Integrand size = 33, antiderivative size = 149 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\frac {2 B \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b d}+\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 d}-\frac {2 a (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 (a+b) d} \] Output:
2*B*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/2 )/b/d+2*(A*b-B*a)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))* sec(d*x+c)^(1/2)/b^2/d-2*a*(A*b-B*a)*cos(d*x+c)^(1/2)*EllipticPi(sin(1/2*d *x+1/2*c),2*b/(a+b),2^(1/2))*sec(d*x+c)^(1/2)/b^2/(a+b)/d
Time = 21.56 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.48 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\frac {\cot (c+d x) \left (-b B \sec ^{\frac {3}{2}}(c+d x)-b B \cos (2 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)+b B \sec ^{\frac {7}{2}}(c+d x)+b B \cos (2 (c+d x)) \sec ^{\frac {7}{2}}(c+d x)-2 b B E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}+2 b B \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}+2 A b \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}-2 a B \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}\right )}{b^2 d} \] Input:
Integrate[(A + B*Cos[c + d*x])/((a + b*Cos[c + d*x])*Sqrt[Sec[c + d*x]]),x ]
Output:
(Cot[c + d*x]*(-(b*B*Sec[c + d*x]^(3/2)) - b*B*Cos[2*(c + d*x)]*Sec[c + d* x]^(3/2) + b*B*Sec[c + d*x]^(7/2) + b*B*Cos[2*(c + d*x)]*Sec[c + d*x]^(7/2 ) - 2*b*B*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 2*b*B*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 2*A*b*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x ]^2] - 2*a*B*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[ c + d*x]^2]))/(b^2*d)
Time = 1.01 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.85, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3439, 3042, 4526, 3042, 4258, 3042, 3119, 4335, 3042, 3282, 3042, 3120, 3284}
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.
\(\displaystyle \int \frac {A+B \cos (c+d x)}{\sqrt {\sec (c+d x)} (a+b \cos (c+d x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3439 |
\(\displaystyle \int \frac {A \sec (c+d x)+B}{\sqrt {\sec (c+d x)} (a \sec (c+d x)+b)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \csc \left (c+d x+\frac {\pi }{2}\right )+B}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+b\right )}dx\) |
\(\Big \downarrow \) 4526 |
\(\displaystyle \frac {(A b-a B) \int \frac {\sqrt {\sec (c+d x)}}{b+a \sec (c+d x)}dx}{b}+\frac {B \int \frac {1}{\sqrt {\sec (c+d x)}}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {B \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {(A b-a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {(A b-a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
\(\Big \downarrow \) 4335 |
\(\displaystyle \frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x)}}{a+b \cos (c+d x)}dx}{b}+\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
\(\Big \downarrow \) 3282 |
\(\displaystyle \frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {a \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}\right )}{b}+\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}+\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}\right )}{b}+\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}\right )}{b}+\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\) |
Input:
Int[(A + B*Cos[c + d*x])/((a + b*Cos[c + d*x])*Sqrt[Sec[c + d*x]]),x]
Output:
(2*B*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b*d ) + ((A*b - a*B)*Sqrt[Cos[c + d*x]]*((2*EllipticF[(c + d*x)/2, 2])/(b*d) - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d))*Sqrt[Sec[c + d*x]])/b
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{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[Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]/((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d/b Int[1/Sqrt[c + d*Sin[e + f*x]], x], x ] + Simp[(b*c - a*d)/b Int[1/((a + b*Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x ]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _)), x_Symbol] :> Simp[Sqrt[d*Sin[e + f*x]]*(Sqrt[d*Csc[e + f*x]]/d) Int[ Sqrt[d*Sin[e + f*x]]/(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[((csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[A/a Int[ (d*Csc[e + f*x])^n, x], x] - Simp[(A*b - a*B)/(a*d) Int[(d*Csc[e + f*x])^ (n + 1)/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(294\) vs. \(2(142)=284\).
Time = 6.44 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.98
method | result | size |
default | \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}-A \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) a b -B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b +B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+B \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) a^{2}\right )}{b^{2} \left (a -b \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(295\) |
Input:
int((A+B*cos(d*x+c))/(a+cos(d*x+c)*b)/sec(d*x+c)^(1/2),x,method=_RETURNVER BOSE)
Output:
-2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/ 2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*(A*EllipticF(cos(1/2*d*x+1 /2*c),2^(1/2))*a*b-A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-A*EllipticP i(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a*b-B*EllipticF(cos(1/2*d*x+1/2*c ),2^(1/2))*a^2+B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-B*EllipticE(cos (1/2*d*x+1/2*c),2^(1/2))*a*b+B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+B *EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^2)/b^2/(a-b)/(-2*sin( 1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2 *d*x+1/2*c)^2-1)^(1/2)/d
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorithm= "fricas")
Output:
Timed out
\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\int \frac {A + B \cos {\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}}\, dx \] Input:
integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))/sec(d*x+c)**(1/2),x)
Output:
Integral((A + B*cos(c + d*x))/((a + b*cos(c + d*x))*sqrt(sec(c + d*x))), x )
\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorithm= "maxima")
Output:
integrate((B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)*sqrt(sec(d*x + c))), x)
\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorithm= "giac")
Output:
integrate((B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)*sqrt(sec(d*x + c))), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \] Input:
int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))),x)
Output:
int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))), x)
\[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )}d x \] Input:
int((A+B*cos(d*x+c))/(a+b*cos(d*x+c))/sec(d*x+c)^(1/2),x)
Output:
int(sqrt(sec(c + d*x))/sec(c + d*x),x)