Integrand size = 23, antiderivative size = 206 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {\sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{a d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{a d \sqrt {a+b \cos (c+d x)}}+\frac {\sqrt {a+b \cos (c+d x)} \tan (c+d x)}{a d} \]
-(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2 *c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/a/d/((a+b*cos(d*x+c))/ (a+b))^(1/2)+(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin (1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/d/ (a+b*cos(d*x+c))^(1/2)-b*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E llipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/ (a+b))^(1/2)/a/d/(a+b*cos(d*x+c))^(1/2)+(a+b*cos(d*x+c))^(1/2)*tan(d*x+c)/ a/d
Result contains complex when optimal does not.
Time = 15.20 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {-\frac {6 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+4 \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 a d} \]
((-6*b*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b) /(a + b)])/Sqrt[a + b*Cos[c + d*x]] - ((2*I)*Sqrt[-((b*(-1 + Cos[c + d*x]) )/(a + b))]*Sqrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*Csc[c + d*x]*(-2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(-2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b* Cos[c + d*x]]], (a + b)/(a - b)] + b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[ -(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)])))/(a*b*Sqrt[-( a + b)^(-1)]) + 4*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(4*a*d)
Time = 1.65 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.06, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 3281, 27, 3042, 3539, 25, 3042, 3134, 3042, 3132, 3481, 3042, 3142, 3042, 3140, 3286, 3042, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3281 |
| \(\displaystyle \frac {\int -\frac {\left (b \cos ^2(c+d x)+b\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a}+\frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\int \frac {\left (b \cos ^2(c+d x)+b\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\int \frac {b \sin \left (c+d x+\frac {\pi }{2}\right )^2+b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}\) |
| \(\Big \downarrow \) 3539 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\int \sqrt {a+b \cos (c+d x)}dx-\frac {\int -\frac {\left (b^2-a b \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{2 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {\left (b^2-a b \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}+\int \sqrt {a+b \cos (c+d x)}dx}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b^2-a b \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b^2-a b \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {\sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b^2-a b \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {\sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b^2-a b \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b^2 \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-a b \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}+\frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b^2 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a b \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b^2 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b^2 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b^2 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {\tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {2 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {2 a b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\) |
-1/2*((2*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/( d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((-2*a*b*Sqrt[(a + b*Cos[c + d*x]) /(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x ]]) + (2*b^2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]))/b)/a + (Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/(a*d)
3.6.28.3.1 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[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(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[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[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[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_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 ))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2* n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 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[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt [c + d*Sin[e + f*x]] Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(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] && !GtQ[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp [C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x], x] - Simp[1/(b*d) Int[Simp[a *c*C - A*b*d + (b*c*C + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*( 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]
Time = 3.06 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.58
| method | result | size |
| default | \(-\frac {\sqrt {-\left (-2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{a \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, b E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{a \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}+\frac {b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \sqrt {-\frac {2 b}{a -b}}\right )}{a \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}\, d}\) | \(532\) |
-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2 *d*x+1/2*c)/a*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2) /(2*cos(1/2*d*x+1/2*c)^2-1)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x +1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1 /2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-(sin(1/2*d *x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/ 2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1 /2*c),(-2*b/(a-b))^(1/2))+1/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d *x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x +1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/a*b* (sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/ (-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(co s(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d *x+1/2*c)^2+a+b)^(1/2)/d
Timed out. \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {a + b \cos {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]