^2 e (C e-4 B f)\right ) x\right ) \sqrt {1-d^2 x^2}}{24 d^4 f}+\frac {\left (C \left (4 d^2 e^2+3 f^2\right )+4 d^2 \left (2 B e f+A \left (2 d^2 e^2+f^2\right )\right )\right ) \int \frac {1}{\sqrt {1-d^2 x^2}} \, dx}{8 d^4}\\ &=\frac {(C e-4 B f) (e+f x)^2 \sqrt {1-d^2 x^2}}{12 d^2 f}-\frac {C (e+f x)^3 \sqrt {1-d^2 x^2}}{4 d^2 f}+\frac {\left (4 \left (C \left (d^2 e^3-8 e f^2\right )-4 f \left (3 A d^2 e f+B \left (d^2 e^2+f^2\right )\right )\right )-f \left (3 \left (3 C+4 A d^2\right ) f^2-2 d^2 e (C e-4 B f)\right ) x\right ) \sqrt {1-d^2 x^2}}{24 d^4 f}+\frac {\left (C \left (4 d^2 e^2+3 f^2\right )+4 d^2 \left (2 B e f+A \left (2 d^2 e^2+f^2\right )\right )\right ) \sin ^{-1}(d x)}{8 d^5}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 160, normalized size = 0.70 \begin {gather*} \frac {3 \sin ^{-1}(d x) \left (4 d^2 \left (A \left (2 d^2 e^2+f^2\right )+2 B e f\right )+C \left (4 d^2 e^2+3 f^2\right )\right )-d \sqrt {1-d^2 x^2} \left (12 A d^2 f (4 e+f x)+8 B \left (d^2 \left (3 e^2+3 e f x+f^2 x^2\right )+2 f^2\right )+C \left (12 d^2 e^2 x+16 e f \left (d^2 x^2+2\right )+3 f^2 x \left (2 d^2 x^2+3\right )\right )\right )}{24 d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(d*Sqrt[1 - d^2*x^2]*(12*A*d^2*f*(4*e + f*x) + C*(12*d^2*e^2*x + 16*e*f*(2 + d^2*x^2) + 3*f^2*x*(3 + 2*d^2*x
^2)) + 8*B*(2*f^2 + d^2*(3*e^2 + 3*e*f*x + f^2*x^2)))) + 3*(C*(4*d^2*e^2 + 3*f^2) + 4*d^2*(2*B*e*f + A*(2*d^2*
e^2 + f^2)))*ArcSin[d*x])/(24*d^5)

