Integrand size = 33, antiderivative size = 168 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=-\frac {B \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {(A+2 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {(A+2 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \] Output:
-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) /a^2/d+1/3*(A+2*B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)) *sec(d*x+c)^(1/2)/a^2/d+1/3*(A+2*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/a^2/d/(1+s ec(d*x+c))-1/3*(A-B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.08 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.52 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (8 (A+2 B) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-i \sin \left (\frac {1}{2} (c+d x)\right )\right )+i \left (B e^{-i (c+d x)} \left (1+e^{i (c+d x)}\right )^3 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+2 \cos (c+d x) (-A-5 B+(A-7 B) \cos (c+d x)-i (A-B) \sin (c+d x))\right )\right ) \left (\cos \left (\frac {1}{2} (c+3 d x)\right )+i \sin \left (\frac {1}{2} (c+3 d x)\right )\right )}{6 a^2 d (1+\cos (c+d x))^2} \] Input:
Integrate[(A + B*Cos[c + d*x])/((a + a*Cos[c + d*x])^2*Sqrt[Sec[c + d*x]]) ,x]
Output:
(Cos[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(8*(A + 2*B)*Cos[(c + d*x)/2]^3*Sqrt[ Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[(c + d*x)/2] - I*Sin[(c + d*x )/2]) + I*((B*(1 + E^(I*(c + d*x)))^3*Sqrt[1 + E^((2*I)*(c + d*x))]*Hyperg eometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x)) + 2*Cos[ c + d*x]*(-A - 5*B + (A - 7*B)*Cos[c + d*x] - I*(A - B)*Sin[c + d*x])))*(C os[(c + 3*d*x)/2] + I*Sin[(c + 3*d*x)/2]))/(6*a^2*d*E^(I*d*x)*(1 + Cos[c + d*x])^2)
Time = 1.08 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3439, 3042, 4507, 27, 3042, 4508, 25, 3042, 4274, 3042, 4258, 3042, 3119, 3120}
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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)}{\sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3439 |
| \(\displaystyle \int \frac {\sqrt {\sec (c+d x)} (A \sec (c+d x)+B)}{(a \sec (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (A \csc \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\int \frac {a (A-B)+3 a (A+B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (A-B)+3 a (A+B) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}dx}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (A-B)+3 a (A+B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\int -\frac {3 a^2 B-a^2 (A+2 B) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}+\frac {2 (A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {2 (A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\int \frac {3 a^2 B-a^2 (A+2 B) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 (A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\int \frac {3 a^2 B-a^2 (A+2 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {2 (A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 B \int \frac {1}{\sqrt {\sec (c+d x)}}dx-a^2 (A+2 B) \int \sqrt {\sec (c+d x)}dx}{a^2}}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 (A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 B \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-a^2 (A+2 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {2 (A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-a^2 (A+2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 (A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^2 (A+2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {2 (A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\frac {6 a^2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-a^2 (A+2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {2 (A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\frac {6 a^2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 a^2 (A+2 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{6 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
Input:
Int[(A + B*Cos[c + d*x])/((a + a*Cos[c + d*x])^2*Sqrt[Sec[c + d*x]]),x]
Output:
-1/3*((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + (-(((6*a^2*B*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d *x]])/d - (2*a^2*(A + 2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sq rt[Sec[c + d*x]])/d)/a^2) + (2*(A + 2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/ (d*(1 + Sec[c + d*x])))/(6*a^2)
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[(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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B , 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(349\) vs. \(2(153)=306\).
Time = 4.87 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.08
| method | result | size |
| default | \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+4 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-20 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-3 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+9 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+A -B \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(350\) |
Input:
int((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^2/sec(d*x+c)^(1/2),x,method=_RETURNV ERBOSE)
Output:
-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*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)*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+12*B*cos(1/2*d*x+1/2*c)^6+4*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)*EllipticF(cos(1 /2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+6*B*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/ 2*d*x+1/2*c),2^(1/2))+2*A*cos(1/2*d*x+1/2*c)^4-20*B*cos(1/2*d*x+1/2*c)^4-3 *A*cos(1/2*d*x+1/2*c)^2+9*B*cos(1/2*d*x+1/2*c)^2+A-B)/a^2/cos(1/2*d*x+1/2* c)^3/(-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
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.93 \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {{\left (\sqrt {2} {\left (-i \, A - 2 i \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (i \, A + 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A + 2 i \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-i \, A - 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (i \, \sqrt {2} B \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} B \cos \left (d x + c\right ) + i \, \sqrt {2} B\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (-i \, \sqrt {2} B \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} B \cos \left (d x + c\right ) - i \, \sqrt {2} B\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, B \cos \left (d x + c\right )^{2} + {\left (A + 2 \, B\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^2/sec(d*x+c)^(1/2),x, algorith m="fricas")
Output:
1/6*((sqrt(2)*(-I*A - 2*I*B)*cos(d*x + c)^2 - 2*sqrt(2)*(I*A + 2*I*B)*cos( d*x + c) + sqrt(2)*(-I*A - 2*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A + 2*I*B)*cos(d*x + c)^2 - 2*sqrt(2)*(-I *A - 2*I*B)*cos(d*x + c) + sqrt(2)*(I*A + 2*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(I*sqrt(2)*B*cos(d*x + c)^2 + 2*I*sq rt(2)*B*cos(d*x + c) + I*sqrt(2)*B)*weierstrassZeta(-4, 0, weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(-I*sqrt(2)*B*cos(d*x + c) ^2 - 2*I*sqrt(2)*B*cos(d*x + c) - I*sqrt(2)*B)*weierstrassZeta(-4, 0, weie rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*B*cos(d*x + c)^2 + (A + 2*B)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos (d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {\int \frac {A}{\cos ^{2}{\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + 2 \cos {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + \sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {B \cos {\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + 2 \cos {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + \sqrt {\sec {\left (c + d x \right )}}}\, dx}{a^{2}} \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))**2/sec(d*x+c)**(1/2),x)
Output:
(Integral(A/(cos(c + d*x)**2*sqrt(sec(c + d*x)) + 2*cos(c + d*x)*sqrt(sec( c + d*x)) + sqrt(sec(c + d*x))), x) + Integral(B*cos(c + d*x)/(cos(c + d*x )**2*sqrt(sec(c + d*x)) + 2*cos(c + d*x)*sqrt(sec(c + d*x)) + sqrt(sec(c + d*x))), x))/a**2
\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^2/sec(d*x+c)^(1/2),x, algorith m="maxima")
Output:
integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^2*sqrt(sec(d*x + c))) , x)
\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^2/sec(d*x+c)^(1/2),x, algorith m="giac")
Output:
integrate((B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^2*sqrt(sec(d*x + c))) , x)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:
int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^2),x )
Output:
int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^2), x)
\[ \int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )+\sec \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )+\sec \left (d x +c \right )}d x \right ) b}{a^{2}} \] Input:
int((A+B*cos(d*x+c))/(a+a*cos(d*x+c))^2/sec(d*x+c)^(1/2),x)
Output:
(int(sqrt(sec(c + d*x))/(cos(c + d*x)**2*sec(c + d*x) + 2*cos(c + d*x)*sec (c + d*x) + sec(c + d*x)),x)*a + int((sqrt(sec(c + d*x))*cos(c + d*x))/(co s(c + d*x)**2*sec(c + d*x) + 2*cos(c + d*x)*sec(c + d*x) + sec(c + d*x)),x )*b)/a**2