Integrand size = 33, antiderivative size = 125 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx=-\frac {(2 b B-a C) \sqrt {a+b x} \sqrt {a c-b c x}}{2 b^3 c}-\frac {C (a+b x)^{3/2} \sqrt {a c-b c x}}{2 b^3 c}+\frac {\left (2 A b^2+a^2 C\right ) \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right )}{b^3 \sqrt {c}} \] Output:
-1/2*(2*B*b-C*a)*(b*x+a)^(1/2)*(-b*c*x+a*c)^(1/2)/b^3/c-1/2*C*(b*x+a)^(3/2 )*(-b*c*x+a*c)^(1/2)/b^3/c+(2*A*b^2+C*a^2)*arctan(c^(1/2)*(b*x+a)^(1/2)/(- b*c*x+a*c)^(1/2))/b^3/c^(1/2)
Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx=\frac {b (-a+b x) \sqrt {a+b x} (2 B+C x)+2 \left (2 A b^2+a^2 C\right ) \sqrt {a-b x} \arctan \left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )}{2 b^3 \sqrt {c (a-b x)}} \] Input:
Integrate[(A + B*x + C*x^2)/(Sqrt[a + b*x]*Sqrt[a*c - b*c*x]),x]
Output:
(b*(-a + b*x)*Sqrt[a + b*x]*(2*B + C*x) + 2*(2*A*b^2 + a^2*C)*Sqrt[a - b*x ]*ArcTan[Sqrt[a + b*x]/Sqrt[a - b*x]])/(2*b^3*Sqrt[c*(a - b*x)])
Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {1189, 83, 646, 45, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx\) |
\(\Big \downarrow \) 1189 |
\(\displaystyle \int \frac {C x^2+A}{\sqrt {a+b x} \sqrt {a c-b c x}}dx+B \int \frac {x}{\sqrt {a+b x} \sqrt {a c-b c x}}dx\) |
\(\Big \downarrow \) 83 |
\(\displaystyle \int \frac {C x^2+A}{\sqrt {a+b x} \sqrt {a c-b c x}}dx-\frac {B \sqrt {a+b x} \sqrt {a c-b c x}}{b^2 c}\) |
\(\Big \downarrow \) 646 |
\(\displaystyle \frac {1}{2} \left (\frac {a^2 C}{b^2}+2 A\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {a c-b c x}}dx-\frac {B \sqrt {a+b x} \sqrt {a c-b c x}}{b^2 c}-\frac {C x \sqrt {a+b x} \sqrt {a c-b c x}}{2 b^2 c}\) |
\(\Big \downarrow \) 45 |
\(\displaystyle \left (\frac {a^2 C}{b^2}+2 A\right ) \int \frac {1}{\frac {c (a+b x) b}{a c-b c x}+b}d\frac {\sqrt {a+b x}}{\sqrt {a c-b c x}}-\frac {B \sqrt {a+b x} \sqrt {a c-b c x}}{b^2 c}-\frac {C x \sqrt {a+b x} \sqrt {a c-b c x}}{2 b^2 c}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\left (\frac {a^2 C}{b^2}+2 A\right ) \arctan \left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a c-b c x}}\right )}{b \sqrt {c}}-\frac {B \sqrt {a+b x} \sqrt {a c-b c x}}{b^2 c}-\frac {C x \sqrt {a+b x} \sqrt {a c-b c x}}{2 b^2 c}\) |
Input:
Int[(A + B*x + C*x^2)/(Sqrt[a + b*x]*Sqrt[a*c - b*c*x]),x]
Output:
-((B*Sqrt[a + b*x]*Sqrt[a*c - b*c*x])/(b^2*c)) - (C*x*Sqrt[a + b*x]*Sqrt[a *c - b*c*x])/(2*b^2*c) + ((2*A + (a^2*C)/b^2)*ArcTan[(Sqrt[c]*Sqrt[a + b*x ])/Sqrt[a*c - b*c*x]])/(b*Sqrt[c])
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && EqQ[a*d*f *(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_) ^2), x_Symbol] :> Simp[b*x*(c + d*x)^(m + 1)*((e + f*x)^(n + 1)/(d*f*(2*m + 3))), x] - Simp[(b*c*e - a*d*f*(2*m + 3))/(d*f*(2*m + 3)) Int[(c + d*x)^ m*(e + f*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[m, n] && EqQ[d*e + c*f, 0] && !LtQ[m, -1]
Int[((d_) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[b Int[x*(d + e*x)^m*(f + g*x)^n, x], x] + Int[(d + e*x)^m*(f + g*x)^n*(a + c*x^2), x] /; FreeQ[{a, b, c, d, e, f , g, m, n}, x] && EqQ[m, n] && EqQ[e*f + d*g, 0]
Time = 0.68 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.01
method | result | size |
