Integrand size = 20, antiderivative size = 76 \[ \int \frac {1}{\sqrt {a+b \cos (c+d x) \sin (c+d x)}} \, dx=\frac {\sqrt {2} \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}{d \sqrt {2 a+b \sin (2 c+2 d x)}} \] Output:
2^(1/2)*InverseJacobiAM(c-1/4*Pi+d*x,2^(1/2)*(b/(2*a+b))^(1/2))*((2*a+b*si n(2*d*x+2*c))/(2*a+b))^(1/2)/d/(2*a+b*sin(2*d*x+2*c))^(1/2)
Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\sqrt {a+b \cos (c+d x) \sin (c+d x)}} \, dx=\frac {\operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}}}{d \sqrt {a+\frac {1}{2} b \sin (2 (c+d x))}} \] Input:
Integrate[1/Sqrt[a + b*Cos[c + d*x]*Sin[c + d*x]],x]
Output:
(EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*(c + d*x)] )/(2*a + b)])/(d*Sqrt[a + (b*Sin[2*(c + d*x)])/2])
Time = 0.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 3145, 3042, 3142, 3042, 3140}
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.
\(\displaystyle \int \frac {1}{\sqrt {a+b \sin (c+d x) \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a+b \sin (c+d x) \cos (c+d x)}}dx\) |
\(\Big \downarrow \) 3145 |
\(\displaystyle \int \frac {1}{\sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}}dx\) |
\(\Big \downarrow \) 3142 |
\(\displaystyle \frac {\sqrt {2} \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} \int \frac {1}{\sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}}dx}{\sqrt {2 a+b \sin (2 c+2 d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {2} \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} \int \frac {1}{\sqrt {\frac {2 a}{2 a+b}+\frac {b \sin (2 c+2 d x)}{2 a+b}}}dx}{\sqrt {2 a+b \sin (2 c+2 d x)}}\) |
\(\Big \downarrow \) 3140 |
\(\displaystyle \frac {\sqrt {2} \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},\frac {2 b}{2 a+b}\right )}{d \sqrt {2 a+b \sin (2 c+2 d x)}}\) |
Input:
Int[1/Sqrt[a + b*Cos[c + d*x]*Sin[c + d*x]],x]
Output:
(Sqrt[2]*EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)])/(d*Sqrt[2*a + b*Sin[2*c + 2*d*x]])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_ Symbol] :> Int[(a + b*(Sin[2*c + 2*d*x]/2))^n, x] /; FreeQ[{a, b, c, d, n}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(164\) vs. \(2(73)=146\).
Time = 0.19 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.17
method | result | size |
default | \(\frac {2 \left (2 a -b \right ) \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {b \left (\sin \left (2 d x +2 c \right )-1\right )}{2 a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (2 d x +2 c \right )\right )}{2 a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right )}{b \cos \left (2 d x +2 c \right ) \sqrt {4 a +2 b \sin \left (2 d x +2 c \right )}\, d}\) | \(165\) |
Input:
int(1/(a+b*cos(d*x+c)*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
2*(2*a-b)*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-b*(sin(2*d*x+2*c)-1)/(2 *a+b))^(1/2)*(-b*(1+sin(2*d*x+2*c))/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2 *d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))/b/cos(2*d*x+2*c)/(4*a+2 *b*sin(2*d*x+2*c))^(1/2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 304, normalized size of antiderivative = 4.00 \[ \int \frac {1}{\sqrt {a+b \cos (c+d x) \sin (c+d x)}} \, dx=-\frac {{\left (b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right )} \sqrt {4 i \, b} \sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 2 i \, a}{b}} F(\arcsin \left (\sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 2 i \, a}{b}} {\left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )}\right )\,|\,\frac {4 i \, a b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 8 \, a^{2} - b^{2}}{b^{2}}) + {\left (\sqrt {-4 i \, b} b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 2 i \, a \sqrt {-4 i \, b}\right )} \sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} - 2 i \, a}{b}} F(\arcsin \left (\sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} - 2 i \, a}{b}} {\left (\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}\right )\,|\,\frac {-4 i \, a b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 8 \, a^{2} - b^{2}}{b^{2}})}{2 \, b^{2} d} \] Input:
integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^(1/2),x, algorithm="fricas")
Output:
-1/2*((b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)*sqrt(4*I*b)*sqrt((b*sqrt(-(4*a^ 2 - b^2)/b^2) + 2*I*a)/b)*elliptic_f(arcsin(sqrt((b*sqrt(-(4*a^2 - b^2)/b^ 2) + 2*I*a)/b)*(cos(d*x + c) + I*sin(d*x + c))), (4*I*a*b*sqrt(-(4*a^2 - b ^2)/b^2) + 8*a^2 - b^2)/b^2) + (sqrt(-4*I*b)*b*sqrt(-(4*a^2 - b^2)/b^2) + 2*I*a*sqrt(-4*I*b))*sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b)*elliptic_ f(arcsin(sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b)*(cos(d*x + c) - I*si n(d*x + c))), (-4*I*a*b*sqrt(-(4*a^2 - b^2)/b^2) + 8*a^2 - b^2)/b^2))/(b^2 *d)
\[ \int \frac {1}{\sqrt {a+b \cos (c+d x) \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {a + b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}}\, dx \] Input:
integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))**(1/2),x)
Output:
Integral(1/sqrt(a + b*sin(c + d*x)*cos(c + d*x)), x)
\[ \int \frac {1}{\sqrt {a+b \cos (c+d x) \sin (c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt(b*cos(d*x + c)*sin(d*x + c) + a), x)
\[ \int \frac {1}{\sqrt {a+b \cos (c+d x) \sin (c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^(1/2),x, algorithm="giac")
Output:
integrate(1/sqrt(b*cos(d*x + c)*sin(d*x + c) + a), x)
Timed out. \[ \int \frac {1}{\sqrt {a+b \cos (c+d x) \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {a+b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}} \,d x \] Input:
int(1/(a + b*cos(c + d*x)*sin(c + d*x))^(1/2),x)
Output:
int(1/(a + b*cos(c + d*x)*sin(c + d*x))^(1/2), x)
\[ \int \frac {1}{\sqrt {a+b \cos (c+d x) \sin (c+d x)}} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right ) \sin \left (d x +c \right ) b +a}}{\cos \left (d x +c \right ) \sin \left (d x +c \right ) b +a}d x \] Input:
int(1/(a+b*cos(d*x+c)*sin(d*x+c))^(1/2),x)
Output:
int(sqrt(cos(c + d*x)*sin(c + d*x)*b + a)/(cos(c + d*x)*sin(c + d*x)*b + a ),x)