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

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

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Rubi [A]  time = 0.45, 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} \begin {gather*} -\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} \end {gather*}

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 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 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 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 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 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]))

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*}

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Mathematica [A]  time = 0.68, size = 200, normalized size = 0.67 \begin {gather*} -\frac {c \left (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 )+\left (a^2-b^2 x^2\right ) \left (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)))\right )\right )}{120 b^4 \sqrt {a+b x} \sqrt {c (a-b x)}} \end {gather*}

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]

-1/120*(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])]))/(b^4*Sqrt[c*(a - b*x)]*Sqrt[a + b*
x])

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IntegrateAlgebraic [B]  time = 0.64, size = 647, normalized size = 2.16 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {a c-b c x}}{\sqrt {c} \sqrt {a+b x}}\right ) \left (-a^4 B \sqrt {c} f+a^4 \left (-\sqrt {c}\right ) C e-4 a^2 A b^2 \sqrt {c} e\right )}{4 b^3}-\frac {a^2 c \sqrt {a c-b c x} \left (\frac {160 a^3 c^3 C f (a c-b c x)}{a+b x}-\frac {64 a^3 c^2 C f (a c-b c x)^2}{(a+b x)^2}+\frac {160 a^3 c C f (a c-b c x)^3}{(a+b x)^3}-15 a^2 b B c^4 f+\frac {90 a^2 b B c^3 f (a c-b c x)}{a+b x}-\frac {90 a^2 b B c f (a c-b c x)^3}{(a+b x)^3}+\frac {15 a^2 b B f (a c-b c x)^4}{(a+b x)^4}-15 a^2 b c^4 C e+\frac {90 a^2 b c^3 C e (a c-b c x)}{a+b x}-\frac {90 a^2 b c C e (a c-b c x)^3}{(a+b x)^3}+\frac {15 a^2 b C e (a c-b c x)^4}{(a+b x)^4}-\frac {120 A b^3 c^3 e (a c-b c x)}{a+b x}+\frac {120 A b^3 c e (a c-b c x)^3}{(a+b x)^3}+\frac {60 A b^3 e (a c-b c x)^4}{(a+b x)^4}+\frac {160 a A b^2 c^3 f (a c-b c x)}{a+b x}+\frac {320 a A b^2 c^2 f (a c-b c x)^2}{(a+b x)^2}+\frac {160 a A b^2 c f (a c-b c x)^3}{(a+b x)^3}+\frac {160 a b^2 B c^3 e (a c-b c x)}{a+b x}+\frac {320 a b^2 B c^2 e (a c-b c x)^2}{(a+b x)^2}+\frac {160 a b^2 B c e (a c-b c x)^3}{(a+b x)^3}-60 A b^3 c^4 e\right )}{60 b^4 \sqrt {a+b x} \left (\frac {a c-b c x}{a+b x}+c\right )^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(a^2*c*Sqrt[a*c - b*c*x]*(-60*A*b^3*c^4*e - 15*a^2*b*c^4*C*e - 15*a^2*b*B*c^4*f - (120*A*b^3*c^3*e*(a*c
- b*c*x))/(a + b*x) + (160*a*b^2*B*c^3*e*(a*c - b*c*x))/(a + b*x) + (90*a^2*b*c^3*C*e*(a*c - b*c*x))/(a + b*x)
+ (160*a*A*b^2*c^3*f*(a*c - b*c*x))/(a + b*x) + (90*a^2*b*B*c^3*f*(a*c - b*c*x))/(a + b*x) + (160*a^3*c^3*C*f
*(a*c - b*c*x))/(a + b*x) + (320*a*b^2*B*c^2*e*(a*c - b*c*x)^2)/(a + b*x)^2 + (320*a*A*b^2*c^2*f*(a*c - b*c*x)
^2)/(a + b*x)^2 - (64*a^3*c^2*C*f*(a*c - b*c*x)^2)/(a + b*x)^2 + (120*A*b^3*c*e*(a*c - b*c*x)^3)/(a + b*x)^3 +
(160*a*b^2*B*c*e*(a*c - b*c*x)^3)/(a + b*x)^3 - (90*a^2*b*c*C*e*(a*c - b*c*x)^3)/(a + b*x)^3 + (160*a*A*b^2*c
*f*(a*c - b*c*x)^3)/(a + b*x)^3 - (90*a^2*b*B*c*f*(a*c - b*c*x)^3)/(a + b*x)^3 + (160*a^3*c*C*f*(a*c - b*c*x)^
3)/(a + b*x)^3 + (60*A*b^3*e*(a*c - b*c*x)^4)/(a + b*x)^4 + (15*a^2*b*C*e*(a*c - b*c*x)^4)/(a + b*x)^4 + (15*a
^2*b*B*f*(a*c - b*c*x)^4)/(a + b*x)^4))/(b^4*Sqrt[a + b*x]*(c + (a*c - b*c*x)/(a + b*x))^5) + ((-4*a^2*A*b^2*S
qrt[c]*e - a^4*Sqrt[c]*C*e - a^4*B*Sqrt[c]*f)*ArcTan[Sqrt[a*c - b*c*x]/(Sqrt[c]*Sqrt[a + b*x])])/(4*b^3)

