3.32 \(\int \frac{A+B x+C x^2}{\sqrt{a+b x} \sqrt{a c-b c x} (e+f x)^2} \, dx\)

Optimal. Leaf size=322 \[ \frac{f \left (a^2-b^2 x^2\right ) \left (A+\frac{e (C e-B f)}{f^2}\right )}{\sqrt{a+b x} (e+f x) \sqrt{a c-b c x} \left (b^2 e^2-a^2 f^2\right )}+\frac{\sqrt{a^2 c-b^2 c x^2} \left (a^2 f^2 (2 C e-B f)-b^2 \left (C e^3-A e f^2\right )\right ) \tan ^{-1}\left (\frac{\sqrt{c} \left (a^2 f+b^2 e x\right )}{\sqrt{a^2 c-b^2 c x^2} \sqrt{b^2 e^2-a^2 f^2}}\right )}{\sqrt{c} f^2 \sqrt{a+b x} \sqrt{a c-b c x} \left (b^2 e^2-a^2 f^2\right )^{3/2}}+\frac{C \sqrt{a^2 c-b^2 c x^2} \tan ^{-1}\left (\frac{b \sqrt{c} x}{\sqrt{a^2 c-b^2 c x^2}}\right )}{b \sqrt{c} f^2 \sqrt{a+b x} \sqrt{a c-b c x}} \]

[Out]

(f*(A + (e*(C*e - B*f))/f^2)*(a^2 - b^2*x^2))/((b^2*e^2 - a^2*f^2)*Sqrt[a + b*x]*Sqrt[a*c - b*c*x]*(e + f*x))
+ (C*Sqrt[a^2*c - b^2*c*x^2]*ArcTan[(b*Sqrt[c]*x)/Sqrt[a^2*c - b^2*c*x^2]])/(b*Sqrt[c]*f^2*Sqrt[a + b*x]*Sqrt[
a*c - b*c*x]) + ((a^2*f^2*(2*C*e - B*f) - b^2*(C*e^3 - A*e*f^2))*Sqrt[a^2*c - b^2*c*x^2]*ArcTan[(Sqrt[c]*(a^2*
f + b^2*e*x))/(Sqrt[b^2*e^2 - a^2*f^2]*Sqrt[a^2*c - b^2*c*x^2])])/(Sqrt[c]*f^2*(b^2*e^2 - a^2*f^2)^(3/2)*Sqrt[
a + b*x]*Sqrt[a*c - b*c*x])

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Rubi [A]  time = 0.530435, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {1610, 1651, 844, 217, 203, 725, 204} \[ \frac{f \left (a^2-b^2 x^2\right ) \left (A+\frac{e (C e-B f)}{f^2}\right )}{\sqrt{a+b x} (e+f x) \sqrt{a c-b c x} \left (b^2 e^2-a^2 f^2\right )}+\frac{\sqrt{a^2 c-b^2 c x^2} \left (a^2 f^2 (2 C e-B f)-b^2 \left (C e^3-A e f^2\right )\right ) \tan ^{-1}\left (\frac{\sqrt{c} \left (a^2 f+b^2 e x\right )}{\sqrt{a^2 c-b^2 c x^2} \sqrt{b^2 e^2-a^2 f^2}}\right )}{\sqrt{c} f^2 \sqrt{a+b x} \sqrt{a c-b c x} \left (b^2 e^2-a^2 f^2\right )^{3/2}}+\frac{C \sqrt{a^2 c-b^2 c x^2} \tan ^{-1}\left (\frac{b \sqrt{c} x}{\sqrt{a^2 c-b^2 c x^2}}\right )}{b \sqrt{c} f^2 \sqrt{a+b x} \sqrt{a c-b c x}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)/(Sqrt[a + b*x]*Sqrt[a*c - b*c*x]*(e + f*x)^2),x]

[Out]

