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

Optimal. Leaf size=246 \[ -\frac {\left (a^2-b^2 x^2\right ) \left (2 \left (2 a^2 C f^2-b^2 \left (C e^2-3 f (A f+B e)\right )\right )-b^2 f x (C e-3 B f)\right )}{6 b^4 f \sqrt {a+b x} \sqrt {a c-b c x}}+\frac {\sqrt {a^2 c-b^2 c x^2} \tan ^{-1}\left (\frac {b \sqrt {c} x}{\sqrt {a^2 c-b^2 c x^2}}\right ) \left (a^2 (B f+C e)+2 A b^2 e\right )}{2 b^3 \sqrt {c} \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {C \left (a^2-b^2 x^2\right ) (e+f x)^2}{3 b^2 f \sqrt {a+b x} \sqrt {a c-b c x}} \]

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Rubi [A]  time = 0.40, antiderivative size = 249, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {1610, 1654, 780, 217, 203} \begin {gather*} -\frac {\left (a^2-b^2 x^2\right ) \left (2 \left (2 a^2 C f^2-\frac {1}{2} b^2 \left (2 C e^2-6 f (A f+B e)\right )\right )-b^2 f x (C e-3 B f)\right )}{6 b^4 f \sqrt {a+b x} \sqrt {a c-b c x}}+\frac {\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 ) \left (a^2 (B f+C e)+2 A b^2 e\right )}{2 b^3 \sqrt {c} \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {C \left (a^2-b^2 x^2\right ) (e+f x)^2}{3 b^2 f \sqrt {a+b x} \sqrt {a c-b c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(C*(e + f*x)^2*(a^2 - b^2*x^2))/(3*b^2*f*Sqrt[a + b*x]*Sqrt[a*c - b*c*x]) - ((2*(2*a^2*C*f^2 - (b^2*(2*C*e^2
- 6*f*(B*e + A*f)))/2) - b^2*f*(C*e - 3*B*f)*x)*(a^2 - b^2*x^2))/(6*b^4*f*Sqrt[a + b*x]*Sqrt[a*c - b*c*x]) + (
(2*A*b^2*e + a^2*(C*e + B*f))*Sqrt[a^2*c - b^2*c*x^2]*ArcTan[(b*Sqrt[c]*x)/Sqrt[a^2*c - b^2*c*x^2]])/(2*b^3*Sq
rt[c]*Sqrt[a + b*x]*Sqrt[a*c - b*c*x])

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 \frac {(e+f x) \left (A+B x+C x^2\right )}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx &=\frac {\sqrt {a^2 c-b^2 c x^2} \int \frac {(e+f x) \left (A+B x+C x^2\right )}{\sqrt {a^2 c-b^2 c x^2}} \, dx}{\sqrt {a+b x} \sqrt {a c-b c x}}\\ &=-\frac {C (e+f x)^2 \left (a^2-b^2 x^2\right )}{3 b^2 f \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {\sqrt {a^2 c-b^2 c x^2} \int \frac {(e+f x) \left (-c \left (3 A b^2+2 a^2 C\right ) f^2+b^2 c f (C e-3 B f) x\right )}{\sqrt {a^2 c-b^2 c x^2}} \, dx}{3 b^2 c f^2 \sqrt {a+b x} \sqrt {a c-b c x}}\\ &=-\frac {C (e+f x)^2 \left (a^2-b^2 x^2\right )}{3 b^2 f \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {\left (2 \left (2 a^2 C f^2-\frac {1}{2} b^2 \left (2 C e^2-6 f (B e+A f)\right )\right )-b^2 f (C e-3 B f) x\right ) \left (a^2-b^2 x^2\right )}{6 b^4 f \sqrt {a+b x} \sqrt {a c-b c x}}+\frac {\left (\left (2 A b^2 e+a^2 (C e+B 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}{2 b^2 \sqrt {a+b x} \sqrt {a c-b c x}}\\ &=-\frac {C (e+f x)^2 \left (a^2-b^2 x^2\right )}{3 b^2 f \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {\left (2 \left (2 a^2 C f^2-\frac {1}{2} b^2 \left (2 C e^2-6 f (B e+A f)\right )\right )-b^2 f (C e-3 B f) x\right ) \left (a^2-b^2 x^2\right )}{6 b^4 f \sqrt {a+b x} \sqrt {a c-b c x}}+\frac {\left (\left (2 A b^2 e+a^2 (C e+B 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 )}{2 b^2 \sqrt {a+b x} \sqrt {a c-b c x}}\\ &=-\frac {C (e+f x)^2 \left (a^2-b^2 x^2\right )}{3 b^2 f \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {\left (2 \left (2 a^2 C f^2-\frac {1}{2} b^2 \left (2 C e^2-6 f (B e+A f)\right )\right )-b^2 f (C e-3 B f) x\right ) \left (a^2-b^2 x^2\right )}{6 b^4 f \sqrt {a+b x} \sqrt {a c-b c x}}+\frac {\left (2 A b^2 e+a^2 (C e+B f)\right ) \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 )}{2 b^3 \sqrt {c} \sqrt {a+b x} \sqrt {a c-b c x}}\\ \end {align*}