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IntegrateAlgebraic [B]  time = 0.47, size = 708, normalized size = 3.11 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {1-d x}}{\sqrt {d x+1}}\right ) \left (-8 A d^4 e^2-4 A d^2 f^2-8 B d^2 e f-4 C d^2 e^2-3 C f^2\right )}{4 d^5}-\frac {\sqrt {1-d x} \left (\frac {144 A d^3 e f (1-d x)}{d x+1}+\frac {144 A d^3 e f (1-d x)^2}{(d x+1)^2}+\frac {48 A d^3 e f (1-d x)^3}{(d x+1)^3}+48 A d^3 e f+\frac {12 A d^2 f^2 (1-d x)}{d x+1}-\frac {12 A d^2 f^2 (1-d x)^2}{(d x+1)^2}-\frac {12 A d^2 f^2 (1-d x)^3}{(d x+1)^3}+12 A d^2 f^2+\frac {72 B d^3 e^2 (1-d x)}{d x+1}+\frac {72 B d^3 e^2 (1-d x)^2}{(d x+1)^2}+\frac {24 B d^3 e^2 (1-d x)^3}{(d x+1)^3}+24 B d^3 e^2+\frac {24 B d^2 e f (1-d x)}{d x+1}-\frac {24 B d^2 e f (1-d x)^2}{(d x+1)^2}-\frac {24 B d^2 e f (1-d x)^3}{(d x+1)^3}+24 B d^2 e f+\frac {40 B d f^2 (1-d x)}{d x+1}+\frac {40 B d f^2 (1-d x)^2}{(d x+1)^2}+\frac {24 B d f^2 (1-d x)^3}{(d x+1)^3}+24 B d f^2+\frac {12 C d^2 e^2 (1-d x)}{d x+1}-\frac {12 C d^2 e^2 (1-d x)^2}{(d x+1)^2}-\frac {12 C d^2 e^2 (1-d x)^3}{(d x+1)^3}+12 C d^2 e^2+\frac {80 C d e f (1-d x)}{d x+1}+\frac {80 C d e f (1-d x)^2}{(d x+1)^2}+\frac {48 C d e f (1-d x)^3}{(d x+1)^3}+48 C d e f-\frac {9 C f^2 (1-d x)}{d x+1}+\frac {9 C f^2 (1-d x)^2}{(d x+1)^2}-\frac {15 C f^2 (1-d x)^3}{(d x+1)^3}+15 C f^2\right )}{12 d^5 \sqrt {d x+1} \left (\frac {1-d x}{d x+1}+1\right )^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(Sqrt[1 - d*x]*(12*C*d^2*e^2 + 24*B*d^3*e^2 + 48*C*d*e*f + 24*B*d^2*e*f + 48*A*d^3*e*f + 15*C*f^2 + 24*B
*d*f^2 + 12*A*d^2*f^2 - (12*C*d^2*e^2*(1 - d*x)^3)/(1 + d*x)^3 + (24*B*d^3*e^2*(1 - d*x)^3)/(1 + d*x)^3 + (48*
C*d*e*f*(1 - d*x)^3)/(1 + d*x)^3 - (24*B*d^2*e*f*(1 - d*x)^3)/(1 + d*x)^3 + (48*A*d^3*e*f*(1 - d*x)^3)/(1 + d*
x)^3 - (15*C*f^2*(1 - d*x)^3)/(1 + d*x)^3 + (24*B*d*f^2*(1 - d*x)^3)/(1 + d*x)^3 - (12*A*d^2*f^2*(1 - d*x)^3)/
(1 + d*x)^3 - (12*C*d^2*e^2*(1 - d*x)^2)/(1 + d*x)^2 + (72*B*d^3*e^2*(1 - d*x)^2)/(1 + d*x)^2 + (80*C*d*e*f*(1
- d*x)^2)/(1 + d*x)^2 - (24*B*d^2*e*f*(1 - d*x)^2)/(1 + d*x)^2 + (144*A*d^3*e*f*(1 - d*x)^2)/(1 + d*x)^2 + (9
*C*f^2*(1 - d*x)^2)/(1 + d*x)^2 + (40*B*d*f^2*(1 - d*x)^2)/(1 + d*x)^2 - (12*A*d^2*f^2*(1 - d*x)^2)/(1 + d*x)^
2 + (12*C*d^2*e^2*(1 - d*x))/(1 + d*x) + (72*B*d^3*e^2*(1 - d*x))/(1 + d*x) + (80*C*d*e*f*(1 - d*x))/(1 + d*x)
+ (24*B*d^2*e*f*(1 - d*x))/(1 + d*x) + (144*A*d^3*e*f*(1 - d*x))/(1 + d*x) - (9*C*f^2*(1 - d*x))/(1 + d*x) +
(40*B*d*f^2*(1 - d*x))/(1 + d*x) + (12*A*d^2*f^2*(1 - d*x))/(1 + d*x)))/(d^5*Sqrt[1 + d*x]*(1 + (1 - d*x)/(1 +
d*x))^4) + ((-4*C*d^2*e^2 - 8*A*d^4*e^2 - 8*B*d^2*e*f - 3*C*f^2 - 4*A*d^2*f^2)*ArcTan[Sqrt[1 - d*x]/Sqrt[1 +
d*x]])/(4*d^5)

