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

Optimal. Leaf size=300 \[ -\frac{\sqrt{a+b x} \left (a^2-b^2 x^2\right ) \sqrt{a c-b c x} \left (4 \left (2 a^2 C f^2-b^2 \left (3 C e^2-5 f (A f+B e)\right )\right )-3 b^2 f x (3 C e-5 B f)\right )}{60 b^4 f}+\frac{a^2 \sqrt{c} \sqrt{a+b x} \sqrt{a c-b c x} \tan ^{-1}\left (\frac{b \sqrt{c} x}{\sqrt{a^2 c-b^2 c x^2}}\right ) \left (a^2 (B f+C e)+4 A b^2 e\right )}{8 b^3 \sqrt{a^2 c-b^2 c x^2}}+\frac{x \sqrt{a+b x} \sqrt{a c-b c x} \left (a^2 (B f+C e)+4 A b^2 e\right )}{8 b^2}-\frac{C \sqrt{a+b x} \left (a^2-b^2 x^2\right ) (e+f x)^2 \sqrt{a c-b c x}}{5 b^2 f} \]

[Out]

((4*A*b^2*e + a^2*(C*e + B*f))*x*Sqrt[a + b*x]*Sqrt[a*c - b*c*x])/(8*b^2) - (C*Sqrt[a + b*x]*Sqrt[a*c - b*c*x]
*(e + f*x)^2*(a^2 - b^2*x^2))/(5*b^2*f) - (Sqrt[a + b*x]*Sqrt[a*c - b*c*x]*(4*(2*a^2*C*f^2 - b^2*(3*C*e^2 - 5*
f*(B*e + A*f))) - 3*b^2*f*(3*C*e - 5*B*f)*x)*(a^2 - b^2*x^2))/(60*b^4*f) + (a^2*Sqrt[c]*(4*A*b^2*e + a^2*(C*e
+ B*f))*Sqrt[a + b*x]*Sqrt[a*c - b*c*x]*ArcTan[(b*Sqrt[c]*x)/Sqrt[a^2*c - b^2*c*x^2]])/(8*b^3*Sqrt[a^2*c - b^2
*c*x^2])

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Rubi [A]  time = 0.445962, antiderivative size = 297, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {1610, 1654, 780, 195, 217, 203} \[ -\frac{\sqrt{a+b x} \left (a^2-b^2 x^2\right ) \sqrt{a c-b c x} \left (4 \left (2 a^2 C f^2-b^2 \left (3 C e^2-5 f (A f+B e)\right )\right )-3 b^2 f x (3 C e-5 B f)\right )}{60 b^4 f}+\frac{a^2 \sqrt{c} \sqrt{a+b x} \sqrt{a c-b c x} \tan ^{-1}\left (\frac{b \sqrt{c} x}{\sqrt{a^2 c-b^2 c x^2}}\right ) \left (a^2 (B f+C e)+4 A b^2 e\right )}{8 b^3 \sqrt{a^2 c-b^2 c x^2}}+\frac{1}{8} x \sqrt{a+b x} \sqrt{a c-b c x} \left (\frac{a^2 (B f+C e)}{b^2}+4 A e\right )-\frac{C \sqrt{a+b x} \left (a^2-b^2 x^2\right ) (e+f x)^2 \sqrt{a c-b c x}}{5 b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A*e + (a^2*(C*e + B*f))/b^2)*x*Sqrt[a + b*x]*Sqrt[a*c - b*c*x])/8 - (C*Sqrt[a + b*x]*Sqrt[a*c - b*c*x]*(e
+ f*x)^2*(a^2 - b^2*x^2))/(5*b^2*f) - (Sqrt[a + b*x]*Sqrt[a*c - b*c*x]*(4*(2*a^2*C*f^2 - b^2*(3*C*e^2 - 5*f*(B
*e + A*f))) - 3*b^2*f*(3*C*e - 5*B*f)*x)*(a^2 - b^2*x^2))/(60*b^4*f) + (a^2*Sqrt[c]*(4*A*b^2*e + a^2*(C*e + B*
f))*Sqrt[a + b*x]*Sqrt[a*c - b*c*x]*ArcTan[(b*Sqrt[c]*x)/Sqrt[a^2*c - b^2*c*x^2]])/(8*b^3*Sqrt[a^2*c - b^2*c*x
^2])

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 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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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])