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Mathematica [A]  time = 1.43, size = 390, normalized size = 1.59 \begin {gather*} -\frac {3 \sqrt {a-b x} \sqrt {a+b x} \left (6 a^{3/2} \sin ^{-1}\left (\frac {\sqrt {a-b x}}{\sqrt {2} \sqrt {a}}\right )+\sqrt {a-b x} (4 a+b x) \sqrt {\frac {b x}{a}+1}\right ) (-3 a C f+b B f+b C e)+6 \sqrt {a-b x} \sqrt {a+b x} \left (\sqrt {a-b x} \sqrt {\frac {b x}{a}+1}+2 \sqrt {a} \sin ^{-1}\left (\frac {\sqrt {a-b x}}{\sqrt {2} \sqrt {a}}\right )\right ) \left (3 a^2 C f-2 a b (B f+C e)+b^2 (A f+B e)\right )+C f \sqrt {a+b x} \left (30 a^{5/2} \sqrt {a-b x} \sin ^{-1}\left (\frac {\sqrt {a-b x}}{\sqrt {2} \sqrt {a}}\right )+(a-b x) \sqrt {\frac {b x}{a}+1} \left (22 a^2+9 a b x+2 b^2 x^2\right )\right )+12 \sqrt {a-b x} \sqrt {\frac {b x}{a}+1} (b e-a f) \tan ^{-1}\left (\frac {\sqrt {a-b x}}{\sqrt {a+b x}}\right ) \left (a (a C-b B)+A b^2\right )}{6 b^4 \sqrt {\frac {b x}{a}+1} \sqrt {c (a-b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(6*(3*a^2*C*f + b^2*(B*e + A*f) - 2*a*b*(C*e + B*f))*Sqrt[a - b*x]*Sqrt[a + b*x]*(Sqrt[a - b*x]*Sqrt[1 +
(b*x)/a] + 2*Sqrt[a]*ArcSin[Sqrt[a - b*x]/(Sqrt[2]*Sqrt[a])]) + 3*(b*C*e + b*B*f - 3*a*C*f)*Sqrt[a - b*x]*Sqrt
[a + b*x]*(Sqrt[a - b*x]*(4*a + b*x)*Sqrt[1 + (b*x)/a] + 6*a^(3/2)*ArcSin[Sqrt[a - b*x]/(Sqrt[2]*Sqrt[a])]) +
C*f*Sqrt[a + b*x]*((a - b*x)*Sqrt[1 + (b*x)/a]*(22*a^2 + 9*a*b*x + 2*b^2*x^2) + 30*a^(5/2)*Sqrt[a - b*x]*ArcSi
n[Sqrt[a - b*x]/(Sqrt[2]*Sqrt[a])]) + 12*(A*b^2 + a*(-(b*B) + a*C))*(b*e - a*f)*Sqrt[a - b*x]*Sqrt[1 + (b*x)/a
]*ArcTan[Sqrt[a - b*x]/Sqrt[a + b*x]])/(b^4*Sqrt[c*(a - b*x)]*Sqrt[1 + (b*x)/a])