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fricas [A]  time = 0.82, size = 192, normalized size = 0.84 \begin {gather*} -\frac {{\left (6 \, C d^{3} f^{2} x^{3} + 24 \, B d^{3} e^{2} + 16 \, B d f^{2} + 16 \, {\left (3 \, A d^{3} + 2 \, C d\right )} e f + 8 \, {\left (2 \, C d^{3} e f + B d^{3} f^{2}\right )} x^{2} + 3 \, {\left (4 \, C d^{3} e^{2} + 8 \, B d^{3} e f + {\left (4 \, A d^{3} + 3 \, C d\right )} f^{2}\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 6 \, {\left (8 \, B d^{2} e f + 4 \, {\left (2 \, A d^{4} + C d^{2}\right )} e^{2} + {\left (4 \, A d^{2} + 3 \, C\right )} f^{2}\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{24 \, d^{5}} \end {gather*}

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]

-1/24*((6*C*d^3*f^2*x^3 + 24*B*d^3*e^2 + 16*B*d*f^2 + 16*(3*A*d^3 + 2*C*d)*e*f + 8*(2*C*d^3*e*f + B*d^3*f^2)*x
^2 + 3*(4*C*d^3*e^2 + 8*B*d^3*e*f + (4*A*d^3 + 3*C*d)*f^2)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 6*(8*B*d^2*e*f +
4*(2*A*d^4 + C*d^2)*e^2 + (4*A*d^2 + 3*C)*f^2)*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/d^5

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giac [A]  time = 1.64, size = 277, normalized size = 1.21 \begin {gather*} -\frac {{\left ({\left (d x + 1\right )} {\left (2 \, {\left (d x + 1\right )} {\left (\frac {3 \, {\left (d x + 1\right )} C f^{2}}{d^{4}} + \frac {4 \, B d^{17} f^{2} + 8 \, C d^{17} f e - 9 \, C d^{16} f^{2}}{d^{20}}\right )} + \frac {12 \, A d^{18} f^{2} + 24 \, B d^{18} f e - 16 \, B d^{17} f^{2} + 12 \, C d^{18} e^{2} - 32 \, C d^{17} f e + 27 \, C d^{16} f^{2}}{d^{20}}\right )} + \frac {3 \, {\left (16 \, A d^{19} f e - 4 \, A d^{18} f^{2} + 8 \, B d^{19} e^{2} - 8 \, B d^{18} f e + 8 \, B d^{17} f^{2} - 4 \, C d^{18} e^{2} + 16 \, C d^{17} f e - 5 \, C d^{16} f^{2}\right )}}{d^{20}}\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - \frac {6 \, {\left (8 \, A d^{4} e^{2} + 4 \, A d^{2} f^{2} + 8 \, B d^{2} f e + 4 \, C d^{2} e^{2} + 3 \, C f^{2}\right )} \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{d^{4}}}{24 \, d} \end {gather*}

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]

-1/24*(((d*x + 1)*(2*(d*x + 1)*(3*(d*x + 1)*C*f^2/d^4 + (4*B*d^17*f^2 + 8*C*d^17*f*e - 9*C*d^16*f^2)/d^20) + (
12*A*d^18*f^2 + 24*B*d^18*f*e - 16*B*d^17*f^2 + 12*C*d^18*e^2 - 32*C*d^17*f*e + 27*C*d^16*f^2)/d^20) + 3*(16*A
*d^19*f*e - 4*A*d^18*f^2 + 8*B*d^19*e^2 - 8*B*d^18*f*e + 8*B*d^17*f^2 - 4*C*d^18*e^2 + 16*C*d^17*f*e - 5*C*d^1
6*f^2)/d^20)*sqrt(d*x + 1)*sqrt(-d*x + 1) - 6*(8*A*d^4*e^2 + 4*A*d^2*f^2 + 8*B*d^2*f*e + 4*C*d^2*e^2 + 3*C*f^2
)*arcsin(1/2*sqrt(2)*sqrt(d*x + 1))/d^4)/d

