Integrand size = 28, antiderivative size = 97 \[ \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{c \sqrt {d} \sqrt {a+b x+c x^2}} \] Output:
2*(-4*a*c+b^2)^(1/4)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*EllipticF((2*c* d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)/c/d^(1/2)/(c*x^2+b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {d (b+2 c x)} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c d \sqrt {a+x (b+c x)}} \] Input:
Integrate[1/(Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2]),x]
Output:
(2*Sqrt[d*(b + 2*c*x)]*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeo metric2F1[1/4, 1/2, 5/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(c*d*Sqrt[a + x*(b + c*x)])
Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1115, 1113, 762}
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.
\(\displaystyle \int \frac {1}{\sqrt {a+b x+c x^2} \sqrt {b d+2 c d x}} \, dx\) |
\(\Big \downarrow \) 1115 |
\(\displaystyle \frac {\sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {b d+2 c x d} \sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{\sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 1113 |
\(\displaystyle \frac {2 \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {1}{\sqrt {1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}}}d\sqrt {b d+2 c x d}}{c d \sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {2 \sqrt [4]{b^2-4 a c} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{c \sqrt {d} \sqrt {a+b x+c x^2}}\) |
Input:
Int[1/(Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2]),x]
Output:
(2*(b^2 - 4*a*c)^(1/4)*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*Ellipt icF[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1])/(c*Sqr t[d]*Sqrt[a + b*x + c*x^2])
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_ Symbol] :> Simp[(4/e)*Sqrt[-c/(b^2 - 4*a*c)] Subst[Int[1/Sqrt[Simp[1 - b^ 2*(x^4/(d^2*(b^2 - 4*a*c))), x]], x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]
Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sym bol] :> Simp[Sqrt[(-c)*((a + b*x + c*x^2)/(b^2 - 4*a*c))]/Sqrt[a + b*x + c* x^2] Int[(d + e*x)^m/Sqrt[(-a)*(c/(b^2 - 4*a*c)) - b*c*(x/(b^2 - 4*a*c)) - c^2*(x^2/(b^2 - 4*a*c))], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c* d - b*e, 0] && EqQ[m^2, 1/4]
Leaf count of result is larger than twice the leaf count of optimal. \(188\) vs. \(2(83)=166\).
Time = 2.67 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.95
method | result | size |
default | \(\frac {\operatorname {EllipticF}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-4 a c +b^{2}}\, \sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}}{c d \left (2 x^{3} c^{2}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right )}\) | \(189\) |
elliptic | \(\frac {2 \sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, \sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\) | \(407\) |
Input:
int(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
EllipticF(1/2*2^(1/2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^ (1/2),2^(1/2))*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*(- (2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b ^2)^(1/2)+b))^(1/2)*(-4*a*c+b^2)^(1/2)*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^( 1/2)/c/d/(2*c^2*x^3+3*b*c*x^2+2*a*c*x+b^2*x+a*b)
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.43 \[ \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx=\frac {\sqrt {2} \sqrt {c^{2} d} {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )}{c^{2} d} \] Input:
integrate(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
Output:
sqrt(2)*sqrt(c^2*d)*weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c)/(c^2*d)
\[ \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx=\int \frac {1}{\sqrt {d \left (b + 2 c x\right )} \sqrt {a + b x + c x^{2}}}\, dx \] Input:
integrate(1/(2*c*d*x+b*d)**(1/2)/(c*x**2+b*x+a)**(1/2),x)
Output:
Integral(1/(sqrt(d*(b + 2*c*x))*sqrt(a + b*x + c*x**2)), x)
\[ \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx=\int { \frac {1}{\sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}} \,d x } \] Input:
integrate(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)), x)
\[ \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx=\int { \frac {1}{\sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}} \,d x } \] Input:
integrate(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
Output:
integrate(1/(sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a)), x)
Timed out. \[ \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx=\int \frac {1}{\sqrt {b\,d+2\,c\,d\,x}\,\sqrt {c\,x^2+b\,x+a}} \,d x \] Input:
int(1/((b*d + 2*c*d*x)^(1/2)*(a + b*x + c*x^2)^(1/2)),x)
Output:
int(1/((b*d + 2*c*d*x)^(1/2)*(a + b*x + c*x^2)^(1/2)), x)
\[ \int \frac {1}{\sqrt {b d+2 c d x} \sqrt {a+b x+c x^2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}}{2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b}d x \right )}{d} \] Input:
int(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^(1/2),x)
Output:
(sqrt(d)*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2))/(a*b + 2*a*c*x + b** 2*x + 3*b*c*x**2 + 2*c**2*x**3),x))/d