Integrand size = 43, antiderivative size = 239 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\frac {\left (A b^2-a (b B-a C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a b \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a b B-a^2 (2 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^4+a^3 b B+a b^3 B+a^4 C-3 a^2 b^2 (A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 (a-b) b (a+b)^2 d}-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \cos (c+d x))} \]
(A*b^2-a*(B*b-C*a))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt icE(sin(1/2*d*x+1/2*c),2^(1/2))/a/b/(a^2-b^2)/d-(A*b^2+B*a*b-a^2*(2*A+C))* (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^2/(a^2-b^2)/d+(A*b^4+B*a^3*b+B*a*b^3+a^4*C-3*a^2*b^2*(A+C))* (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^2/(a-b)/b/(a+b)^2/d-(A*b^2-a*(B*b-C*a))*sin(d*x+c )*cos(d*x+c)^(1/2)/b/(a^2-b^2)/d/(b+a*cos(d*x+c))
Time = 5.03 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.25 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=-\frac {\frac {4 \left (A b^2+a (-b B+a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (b+a \cos (c+d x))}+\frac {\frac {2 \left (-A b^2+a b B+3 a^2 C-4 b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 b (-b B+a (A+C)) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a (a+b)}+\frac {2 \left (A b^2+a (-b B+a C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(-a+b) (a+b)}}{4 b d} \]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^2),x]
-1/4*((4*(A*b^2 + a*(-(b*B) + a*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(b + a*Cos[c + d*x])) + ((2*(-(A*b^2) + a*b*B + 3*a^2*C - 4*b^2*C) *EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*b*(-(b*B) + a*(A + C))*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a*(a + b)) + (2*(A*b^2 + a*(-(b*B) + a*C))*(-2*a*b*Ellipti cE[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^2*b*Sqrt[Sin[c + d*x]^2]))/((-a + b)*(a + b))) /(b*d)
Time = 1.62 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.97, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4600, 3042, 3534, 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.
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2}dx\) |
\(\Big \downarrow \) 4600 |
\(\displaystyle \int \frac {A \cos ^2(c+d x)+B \cos (c+d x)+C}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+b)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2}dx\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle -\frac {\int \frac {-C a^2-b B a+A b^2-\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)+2 b^2 C+2 b (b B-a (A+C)) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {-C a^2-b B a+A b^2-\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)+2 b^2 C+2 b (b B-a (A+C)) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {-C a^2-b B a+A b^2+\left (a (b B-a C)-A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b^2 C+2 b (b B-a (A+C)) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle -\frac {-\frac {\int -\frac {a \left (-C a^2-b B a+A b^2+2 b^2 C\right )+b \left (-\left ((2 A+C) a^2\right )+b B a+A b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {\left (A b^2-a (b B-a C)\right ) \int \sqrt {\cos (c+d x)}dx}{a}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\frac {\int \frac {a \left (-C a^2-b B a+A b^2+2 b^2 C\right )+b \left (-\left ((2 A+C) a^2\right )+b B a+A b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {\left (A b^2-a (b B-a C)\right ) \int \sqrt {\cos (c+d x)}dx}{a}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\int \frac {a \left (-C a^2-b B a+A b^2+2 b^2 C\right )+b \left (-\left ((2 A+C) a^2\right )+b B a+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {\left (A b^2-a (b B-a C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {\frac {\int \frac {a \left (-C a^2-b B a+A b^2+2 b^2 C\right )+b \left (-\left ((2 A+C) a^2\right )+b B a+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{a d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle -\frac {\frac {\frac {b \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}-\frac {\left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{a d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\frac {b \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {\left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{a d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {\frac {\frac {2 b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right )}{a d}-\frac {\left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{a d}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle -\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {\frac {\frac {2 b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right )}{a d}-\frac {2 \left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{a d}}{2 b \left (a^2-b^2\right )}\) |
-1/2*((-2*(A*b^2 - a*(b*B - a*C))*EllipticE[(c + d*x)/2, 2])/(a*d) + ((2*b *(A*b^2 + a*b*B - a^2*(2*A + C))*EllipticF[(c + d*x)/2, 2])/(a*d) - (2*(A* b^4 + a^3*b*B + a*b^3*B + a^4*C - 3*a^2*b^2*(A + C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a*(a + b)*d))/a)/(b*(a^2 - b^2)) - ((A*b^2 - a*(b*B - a*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b*(a^2 - b^2)*d*(b + a*Cos[c + d*x]))
3.14.26.3.1 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_.) + (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_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*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[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 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]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x _)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) *(x_)]^2), x_Symbol] :> Simp[d^(m + 2) Int[(b + a*Cos[e + f*x])^m*(d*Cos[ e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr eeQ[{a, b, d, e, f, A, B, C, n}, x] && !IntegerQ[n] && IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(808\) vs. \(2(315)=630\).
Time = 3.98 (sec) , antiderivative size = 809, normalized size of antiderivative = 3.38
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),x, method=_RETURNVERBOSE)
-(-(-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*(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))- 2*(-2*A*b+B*a)/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)*Ellip ticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2/a^2*(A*b^2-B*a*b+C*a^2)*(a^2 /b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c )^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-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)*Elli pticE(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)*EllipticPi(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*(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))))/sin(1/2*d*x+1/2*c...
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2/cos(d*x+c)^(1 /2),x, algorithm="fricas")
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2} \sqrt {\cos {\left (c + d x \right )}}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)/((a + b*sec(c + d*x))**2 *sqrt(cos(c + d*x))), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2/cos(d*x+c)^(1 /2),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^2* sqrt(cos(d*x + c))), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2/cos(d*x+c)^(1 /2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^2* sqrt(cos(d*x + c))), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]