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

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

[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)

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Rubi [A]  time = 0.492606, antiderivative size = 228, normalized size of antiderivative = 1., 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} \[ \frac{\sqrt{1-d^2 x^2} \left (4 \left (C \left (d^2 e^3-8 e f^2\right )-4 f \left (3 A d^2 e f+B \left (d^2 e^2+f^2\right )\right )\right )-f x \left (3 f^2 \left (4 A d^2+3 C\right )-2 d^2 e (C e-4 B f)\right )\right )}{24 d^4 f}+\frac{\sin ^{-1}(d x) \left (4 d^2 \left (A \left (2 d^2 e^2+f^2\right )+2 B e f\right )+C \left (4 d^2 e^2+3 f^2\right )\right )}{8 d^5}+\frac{\sqrt{1-d^2 x^2} (e+f x)^2 (C e-4 B f)}{12 d^2 f}-\frac{C \sqrt{1-d^2 x^2} (e+f x)^3}{4 d^2 f} \]

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 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)^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 (3 C+4 A d^2\right ) f^2+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*}

Mathematica [A]  time = 0.207138, size = 160, normalized size = 0.7 \[ \frac{3 \sin ^{-1}(d x) \left (4 d^2 \left (A \left (2 d^2 e^2+f^2\right )+2 B e f\right )+C \left (4 d^2 e^2+3 f^2\right )\right )-d \sqrt{1-d^2 x^2} \left (12 A d^2 f (4 e+f x)+8 B \left (d^2 \left (3 e^2+3 e f x+f^2 x^2\right )+2 f^2\right )+C \left (12 d^2 e^2 x+16 e f \left (d^2 x^2+2\right )+3 f^2 x \left (2 d^2 x^2+3\right )\right )\right )}{24 d^5} \]

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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Maple [C]  time = 0.024, size = 423, normalized size = 1.9 \begin{align*} -{\frac{{\it csgn} \left ( d \right ) }{24\,{d}^{5}}\sqrt{-dx+1}\sqrt{dx+1} \left ( 6\,C{\it csgn} \left ( d \right ){d}^{3}\sqrt{-{d}^{2}{x}^{2}+1}{x}^{3}{f}^{2}+8\,B{\it csgn} \left ( d \right ){d}^{3}\sqrt{-{d}^{2}{x}^{2}+1}{x}^{2}{f}^{2}+16\,C{\it csgn} \left ( d \right ){d}^{3}\sqrt{-{d}^{2}{x}^{2}+1}{x}^{2}ef+12\,A{\it csgn} \left ( d \right ){d}^{3}\sqrt{-{d}^{2}{x}^{2}+1}x{f}^{2}+24\,B{\it csgn} \left ( d \right ){d}^{3}\sqrt{-{d}^{2}{x}^{2}+1}xef+12\,C{\it csgn} \left ( d \right ){d}^{3}\sqrt{-{d}^{2}{x}^{2}+1}x{e}^{2}+48\,A{\it csgn} \left ( d \right ){d}^{3}\sqrt{-{d}^{2}{x}^{2}+1}ef-24\,A\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ){d}^{4}{e}^{2}+24\,B{\it csgn} \left ( d \right ){d}^{3}\sqrt{-{d}^{2}{x}^{2}+1}{e}^{2}+9\,C{\it csgn} \left ( d \right ) d\sqrt{-{d}^{2}{x}^{2}+1}x{f}^{2}-12\,A\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ){d}^{2}{f}^{2}+16\,B{\it csgn} \left ( d \right ) d\sqrt{-{d}^{2}{x}^{2}+1}{f}^{2}-24\,B\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ){d}^{2}ef+32\,C{\it csgn} \left ( d \right ) d\sqrt{-{d}^{2}{x}^{2}+1}ef-12\,C\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ){d}^{2}{e}^{2}-9\,C\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ){f}^{2} \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)^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*C*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*x^3*f^2+8*B*csgn(d)*d^3*(-d^2*x^2+1)^(1
/2)*x^2*f^2+16*C*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*x^2*e*f+12*A*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*x*f^2+24*B*csgn(d)
*d^3*(-d^2*x^2+1)^(1/2)*x*e*f+12*C*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*x*e^2+48*A*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*e*
f-24*A*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d^4*e^2+24*B*csgn(d)*d^3*(-d^2*x^2+1)^(1/2)*e^2+9*C*csgn(d)*d*(-
d^2*x^2+1)^(1/2)*x*f^2-12*A*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d^2*f^2+16*B*csgn(d)*d*(-d^2*x^2+1)^(1/2)*f
^2-24*B*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d^2*e*f+32*C*csgn(d)*d*(-d^2*x^2+1)^(1/2)*e*f-12*C*arctan(csgn(
d)*d*x/(-d^2*x^2+1)^(1/2))*d^2*e^2-9*C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*f^2)*csgn(d)/d^5/(-d^2*x^2+1)^(1
/2)