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maple [C]  time = 0.03, size = 423, normalized size = 1.86 \begin {gather*} -\frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \left (6 \sqrt {-d^{2} x^{2}+1}\, C \,d^{3} f^{2} x^{3} \mathrm {csgn}\relax (d )+8 \sqrt {-d^{2} x^{2}+1}\, B \,d^{3} f^{2} x^{2} \mathrm {csgn}\relax (d )+16 \sqrt {-d^{2} x^{2}+1}\, C \,d^{3} e f \,x^{2} \mathrm {csgn}\relax (d )-24 A \,d^{4} e^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )+12 \sqrt {-d^{2} x^{2}+1}\, A \,d^{3} f^{2} x \,\mathrm {csgn}\relax (d )+24 \sqrt {-d^{2} x^{2}+1}\, B \,d^{3} e f x \,\mathrm {csgn}\relax (d )+12 \sqrt {-d^{2} x^{2}+1}\, C \,d^{3} e^{2} x \,\mathrm {csgn}\relax (d )+48 \sqrt {-d^{2} x^{2}+1}\, A \,d^{3} e f \,\mathrm {csgn}\relax (d )+24 \sqrt {-d^{2} x^{2}+1}\, B \,d^{3} e^{2} \mathrm {csgn}\relax (d )-12 A \,d^{2} f^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )-24 B \,d^{2} e f \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )-12 C \,d^{2} e^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )+9 \sqrt {-d^{2} x^{2}+1}\, C d \,f^{2} x \,\mathrm {csgn}\relax (d )+16 \sqrt {-d^{2} x^{2}+1}\, B d \,f^{2} \mathrm {csgn}\relax (d )+32 \sqrt {-d^{2} x^{2}+1}\, C d e f \,\mathrm {csgn}\relax (d )-9 C \,f^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )\right ) \mathrm {csgn}\relax (d )}{24 \sqrt {-d^{2} x^{2}+1}\, d^{5}} \end {gather*}

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]

-1/24*(-d*x+1)^(1/2)*(d*x+1)^(1/2)*(6*(-d^2*x^2+1)^(1/2)*C*d^3*f^2*x^3*csgn(d)+8*(-d^2*x^2+1)^(1/2)*B*d^3*f^2*
x^2*csgn(d)+16*(-d^2*x^2+1)^(1/2)*C*d^3*e*f*x^2*csgn(d)+12*(-d^2*x^2+1)^(1/2)*A*d^3*f^2*x*csgn(d)+24*(-d^2*x^2
+1)^(1/2)*B*d^3*e*f*x*csgn(d)+12*(-d^2*x^2+1)^(1/2)*C*d^3*e^2*x*csgn(d)+48*(-d^2*x^2+1)^(1/2)*A*d^3*e*f*csgn(d
)-24*A*d^4*e^2*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d))+24*(-d^2*x^2+1)^(1/2)*B*d^3*e^2*csgn(d)+9*(-d^2*x^2+1)
^(1/2)*C*d*f^2*x*csgn(d)-12*A*d^2*f^2*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d))+16*(-d^2*x^2+1)^(1/2)*B*d*f^2*c
sgn(d)-24*B*d^2*e*f*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d))+32*(-d^2*x^2+1)^(1/2)*C*d*e*f*csgn(d)-12*C*d^2*e^
2*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d))-9*C*f^2*arctan(1/(-d^2*x^2+1)^(1/2)*d*x*csgn(d)))*csgn(d)/d^5/(-d^2
*x^2+1)^(1/2)

