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

Optimal. Leaf size=248 \[ \frac{\sqrt{1-d^2 x^2} \left (A f^2-B e f+C e^2\right )}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}-\frac{\sqrt{1-d^2 x^2} \left (-3 A d^2 e f^2+B d^2 e^2 f+2 B f^3+C d^2 e^3-4 C e f^2\right )}{2 f \left (d^2 e^2-f^2\right )^2 (e+f x)}+\frac{\tan ^{-1}\left (\frac{d^2 e x+f}{\sqrt{1-d^2 x^2} \sqrt{d^2 e^2-f^2}}\right ) \left (C \left (d^2 e^2+2 f^2\right )-d^2 \left (3 B e f-A \left (2 d^2 e^2+f^2\right )\right )\right )}{2 \left (d^2 e^2-f^2\right )^{5/2}} \]

[Out]

((C*e^2 - B*e*f + A*f^2)*Sqrt[1 - d^2*x^2])/(2*f*(d^2*e^2 - f^2)*(e + f*x)^2) - ((C*d^2*e^3 + B*d^2*e^2*f - 4*
C*e*f^2 - 3*A*d^2*e*f^2 + 2*B*f^3)*Sqrt[1 - d^2*x^2])/(2*f*(d^2*e^2 - f^2)^2*(e + f*x)) + ((C*(d^2*e^2 + 2*f^2
) - d^2*(3*B*e*f - A*(2*d^2*e^2 + f^2)))*ArcTan[(f + d^2*e*x)/(Sqrt[d^2*e^2 - f^2]*Sqrt[1 - d^2*x^2])])/(2*(d^
2*e^2 - f^2)^(5/2))

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Rubi [A]  time = 0.328948, antiderivative size = 248, 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, 1651, 807, 725, 204} \[ \frac{\sqrt{1-d^2 x^2} \left (A f^2-B e f+C e^2\right )}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}-\frac{\sqrt{1-d^2 x^2} \left (-3 A d^2 e f^2+B d^2 e^2 f+2 B f^3+C d^2 e^3-4 C e f^2\right )}{2 f \left (d^2 e^2-f^2\right )^2 (e+f x)}+\frac{\tan ^{-1}\left (\frac{d^2 e x+f}{\sqrt{1-d^2 x^2} \sqrt{d^2 e^2-f^2}}\right ) \left (C \left (d^2 e^2+2 f^2\right )-d^2 \left (3 B e f-A \left (2 d^2 e^2+f^2\right )\right )\right )}{2 \left (d^2 e^2-f^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((C*e^2 - B*e*f + A*f^2)*Sqrt[1 - d^2*x^2])/(2*f*(d^2*e^2 - f^2)*(e + f*x)^2) - ((C*d^2*e^3 + B*d^2*e^2*f - 4*
C*e*f^2 - 3*A*d^2*e*f^2 + 2*B*f^3)*Sqrt[1 - d^2*x^2])/(2*f*(d^2*e^2 - f^2)^2*(e + f*x)) + ((C*(d^2*e^2 + 2*f^2
) - d^2*(3*B*e*f - A*(2*d^2*e^2 + f^2)))*ArcTan[(f + d^2*e*x)/(Sqrt[d^2*e^2 - f^2]*Sqrt[1 - d^2*x^2])])/(2*(d^
2*e^2 - f^2)^(5/2))