Rubi steps

\begin{align*} \int \sqrt{a+b x} \sqrt{a c-b c x} (e+f x) \left (A+B x+C x^2\right ) \, dx &=\frac{\left (\sqrt{a+b x} \sqrt{a c-b c x}\right ) \int (e+f x) \sqrt{a^2 c-b^2 c x^2} \left (A+B x+C x^2\right ) \, dx}{\sqrt{a^2 c-b^2 c x^2}}\\ &=-\frac{C \sqrt{a+b x} \sqrt{a c-b c x} (e+f x)^2 \left (a^2-b^2 x^2\right )}{5 b^2 f}-\frac{\left (\sqrt{a+b x} \sqrt{a c-b c x}\right ) \int (e+f x) \left (-c \left (5 A b^2+2 a^2 C\right ) f^2+b^2 c f (3 C e-5 B f) x\right ) \sqrt{a^2 c-b^2 c x^2} \, dx}{5 b^2 c f^2 \sqrt{a^2 c-b^2 c x^2}}\\ &=-\frac{C \sqrt{a+b x} \sqrt{a c-b c x} (e+f x)^2 \left (a^2-b^2 x^2\right )}{5 b^2 f}-\frac{\sqrt{a+b x} \sqrt{a c-b c x} \left (4 \left (2 a^2 C f^2-b^2 \left (3 C e^2-5 f (B e+A f)\right )\right )-3 b^2 f (3 C e-5 B f) x\right ) \left (a^2-b^2 x^2\right )}{60 b^4 f}+\frac{\left (\left (4 A b^2 e+a^2 (C e+B f)\right ) \sqrt{a+b x} \sqrt{a c-b c x}\right ) \int \sqrt{a^2 c-b^2 c x^2} \, dx}{4 b^2 \sqrt{a^2 c-b^2 c x^2}}\\ &=\frac{1}{8} \left (4 A e+\frac{a^2 (C e+B f)}{b^2}\right ) x \sqrt{a+b x} \sqrt{a c-b c x}-\frac{C \sqrt{a+b x} \sqrt{a c-b c x} (e+f x)^2 \left (a^2-b^2 x^2\right )}{5 b^2 f}-\frac{\sqrt{a+b x} \sqrt{a c-b c x} \left (4 \left (2 a^2 C f^2-b^2 \left (3 C e^2-5 f (B e+A f)\right )\right )-3 b^2 f (3 C e-5 B f) x\right ) \left (a^2-b^2 x^2\right )}{60 b^4 f}+\frac{\left (a^2 c \left (4 A b^2 e+a^2 (C e+B f)\right ) \sqrt{a+b x} \sqrt{a c-b c x}\right ) \int \frac{1}{\sqrt{a^2 c-b^2 c x^2}} \, dx}{8 b^2 \sqrt{a^2 c-b^2 c x^2}}\\ &=\frac{1}{8} \left (4 A e+\frac{a^2 (C e+B f)}{b^2}\right ) x \sqrt{a+b x} \sqrt{a c-b c x}-\frac{C \sqrt{a+b x} \sqrt{a c-b c x} (e+f x)^2 \left (a^2-b^2 x^2\right )}{5 b^2 f}-\frac{\sqrt{a+b x} \sqrt{a c-b c x} \left (4 \left (2 a^2 C f^2-b^2 \left (3 C e^2-5 f (B e+A f)\right )\right )-3 b^2 f (3 C e-5 B f) x\right ) \left (a^2-b^2 x^2\right )}{60 b^4 f}+\frac{\left (a^2 c \left (4 A b^2 e+a^2 (C e+B f)\right ) \sqrt{a+b x} \sqrt{a c-b c x}\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 )}{8 b^2 \sqrt{a^2 c-b^2 c x^2}}\\ &=\frac{1}{8} \left (4 A e+\frac{a^2 (C e+B f)}{b^2}\right ) x \sqrt{a+b x} \sqrt{a c-b c x}-\frac{C \sqrt{a+b x} \sqrt{a c-b c x} (e+f x)^2 \left (a^2-b^2 x^2\right )}{5 b^2 f}-\frac{\sqrt{a+b x} \sqrt{a c-b c x} \left (4 \left (2 a^2 C f^2-b^2 \left (3 C e^2-5 f (B e+A f)\right )\right )-3 b^2 f (3 C e-5 B f) x\right ) \left (a^2-b^2 x^2\right )}{60 b^4 f}+\frac{a^2 \sqrt{c} \left (4 A b^2 e+a^2 (C e+B f)\right ) \sqrt{a+b x} \sqrt{a c-b c x} \tan ^{-1}\left (\frac{b \sqrt{c} x}{\sqrt{a^2 c-b^2 c x^2}}\right )}{8 b^3 \sqrt{a^2 c-b^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.646006, size = 200, normalized size = 0.67 \[ -\frac{c \left (\left (a^2-b^2 x^2\right ) \left (a^2 b^2 (40 A f+5 B (8 e+3 f x)+C x (15 e+8 f x))+16 a^4 C f-2 b^4 x (10 A (3 e+2 f x)+x (5 B (4 e+3 f x)+3 C x (5 e+4 f x)))\right )+30 a^{5/2} b \sqrt{a-b x} \sqrt{\frac{b x}{a}+1} \sin ^{-1}\left (\frac{\sqrt{a-b x}}{\sqrt{2} \sqrt{a}}\right ) \left (a^2 (B f+C e)+4 A b^2 e\right )\right )}{120 b^4 \sqrt{a+b x} \sqrt{c (a-b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*((a^2 - b^2*x^2)*(16*a^4*C*f + a^2*b^2*(40*A*f + 5*B*(8*e + 3*f*x) + C*x*(15*e + 8*f*x)) - 2*b^4*x*(10*A*(
3*e + 2*f*x) + x*(5*B*(4*e + 3*f*x) + 3*C*x*(5*e + 4*f*x)))) + 30*a^(5/2)*b*(4*A*b^2*e + a^2*(C*e + B*f))*Sqrt
[a - b*x]*Sqrt[1 + (b*x)/a]*ArcSin[Sqrt[a - b*x]/(Sqrt[2]*Sqrt[a])]))/(120*b^4*Sqrt[c*(a - b*x)]*Sqrt[a + b*x]
)