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maxima [A]  time = 1.27, size = 231, normalized size = 1.01 \begin {gather*} -\frac {\sqrt {-d^{2} x^{2} + 1} C f^{2} x^{3}}{4 \, d^{2}} + \frac {A e^{2} \arcsin \left (d x\right )}{d} - \frac {\sqrt {-d^{2} x^{2} + 1} B e^{2}}{d^{2}} - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} A e f}{d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} {\left (2 \, C e f + B f^{2}\right )} x^{2}}{3 \, d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} {\left (C e^{2} + 2 \, B e f + A f^{2}\right )} x}{2 \, d^{2}} - \frac {3 \, \sqrt {-d^{2} x^{2} + 1} C f^{2} x}{8 \, d^{4}} + \frac {{\left (C e^{2} + 2 \, B e f + A f^{2}\right )} \arcsin \left (d x\right )}{2 \, d^{3}} + \frac {3 \, C f^{2} \arcsin \left (d x\right )}{8 \, d^{5}} - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} {\left (2 \, C e f + B f^{2}\right )}}{3 \, d^{4}} \end {gather*}

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]

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

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mupad [B]  time = 33.64, size = 1732, normalized size = 7.60

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((14*A*f^2*((1 - d*x)^(1/2) - 1)^3)/((d*x + 1)^(1/2) - 1)^3 - (2*A*f^2*((1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/
2) - 1) - (14*A*f^2*((1 - d*x)^(1/2) - 1)^5)/((d*x + 1)^(1/2) - 1)^5 + (2*A*f^2*((1 - d*x)^(1/2) - 1)^7)/((d*x
+ 1)^(1/2) - 1)^7 + (16*A*d*e*f*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 + (32*A*d*e*f*((1 - d*x)^(1/
2) - 1)^4)/((d*x + 1)^(1/2) - 1)^4 + (16*A*d*e*f*((1 - d*x)^(1/2) - 1)^6)/((d*x + 1)^(1/2) - 1)^6)/(d^3 + (4*d
^3*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 + (6*d^3*((1 - d*x)^(1/2) - 1)^4)/((d*x + 1)^(1/2) - 1)^4
+ (4*d^3*((1 - d*x)^(1/2) - 1)^6)/((d*x + 1)^(1/2) - 1)^6 + (d^3*((1 - d*x)^(1/2) - 1)^8)/((d*x + 1)^(1/2) - 1
)^8) - ((((1 - d*x)^(1/2) - 1)^4*(64*B*f^2 + 32*B*d^2*e^2))/((d*x + 1)^(1/2) - 1)^4 + (((1 - d*x)^(1/2) - 1)^8
*(64*B*f^2 + 32*B*d^2*e^2))/((d*x + 1)^(1/2) - 1)^8 - (((1 - d*x)^(1/2) - 1)^6*((128*B*f^2)/3 - 48*B*d^2*e^2))
/((d*x + 1)^(1/2) - 1)^6 + (8*B*d^2*e^2*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 + (8*B*d^2*e^2*((1 -
d*x)^(1/2) - 1)^10)/((d*x + 1)^(1/2) - 1)^10 + (20*B*d*e*f*((1 - d*x)^(1/2) - 1)^3)/((d*x + 1)^(1/2) - 1)^3 +
(24*B*d*e*f*((1 - d*x)^(1/2) - 1)^5)/((d*x + 1)^(1/2) - 1)^5 - (24*B*d*e*f*((1 - d*x)^(1/2) - 1)^7)/((d*x + 1)
^(1/2) - 1)^7 - (20*B*d*e*f*((1 - d*x)^(1/2) - 1)^9)/((d*x + 1)^(1/2) - 1)^9 + (4*B*d*e*f*((1 - d*x)^(1/2) - 1
