3.13 \(\int \frac{A+B x+C x^2}{\sqrt{1-d x} \sqrt{1+d x} (e+f x)^2} \, dx\)
Optimal. Leaf size=163 \[ \frac{\sqrt{1-d^2 x^2} \left (A f^2-B e f+C e^2\right )}{f \left (d^2 e^2-f^2\right ) (e+f x)}-\frac{\tan ^{-1}\left (\frac{d^2 e x+f}{\sqrt{1-d^2 x^2} \sqrt{d^2 e^2-f^2}}\right ) \left (-A d^2 e f^2+B f^3+C d^2 e^3-2 C e f^2\right )}{f^2 \left (d^2 e^2-f^2\right )^{3/2}}+\frac{C \sin ^{-1}(d x)}{d f^2} \]
[Out]
((C*e^2 - B*e*f + A*f^2)*Sqrt[1 - d^2*x^2])/(f*(d^2*e^2 - f^2)*(e + f*x)) + (C*ArcSin[d*x])/(d*f^2) - ((C*d^2*
e^3 - 2*C*e*f^2 - A*d^2*e*f^2 + B*f^3)*ArcTan[(f + d^2*e*x)/(Sqrt[d^2*e^2 - f^2]*Sqrt[1 - d^2*x^2])])/(f^2*(d^
2*e^2 - f^2)^(3/2))
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Rubi [A] time = 0.295465, antiderivative size = 163, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used =
{1609, 1651, 844, 216, 725, 204} \[ \frac{\sqrt{1-d^2 x^2} \left (A f^2-B e f+C e^2\right )}{f \left (d^2 e^2-f^2\right ) (e+f x)}-\frac{\tan ^{-1}\left (\frac{d^2 e x+f}{\sqrt{1-d^2 x^2} \sqrt{d^2 e^2-f^2}}\right ) \left (-A d^2 e f^2+B f^3+C d^2 e^3-2 C e f^2\right )}{f^2 \left (d^2 e^2-f^2\right )^{3/2}}+\frac{C \sin ^{-1}(d x)}{d f^2} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*x + C*x^2)/(Sqrt[1 - d*x]*Sqrt[1 + d*x]*(e + f*x)^2),x]
[Out]
((C*e^2 - B*e*f + A*f^2)*Sqrt[1 - d^2*x^2])/(f*(d^2*e^2 - f^2)*(e + f*x)) + (C*ArcSin[d*x])/(d*f^2) - ((C*d^2*
e^3 - 2*C*e*f^2 - A*d^2*e*f^2 + B*f^3)*ArcTan[(f + d^2*e*x)/(Sqrt[d^2*e^2 - f^2]*Sqrt[1 - d^2*x^2])])/(f^2*(d^
2*e^2 - f^2)^(3/2))
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 844
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] && !IGtQ[m, 0]
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 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)^2} \, dx &=\int \frac{A+B x+C x^2}{(e+f x)^2 \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}}{f \left (d^2 e^2-f^2\right ) (e+f x)}+\frac{\int \frac{C e+A d^2 e-B f+C \left (\frac{d^2 e^2}{f}-f\right ) x}{(e+f x) \sqrt{1-d^2 x^2}} \, dx}{d^2 e^2-f^2}\\ &=\frac{\left (C e^2-B e f+A f^2\right ) \sqrt{1-d^2 x^2}}{f \left (d^2 e^2-f^2\right ) (e+f x)}+\frac{C \int \frac{1}{\sqrt{1-d^2 x^2}} \, dx}{f^2}+\frac{\left (2 C e+A d^2 e-\frac{C d^2 e^3}{f^2}-B f\right ) \int \frac{1}{(e+f x) \sqrt{1-d^2 x^2}} \, dx}{d^2 e^2-f^2}\\ &=\frac{\left (C e^2-B e f+A f^2\right ) \sqrt{1-d^2 x^2}}{f \left (d^2 e^2-f^2\right ) (e+f x)}+\frac{C \sin ^{-1}(d x)}{d f^2}-\frac{\left (2 C e+A d^2 e-\frac{C d^2 e^3}{f^2}-B f\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 )}{d^2 e^2-f^2}\\ &=\frac{\left (C e^2-B e f+A f^2\right ) \sqrt{1-d^2 x^2}}{f \left (d^2 e^2-f^2\right ) (e+f x)}+\frac{C \sin ^{-1}(d x)}{d f^2}+\frac{\left (2 C e+A d^2 e-\frac{C d^2 e^3}{f^2}-B f\right ) \tan ^{-1}\left (\frac{f+d^2 e x}{\sqrt{d^2 e^2-f^2} \sqrt{1-d^2 x^2}}\right )}{\left (d^2 e^2-f^2\right )^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.412729, size = 211, normalized size = 1.29 \[ \frac{-\frac{f \sqrt{1-d^2 x^2} \left (f (A f-B e)+C e^2\right )}{\left (f^2-d^2 e^2\right ) (e+f x)}-\frac{\log \left (\sqrt{1-d^2 x^2} \sqrt{f^2-d^2 e^2}+d^2 e x+f\right ) \left (-A d^2 e f^2+B f^3+C d^2 e^3-2 C e f^2\right )}{\left (f^2-d^2 e^2\right )^{3/2}}+\frac{\log (e+f x) \left (-A d^2 e f^2+B f^3+C d^2 e^3-2 C e f^2\right )}{\left (f^2-d^2 e^2\right )^{3/2}}+\frac{C \sin ^{-1}(d x)}{d}}{f^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*x + C*x^2)/(Sqrt[1 - d*x]*Sqrt[1 + d*x]*(e + f*x)^2),x]
[Out]
(-((f*(C*e^2 + f*(-(B*e) + A*f))*Sqrt[1 - d^2*x^2])/((-(d^2*e^2) + f^2)*(e + f*x))) + (C*ArcSin[d*x])/d + ((C*
