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

Optimal. Leaf size=130 \[ -\frac{\sqrt{1-d^2 x^2} \left (2 \left (3 d^2 f (A f+B e)-C \left (d^2 e^2-2 f^2\right )\right )-d^2 f x (C e-3 B f)\right )}{6 d^4 f}+\frac{\sin ^{-1}(d x) \left (2 A d^2 e+B f+C e\right )}{2 d^3}-\frac{C \sqrt{1-d^2 x^2} (e+f x)^2}{3 d^2 f} \]

[Out]

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

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Rubi [A]  time = 0.229777, antiderivative size = 133, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {1609, 1654, 780, 216} \[ -\frac{\sqrt{1-d^2 x^2} \left (2 \left (3 d^2 f (A f+B e)-\frac{1}{2} C \left (2 d^2 e^2-4 f^2\right )\right )-d^2 f x (C e-3 B f)\right )}{6 d^4 f}+\frac{\sin ^{-1}(d x) \left (2 A d^2 e+B f+C e\right )}{2 d^3}-\frac{C \sqrt{1-d^2 x^2} (e+f x)^2}{3 d^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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) \left (A+B x+C x^2\right )}{\sqrt{1-d x} \sqrt{1+d x}} \, dx &=\int \frac{(e+f x) \left (A+B x+C x^2\right )}{\sqrt{1-d^2 x^2}} \, dx\\ &=-\frac{C (e+f x)^2 \sqrt{1-d^2 x^2}}{3 d^2 f}-\frac{\int \frac{(e+f x) \left (-\left (2 C+3 A d^2\right ) f^2+d^2 f (C e-3 B f) x\right )}{\sqrt{1-d^2 x^2}} \, dx}{3 d^2 f^2}\\ &=-\frac{C (e+f x)^2 \sqrt{1-d^2 x^2}}{3 d^2 f}-\frac{\left (2 \left (3 d^2 f (B e+A f)-\frac{1}{2} C \left (2 d^2 e^2-4 f^2\right )\right )-d^2 f (C e-3 B f) x\right ) \sqrt{1-d^2 x^2}}{6 d^4 f}+\frac{\left (C e+2 A d^2 e+B f\right ) \int \frac{1}{\sqrt{1-d^2 x^2}} \, dx}{2 d^2}\\ &=-\frac{C (e+f x)^2 \sqrt{1-d^2 x^2}}{3 d^2 f}-\frac{\left (2 \left (3 d^2 f (B e+A f)-\frac{1}{2} C \left (2 d^2 e^2-4 f^2\right )\right )-d^2 f (C e-3 B f) x\right ) \sqrt{1-d^2 x^2}}{6 d^4 f}+\frac{\left (C e+2 A d^2 e+B f\right ) \sin ^{-1}(d x)}{2 d^3}\\ \end{align*}

Mathematica [A]  time = 0.100861, size = 88, normalized size = 0.68 \[ \frac{3 d \sin ^{-1}(d x) \left (2 A d^2 e+B f+C e\right )-\sqrt{1-d^2 x^2} \left (6 A d^2 f+3 B d^2 (2 e+f x)+C \left (3 d^2 e x+2 d^2 f x^2+4 f\right )\right )}{6 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.018, size = 235, normalized size = 1.8 \begin{align*} -{\frac{{\it csgn} \left ( d \right ) }{6\,{d}^{4}}\sqrt{-dx+1}\sqrt{dx+1} \left ( 2\,C{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}{x}^{2}{d}^{2}f+3\,B{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}x{d}^{2}f+3\,C{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}x{d}^{2}e+6\,A{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}{d}^{2}f-6\,A\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ){d}^{3}e+6\,B{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}{d}^{2}e-3\,B\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ) df+4\,C{\it csgn} \left ( d \right ) \sqrt{-{d}^{2}{x}^{2}+1}f-3\,C\arctan \left ({\frac{{\it csgn} \left ( d \right ) dx}{\sqrt{-{d}^{2}{x}^{2}+1}}} \right ) de \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)*(C*x^2+B*x+A)/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x)

[Out]

-1/6*(-d*x+1)^(1/2)*(d*x+1)^(1/2)*(2*C*csgn(d)*(-d^2*x^2+1)^(1/2)*x^2*d^2*f+3*B*csgn(d)*(-d^2*x^2+1)^(1/2)*x*d
^2*f+3*C*csgn(d)*(-d^2*x^2+1)^(1/2)*x*d^2*e+6*A*csgn(d)*(-d^2*x^2+1)^(1/2)*d^2*f-6*A*arctan(csgn(d)*d*x/(-d^2*
x^2+1)^(1/2))*d^3*e+6*B*csgn(d)*(-d^2*x^2+1)^(1/2)*d^2*e-3*B*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d*f+4*C*cs
gn(d)*(-d^2*x^2+1)^(1/2)*f-3*C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*d*e)*csgn(d)/d^4/(-d^2*x^2+1)^(1/2)