)^11)/((d*x + 1)^(1/2) - 1)^11 - (4*B*d*e*f*((1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/2) - 1))/(d^4 + (6*d^4*((1 -
d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 + (15*d^4*((1 - d*x)^(1/2) - 1)^4)/((d*x + 1)^(1/2) - 1)^4 + (20*d^
4*((1 - d*x)^(1/2) - 1)^6)/((d*x + 1)^(1/2) - 1)^6 + (15*d^4*((1 - d*x)^(1/2) - 1)^8)/((d*x + 1)^(1/2) - 1)^8
+ (6*d^4*((1 - d*x)^(1/2) - 1)^10)/((d*x + 1)^(1/2) - 1)^10 + (d^4*((1 - d*x)^(1/2) - 1)^12)/((d*x + 1)^(1/2)
- 1)^12) - ((((1 - d*x)^(1/2) - 1)^15*((3*C*f^2)/2 + 2*C*d^2*e^2))/((d*x + 1)^(1/2) - 1)^15 - (((1 - d*x)^(1/2
) - 1)^3*((23*C*f^2)/2 - 6*C*d^2*e^2))/((d*x + 1)^(1/2) - 1)^3 - (((1 - d*x)^(1/2) - 1)*((3*C*f^2)/2 + 2*C*d^2
*e^2))/((d*x + 1)^(1/2) - 1) + (((1 - d*x)^(1/2) - 1)^13*((23*C*f^2)/2 - 6*C*d^2*e^2))/((d*x + 1)^(1/2) - 1)^1
3 + (((1 - d*x)^(1/2) - 1)^5*((333*C*f^2)/2 + 30*C*d^2*e^2))/((d*x + 1)^(1/2) - 1)^5 - (((1 - d*x)^(1/2) - 1)^
11*((333*C*f^2)/2 + 30*C*d^2*e^2))/((d*x + 1)^(1/2) - 1)^11 - (((1 - d*x)^(1/2) - 1)^7*((671*C*f^2)/2 - 22*C*d
^2*e^2))/((d*x + 1)^(1/2) - 1)^7 + (((1 - d*x)^(1/2) - 1)^9*((671*C*f^2)/2 - 22*C*d^2*e^2))/((d*x + 1)^(1/2) -
1)^9 + (128*C*d*e*f*((1 - d*x)^(1/2) - 1)^4)/((d*x + 1)^(1/2) - 1)^4 + (512*C*d*e*f*((1 - d*x)^(1/2) - 1)^6)/
(3*((d*x + 1)^(1/2) - 1)^6) + (256*C*d*e*f*((1 - d*x)^(1/2) - 1)^8)/(3*((d*x + 1)^(1/2) - 1)^8) + (512*C*d*e*f
*((1 - d*x)^(1/2) - 1)^10)/(3*((d*x + 1)^(1/2) - 1)^10) + (128*C*d*e*f*((1 - d*x)^(1/2) - 1)^12)/((d*x + 1)^(1
/2) - 1)^12)/(d^5 + (8*d^5*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 + (28*d^5*((1 - d*x)^(1/2) - 1)^4)
/((d*x + 1)^(1/2) - 1)^4 + (56*d^5*((1 - d*x)^(1/2) - 1)^6)/((d*x + 1)^(1/2) - 1)^6 + (70*d^5*((1 - d*x)^(1/2)
- 1)^8)/((d*x + 1)^(1/2) - 1)^8 + (56*d^5*((1 - d*x)^(1/2) - 1)^10)/((d*x + 1)^(1/2) - 1)^10 + (28*d^5*((1 -
d*x)^(1/2) - 1)^12)/((d*x + 1)^(1/2) - 1)^12 + (8*d^5*((1 - d*x)^(1/2) - 1)^14)/((d*x + 1)^(1/2) - 1)^14 + (d^
5*((1 - d*x)^(1/2) - 1)^16)/((d*x + 1)^(1/2) - 1)^16) - (C*atan((C*((1 - d*x)^(1/2) - 1)*(3*f^2 + 4*d^2*e^2))/
(((d*x + 1)^(1/2) - 1)*(3*C*f^2 + 4*C*d^2*e^2)))*(3*f^2 + 4*d^2*e^2))/(2*d^5) - (2*A*atan((A*(f^2 + 2*d^2*e^2)
*((1 - d*x)^(1/2) - 1))/(((d*x + 1)^(1/2) - 1)*(A*f^2 + 2*A*d^2*e^2)))*(f^2 + 2*d^2*e^2))/d^3 - (4*B*e*f*atan(
((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/d^3

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

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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