Integrand size = 25, antiderivative size = 431 \[ \int \frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=-\frac {(a-b) \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{d \sqrt {\sec (c+d x)}}-\frac {a \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]
sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)/d-(a-b)*csc(d*x+c)*Elli pticE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^( 1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+se c(d*x+c))/(a-b))^(1/2)/a/d/sec(d*x+c)^(1/2)+csc(d*x+c)*EllipticF((a+b*cos( d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b)^(1/ 2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b) )^(1/2)/d/sec(d*x+c)^(1/2)-a*csc(d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/2)/ (a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos (d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2 )/b/d/sec(d*x+c)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(2995\) vs. \(2(431)=862\).
Time = 19.00 (sec) , antiderivative size = 2995, normalized size of antiderivative = 6.95 \[ \int \frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\text {Result too large to show} \]
(Sec[(c + d*x)/2]^2*Sqrt[1 + Sec[c + d*x]]*(8*(a + b)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]* EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 16*a*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d *x]))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] - 16*a*E llipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x ])^(-1)]*Sqrt[(b + a*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + 2*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^3*(-Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/ 2])))/(16*d*((1 + Cos[c + d*x])^(-1))^(3/2)*Sqrt[Sec[c + d*x]]*((b*Sec[(c + d*x)/2]^2*Sqrt[1 + Sec[c + d*x]]*Sin[c + d*x]*(8*(a + b)*Sqrt[Cos[c + d* x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x ]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 16*a*Sqrt[Cos [c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[ c + d*x]))]*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] - 1 6*a*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(b + a*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + 2* (a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^3*(-Sin[(c + d*x)/2] + Sin[(3*(c + d *x))/2])))/(32*((1 + Cos[c + d*x])^(-1))^(3/2)*(a + b*Cos[c + d*x])^(3/2)) - (3*Sec[(c + d*x)/2]^2*Sqrt[1 + Sec[c + d*x]]*Sin[c + d*x]*(8*(a + b)*Sq rt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*...
Time = 1.51 (sec) , antiderivative size = 405, normalized size of antiderivative = 0.94, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4710, 3042, 3300, 27, 3042, 3533, 27, 3042, 3280, 3042, 3288, 3295, 3473}
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.
\(\displaystyle \int \frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4710 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3300 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int -\frac {a b-a b \cos ^2(c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{b}+\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a b-a b \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a b-a b \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )\) |
\(\Big \downarrow \) 3533 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a b}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a b \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{2 b}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {a b \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a b \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{2 b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {a b \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a b \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )\) |
\(\Big \downarrow \) 3280 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {a b \left (\int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx\right )-a b \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {a b \left (\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-a b \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )\) |
\(\Big \downarrow \) 3288 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {a b \left (\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 a \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}\right )\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {a b \left (\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}\right )+\frac {2 a \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}\right )\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {a b \left (\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}\right )+\frac {2 a \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((2*a*Sqrt[a + b]*Cot[c + d*x] *EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[C os[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*S qrt[(a*(1 + Sec[c + d*x]))/(a - b)])/d + a*b*((2*(a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[( a*(1 + Sec[c + d*x]))/(a - b)])/(a^2*d) - (2*Sqrt[a + b]*Cot[c + d*x]*Elli pticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], - ((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d)))/b + (Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(d *Sqrt[Cos[c + d*x]]))
3.8.34.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin [(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[1/(a - b) Int[1/(Sqrt[a + b*Sin [e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), x], x] - Simp[b/(a - b) Int[(1 + Si n[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*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[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c *((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti cPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x] )^(m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) I nt[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a^2*c*d*( m + n) + b*d*(b*c*(m - 1) + a*d*n) + (a*d*(2*b*c + a*d)*(m + n) - b*d*(a*c - b*d*(m + n - 1)))*Sin[e + f*x] + b*d*(b*c*n + a*d*(2*m + n - 1))*Sin[e + f*x]^2, 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] && LtQ[0, m, 2] && LtQ[-1, n, 2] && NeQ[m + n, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_.) + (b_.)*sin[(e_.) + ( f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] : > Simp[C/b^2 Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x ] + Simp[1/b^2 Int[(A*b^2 - a^2*C - 2*a*b*C*Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1068\) vs. \(2(391)=782\).
Time = 6.16 (sec) , antiderivative size = 1069, normalized size of antiderivative = 2.48
1/d/(1+cos(d*x+c))/(a+cos(d*x+c)*b)^(1/2)/sec(d*x+c)^(1/2)*(2*EllipticF(co t(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2 )*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a*cos(d*x+c)-EllipticE(cot (d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2) *((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a*cos(d*x+c)-EllipticE(cot( d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)* ((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*b*cos(d*x+c)-2*EllipticPi(co t(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^( 1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a*cos(d*x+c)+4*(cos(d*x +c)/(1+cos(d*x+c)))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^( 1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a-2*(cos(d*x+c)/(1+cos (d*x+c)))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+ cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a-2*(cos(d*x+c)/(1+cos(d*x+c)))^ (1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c) *b)/(1+cos(d*x+c))/(a+b))^(1/2)*b-4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*Elli pticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1 +cos(d*x+c))/(a+b))^(1/2)*a+2*sec(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2) *EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/( 1+cos(d*x+c))/(a+b))^(1/2)*a-sec(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)* EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)...
\[ \int \frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {\sqrt {b \cos \left (d x + c\right ) + a}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
\[ \int \frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\sqrt {a + b \cos {\left (c + d x \right )}}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {\sqrt {b \cos \left (d x + c\right ) + a}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
\[ \int \frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {\sqrt {b \cos \left (d x + c\right ) + a}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\sqrt {a+b\,\cos \left (c+d\,x\right )}}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]