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 1651

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, d
+ e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1
)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p*ExpandToSum[(m
+ 1)*(c*d^2 + a*e^2)*Q + c*d*R*(m + 1) - c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e, p}, x] && Po
lyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{A+B x+C x^2}{\sqrt{1-d x} \sqrt{1+d x} (e+f x)^3} \, dx &=\int \frac{A+B x+C x^2}{(e+f x)^3 \sqrt{1-d^2 x^2}} \, dx\\ &=\frac{\left (C e^2-B e f+A f^2\right ) \sqrt{1-d^2 x^2}}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}+\frac{\int \frac{2 \left (C e+A d^2 e-B f\right )+\left (B d^2 e+\frac{C d^2 e^2}{f}-2 C f-A d^2 f\right ) x}{(e+f x)^2 \sqrt{1-d^2 x^2}} \, dx}{2 \left (d^2 e^2-f^2\right )}\\ &=\frac{\left (C e^2-B e f+A f^2\right ) \sqrt{1-d^2 x^2}}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}-\frac{\left (C d^2 e^3+B d^2 e^2 f-4 C e f^2-3 A d^2 e f^2+2 B f^3\right ) \sqrt{1-d^2 x^2}}{2 f \left (d^2 e^2-f^2\right )^2 (e+f x)}+\frac{\left (C \left (d^2 e^2+2 f^2\right )-d^2 \left (3 B e f-A \left (2 d^2 e^2+f^2\right )\right )\right ) \int \frac{1}{(e+f x) \sqrt{1-d^2 x^2}} \, dx}{2 \left (d^2 e^2-f^2\right )^2}\\ &=\frac{\left (C e^2-B e f+A f^2\right ) \sqrt{1-d^2 x^2}}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}-\frac{\left (C d^2 e^3+B d^2 e^2 f-4 C e f^2-3 A d^2 e f^2+2 B f^3\right ) \sqrt{1-d^2 x^2}}{2 f \left (d^2 e^2-f^2\right )^2 (e+f x)}-\frac{\left (C \left (d^2 e^2+2 f^2\right )-d^2 \left (3 B e f-A \left (2 d^2 e^2+f^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-d^2 e^2+f^2-x^2} \, dx,x,\frac{f+d^2 e x}{\sqrt{1-d^2 x^2}}\right )}{2 \left (d^2 e^2-f^2\right )^2}\\ &=\frac{\left (C e^2-B e f+A f^2\right ) \sqrt{1-d^2 x^2}}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}-\frac{\left (C d^2 e^3+B d^2 e^2 f-4 C e f^2-3 A d^2 e f^2+2 B f^3\right ) \sqrt{1-d^2 x^2}}{2 f \left (d^2 e^2-f^2\right )^2 (e+f x)}+\frac{\left (C \left (d^2 e^2+2 f^2\right )-d^2 \left (3 B e f-A \left (2 d^2 e^2+f^2\right )\right )\right ) \tan ^{-1}\left (\frac{f+d^2 e x}{\sqrt{d^2 e^2-f^2} \sqrt{1-d^2 x^2}}\right )}{2 \left (d^2 e^2-f^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.178906, size = 273, normalized size = 1.1 \[ \frac{1}{2} \left (-\frac{\sqrt{1-d^2 x^2} \left (-A d^2 e f (4 e+3 f x)+A f^3+B d^2 e^2 (2 e+f x)+B f^2 (e+2 f x)+C e \left (d^2 e^2 x-3 e f-4 f^2 x\right )\right )}{\left (f^2-d^2 e^2\right )^2 (e+f x)^2}-\frac{\log \left (\sqrt{1-d^2 x^2} \sqrt{f^2-d^2 e^2}+d^2 e x+f\right ) \left (d^2 \left (A \left (2 d^2 e^2+f^2\right )-3 B e f\right )+C \left (d^2 e^2+2 f^2\right )\right )}{\left (f^2-d^2 e^2\right )^{5/2}}+\frac{\log (e+f x) \left (d^2 \left (A \left (2 d^2 e^2+f^2\right )-3 B e f\right )+C \left (d^2 e^2+2 f^2\right )\right )}{\left (f^2-d^2 e^2\right )^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((Sqrt[1 - d^2*x^2]*(A*f^3 + B*d^2*e^2*(2*e + f*x) + B*f^2*(e + 2*f*x) - A*d^2*e*f*(4*e + 3*f*x) + C*e*(-3*e
*f + d^2*e^2*x - 4*f^2*x)))/((-(d^2*e^2) + f^2)^2*(e + f*x)^2)) + ((C*(d^2*e^2 + 2*f^2) + d^2*(-3*B*e*f + A*(2
*d^2*e^2 + f^2)))*Log[e + f*x])/(-(d^2*e^2) + f^2)^(5/2) - ((C*(d^2*e^2 + 2*f^2) + d^2*(-3*B*e*f + A*(2*d^2*e^
2 + f^2)))*Log[f + d^2*e*x + Sqrt[-(d^2*e^2) + f^2]*Sqrt[1 - d^2*x^2]])/(-(d^2*e^2) + f^2)^(5/2))/2

