Integrand size = 21, antiderivative size = 75 \[ \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right ),2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}{d \sqrt {a \cos (c+d x)+b \sin (c+d x)}} \] Output:
2*InverseJacobiAM(1/2*c+1/2*d*x-1/2*arctan(b,a),2^(1/2))*((a*cos(d*x+c)+b* sin(d*x+c))/(a^2+b^2)^(1/2))^(1/2)/d/(a*cos(d*x+c)+b*sin(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.23 \[ \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}} \, dx=\frac {2 \sqrt {\cos ^2\left (c+d x+\arctan \left (\frac {a}{b}\right )\right )} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (c+d x+\arctan \left (\frac {a}{b}\right )\right )\right ) \tan \left (c+d x+\arctan \left (\frac {a}{b}\right )\right )}{d \sqrt {\sqrt {1+\frac {a^2}{b^2}} b \sin \left (c+d x+\arctan \left (\frac {a}{b}\right )\right )}} \] Input:
Integrate[1/Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]],x]
Output:
(2*Sqrt[Cos[c + d*x + ArcTan[a/b]]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[c + d*x + ArcTan[a/b]]^2]*Tan[c + d*x + ArcTan[a/b]])/(d*Sqrt[Sqrt[1 + a^2/b^2]*b*Sin[c + d*x + ArcTan[a/b]]])
Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3557, 3042, 3120}
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.
\(\displaystyle \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}}dx\) |
\(\Big \downarrow \) 3557 |
\(\displaystyle \frac {\sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} \int \frac {1}{\sqrt {\cos \left (c+d x-\tan ^{-1}(a,b)\right )}}dx}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} \int \frac {1}{\sqrt {\sin \left (c+d x-\tan ^{-1}(a,b)+\frac {\pi }{2}\right )}}dx}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right ),2\right )}{d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}\) |
Input:
Int[1/Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]],x]
Output:
(2*EllipticF[(c + d*x - ArcTan[a, b])/2, 2]*Sqrt[(a*Cos[c + d*x] + b*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(d*Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]])
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(a*Cos[c + d*x] + b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*S in[c + d*x])/Sqrt[a^2 + b^2])^n Int[Cos[c + d*x - ArcTan[a, b]]^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && !(GeQ[n, 1] || LeQ[n, -1]) && !(GtQ[a^2 + b^2, 0] || EqQ[a^2 + b^2, 0])
Time = 0.36 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.65
method | result | size |
default | \(-\frac {\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )}{\cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d}\) | \(124\) |
Input:
int(1/(cos(d*x+c)*a+b*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
-(-sin(d*x+c-arctan(-a,b))+1)^(1/2)*(2*sin(d*x+c-arctan(-a,b))+2)^(1/2)*si n(d*x+c-arctan(-a,b))^(1/2)*EllipticF((-sin(d*x+c-arctan(-a,b))+1)^(1/2),1 /2*2^(1/2))/cos(d*x+c-arctan(-a,b))/(sin(d*x+c-arctan(-a,b))*(a^2+b^2)^(1/ 2))^(1/2)/d
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.75 \[ \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}} \, dx=-\frac {2 \, {\left (\sqrt {\frac {1}{2} \, a - \frac {1}{2} i \, b} {\left (i \, a - b\right )} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {\frac {1}{2} \, a + \frac {1}{2} i \, b} {\left (-i \, a - b\right )} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}}{{\left (a^{2} + b^{2}\right )} d} \] Input:
integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(1/2),x, algorithm="fricas")
Output:
-2*(sqrt(1/2*a - 1/2*I*b)*(I*a - b)*weierstrassPInverse(-4*(a^2 + 2*I*a*b - b^2)/(a^2 + b^2), 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(1/2*a + 1/2*I *b)*(-I*a - b)*weierstrassPInverse(-4*(a^2 - 2*I*a*b - b^2)/(a^2 + b^2), 0 , cos(d*x + c) - I*sin(d*x + c)))/((a^2 + b^2)*d)
\[ \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}}}\, dx \] Input:
integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))**(1/2),x)
Output:
Integral(1/sqrt(a*cos(c + d*x) + b*sin(c + d*x)), x)
\[ \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}} \, dx=\int { \frac {1}{\sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}} \,d x } \] Input:
integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt(a*cos(d*x + c) + b*sin(d*x + c)), x)
\[ \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}} \, dx=\int { \frac {1}{\sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}} \,d x } \] Input:
integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(1/2),x, algorithm="giac")
Output:
integrate(1/sqrt(a*cos(d*x + c) + b*sin(d*x + c)), x)
Timed out. \[ \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {a\,\cos \left (c+d\,x\right )+b\,\sin \left (c+d\,x\right )}} \,d x \] Input:
int(1/(a*cos(c + d*x) + b*sin(c + d*x))^(1/2),x)
Output:
int(1/(a*cos(c + d*x) + b*sin(c + d*x))^(1/2), x)
\[ \int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}}{\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) b}d x \] Input:
int(1/(a*cos(d*x+c)+b*sin(d*x+c))^(1/2),x)
Output:
int(sqrt(cos(c + d*x)*a + sin(c + d*x)*b)/(cos(c + d*x)*a + sin(c + d*x)*b ),x)