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fricas [A]  time = 1.20, size = 441, normalized size = 1.47 \begin {gather*} \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 {gather*}

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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maple [B]  time = 0.01, size = 588, normalized size = 1.96 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {-\left (b x -a \right ) c}\, \left (60 A \,a^{2} b^{4} c e \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}}\right )+15 B \,a^{4} b^{2} c f \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}}\right )+15 C \,a^{4} b^{2} c e \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}}\right )+24 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, C \,b^{4} f \,x^{4}+30 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, B \,b^{4} f \,x^{3}+30 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, C \,b^{4} e \,x^{3}+40 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, A \,b^{4} f \,x^{2}+40 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, B \,b^{4} e \,x^{2}-8 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, C \,a^{2} b^{2} f \,x^{2}+60 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, A \,b^{4} e x -15 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, B \,a^{2} b^{2} f x -15 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, C \,a^{2} b^{2} e x -40 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, A \,a^{2} b^{2} f -40 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, B \,a^{2} b^{2} e -16 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, C \,a^{4} f \right )}{120 \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, \sqrt {b^{2} c}\, b^{4}} \end {gather*}

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

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maxima [A]  time = 2.25, size = 248, normalized size = 0.83 \begin {gather*} \frac {A a^{2} \sqrt {c} e \arcsin \left (\frac {b x}{a}\right )}{2 \, b} + \frac {1}{2} \, \sqrt {-b^{2} c x^{2} + a^{2} c} A e x + \frac {{\left (C e + B f\right )} a^{4} \sqrt {c} \arcsin \left (\frac {b x}{a}\right )}{8 \, b^{3}} + \frac {\sqrt {-b^{2} c x^{2} + a^{2} c} {\left (C e + B f\right )} a^{2} x}{8 \, b^{2}} - \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} C f x^{2}}{5 \, b^{2} c} - \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} B e}{3 \, b^{2} c} - \frac {2 \, {\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} C a^{2} f}{15 \, b^{4} c} - \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} A f}{3 \, b^{2} c} - \frac {{\left (-b^{2} c x^{2} + a^{2} c\right )}^{\frac {3}{2}} {\left (C e + B f\right )} x}{4 \, b^{2} c} \end {gather*}

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]

1/2*A*a^2*sqrt(c)*e*arcsin(b*x/a)/b + 1/2*sqrt(-b^2*c*x^2 + a^2*c)*A*e*x + 1/8*(C*e + B*f)*a^4*sqrt(c)*arcsin(
b*x/a)/b^3 + 1/8*sqrt(-b^2*c*x^2 + a^2*c)*(C*e + B*f)*a^2*x/b^2 - 1/5*(-b^2*c*x^2 + a^2*c)^(3/2)*C*f*x^2/(b^2*
c) - 1/3*(-b^2*c*x^2 + a^2*c)^(3/2)*B*e/(b^2*c) - 2/15*(-b^2*c*x^2 + a^2*c)^(3/2)*C*a^2*f/(b^4*c) - 1/3*(-b^2*
c*x^2 + a^2*c)^(3/2)*A*f/(b^2*c) - 1/4*(-b^2*c*x^2 + a^2*c)^(3/2)*(C*e + B*f)*x/(b^2*c)

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mupad [B]  time = 30.58, size = 1765, normalized size = 5.88