risch | \(-\frac {\left (C x +2 B \right ) \left (-b x +a \right ) \sqrt {b x +a}}{2 b^{2} \sqrt {-c \left (b x -a \right )}}+\frac {\left (2 b^{2} A +a^{2} C \right ) \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-b^{2} c \,x^{2}+a^{2} c}}\right ) \sqrt {-\left (b x +a \right ) c \left (b x -a \right )}}{2 b^{2} \sqrt {b^{2} c}\, \sqrt {b x +a}\, \sqrt {-c \left (b x -a \right )}}\) | \(126\) |
default | \(\frac {\sqrt {b x +a}\, \sqrt {c \left (-b x +a \right )}\, \left (2 A \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}\right ) b^{2} c +C \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}}\right ) a^{2} c -C \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, \sqrt {b^{2} c}\, x -2 B \sqrt {b^{2} c}\, \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\right )}{2 b^{2} \sqrt {c \left (-b^{2} x^{2}+a^{2}\right )}\, c \sqrt {b^{2} c}}\) | \(168\) |
Input:
int((C*x^2+B*x+A)/(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2),x,method=_RETURNVERBOSE )
Output:
-1/2*(C*x+2*B)/b^2*(-b*x+a)*(b*x+a)^(1/2)/(-c*(b*x-a))^(1/2)+1/2*(2*A*b^2+ C*a^2)/b^2/(b^2*c)^(1/2)*arctan((b^2*c)^(1/2)*x/(-b^2*c*x^2+a^2*c)^(1/2))* (-(b*x+a)*c*(b*x-a))^(1/2)/(b*x+a)^(1/2)/(-c*(b*x-a))^(1/2)
Time = 0.10 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.57 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx=\left [-\frac {{\left (C a^{2} + 2 \, A b^{2}\right )} \sqrt {-c} \log \left (2 \, b^{2} c x^{2} - 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {-c} x - a^{2} c\right ) + 2 \, {\left (C b x + 2 \, B b\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{4 \, b^{3} c}, -\frac {{\left (C a^{2} + 2 \, A b^{2}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {c} x}{b^{2} c x^{2} - a^{2} c}\right ) + {\left (C b x + 2 \, B b\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{2 \, b^{3} c}\right ] \] Input:
integrate((C*x^2+B*x+A)/(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2),x, algorithm="fri cas")
Output:
[-1/4*((C*a^2 + 2*A*b^2)*sqrt(-c)*log(2*b^2*c*x^2 - 2*sqrt(-b*c*x + a*c)*s qrt(b*x + a)*b*sqrt(-c)*x - a^2*c) + 2*(C*b*x + 2*B*b)*sqrt(-b*c*x + a*c)* sqrt(b*x + a))/(b^3*c), -1/2*((C*a^2 + 2*A*b^2)*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)) + (C*b*x + 2*B*b)*s qrt(-b*c*x + a*c)*sqrt(b*x + a))/(b^3*c)]
Timed out. \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx=\text {Timed out} \] Input:
integrate((C*x**2+B*x+A)/(b*x+a)**(1/2)/(-b*c*x+a*c)**(1/2),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.70 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx=\frac {C a^{2} \arcsin \left (\frac {b x}{a}\right )}{2 \, b^{3} \sqrt {c}} + \frac {A \arcsin \left (\frac {b x}{a}\right )}{b \sqrt {c}} - \frac {\sqrt {-b^{2} c x^{2} + a^{2} c} C x}{2 \, b^{2} c} - \frac {\sqrt {-b^{2} c x^{2} + a^{2} c} B}{b^{2} c} \] Input:
integrate((C*x^2+B*x+A)/(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2),x, algorithm="max ima")
Output:
1/2*C*a^2*arcsin(b*x/a)/(b^3*sqrt(c)) + A*arcsin(b*x/a)/(b*sqrt(c)) - 1/2* sqrt(-b^2*c*x^2 + a^2*c)*C*x/(b^2*c) - sqrt(-b^2*c*x^2 + a^2*c)*B/(b^2*c)