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Maxima [A]  time = 4.34213, size = 356, normalized size = 1.56 \begin{align*} -\frac{\sqrt{-d^{2} x^{2} + 1} C f^{2} x^{3}}{4 \, d^{2}} + \frac{A e^{2} \arcsin \left (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{\sqrt{d^{2}}} - \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 (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{2 \, \sqrt{d^{2}} d^{2}} + \frac{3 \, C f^{2} \arcsin \left (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{8 \, \sqrt{d^{2}} d^{4}} - \frac{2 \, \sqrt{-d^{2} x^{2} + 1}{\left (2 \, C e f + B f^{2}\right )}}{3 \, 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]

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

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Fricas [A]  time = 1.14281, size = 435, normalized size = 1.91 \begin{align*} -\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{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]

-1/24*((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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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 [A]  time = 2.82415, size = 352, normalized size = 1.54 \begin{align*} -\frac{{\left (48 \, A d^{19} f e - 12 \, A d^{18} f^{2} + 24 \, B d^{19} e^{2} - 24 \, B d^{18} f e + 24 \, B d^{17} f^{2} - 12 \, C d^{18} e^{2} + 48 \, C d^{17} f e - 15 \, C d^{16} f^{2} +{\left (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} + 2 \,{\left (3 \,{\left (d x + 1\right )} C d^{16} f^{2} + 4 \, B d^{17} f^{2} + 8 \, C d^{17} f e - 9 \, C d^{16} f^{2}\right )}{\left (d x + 1\right )}\right )}{\left (d x + 1\right )}\right )} \sqrt{d x + 1} \sqrt{-d x + 1} - 6 \,{\left (8 \, A d^{20} e^{2} + 4 \, A d^{18} f^{2} + 8 \, B d^{18} f e + 4 \, C d^{18} e^{2} + 3 \, C d^{16} f^{2}\right )} \arcsin \left (\frac{1}{2} \, \sqrt{2} \sqrt{d x + 1}\right )}{86016 \, d} \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]

-1/86016*((48*A*d^19*f*e - 12*A*d^18*f^2 + 24*B*d^19*e^2 - 24*B*d^18*f*e + 24*B*d^17*f^2 - 12*C*d^18*e^2 + 48*
C*d^17*f*e - 15*C*d^16*f^2 + (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 + 2*(3*(d*x + 1)*C*d^16*f^2 + 4*B*d^17*f^2 + 8*C*d^17*f*e - 9*C*d^16*f^2)*(d*x + 1))*(d*x + 1))*
sqrt(d*x + 1)*sqrt(-d*x + 1) - 6*(8*A*d^20*e^2 + 4*A*d^18*f^2 + 8*B*d^18*f*e + 4*C*d^18*e^2 + 3*C*d^16*f^2)*ar
csin(1/2*sqrt(2)*sqrt(d*x + 1)))/d