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

Optimal. Leaf size=122 \[ \frac{\left (A f^2-B e f+C e^2\right ) \tan ^{-1}\left (\frac{d^2 e x+f}{\sqrt{1-d^2 x^2} \sqrt{d^2 e^2-f^2}}\right )}{f^2 \sqrt{d^2 e^2-f^2}}-\frac{\sin ^{-1}(d x) (C e-B f)}{d f^2}-\frac{C \sqrt{1-d^2 x^2}}{d^2 f} \]

[Out]

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

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Rubi [A]  time = 0.282661, antiderivative size = 122, 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, 1654, 844, 216, 725, 204} \[ \frac{\left (A f^2-B e f+C e^2\right ) \tan ^{-1}\left (\frac{d^2 e x+f}{\sqrt{1-d^2 x^2} \sqrt{d^2 e^2-f^2}}\right )}{f^2 \sqrt{d^2 e^2-f^2}}-\frac{\sin ^{-1}(d x) (C e-B f)}{d f^2}-\frac{C \sqrt{1-d^2 x^2}}{d^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((C*Sqrt[1 - d^2*x^2])/(d^2*f)) - ((C*e - B*f)*ArcSin[d*x])/(d*f^2) + ((C*e^2 - B*e*f + A*f^2)*ArcTan[(f + d^
2*e*x)/(Sqrt[d^2*e^2 - f^2]*Sqrt[1 - d^2*x^2])])/(f^2*Sqrt[d^2*e^2 - f^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 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 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)} \, dx &=\int \frac{A+B x+C x^2}{(e+f x) \sqrt{1-d^2 x^2}} \, dx\\ &=-\frac{C \sqrt{1-d^2 x^2}}{d^2 f}-\frac{\int \frac{-A d^2 f^2+d^2 f (C e-B f) x}{(e+f x) \sqrt{1-d^2 x^2}} \, dx}{d^2 f^2}\\ &=-\frac{C \sqrt{1-d^2 x^2}}{d^2 f}-\frac{(C e-B f) \int \frac{1}{\sqrt{1-d^2 x^2}} \, dx}{f^2}+\frac{\left (C e^2-B e f+A f^2\right ) \int \frac{1}{(e+f x) \sqrt{1-d^2 x^2}} \, dx}{f^2}\\ &=-\frac{C \sqrt{1-d^2 x^2}}{d^2 f}-\frac{(C e-B f) \sin ^{-1}(d x)}{d f^2}-\frac{\left (C e^2-B e f+A f^2\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 )}{f^2}\\ &=-\frac{C \sqrt{1-d^2 x^2}}{d^2 f}-\frac{(C e-B f) \sin ^{-1}(d x)}{d f^2}+\frac{\left (C e^2-B e f+A f^2\right ) \tan ^{-1}\left (\frac{f+d^2 e x}{\sqrt{d^2 e^2-f^2} \sqrt{1-d^2 x^2}}\right )}{f^2 \sqrt{d^2 e^2-f^2}}\\ \end{align*}