(f*(A + (e*(C*e - B*f))/f^2)*(a^2 - b^2*x^2))/((b^2*e^2 - a^2*f^2)*Sqrt[a + b*x]*Sqrt[a*c - b*c*x]*(e + f*x))
+ (C*Sqrt[a^2*c - b^2*c*x^2]*ArcTan[(b*Sqrt[c]*x)/Sqrt[a^2*c - b^2*c*x^2]])/(b*Sqrt[c]*f^2*Sqrt[a + b*x]*Sqrt[
a*c - b*c*x]) + ((a^2*f^2*(2*C*e - B*f) - b^2*(C*e^3 - A*e*f^2))*Sqrt[a^2*c - b^2*c*x^2]*ArcTan[(Sqrt[c]*(a^2*
f + b^2*e*x))/(Sqrt[b^2*e^2 - a^2*f^2]*Sqrt[a^2*c - b^2*c*x^2])])/(Sqrt[c]*f^2*(b^2*e^2 - a^2*f^2)^(3/2)*Sqrt[
a + b*x]*Sqrt[a*c - b*c*x])

Rule 1610

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Dist[((
a + b*x)^FracPart[m]*(c + d*x)^FracPart[m])/(a*c + b*d*x^2)^FracPart[m], Int[Px*(a*c + b*d*x^2)^m*(e + f*x)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && EqQ[b*c + a*d, 0] && EqQ[m, n] &&  !Intege
rQ[m]

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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{a+b x} \sqrt{a c-b c x} (e+f x)^2} \, dx &=\frac{\sqrt{a^2 c-b^2 c x^2} \int \frac{A+B x+C x^2}{(e+f x)^2 \sqrt{a^2 c-b^2 c x^2}} \, dx}{\sqrt{a+b x} \sqrt{a c-b c x}}\\ &=\frac{f \left (A+\frac{e (C e-B f)}{f^2}\right ) \left (a^2-b^2 x^2\right )}{\left (b^2 e^2-a^2 f^2\right ) \sqrt{a+b x} \sqrt{a c-b c x} (e+f x)}+\frac{\sqrt{a^2 c-b^2 c x^2} \int \frac{c \left (A b^2 e+a^2 (C e-B f)\right )+c C \left (\frac{b^2 e^2}{f}-a^2 f\right ) x}{(e+f x) \sqrt{a^2 c-b^2 c x^2}} \, dx}{c \left (b^2 e^2-a^2 f^2\right ) \sqrt{a+b x} \sqrt{a c-b c x}}\\ &=\frac{f \left (A+\frac{e (C e-B f)}{f^2}\right ) \left (a^2-b^2 x^2\right )}{\left (b^2 e^2-a^2 f^2\right ) \sqrt{a+b x} \sqrt{a c-b c x} (e+f x)}+\frac{\left (C \left (\frac{b^2 e^2}{f}-a^2 f\right ) \sqrt{a^2 c-b^2 c x^2}\right ) \int \frac{1}{\sqrt{a^2 c-b^2 c x^2}} \, dx}{f \left (b^2 e^2-a^2 f^2\right ) \sqrt{a+b x} \sqrt{a c-b c x}}+\frac{\left (\left (-c C e \left (\frac{b^2 e^2}{f}-a^2 f\right )+c f \left (A b^2 e+a^2 (C e-B f)\right )\right ) \sqrt{a^2 c-b^2 c x^2}\right ) \int \frac{1}{(e+f x) \sqrt{a^2 c-b^2 c x^2}} \, dx}{c f \left (b^2 e^2-a^2 f^2\right ) \sqrt{a+b x} \sqrt{a c-b c x}}\\ &=\frac{f \left (A+\frac{e (C e-B f)}{f^2}\right ) \left (a^2-b^2 x^2\right )}{\left (b^2 e^2-a^2 f^2\right ) \sqrt{a+b x} \sqrt{a c-b c x} (e+f x)}+\frac{\left (C \left (\frac{b^2 e^2}{f}-a^2 f\right ) \sqrt{a^2 c-b^2 c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{1+b^2 c x^2} \, dx,x,\frac{x}{\sqrt{a^2 c-b^2 c x^2}}\right )}{f \left (b^2 e^2-a^2 f^2\right ) \sqrt{a+b x} \sqrt{a c-b c x}}-\frac{\left (\left (-c C e \left (\frac{b^2 e^2}{f}-a^2 f\right )+c f \left (A b^2 e+a^2 (C e-B f)\right )\right ) \sqrt{a^2 c-b^2 c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{-b^2 c e^2+a^2 c f^2-x^2} \, dx,x,\frac{a^2 c f+b^2 c e x}{\sqrt{a^2 c-b^2 c x^2}}\right )}{c f \left (b^2 e^2-a^2 f^2\right ) \sqrt{a+b x} \sqrt{a c-b c x}}\\ &=\frac{f \left (A+\frac{e (C e-B f)}{f^2}\right ) \left (a^2-b^2 x^2\right )}{\left (b^2 e^2-a^2 f^2\right ) \sqrt{a+b x} \sqrt{a c-b c x} (e+f x)}+\frac{C \sqrt{a^2 c-b^2 c x^2} \tan ^{-1}\left (\frac{b \sqrt{c} x}{\sqrt{a^2 c-b^2 c x^2}}\right )}{b \sqrt{c} f^2 \sqrt{a+b x} \sqrt{a c-b c x}}+\frac{\left (a^2 f^2 (2 C e-B f)-b^2 \left (C e^3-A e f^2\right )\right ) \sqrt{a^2 c-b^2 c x^2} \tan ^{-1}\left (\frac{\sqrt{c} \left (a^2 f+b^2 e x\right )}{\sqrt{b^2 e^2-a^2 f^2} \sqrt{a^2 c-b^2 c x^2}}\right )}{\sqrt{c} f^2 \left (b^2 e^2-a^2 f^2\right )^{3/2} \sqrt{a+b x} \sqrt{a c-b c x}}\\ \end{align*}