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IntegrateAlgebraic [A]  time = 0.41, size = 356, normalized size = 1.45 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {a c-b c x}}{\sqrt {c} \sqrt {a+b x}}\right ) \left (-a^2 B f+a^2 (-C) e-2 A b^2 e\right )}{b^3 \sqrt {c}}-\frac {a \sqrt {a c-b c x} \left (\frac {6 a^2 C f (a c-b c x)^2}{(a+b x)^2}+\frac {4 a^2 c C f (a c-b c x)}{a+b x}+6 a^2 c^2 C f+\frac {6 A b^2 f (a c-b c x)^2}{(a+b x)^2}+\frac {12 A b^2 c f (a c-b c x)}{a+b x}+\frac {6 b^2 B e (a c-b c x)^2}{(a+b x)^2}+\frac {12 b^2 B c e (a c-b c x)}{a+b x}+3 a b B c^2 f-\frac {3 a b B f (a c-b c x)^2}{(a+b x)^2}+3 a b c^2 C e-\frac {3 a b C e (a c-b c x)^2}{(a+b x)^2}+6 A b^2 c^2 f+6 b^2 B c^2 e\right )}{3 b^4 \sqrt {a+b x} \left (\frac {a c-b c x}{a+b x}+c\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a*Sqrt[a*c - b*c*x]*(6*b^2*B*c^2*e + 3*a*b*c^2*C*e + 6*A*b^2*c^2*f + 3*a*b*B*c^2*f + 6*a^2*c^2*C*f + (12
*b^2*B*c*e*(a*c - b*c*x))/(a + b*x) + (12*A*b^2*c*f*(a*c - b*c*x))/(a + b*x) + (4*a^2*c*C*f*(a*c - b*c*x))/(a
+ b*x) + (6*b^2*B*e*(a*c - b*c*x)^2)/(a + b*x)^2 - (3*a*b*C*e*(a*c - b*c*x)^2)/(a + b*x)^2 + (6*A*b^2*f*(a*c -
b*c*x)^2)/(a + b*x)^2 - (3*a*b*B*f*(a*c - b*c*x)^2)/(a + b*x)^2 + (6*a^2*C*f*(a*c - b*c*x)^2)/(a + b*x)^2))/(
b^4*Sqrt[a + b*x]*(c + (a*c - b*c*x)/(a + b*x))^3) + ((-2*A*b^2*e - a^2*C*e - a^2*B*f)*ArcTan[Sqrt[a*c - b*c*x
]/(Sqrt[c]*Sqrt[a + b*x])])/(b^3*Sqrt[c])

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fricas [A]  time = 0.71, size = 302, normalized size = 1.23 \begin {gather*} \left [-\frac {3 \, {\left (B a^{2} b f + {\left (C a^{2} b + 2 \, A 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 (2 \, C b^{2} f x^{2} + 6 \, B b^{2} e + 2 \, {\left (2 \, C a^{2} + 3 \, A b^{2}\right )} f + 3 \, {\left (C b^{2} e + B b^{2} f\right )} x\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{12 \, b^{4} c}, -\frac {3 \, {\left (B a^{2} b f + {\left (C a^{2} b + 2 \, A 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 (2 \, C b^{2} f x^{2} + 6 \, B b^{2} e + 2 \, {\left (2 \, C a^{2} + 3 \, A b^{2}\right )} f + 3 \, {\left (C b^{2} e + B b^{2} f\right )} x\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{6 \, b^{4} c}\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]

[-1/12*(3*(B*a^2*b*f + (C*a^2*b + 2*A*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*(2*C*b^2*f*x^2 + 6*B*b^2*e + 2*(2*C*a^2 + 3*A*b^2)*f + 3*(C*b^2*e + B*b^2*f)*x)*sqrt(-
b*c*x + a*c)*sqrt(b*x + a))/(b^4*c), -1/6*(3*(B*a^2*b*f + (C*a^2*b + 2*A*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)) + (2*C*b^2*f*x^2 + 6*B*b^2*e + 2*(2*C*a^2 + 3*A*b^2)*f + 3
*(C*b^2*e + B*b^2*f)*x)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/(b^4*c)]