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Maxima [A]  time = 3.60375, size = 205, normalized size = 1.58 \begin{align*} -\frac{\sqrt{-d^{2} x^{2} + 1} C f x^{2}}{3 \, d^{2}} + \frac{A e \arcsin \left (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{\sqrt{d^{2}}} - \frac{\sqrt{-d^{2} x^{2} + 1} B e}{d^{2}} - \frac{\sqrt{-d^{2} x^{2} + 1} A f}{d^{2}} - \frac{\sqrt{-d^{2} x^{2} + 1}{\left (C e + B f\right )} x}{2 \, d^{2}} + \frac{{\left (C e + B f\right )} \arcsin \left (\frac{d^{2} x}{\sqrt{d^{2}}}\right )}{2 \, \sqrt{d^{2}} d^{2}} - \frac{2 \, \sqrt{-d^{2} x^{2} + 1} C f}{3 \, d^{4}} \end{align*}

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]

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

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Fricas [A]  time = 1.08703, size = 267, normalized size = 2.05 \begin{align*} -\frac{{\left (2 \, C d^{2} f x^{2} + 6 \, B d^{2} e + 2 \,{\left (3 \, A d^{2} + 2 \, C\right )} f + 3 \,{\left (C d^{2} e + B d^{2} f\right )} x\right )} \sqrt{d x + 1} \sqrt{-d x + 1} + 6 \,{\left (B d f +{\left (2 \, A d^{3} + C d\right )} e\right )} \arctan \left (\frac{\sqrt{d x + 1} \sqrt{-d x + 1} - 1}{d x}\right )}{6 \, d^{4}} \end{align*}

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]

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

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Sympy [C]  time = 112.876, size = 617, normalized size = 4.75 \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)*(C*x**2+B*x+A)/(-d*x+1)**(1/2)/(d*x+1)**(1/2),x)

[Out]

-I*A*e*meijerg(((1/4, 3/4), (1/2, 1/2, 1, 1)), ((0, 1/4, 1/2, 3/4, 1, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d)
+ A*e*meijerg(((-1/2, -1/4, 0, 1/4, 1/2, 1), ()), ((-1/4, 1/4), (-1/2, 0, 0, 0)), exp_polar(-2*I*pi)/(d**2*x**
2))/(4*pi**(3/2)*d) - I*A*f*meijerg(((-1/4, 1/4), (0, 0, 1/2, 1)), ((-1/2, -1/4, 0, 1/4, 1/2, 0), ()), 1/(d**2
*x**2))/(4*pi**(3/2)*d**2) - A*f*meijerg(((-1, -3/4, -1/2, -1/4, 0, 1), ()), ((-3/4, -1/4), (-1, -1/2, -1/2, 0
)), exp_polar(-2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**2) - I*B*e*meijerg(((-1/4, 1/4), (0, 0, 1/2, 1)), ((-1/2,
-1/4, 0, 1/4, 1/2, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d**2) - B*e*meijerg(((-1, -3/4, -1/2, -1/4, 0, 1), ())
, ((-3/4, -1/4), (-1, -1/2, -1/2, 0)), exp_polar(-2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**2) - I*B*f*meijerg(((-3
/4, -1/4), (-1/2, -1/2, 0, 1)), ((-1, -3/4, -1/2, -1/4, 0, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d**3) + B*f*me
ijerg(((-3/2, -5/4, -1, -3/4, -1/2, 1), ()), ((-5/4, -3/4), (-3/2, -1, -1, 0)), exp_polar(-2*I*pi)/(d**2*x**2)
)/(4*pi**(3/2)*d**3) - I*C*e*meijerg(((-3/4, -1/4), (-1/2, -1/2, 0, 1)), ((-1, -3/4, -1/2, -1/4, 0, 0), ()), 1
/(d**2*x**2))/(4*pi**(3/2)*d**3) + C*e*meijerg(((-3/2, -5/4, -1, -3/4, -1/2, 1), ()), ((-5/4, -3/4), (-3/2, -1
, -1, 0)), exp_polar(-2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**3) - I*C*f*meijerg(((-5/4, -3/4), (-1, -1, -1/2, 1)
), ((-3/2, -5/4, -1, -3/4, -1/2, 0), ()), 1/(d**2*x**2))/(4*pi**(3/2)*d**4) - C*f*meijerg(((-2, -7/4, -3/2, -5
/4, -1, 1), ()), ((-7/4, -5/4), (-2, -3/2, -3/2, 0)), exp_polar(-2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**4)

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Giac [A]  time = 1.98318, size = 186, normalized size = 1.43 \begin{align*} -\frac{{\left (6 \, A d^{11} f + 6 \, B d^{11} e - 3 \, B d^{10} f - 3 \, C d^{10} e + 6 \, C d^{9} f +{\left (2 \,{\left (d x + 1\right )} C d^{9} f + 3 \, B d^{10} f + 3 \, C d^{10} e - 4 \, C d^{9} f\right )}{\left (d x + 1\right )}\right )} \sqrt{d x + 1} \sqrt{-d x + 1} - 6 \,{\left (2 \, A d^{12} e + B d^{10} f + C d^{10} e\right )} \arcsin \left (\frac{1}{2} \, \sqrt{2} \sqrt{d x + 1}\right )}{3840 \, d} \end{align*}

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]

-1/3840*((6*A*d^11*f + 6*B*d^11*e - 3*B*d^10*f - 3*C*d^10*e + 6*C*d^9*f + (2*(d*x + 1)*C*d^9*f + 3*B*d^10*f +
3*C*d^10*e - 4*C*d^9*f)*(d*x + 1))*sqrt(d*x + 1)*sqrt(-d*x + 1) - 6*(2*A*d^12*e + B*d^10*f + C*d^10*e)*arcsin(
1/2*sqrt(2)*sqrt(d*x + 1)))/d