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Maple [B]  time = 0.013, size = 588, normalized size = 2. \begin{align*}{\frac{1}{120\,{b}^{4}}\sqrt{bx+a}\sqrt{-c \left ( bx-a \right ) } \left ( 24\,C{x}^{4}{b}^{4}f\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }+30\,B{x}^{3}{b}^{4}f\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }+30\,C{x}^{3}{b}^{4}e\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }+60\,A\arctan \left ({\frac{\sqrt{{b}^{2}c}x}{\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }}} \right ){a}^{2}{b}^{4}ce+40\,A{x}^{2}{b}^{4}f\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }+15\,B\arctan \left ({\frac{\sqrt{{b}^{2}c}x}{\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }}} \right ){a}^{4}{b}^{2}cf+40\,B{x}^{2}{b}^{4}e\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }+15\,C\arctan \left ({\frac{\sqrt{{b}^{2}c}x}{\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }}} \right ){a}^{4}{b}^{2}ce-8\,C{x}^{2}{a}^{2}{b}^{2}f\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }+60\,A\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }x{b}^{4}e-15\,B\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }x{a}^{2}{b}^{2}f-15\,C\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }x{a}^{2}{b}^{2}e-40\,A{a}^{2}{b}^{2}f\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }-40\,B{a}^{2}{b}^{2}e\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }-16\,C{a}^{4}f\sqrt{{b}^{2}c}\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) } \right ){\frac{1}{\sqrt{-c \left ({b}^{2}{x}^{2}-{a}^{2} \right ) }}}{\frac{1}{\sqrt{{b}^{2}c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*(C*x^2+B*x+A)*(b*x+a)^(1/2)*(-b*c*x+a*c)^(1/2),x)

[Out]