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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 [A]  time = 0.03, size = 365, normalized size = 1.48 \begin {gather*} \frac {\sqrt {b x +a}\, \sqrt {-\left (b x -a \right ) c}\, \left (6 A \,b^{4} c e \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}}\right )+3 B \,a^{2} b^{2} c f \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}}\right )+3 C \,a^{2} b^{2} c e \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}}\right )-2 \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, \sqrt {b^{2} c}\, C \,b^{2} f \,x^{2}-3 \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, \sqrt {b^{2} c}\, B \,b^{2} f x -3 \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, \sqrt {b^{2} c}\, C \,b^{2} e x -6 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, A \,b^{2} f -6 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, B \,b^{2} e -4 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, C \,a^{2} f \right )}{6 \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, \sqrt {b^{2} c}\, b^{4} c} \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]

1/6*(b*x+a)^(1/2)*(-(b*x-a)*c)^(1/2)/c*(6*A*arctan((b^2*c)^(1/2)/(-(b^2*x^2-a^2)*c)^(1/2)*x)*b^4*c*e+3*B*arcta
n((b^2*c)^(1/2)/(-(b^2*x^2-a^2)*c)^(1/2)*x)*a^2*b^2*c*f+3*C*arctan((b^2*c)^(1/2)/(-(b^2*x^2-a^2)*c)^(1/2)*x)*a
^2*b^2*c*e-2*C*x^2*b^2*f*(-(b^2*x^2-a^2)*c)^(1/2)*(b^2*c)^(1/2)-3*B*(-(b^2*x^2-a^2)*c)^(1/2)*(b^2*c)^(1/2)*x*b
^2*f-3*C*(-(b^2*x^2-a^2)*c)^(1/2)*(b^2*c)^(1/2)*x*b^2*e-6*A*(b^2*c)^(1/2)*(-(b^2*x^2-a^2)*c)^(1/2)*b^2*f-6*B*(
b^2*c)^(1/2)*(-(b^2*x^2-a^2)*c)^(1/2)*b^2*e-4*C*(b^2*c)^(1/2)*(-(b^2*x^2-a^2)*c)^(1/2)*a^2*f)/(-(b^2*x^2-a^2)*
c)^(1/2)/b^4/(b^2*c)^(1/2)

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maxima [A]  time = 2.05, size = 189, normalized size = 0.77 \begin {gather*} -\frac {\sqrt {-b^{2} c x^{2} + a^{2} c} C f x^{2}}{3 \, b^{2} c} + \frac {A e \arcsin \left (\frac {b x}{a}\right )}{b \sqrt {c}} + \frac {{\left (C e + B f\right )} a^{2} \arcsin \left (\frac {b x}{a}\right )}{2 \, b^{3} \sqrt {c}} - \frac {\sqrt {-b^{2} c x^{2} + a^{2} c} B e}{b^{2} c} - \frac {2 \, \sqrt {-b^{2} c x^{2} + a^{2} c} C a^{2} f}{3 \, b^{4} c} - \frac {\sqrt {-b^{2} c x^{2} + a^{2} c} A f}{b^{2} c} - \frac {\sqrt {-b^{2} c x^{2} + a^{2} c} {\left (C e + B f\right )} x}{2 \, 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]

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

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mupad [B]  time = 30.74, size = 1011, normalized size = 4.11 \begin {gather*} -\frac {\frac {2\,B\,a^2\,f\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^7}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7}-\frac {2\,B\,a^2\,c^3\,f\,\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}{\sqrt {a+b\,x}-\sqrt {a}}-\frac {14\,B\,a^2\,c\,f\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^5}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5}+\frac {14\,B\,a^2\,c^2\,f\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^3}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}}{b^3\,c^4+\frac {b^3\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^8}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}+\frac {4\,b^3\,c^3\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^2}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}+\frac {6\,b^3\,c^2\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^4}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}+\frac {4\,b^3\,c\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^6}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}}-\frac {\frac {2\,C\,a^2\,e\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^7}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7}-\frac {2\,C\,a^2\,c^3\,e\,\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}{\sqrt {a+b\,x}-\sqrt {a}}-\frac {14\,C\,a^2\,c\,e\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^5}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5}+\frac {14\,C\,a^2\,c^2\,e\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^3}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}}{b^3\,c^4+\frac {b^3\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^8}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}+\frac {4\,b^3\,c^3\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^2}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}+\frac {6\,b^3\,c^2\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^4}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}+\frac {4\,b^3\,c\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^6}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}}-\frac {\sqrt {a\,c-b\,c\,x}\,\left (\frac {2\,C\,a^3\,f}{3\,b^4\,c}+\frac {C\,f\,x^3}{3\,b\,c}+\frac {C\,a\,f\,x^2}{3\,b^2\,c}+\frac {2\,C\,a^2\,f\,x}{3\,b^3\,c}\right )}{\sqrt {a+b\,x}}-\frac {4\,A\,e\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}{\sqrt {b^2\,c}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{\sqrt {b^2\,c}}-\frac {A\,f\,\sqrt {a\,c-b\,c\,x}\,\sqrt {a+b\,x}}{b^2\,c}-\frac {B\,e\,\sqrt {a\,c-b\,c\,x}\,\sqrt {a+b\,x}}{b^2\,c}-\frac {2\,B\,a^2\,f\,\mathrm {atan}\left (\frac {\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}}{\sqrt {c}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{b^3\,\sqrt {c}}-\frac {2\,C\,a^2\,e\,\mathrm {atan}\left (\frac {\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}}{\sqrt {c}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{b^3\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*B*a^2*f*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^7)/((a + b*x)^(1/2) - a^(1/2))^7 - (2*B*a^2*c^3*f*((a*c - b*
c*x)^(1/2) - (a*c)^(1/2)))/((a + b*x)^(1/2) - a^(1/2)) - (14*B*a^2*c*f*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^5)/
((a + b*x)^(1/2) - a^(1/2))^5 + (14*B*a^2*c^2*f*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^3)/((a + b*x)^(1/2) - a^(1
/2))^3)/(b^3*c^4 + (b^3*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^8)/((a + b*x)^(1/2) - a^(1/2))^8 + (4*b^3*c^3*((a*
c - b*c*x)^(1/2) - (a*c)^(1/2))^2)/((a + b*x)^(1/2) - a^(1/2))^2 + (6*b^3*c^2*((a*c - b*c*x)^(1/2) - (a*c)^(1/
2))^4)/((a + b*x)^(1/2) - a^(1/2))^4 + (4*b^3*c*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^6)/((a + b*x)^(1/2) - a^(1
/2))^6) - ((2*C*a^2*e*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^7)/((a + b*x)^(1/2) - a^(1/2))^7 - (2*C*a^2*c^3*e*((
a*c - b*c*x)^(1/2) - (a*c)^(1/2)))/((a + b*x)^(1/2) - a^(1/2)) - (14*C*a^2*c*e*((a*c - b*c*x)^(1/2) - (a*c)^(1
/2))^5)/((a + b*x)^(1/2) - a^(1/2))^5 + (14*C*a^2*c^2*e*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^3)/((a + b*x)^(1/2
) - a^(1/2))^3)/(b^3*c^4 + (b^3*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^8)/((a + b*x)^(1/2) - a^(1/2))^8 + (4*b^3*
c^3*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^2)/((a + b*x)^(1/2) - a^(1/2))^2 + (6*b^3*c^2*((a*c - b*c*x)^(1/2) - (
a*c)^(1/2))^4)/((a + b*x)^(1/2) - a^(1/2))^4 + (4*b^3*c*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^6)/((a + b*x)^(1/2
) - a^(1/2))^6) - ((a*c - b*c*x)^(1/2)*((2*C*a^3*f)/(3*b^4*c) + (C*f*x^3)/(3*b*c) + (C*a*f*x^2)/(3*b^2*c) + (2
*C*a^2*f*x)/(3*b^3*c)))/(a + b*x)^(1/2) - (4*A*e*atan((b*((a*c - b*c*x)^(1/2) - (a*c)^(1/2)))/((b^2*c)^(1/2)*(
(a + b*x)^(1/2) - a^(1/2)))))/(b^2*c)^(1/2) - (A*f*(a*c - b*c*x)^(1/2)*(a + b*x)^(1/2))/(b^2*c) - (B*e*(a*c -
b*c*x)^(1/2)*(a + b*x)^(1/2))/(b^2*c) - (2*B*a^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)))))/(b^3*c^(1/2)) - (2*C*a^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)))))/(b^3*c^(1/2))

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sympy [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)

[Out]

Timed out

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