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((B*a^4*c^8*f*((a*c - b*c*x)^(1/2) - (a*c)^(1/2)))/(2*((a + b*x)^(1/2) - a^(1/2))) - (B*a^4*c*f*((a*c - b*c*x)
^(1/2) - (a*c)^(1/2))^15)/(2*((a + b*x)^(1/2) - a^(1/2))^15) - (35*B*a^4*c^7*f*((a*c - b*c*x)^(1/2) - (a*c)^(1
/2))^3)/(2*((a + b*x)^(1/2) - a^(1/2))^3) + (273*B*a^4*c^6*f*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^5)/(2*((a + b
*x)^(1/2) - a^(1/2))^5) - (715*B*a^4*c^5*f*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^7)/(2*((a + b*x)^(1/2) - a^(1/2
))^7) + (715*B*a^4*c^4*f*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^9)/(2*((a + b*x)^(1/2) - a^(1/2))^9) - (273*B*a^4
*c^3*f*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^11)/(2*((a + b*x)^(1/2) - a^(1/2))^11) + (35*B*a^4*c^2*f*((a*c - b*
c*x)^(1/2) - (a*c)^(1/2))^13)/(2*((a + b*x)^(1/2) - a^(1/2))^13))/(b^3*c^8 + (b^3*((a*c - b*c*x)^(1/2) - (a*c)
^(1/2))^16)/((a + b*x)^(1/2) - a^(1/2))^16 + (8*b^3*c^7*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^2)/((a + b*x)^(1/2
) - a^(1/2))^2 + (28*b^3*c^6*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^4)/((a + b*x)^(1/2) - a^(1/2))^4 + (56*b^3*c^
5*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^6)/((a + b*x)^(1/2) - a^(1/2))^6 + (70*b^3*c^4*((a*c - b*c*x)^(1/2) - (a
*c)^(1/2))^8)/((a + b*x)^(1/2) - a^(1/2))^8 + (56*b^3*c^3*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^10)/((a + b*x)^(
1/2) - a^(1/2))^10 + (28*b^3*c^2*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^12)/((a + b*x)^(1/2) - a^(1/2))^12 + (8*b
^3*c*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^14)/((a + b*x)^(1/2) - a^(1/2))^14) - (a*c - b*c*x)^(1/2)*((2*C*a^4*f
*(a + b*x)^(1/2))/(15*b^4) - (C*f*x^4*(a + b*x)^(1/2))/5 + (C*a^2*f*x^2*(a + b*x)^(1/2))/(15*b^2)) + ((C*a^4*c
^8*e*((a*c - b*c*x)^(1/2) - (a*c)^(1/2)))/(2*((a + b*x)^(1/2) - a^(1/2))) - (C*a^4*c*e*((a*c - b*c*x)^(1/2) -
(a*c)^(1/2))^15)/(2*((a + b*x)^(1/2) - a^(1/2))^15) - (35*C*a^4*c^7*e*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^3)/(
2*((a + b*x)^(1/2) - a^(1/2))^3) + (273*C*a^4*c^6*e*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^5)/(2*((a + b*x)^(1/2)
- a^(1/2))^5) - (715*C*a^4*c^5*e*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^7)/(2*((a + b*x)^(1/2) - a^(1/2))^7) + (
715*C*a^4*c^4*e*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^9)/(2*((a + b*x)^(1/2) - a^(1/2))^9) - (273*C*a^4*c^3*e*((
a*c - b*c*x)^(1/2) - (a*c)^(1/2))^11)/(2*((a + b*x)^(1/2) - a^(1/2))^11) + (35*C*a^4*c^2*e*((a*c - b*c*x)^(1/2
) - (a*c)^(1/2))^13)/(2*((a + b*x)^(1/2) - a^(1/2))^13))/(b^3*c^8 + (b^3*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^1
6)/((a + b*x)^(1/2) - a^(1/2))^16 + (8*b^3*c^7*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^2)/((a + b*x)^(1/2) - a^(1/
2))^2 + (28*b^3*c^6*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^4)/((a + b*x)^(1/2) - a^(1/2))^4 + (56*b^3*c^5*((a*c -
b*c*x)^(1/2) - (a*c)^(1/2))^6)/((a + b*x)^(1/2) - a^(1/2))^6 + (70*b^3*c^4*((a*c - b*c*x)^(1/2) - (a*c)^(1/2)
)^8)/((a + b*x)^(1/2) - a^(1/2))^8 + (56*b^3*c^3*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^10)/((a + b*x)^(1/2) - a^
(1/2))^10 + (28*b^3*c^2*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^12)/((a + b*x)^(1/2) - a^(1/2))^12 + (8*b^3*c*((a*
c - b*c*x)^(1/2) - (a*c)^(1/2))^14)/((a + b*x)^(1/2) - a^(1/2))^14) + (A*e*x*(a*c - b*c*x)^(1/2)*(a + b*x)^(1/
2))/2 - (A*f*(a^2 - b^2*x^2)*(a*c - b*c*x)^(1/2)*(a + b*x)^(1/2))/(3*b^2) - (B*e*(a^2 - b^2*x^2)*(a*c - b*c*x)
^(1/2)*(a + b*x)^(1/2))/(3*b^2) - (B*a^4*c^(1/2)*f*atan(((a*c - b*c*x)^(1/2) - (a*c)^(1/2))/(c^(1/2)*((a + b*x
)^(1/2) - a^(1/2)))))/(2*b^3) - (C*a^4*c^(1/2)*e*atan(((a*c - b*c*x)^(1/2) - (a*c)^(1/2))/(c^(1/2)*((a + b*x)^
(1/2) - a^(1/2)))))/(2*b^3) - (A*a^2*b^(1/2)*c^2*e*log((-b*c)^(1/2)*(c*(a - b*x))^(1/2)*(a + b*x)^(1/2) - b^(3
/2)*c*x))/(2*(-b*c)^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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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