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Maple [C]  time = 0., size = 1449, normalized size = 5.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)/(f*x+e)^3/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.42969, size = 3105, normalized size = 12.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*B*d^4*e^7 - B*d^2*e^5*f^2 - (4*A*d^4 + 3*C*d^2)*e^6*f + (5*A*d^2 + 3*C)*e^4*f^3 - B*e^3*f^4 - A*e^2*f
^5 + (2*B*d^4*e^5*f^2 - B*d^2*e^3*f^4 - (4*A*d^4 + 3*C*d^2)*e^4*f^3 + (5*A*d^2 + 3*C)*e^2*f^5 - B*e*f^6 - A*f^
7)*x^2 - (3*B*d^2*e^5*f - (2*A*d^4 + C*d^2)*e^6 - (A*d^2 + 2*C)*e^4*f^2 + (3*B*d^2*e^3*f^3 - (2*A*d^4 + C*d^2)
*e^4*f^2 - (A*d^2 + 2*C)*e^2*f^4)*x^2 + 2*(3*B*d^2*e^4*f^2 - (2*A*d^4 + C*d^2)*e^5*f - (A*d^2 + 2*C)*e^3*f^3)*
x)*sqrt(-d^2*e^2 + f^2)*log((d^2*e*f*x + f^2 - sqrt(-d^2*e^2 + f^2)*(d^2*e*x + f) - (sqrt(-d^2*e^2 + f^2)*sqrt
(-d*x + 1)*f + (d^2*e^2 - f^2)*sqrt(-d*x + 1))*sqrt(d*x + 1))/(f*x + e)) + (2*B*d^4*e^7 - B*d^2*e^5*f^2 - (4*A
*d^4 + 3*C*d^2)*e^6*f + (5*A*d^2 + 3*C)*e^4*f^3 - B*e^3*f^4 - A*e^2*f^5 + (C*d^4*e^7 + B*d^4*e^6*f + B*d^2*e^4
*f^3 - (3*A*d^4 + 5*C*d^2)*e^5*f^2 + (3*A*d^2 + 4*C)*e^3*f^4 - 2*B*e^2*f^5)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) +
2*(2*B*d^4*e^6*f - B*d^2*e^4*f^3 - (4*A*d^4 + 3*C*d^2)*e^5*f^2 + (5*A*d^2 + 3*C)*e^3*f^4 - B*e^2*f^5 - A*e*f^6
)*x)/(d^6*e^10 - 3*d^4*e^8*f^2 + 3*d^2*e^6*f^4 - e^4*f^6 + (d^6*e^8*f^2 - 3*d^4*e^6*f^4 + 3*d^2*e^4*f^6 - e^2*
f^8)*x^2 + 2*(d^6*e^9*f - 3*d^4*e^7*f^3 + 3*d^2*e^5*f^5 - e^3*f^7)*x), -1/2*(2*B*d^4*e^7 - B*d^2*e^5*f^2 - (4*
A*d^4 + 3*C*d^2)*e^6*f + (5*A*d^2 + 3*C)*e^4*f^3 - B*e^3*f^4 - A*e^2*f^5 + (2*B*d^4*e^5*f^2 - B*d^2*e^3*f^4 -
(4*A*d^4 + 3*C*d^2)*e^4*f^3 + (5*A*d^2 + 3*C)*e^2*f^5 - B*e*f^6 - A*f^7)*x^2 + 2*(3*B*d^2*e^5*f - (2*A*d^4 + C
*d^2)*e^6 - (A*d^2 + 2*C)*e^4*f^2 + (3*B*d^2*e^3*f^3 - (2*A*d^4 + C*d^2)*e^4*f^2 - (A*d^2 + 2*C)*e^2*f^4)*x^2
+ 2*(3*B*d^2*e^4*f^2 - (2*A*d^4 + C*d^2)*e^5*f - (A*d^2 + 2*C)*e^3*f^3)*x)*sqrt(d^2*e^2 - f^2)*arctan(-(sqrt(d
^2*e^2 - f^2)*sqrt(d*x + 1)*sqrt(-d*x + 1)*e - sqrt(d^2*e^2 - f^2)*(f*x + e))/((d^2*e^2 - f^2)*x)) + (2*B*d^4*
e^7 - B*d^2*e^5*f^2 - (4*A*d^4 + 3*C*d^2)*e^6*f + (5*A*d^2 + 3*C)*e^4*f^3 - B*e^3*f^4 - A*e^2*f^5 + (C*d^4*e^7
+ B*d^4*e^6*f + B*d^2*e^4*f^3 - (3*A*d^4 + 5*C*d^2)*e^5*f^2 + (3*A*d^2 + 4*C)*e^3*f^4 - 2*B*e^2*f^5)*x)*sqrt(
d*x + 1)*sqrt(-d*x + 1) + 2*(2*B*d^4*e^6*f - B*d^2*e^4*f^3 - (4*A*d^4 + 3*C*d^2)*e^5*f^2 + (5*A*d^2 + 3*C)*e^3
*f^4 - B*e^2*f^5 - A*e*f^6)*x)/(d^6*e^10 - 3*d^4*e^8*f^2 + 3*d^2*e^6*f^4 - e^4*f^6 + (d^6*e^8*f^2 - 3*d^4*e^6*
f^4 + 3*d^2*e^4*f^6 - e^2*f^8)*x^2 + 2*(d^6*e^9*f - 3*d^4*e^7*f^3 + 3*d^2*e^5*f^5 - e^3*f^7)*x)]

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)/(f*x+e)**3/(-d*x+1)**(1/2)/(d*x+1)**(1/2),x)

[Out]

Exception raised: ValueError

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError