Integrand size = 33, antiderivative size = 197 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=-\frac {(A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b \left (a^2-b^2\right ) d}-\frac {(A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a \left (a^2-b^2\right ) d}+\frac {\left (a^2 A b+A b^3+a^3 B-3 a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a (a-b) b (a+b)^2 d}+\frac {a (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \cos (c+d x))} \]
-(A*b-B*a)*(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^2-b^2)/d-(A*b-B*a)*(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))/a/(a^2-b^2)/d+( A*a^2*b+A*b^3+B*a^3-3*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/ 2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a/(a-b)/b/(a+b)^2/d+ a*(A*b-B*a)*sin(d*x+c)*cos(d*x+c)^(1/2)/b/(a^2-b^2)/d/(b+a*cos(d*x+c))
Time = 3.04 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.39 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\frac {-\frac {4 a (-A b+a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (b+a \cos (c+d x))}+\frac {\frac {2 \left (a A b+3 a^2 B-4 b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {4 b (-A b+a B) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{a}+\frac {2 (-A b+a B) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{4 b d} \]
((-4*a*(-(A*b) + a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(b + a *Cos[c + d*x])) + ((2*(a*A*b + 3*a^2*B - 4*b^2*B)*EllipticPi[(2*a)/(a + b) , (c + d*x)/2, 2])/(a + b) + (4*b*(-(A*b) + a*B)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)))/a + (2*(-( A*b) + a*B)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b )*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a /b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x] ^2]))/((a - b)*(a + b)))/(4*b*d)
Time = 1.39 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.92, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3433, 3042, 3479, 27, 3042, 3538, 25, 3042, 3119, 3481, 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.
\(\displaystyle \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3433 |
\(\displaystyle \int \frac {A \cos (c+d x)+B}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+b)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )+B}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2}dx\) |
\(\Big \downarrow \) 3479 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {\int -\frac {B a^2-(A b-a B) \cos ^2(c+d x) a+A b a-2 b^2 B-2 b (A b-a B) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {B a^2-(A b-a B) \cos ^2(c+d x) a+A b a-2 b^2 B-2 b (A b-a B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {B a^2-(A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a+A b a-2 b^2 B-2 b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {-\frac {\int -\frac {a \left (B a^2+A b a-2 b^2 B\right )-a b (A b-a B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\left ((A b-a B) \int \sqrt {\cos (c+d x)}dx\right )}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {a \left (B a^2+A b a-2 b^2 B\right )-a b (A b-a B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-(A b-a B) \int \sqrt {\cos (c+d x)}dx}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {a \left (B a^2+A b a-2 b^2 B\right )-a b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-(A b-a B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\int \frac {a \left (B a^2+A b a-2 b^2 B\right )-a b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\frac {\left (a^3 B+a^2 A b-3 a b^2 B+A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx-b (A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (a^3 B+a^2 A b-3 a b^2 B+A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-b (A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\left (a^3 B+a^2 A b-3 a b^2 B+A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-\frac {2 b (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a}-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {\frac {\frac {2 \left (a^3 B+a^2 A b-3 a b^2 B+A b^3\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}-\frac {2 b (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a}-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 b \left (a^2-b^2\right )}\) |
((-2*(A*b - a*B)*EllipticE[(c + d*x)/2, 2])/d + ((-2*b*(A*b - a*B)*Ellipti cF[(c + d*x)/2, 2])/d + (2*(a^2*A*b + A*b^3 + a^3*B - 3*a*b^2*B)*EllipticP i[(2*a)/(a + b), (c + d*x)/2, 2])/((a + b)*d))/a)/(2*b*(a^2 - b^2)) + (a*( A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b*(a^2 - b^2)*d*(b + a*Cos[c + d*x]))
3.6.84.3.1 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[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[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[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* (d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + c*Sin[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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin [e + f*x])^(1 + n)/(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*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) *(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n }, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(I ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (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 - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(714\) vs. \(2(273)=546\).
Time = 14.71 (sec) , antiderivative size = 715, normalized size of antiderivative = 3.63
method | result | size |
default | \(-\frac {\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}}\, \left (-\frac {2 A \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}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right )}{\left (a^{2}-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}}}+\frac {2 \left (-A b +B a \right ) \left (\frac {a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\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}}}{b \left (a^{2}-b^{2}\right ) \left (2 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a +b \right )}-\frac {\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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{2 \left (a +b \right ) b \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {a \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}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{2 b \left (a^{2}-b^{2}\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}}}-\frac {a \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}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{2 b \left (a^{2}-b^{2}\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}}}-\frac {a^{3} \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}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right )}{2 b \left (a^{2}-b^{2}\right ) \left (a^{2}-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}}}+\frac {3 b a \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}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right )}{2 \left (a^{2}-b^{2}\right ) \left (a^{2}-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}}}\right )}{a}\right )}{\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}\) | \(715\) |
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*A/(a^2-a*b) *(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/ 2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2 *a/(a-b),2^(1/2))+2*(-A*b+B*a)/a*(a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*s in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*a*cos(1/2*d*x+1/2*c)^2- a+b)-1/2/(a+b)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^ (1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1 /2*d*x+1/2*c),2^(1/2))+1/2*a/b/(a^2-b^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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^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)/(-2*sin(1/2*d*x+1/2* c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2 /b/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/ 2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip ticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(s in(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d *x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/ (a-b),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. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]