d^2*e^3 - 2*C*e*f^2 - A*d^2*e*f^2 + B*f^3)*Log[e + f*x])/(-(d^2*e^2) + f^2)^(3/2) - ((C*d^2*e^3 - 2*C*e*f^2 -
A*d^2*e*f^2 + B*f^3)*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)^(3/2))/f^
2
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Maple [C] time = 0., size = 899, normalized size = 5.5 \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)^2/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x)
[Out]
(-A*csgn(d)*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^3*e*f^3+C*csgn(d)*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^3*e^3*f-A*csgn(d)*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^3*e^2*f^2+C*csgn(d)*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^3*e^4+C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*x*d^2*e^2*f^2*(-(
d^2*e^2-f^2)/f^2)^(1/2)+A*csgn(d)*d*f^4*(-(d^2*e^2-f^2)/f^2)^(1/2)*(-d^2*x^2+1)^(1/2)+B*csgn(d)*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*f^4-B*csgn(d)*d*e*f^3*(-(d^2*e^2-f^2)/f^2)^(1/
2)*(-d^2*x^2+1)^(1/2)-2*C*csgn(d)*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*e*f^3+C*csgn(d)*d*e^2*f^2*(-(d^2*e^2-f^2)/f^2)^(1/2)*(-d^2*x^2+1)^(1/2)+C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1
/2))*d^2*e^3*f*(-(d^2*e^2-f^2)/f^2)^(1/2)+B*csgn(d)*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*e*f^3-2*C*csgn(d)*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*e^2*f^2-C*arctan(csgn(d)*d*x/(-d^2*x^2+1)^(1/2))*x*f^4*(-(d^2*e^2-f^2)/f^2)^(1/2)-C*arctan(csgn(d)*d*x/(-d^2
*x^2+1)^(1/2))*e*f^3*(-(d^2*e^2-f^2)/f^2)^(1/2))*csgn(d)*(d*x+1)^(1/2)*(-d*x+1)^(1/2)/(-d^2*x^2+1)^(1/2)/(d*e+
f)/(d*e-f)/(f*x+e)/d/(-(d^2*e^2-f^2)/f^2)^(1/2)/f^3
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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)^2/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 117.112, size = 2082, normalized size = 12.77 \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)^2/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="fricas")
[Out]
[(C*d^3*e^5*f - B*d^3*e^4*f^2 + B*d*e^2*f^4 - A*d*e*f^5 + (A*d^3 - C*d)*e^3*f^3 - (C*d^3*e^5 + B*d*e^2*f^3 - (
A*d^3 + 2*C*d)*e^3*f^2 + (C*d^3*e^4*f + B*d*e*f^4 - (A*d^3 + 2*C*d)*e^2*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)*sq
rt(-d*x + 1))*sqrt(d*x + 1))/(f*x + e)) + (C*d^3*e^5*f - B*d^3*e^4*f^2 + B*d*e^2*f^4 - A*d*e*f^5 + (A*d^3 - C*
d)*e^3*f^3)*sqrt(d*x + 1)*sqrt(-d*x + 1) + (C*d^3*e^4*f^2 - B*d^3*e^3*f^3 + B*d*e*f^5 - A*d*f^6 + (A*d^3 - C*d
)*e^2*f^4)*x - 2*(C*d^4*e^6 - 2*C*d^2*e^4*f^2 + C*e^2*f^4 + (C*d^4*e^5*f - 2*C*d^2*e^3*f^3 + C*e*f^5)*x)*arcta
n((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/(d^5*e^6*f^2 - 2*d^3*e^4*f^4 + d*e^2*f^6 + (d^5*e^5*f^3 - 2*d^3*e
^3*f^5 + d*e*f^7)*x), (C*d^3*e^5*f - B*d^3*e^4*f^2 + B*d*e^2*f^4 - A*d*e*f^5 + (A*d^3 - C*d)*e^3*f^3 - 2*(C*d^
3*e^5 + B*d*e^2*f^3 - (A*d^3 + 2*C*d)*e^3*f^2 + (C*d^3*e^4*f + B*d*e*f^4 - (A*d^3 + 2*C*d)*e^2*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)) + (C*d^3*e^5*f - B*d^3*e^4*f^2 + B*d*e^2*f^4 - A*d*e*f^5 + (A*d^3 - C*d)*e^3*f^3)*sqrt(d*x +
1)*sqrt(-d*x + 1) + (C*d^3*e^4*f^2 - B*d^3*e^3*f^3 + B*d*e*f^5 - A*d*f^6 + (A*d^3 - C*d)*e^2*f^4)*x - 2*(C*d^4
*e^6 - 2*C*d^2*e^4*f^2 + C*e^2*f^4 + (C*d^4*e^5*f - 2*C*d^2*e^3*f^3 + C*e*f^5)*x)*arctan((sqrt(d*x + 1)*sqrt(-
d*x + 1) - 1)/(d*x)))/(d^5*e^6*f^2 - 2*d^3*e^4*f^4 + d*e^2*f^6 + (d^5*e^5*f^3 - 2*d^3*e^3*f^5 + d*e*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)**2/(-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)^2/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="giac")
[Out]
Exception raised: TypeError