Mathematica [A]  time = 0.968873, size = 309, normalized size = 0.96 \[ \frac{\frac{2 b^2 e \sqrt{a-b x} \left (f (A f-B e)+C e^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b x} \sqrt{a f-b e}}{\sqrt{a+b x} \sqrt{-a f-b e}}\right )}{(-a f-b e)^{3/2} (a f-b e)^{3/2}}+\frac{f (b x-a) \sqrt{a+b x} \left (f (A f-B e)+C e^2\right )}{(e+f x) (a f-b e) (a f+b e)}-\frac{2 \sqrt{a-b x} (2 C e-B f) \tan ^{-1}\left (\frac{\sqrt{a-b x} \sqrt{a f-b e}}{\sqrt{a+b x} \sqrt{-a f-b e}}\right )}{\sqrt{-a f-b e} \sqrt{a f-b e}}-\frac{2 C \sqrt{a-b x} \tan ^{-1}\left (\frac{\sqrt{a-b x}}{\sqrt{a+b x}}\right )}{b}}{f^2 \sqrt{c (a-b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x + C*x^2)/(Sqrt[a + b*x]*Sqrt[a*c - b*c*x]*(e + f*x)^2),x]

[Out]

((f*(C*e^2 + f*(-(B*e) + A*f))*(-a + b*x)*Sqrt[a + b*x])/((-(b*e) + a*f)*(b*e + a*f)*(e + f*x)) - (2*C*Sqrt[a
- b*x]*ArcTan[Sqrt[a - b*x]/Sqrt[a + b*x]])/b - (2*(2*C*e - B*f)*Sqrt[a - b*x]*ArcTan[(Sqrt[-(b*e) + a*f]*Sqrt
[a - b*x])/(Sqrt[-(b*e) - a*f]*Sqrt[a + b*x])])/(Sqrt[-(b*e) - a*f]*Sqrt[-(b*e) + a*f]) + (2*b^2*e*(C*e^2 + f*
(-(B*e) + A*f))*Sqrt[a - b*x]*ArcTan[(Sqrt[-(b*e) + a*f]*Sqrt[a - b*x])/(Sqrt[-(b*e) - a*f]*Sqrt[a + b*x])])/(
(-(b*e) - a*f)^(3/2)*(-(b*e) + a*f)^(3/2)))/(f^2*Sqrt[c*(a - b*x)])

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Maple [B]  time = 0., size = 1200, normalized size = 3.7 \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/(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2),x)

[Out]

(A*ln(2*(b^2*c*e*x+a^2*c*f+(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)*f)/(f*x+e))*x*b^2*c*e*f^3*
(b^2*c)^(1/2)-B*ln(2*(b^2*c*e*x+a^2*c*f+(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)*f)/(f*x+e))*x
*a^2*c*f^4*(b^2*c)^(1/2)+2*C*ln(2*(b^2*c*e*x+a^2*c*f+(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)*
f)/(f*x+e))*x*a^2*c*e*f^3*(b^2*c)^(1/2)-C*ln(2*(b^2*c*e*x+a^2*c*f+(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2)*(-c*(b^2*x^2
-a^2))^(1/2)*f)/(f*x+e))*x*b^2*c*e^3*f*(b^2*c)^(1/2)+C*arctan((b^2*c)^(1/2)*x/(-c*(b^2*x^2-a^2))^(1/2))*x*a^2*
c*f^4*(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2)-C*arctan((b^2*c)^(1/2)*x/(-c*(b^2*x^2-a^2))^(1/2))*x*b^2*c*e^2*f^2*(c*(a
^2*f^2-b^2*e^2)/f^2)^(1/2)+A*ln(2*(b^2*c*e*x+a^2*c*f+(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)*
f)/(f*x+e))*b^2*c*e^2*f^2*(b^2*c)^(1/2)-B*ln(2*(b^2*c*e*x+a^2*c*f+(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2)*(-c*(b^2*x^2
-a^2))^(1/2)*f)/(f*x+e))*a^2*c*e*f^3*(b^2*c)^(1/2)+2*C*ln(2*(b^2*c*e*x+a^2*c*f+(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2)
*(-c*(b^2*x^2-a^2))^(1/2)*f)/(f*x+e))*a^2*c*e^2*f^2*(b^2*c)^(1/2)-C*ln(2*(b^2*c*e*x+a^2*c*f+(c*(a^2*f^2-b^2*e^
2)/f^2)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)*f)/(f*x+e))*b^2*c*e^4*(b^2*c)^(1/2)+C*arctan((b^2*c)^(1/2)*x/(-c*(b^2*x
^2-a^2))^(1/2))*a^2*c*e*f^3*(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2)-C*arctan((b^2*c)^(1/2)*x/(-c*(b^2*x^2-a^2))^(1/2))
*b^2*c*e^3*f*(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2)-A*f^4*(-c*(b^2*x^2-a^2))^(1/2)*(b^2*c)^(1/2)*(c*(a^2*f^2-b^2*e^2)
/f^2)^(1/2)+B*e*f^3*(-c*(b^2*x^2-a^2))^(1/2)*(b^2*c)^(1/2)*(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2)-C*e^2*f^2*(-c*(b^2*
x^2-a^2))^(1/2)*(b^2*c)^(1/2)*(c*(a^2*f^2-b^2*e^2)/f^2)^(1/2))/c*(-c*(b*x-a))^(1/2)*(b*x+a)^(1/2)/(-c*(b^2*x^2
-a^2))^(1/2)/(a*f+b*e)/(a*f-b*e)/(f*x+e)/(b^2*c)^(1/2)/(c*(a^2*f^2-b^2*e^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/(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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((C*x^2+B*x+A)/(f*x+e)^2/(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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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/(b*x+a)**(1/2)/(-b*c*x+a*c)**(1/2),x)

[Out]

Exception raised: ValueError

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Giac [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((C*x^2+B*x+A)/(f*x+e)^2/(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2),x, algorithm="giac")

[Out]

Timed out