Time = 0.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx=-\frac {\sqrt {-{\left (b x + a\right )} c + 2 \, a c} \sqrt {b x + a} {\left (\frac {{\left (b x + a\right )} C}{c} - \frac {C a c - 2 \, B b c}{c^{2}}\right )} + \frac {2 \, {\left (C a^{2} + 2 \, A b^{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {-c} + \sqrt {-{\left (b x + a\right )} c + 2 \, a c} \right |}\right )}{\sqrt {-c}}}{2 \, b^{3}} \] Input:
integrate((C*x^2+B*x+A)/(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2),x, algorithm="gia c")
Output:
-1/2*(sqrt(-(b*x + a)*c + 2*a*c)*sqrt(b*x + a)*((b*x + a)*C/c - (C*a*c - 2 *B*b*c)/c^2) + 2*(C*a^2 + 2*A*b^2)*log(abs(-sqrt(b*x + a)*sqrt(-c) + sqrt( -(b*x + a)*c + 2*a*c)))/sqrt(-c))/b^3
Time = 11.52 (sec) , antiderivative size = 489, normalized size of antiderivative = 3.91 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx=-\frac {\frac {2\,C\,a^2\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^7}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7}-\frac {2\,C\,a^2\,c^3\,\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}{\sqrt {a+b\,x}-\sqrt {a}}-\frac {14\,C\,a^2\,c\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^5}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5}+\frac {14\,C\,a^2\,c^2\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^3}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}}{b^3\,c^4+\frac {b^3\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^8}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}+\frac {4\,b^3\,c^3\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^2}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}+\frac {6\,b^3\,c^2\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^4}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}+\frac {4\,b^3\,c\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^6}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}}-\frac {4\,A\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}{\sqrt {b^2\,c}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{\sqrt {b^2\,c}}-\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}}{\sqrt {c}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{b^3\,\sqrt {c}}-\frac {B\,\sqrt {a\,c-b\,c\,x}\,\sqrt {a+b\,x}}{b^2\,c} \] Input:
int((A + B*x + C*x^2)/((a*c - b*c*x)^(1/2)*(a + b*x)^(1/2)),x)
Output:
- ((2*C*a^2*((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*((a*c - b*c*x)^(1/2) - (a*c)^(1/2)))/((a + b*x)^(1/2 ) - a^(1/2)) - (14*C*a^2*c*((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*((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) - (4*A *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) - (2*C*a^2*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)) - (B*(a*c - b*c*x)^(1/2)*(a + b*x)^(1/2))/(b^2*c)
Time = 0.15 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx=\frac {\sqrt {c}\, \left (-2 \mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right ) a^{2} c -4 \mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right ) a \,b^{2}-2 \sqrt {b x +a}\, \sqrt {-b x +a}\, b^{2}-\sqrt {b x +a}\, \sqrt {-b x +a}\, b c x \right )}{2 b^{3} c} \] Input:
int((C*x^2+B*x+A)/(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2),x)
Output:
(sqrt(c)*( - 2*asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))*a**2*c - 4*asin(sqrt( a - b*x)/(sqrt(a)*sqrt(2)))*a*b**2 - 2*sqrt(a + b*x)*sqrt(a - b*x)*b**2 - sqrt(a + b*x)*sqrt(a - b*x)*b*c*x))/(2*b**3*c)