1/120*(b*x+a)^(1/2)*(-c*(b*x-a))^(1/2)*(24*C*x^4*b^4*f*(b^2*c)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)+30*B*x^3*b^4*f*(
b^2*c)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)+30*C*x^3*b^4*e*(b^2*c)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)+60*A*arctan((b^2*c
)^(1/2)*x/(-c*(b^2*x^2-a^2))^(1/2))*a^2*b^4*c*e+40*A*x^2*b^4*f*(b^2*c)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)+15*B*arc
tan((b^2*c)^(1/2)*x/(-c*(b^2*x^2-a^2))^(1/2))*a^4*b^2*c*f+40*B*x^2*b^4*e*(b^2*c)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2
)+15*C*arctan((b^2*c)^(1/2)*x/(-c*(b^2*x^2-a^2))^(1/2))*a^4*b^2*c*e-8*C*x^2*a^2*b^2*f*(b^2*c)^(1/2)*(-c*(b^2*x
^2-a^2))^(1/2)+60*A*(b^2*c)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)*x*b^4*e-15*B*(b^2*c)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)
*x*a^2*b^2*f-15*C*(b^2*c)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)*x*a^2*b^2*e-40*A*a^2*b^2*f*(b^2*c)^(1/2)*(-c*(b^2*x^2
-a^2))^(1/2)-40*B*a^2*b^2*e*(b^2*c)^(1/2)*(-c*(b^2*x^2-a^2))^(1/2)-16*C*a^4*f*(b^2*c)^(1/2)*(-c*(b^2*x^2-a^2))
^(1/2))/(-c*(b^2*x^2-a^2))^(1/2)/b^4/(b^2*c)^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27366, size = 980, normalized size = 3.27 \begin{align*} \left [\frac{15 \,{\left (B a^{4} b f +{\left (C a^{4} b + 4 \, A a^{2} b^{3}\right )} e\right )} \sqrt{-c} \log \left (2 \, b^{2} c x^{2} + 2 \, \sqrt{-b c x + a c} \sqrt{b x + a} b \sqrt{-c} x - a^{2} c\right ) + 2 \,{\left (24 \, C b^{4} f x^{4} - 40 \, B a^{2} b^{2} e + 30 \,{\left (C b^{4} e + B b^{4} f\right )} x^{3} + 8 \,{\left (5 \, B b^{4} e -{\left (C a^{2} b^{2} - 5 \, A b^{4}\right )} f\right )} x^{2} - 8 \,{\left (2 \, C a^{4} + 5 \, A a^{2} b^{2}\right )} f - 15 \,{\left (B a^{2} b^{2} f +{\left (C a^{2} b^{2} - 4 \, A b^{4}\right )} e\right )} x\right )} \sqrt{-b c x + a c} \sqrt{b x + a}}{240 \, b^{4}}, -\frac{15 \,{\left (B a^{4} b f +{\left (C a^{4} b + 4 \, A a^{2} b^{3}\right )} e\right )} \sqrt{c} \arctan \left (\frac{\sqrt{-b c x + a c} \sqrt{b x + a} b \sqrt{c} x}{b^{2} c x^{2} - a^{2} c}\right ) -{\left (24 \, C b^{4} f x^{4} - 40 \, B a^{2} b^{2} e + 30 \,{\left (C b^{4} e + B b^{4} f\right )} x^{3} + 8 \,{\left (5 \, B b^{4} e -{\left (C a^{2} b^{2} - 5 \, A b^{4}\right )} f\right )} x^{2} - 8 \,{\left (2 \, C a^{4} + 5 \, A a^{2} b^{2}\right )} f - 15 \,{\left (B a^{2} b^{2} f +{\left (C a^{2} b^{2} - 4 \, A b^{4}\right )} e\right )} x\right )} \sqrt{-b c x + a c} \sqrt{b x + a}}{120 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/240*(15*(B*a^4*b*f + (C*a^4*b + 4*A*a^2*b^3)*e)*sqrt(-c)*log(2*b^2*c*x^2 + 2*sqrt(-b*c*x + a*c)*sqrt(b*x +
a)*b*sqrt(-c)*x - a^2*c) + 2*(24*C*b^4*f*x^4 - 40*B*a^2*b^2*e + 30*(C*b^4*e + B*b^4*f)*x^3 + 8*(5*B*b^4*e - (C
*a^2*b^2 - 5*A*b^4)*f)*x^2 - 8*(2*C*a^4 + 5*A*a^2*b^2)*f - 15*(B*a^2*b^2*f + (C*a^2*b^2 - 4*A*b^4)*e)*x)*sqrt(
-b*c*x + a*c)*sqrt(b*x + a))/b^4, -1/120*(15*(B*a^4*b*f + (C*a^4*b + 4*A*a^2*b^3)*e)*sqrt(c)*arctan(sqrt(-b*c*
x + a*c)*sqrt(b*x + a)*b*sqrt(c)*x/(b^2*c*x^2 - a^2*c)) - (24*C*b^4*f*x^4 - 40*B*a^2*b^2*e + 30*(C*b^4*e + B*b
^4*f)*x^3 + 8*(5*B*b^4*e - (C*a^2*b^2 - 5*A*b^4)*f)*x^2 - 8*(2*C*a^4 + 5*A*a^2*b^2)*f - 15*(B*a^2*b^2*f + (C*a
^2*b^2 - 4*A*b^4)*e)*x)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/b^4]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-c*(-a + b*x))*sqrt(a + b*x)*(e + f*x)*(A + B*x + C*x**2), x)

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

[Out]

Timed out