Mathematica [A]  time = 0.127119, size = 117, normalized size = 0.96 \[ \frac{\frac{\left (f (A f-B e)+C e^2\right ) \tan ^{-1}\left (\frac{d^2 e x+f}{\sqrt{1-d^2 x^2} \sqrt{d^2 e^2-f^2}}\right )}{\sqrt{d^2 e^2-f^2}}+\frac{\sin ^{-1}(d x) (B f-C e)}{d}-\frac{C f \sqrt{1-d^2 x^2}}{d^2}}{f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0., size = 373, normalized size = 3.1 \begin{align*}{\frac{{\it csgn} \left ( d \right ) }{{f}^{3}{d}^{2}} \left ( -A{\it csgn} \left ( d \right ) \ln \left ( 2\,{\frac{1}{fx+e} \left ({d}^{2}ex+\sqrt{-{\frac{{d}^{2}{e}^{2}-{f}^{2}}{{f}^{2}}}}\sqrt{-{d}^{2}{x}^{2}+1}f+f \right ) } \right ){d}^{2}{f}^{2}+B{\it csgn} \left ( d \right ) \ln \left ( 2\,{\frac{1}{fx+e} \left ({d}^{2}ex+\sqrt{-{\frac{{d}^{2}{e}^{2}-{f}^{2}}{{f}^{2}}}}\sqrt{-{d}^{2}{x}^{2}+1}f+f \right ) } \right ){d}^{2}ef-C{\it csgn} \left ( d \right ) \ln \left ( 2\,{\frac{1}{fx+e} \left ({d}^{2}ex+\sqrt{-{\frac{{d}^{2}{e}^{2}-{f}^{2}}{{f}^{2}}}}\sqrt{-{d}^{2}{x}^{2}+1}f+f \right ) } \right ){d}^{2}{e}^{2}+B\arctan \left ({{\it csgn} \left ( d \right ) dx{\frac{1}{\sqrt{-{d}^{2}{x}^{2}+1}}}} \right ) d{f}^{2}\sqrt{-{\frac{{d}^{2}{e}^{2}-{f}^{2}}{{f}^{2}}}}-C{\it csgn} \left ( d \right ){f}^{2}\sqrt{-{d}^{2}{x}^{2}+1}\sqrt{-{\frac{{d}^{2}{e}^{2}-{f}^{2}}{{f}^{2}}}}-C\arctan \left ({{\it csgn} \left ( d \right ) dx{\frac{1}{\sqrt{-{d}^{2}{x}^{2}+1}}}} \right ) def\sqrt{-{\frac{{d}^{2}{e}^{2}-{f}^{2}}{{f}^{2}}}} \right ) \sqrt{-dx+1}\sqrt{dx+1}{\frac{1}{\sqrt{-{\frac{{d}^{2}{e}^{2}-{f}^{2}}{{f}^{2}}}}}}{\frac{1}{\sqrt{-{d}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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)/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 30.0132, size = 1019, normalized size = 8.35 \begin{align*} \left [-\frac{{\left (C d^{2} e^{2} - B d^{2} e f + A d^{2} f^{2}\right )} \sqrt{-d^{2} e^{2} + f^{2}} \log \left (\frac{d^{2} e f x + f^{2} - \sqrt{-d^{2} e^{2} + f^{2}}{\left (d^{2} e x + f\right )} -{\left (\sqrt{-d^{2} e^{2} + f^{2}} \sqrt{-d x + 1} f +{\left (d^{2} e^{2} - f^{2}\right )} \sqrt{-d x + 1}\right )} \sqrt{d x + 1}}{f x + e}\right ) +{\left (C d^{2} e^{2} f - C f^{3}\right )} \sqrt{d x + 1} \sqrt{-d x + 1} - 2 \,{\left (C d^{3} e^{3} - B d^{3} e^{2} f - C d e f^{2} + B d f^{3}\right )} \arctan \left (\frac{\sqrt{d x + 1} \sqrt{-d x + 1} - 1}{d x}\right )}{d^{4} e^{2} f^{2} - d^{2} f^{4}}, \frac{2 \,{\left (C d^{2} e^{2} - B d^{2} e f + A d^{2} f^{2}\right )} \sqrt{d^{2} e^{2} - f^{2}} \arctan \left (-\frac{\sqrt{d^{2} e^{2} - f^{2}} \sqrt{d x + 1} \sqrt{-d x + 1} e - \sqrt{d^{2} e^{2} - f^{2}}{\left (f x + e\right )}}{{\left (d^{2} e^{2} - f^{2}\right )} x}\right ) -{\left (C d^{2} e^{2} f - C f^{3}\right )} \sqrt{d x + 1} \sqrt{-d x + 1} + 2 \,{\left (C d^{3} e^{3} - B d^{3} e^{2} f - C d e f^{2} + B d f^{3}\right )} \arctan \left (\frac{\sqrt{d x + 1} \sqrt{-d x + 1} - 1}{d x}\right )}{d^{4} e^{2} f^{2} - d^{2} f^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-((C*d^2*e^2 - B*d^2*e*f + A*d^2*f^2)*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)) +
(C*d^2*e^2*f - C*f^3)*sqrt(d*x + 1)*sqrt(-d*x + 1) - 2*(C*d^3*e^3 - B*d^3*e^2*f - C*d*e*f^2 + B*d*f^3)*arctan
((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/(d^4*e^2*f^2 - d^2*f^4), (2*(C*d^2*e^2 - B*d^2*e*f + A*d^2*f^2)*sq
rt(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^2*e^2*f - C*f^3)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 2*(C*d^3*e^3 - B*d^3*e^2*f - C*d*
e*f^2 + B*d*f^3)*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(d*x)))/(d^4*e^2*f^2 - d^2*f^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x + C x^{2}}{\left (e + f x\right ) \sqrt{- d x + 1} \sqrt{d x + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(-d*x+1)^(1/2)/(d*x+1)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError