3.1.25 \(\int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x} (e+f x)^2} \, dx\) [25]

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

[Out]

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

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Rubi [A]
time = 0.38, antiderivative size = 322, normalized size of antiderivative = 1.00, 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 = {1624, 1665, 858, 223, 209, 739, 210} \begin {gather*} \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 ) \text {ArcTan}\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 {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 {C \sqrt {a^2 c-b^2 c x^2} \text {ArcTan}\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}} \end {gather*}

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 223

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 739

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 858

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 1624

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 1665

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]

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.97, size = 229, normalized size = 0.71 \begin {gather*} \frac {2 \left (\frac {f \left (C e^2+f (-B e+A f)\right ) (-a+b x) \sqrt {a+b x}}{2 (-b e+a f) (b e+a f) (e+f x)}+\frac {C \sqrt {a-b x} \tan ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )}{b}-\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-b x} \tan ^{-1}\left (\frac {\sqrt {b e+a f} \sqrt {a+b x}}{\sqrt {b e-a f} \sqrt {a-b x}}\right )}{(b e-a f)^{3/2} (b e+a f)^{3/2}}\right )}{f^2 \sqrt {c (a-b x)}} \end {gather*}

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]

(2*((f*(C*e^2 + f*(-(B*e) + A*f))*(-a + b*x)*Sqrt[a + b*x])/(2*(-(b*e) + a*f)*(b*e + a*f)*(e + f*x)) + (C*Sqrt
[a - b*x]*ArcTan[Sqrt[a + b*x]/Sqrt[a - b*x]])/b - ((a^2*f^2*(-2*C*e + B*f) + b^2*(C*e^3 - A*e*f^2))*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] Leaf count of result is larger than twice the leaf count of optimal. \(1165\) vs. \(2(290)=580\).
time = 0.11, size = 1166, normalized size = 3.62

method	result	size
default	\(\frac {\left (A \ln \left (\frac {2 b^{2} c e x +2 a^{2} c f +2 \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, f}{f x +e}\right ) b^{2} c e \,f^{3} x \sqrt {b^{2} c}-B \ln \left (\frac {2 b^{2} c e x +2 a^{2} c f +2 \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, f}{f x +e}\right ) a^{2} c \,f^{4} x \sqrt {b^{2} c}+2 C \ln \left (\frac {2 b^{2} c e x +2 a^{2} c f +2 \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, f}{f x +e}\right ) a^{2} c e \,f^{3} x \sqrt {b^{2} c}-C \ln \left (\frac {2 b^{2} c e x +2 a^{2} c f +2 \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, f}{f x +e}\right ) b^{2} c \,e^{3} f x \sqrt {b^{2} c}+C \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}\right ) a^{2} c \,f^{4} x \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}-C \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}\right ) b^{2} c \,e^{2} f^{2} x \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}+A \ln \left (\frac {2 b^{2} c e x +2 a^{2} c f +2 \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, f}{f x +e}\right ) b^{2} c \,e^{2} f^{2} \sqrt {b^{2} c}-B \ln \left (\frac {2 b^{2} c e x +2 a^{2} c f +2 \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, f}{f x +e}\right ) a^{2} c e \,f^{3} \sqrt {b^{2} c}+2 C \ln \left (\frac {2 b^{2} c e x +2 a^{2} c f +2 \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, f}{f x +e}\right ) a^{2} c \,e^{2} f^{2} \sqrt {b^{2} c}-C \ln \left (\frac {2 b^{2} c e x +2 a^{2} c f +2 \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, f}{f x +e}\right ) b^{2} c \,e^{4} \sqrt {b^{2} c}+C \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}\right ) a^{2} c e \,f^{3} \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}-C \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}\right ) b^{2} c \,e^{3} f \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}-A \,f^{4} \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {b^{2} c}+B e \,f^{3} \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {b^{2} c}-C \,e^{2} f^{2} \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {b^{2} c}\right ) \sqrt {b x +a}\, \sqrt {c \left (-b x +a \right )}}{c \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \left (a f -b e \right ) \sqrt {b^{2} c}\, \left (a f +b e \right ) \left (f x +e \right ) \sqrt {\frac {c \left (a^{2} f^{2}-b^{2} e^{2}\right )}{f^{2}}}\, f^{3}}\)	\(1166\)

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,method=_RETURNVERBOSE)

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume((4*b^2*c>0)', see `assume?` fo
r more detai

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Fricas [F(-1)] Timed out
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((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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x + C x^{2}}{\sqrt {- c \left (- a + b x\right )} \sqrt {a + b x} \left (e + f x\right )^{2}}\, dx \end {gather*}

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]

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

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Giac [A]
time = 1.34, size = 523, normalized size = 1.62 \begin {gather*} -\frac {\frac {2 \, {\left (B a^{2} b \sqrt {-c} f^{3} - 2 \, C a^{2} b \sqrt {-c} f^{2} e - A b^{3} \sqrt {-c} f^{2} e + C b^{3} \sqrt {-c} e^{3}\right )} \arctan \left (\frac {{\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} f - 2 \, b c e}{2 \, \sqrt {a^{2} f^{2} - b^{2} e^{2}} c}\right )}{{\left (a^{2} f^{4} - b^{2} f^{2} e^{2}\right )} \sqrt {a^{2} f^{2} - b^{2} e^{2}} c} + \frac {C \log \left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2}\right )}{\sqrt {-c} f^{2}} + \frac {4 \, {\left (2 \, A a^{2} b^{2} \sqrt {-c} c f^{3} - A b^{3} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} \sqrt {-c} f^{2} e - 2 \, B a^{2} b^{2} \sqrt {-c} c f^{2} e + B b^{3} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} \sqrt {-c} f e^{2} + 2 \, C a^{2} b^{2} \sqrt {-c} c f e^{2} - C b^{3} {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} \sqrt {-c} e^{3}\right )}}{{\left (a^{2} f^{4} - b^{2} f^{2} e^{2}\right )} {\left ({\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{4} f + 4 \, a^{2} c^{2} f - 4 \, b {\left (\sqrt {b x + a} \sqrt {-c} - \sqrt {-{\left (b x + a\right )} c + 2 \, a c}\right )}^{2} c e\right )}}}{b} \end {gather*}

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]

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

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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