Integrand size = 33, antiderivative size = 171 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sqrt {\sec (c+d x)} \, dx=\frac {2 \left (10 a A b+5 a^2 B+3 b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (3 a^2 A+A b^2+2 a b B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \] Output:
2/5*(10*A*a*b+5*B*a^2+3*B*b^2)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2* c),2^(1/2))*sec(d*x+c)^(1/2)/d+2/3*(3*A*a^2+A*b^2+2*B*a*b)*cos(d*x+c)^(1/2 )*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+2/5*b^2*B*sin( d*x+c)/d/sec(d*x+c)^(3/2)+2/3*b*(A*b+2*B*a)*sin(d*x+c)/d/sec(d*x+c)^(1/2)
Time = 6.72 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.75 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sqrt {\sec (c+d x)} \, dx=\frac {\sqrt {\sec (c+d x)} \left (6 \left (10 a A b+5 a^2 B+3 b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (3 a^2 A+A b^2+2 a b B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+b (5 A b+10 a B+3 b B \cos (c+d x)) \sin (2 (c+d x))\right )}{15 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sqrt[Sec[c + d*x]],x ]
Output:
(Sqrt[Sec[c + d*x]]*(6*(10*a*A*b + 5*a^2*B + 3*b^2*B)*Sqrt[Cos[c + d*x]]*E llipticE[(c + d*x)/2, 2] + 10*(3*a^2*A + A*b^2 + 2*a*b*B)*Sqrt[Cos[c + d*x ]]*EllipticF[(c + d*x)/2, 2] + b*(5*A*b + 10*a*B + 3*b*B*Cos[c + d*x])*Sin [2*(c + d*x)]))/(15*d)
Time = 1.17 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.02, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 3439, 3042, 4512, 27, 3042, 4535, 3042, 4258, 3042, 3119, 4533, 3042, 4258, 3042, 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.
| \(\displaystyle \int \sqrt {\sec (c+d x)} (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3439 |
| \(\displaystyle \int \frac {(a \sec (c+d x)+b)^2 (A \sec (c+d x)+B)}{\sec ^{\frac {5}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+b\right )^2 \left (A \csc \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4512 |
| \(\displaystyle \frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2}{5} \int -\frac {5 a^2 A \sec ^2(c+d x)+\left (5 B a^2+10 A b a+3 b^2 B\right ) \sec (c+d x)+5 b (A b+2 a B)}{2 \sec ^{\frac {3}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {5 a^2 A \sec ^2(c+d x)+\left (5 B a^2+10 A b a+3 b^2 B\right ) \sec (c+d x)+5 b (A b+2 a B)}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {5 a^2 A \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (5 B a^2+10 A b a+3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+5 b (A b+2 a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (\left (5 a^2 B+10 a A b+3 b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\int \frac {5 a^2 A \sec ^2(c+d x)+5 b (A b+2 a B)}{\sec ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\left (5 a^2 B+10 a A b+3 b^2 B\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {5 a^2 A \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 b (A b+2 a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\left (5 a^2 B+10 a A b+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\int \frac {5 a^2 A \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 b (A b+2 a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\left (5 a^2 B+10 a A b+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\int \frac {5 a^2 A \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 b (A b+2 a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a^2 A \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 b (A b+2 a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (3 a^2 A+2 a b B+A b^2\right ) \int \sqrt {\sec (c+d x)}dx+\frac {2 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (3 a^2 A+2 a b B+A b^2\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (3 a^2 A+2 a b B+A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (3 a^2 A+2 a b B+A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {10 \left (3 a^2 A+2 a b B+A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \left (5 a^2 B+10 a A b+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {10 b (2 a B+A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b^2 B \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sqrt[Sec[c + d*x]],x]
Output:
(2*b^2*B*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ((2*(10*a*A*b + 5*a^2*B + 3*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] )/d + (10*(3*a^2*A + A*b^2 + 2*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d* x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (10*b*(A*b + 2*a*B)*Sin[c + d*x])/(3* d*Sqrt[Sec[c + d*x]]))/5
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_))^2*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a^2*A*Cos[ e + f*x]*((d*Csc[e + f*x])^(n + 1)/(d*f*n)), x] + Simp[1/(d*n) Int[(d*Csc [e + f*x])^(n + 1)*(a*(2*A*b + a*B)*n + (2*a*b*B*n + A*(b^2*n + a^2*(n + 1) ))*Csc[e + f*x] + b^2*B*n*Csc[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(486\) vs. \(2(154)=308\).
Time = 12.23 (sec) , antiderivative size = 487, normalized size of antiderivative = 2.85
| method | result | size |
| default | \(-\frac {2 \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 (-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} b^{2}+\left (20 A \,b^{2}+40 B a b +24 B \,b^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-10 A \,b^{2}-20 B a b -6 B \,b^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 a^{2} A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \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 )+5 A \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \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 )-30 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \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 ) a b +10 B a b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \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 )-15 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \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 ) a^{2}-9 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \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 ) b^{2}\right )}{15 \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}\) | \(487\) |
| parts | \(-\frac {2 a^{2} A \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}}\, \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 )}{\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}-\frac {2 \left (A \,b^{2}+2 B a b \right ) \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 (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \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 )\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}+\frac {2 \left (2 A a b +a^{2} B \right ) \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}}\, \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 )}{\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}-\frac {2 B \,b^{2} \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 (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \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 \sin \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 )\right )}{5 \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}\) | \(675\) |
Input:
int((a+cos(d*x+c)*b)^2*(A+B*cos(d*x+c))*sec(d*x+c)^(1/2),x,method=_RETURNV ERBOSE)
Output:
-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*B*cos(1 /2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*b^2+(20*A*b^2+40*B*a*b+24*B*b^2)*sin(1/ 2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*A*b^2-20*B*a*b-6*B*b^2)*sin(1/2*d*x +1/2*c)^2*cos(1/2*d*x+1/2*c)+15*a^2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin( 1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+5*A*b^2*(s in(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos( 1/2*d*x+1/2*c),2^(1/2))-30*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1 /2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b+10*B*a*b*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2* d*x+1/2*c),2^(1/2))-15*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c )^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-9*B*(sin(1/2*d*x+1/ 2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c ),2^(1/2))*b^2)/(-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 = 0.09 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.32 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sqrt {\sec (c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, A a^{2} + 2 i \, B a b + i \, A b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, A a^{2} - 2 i \, B a b - i \, A b^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a^{2} - 10 i \, A a b - 3 i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a^{2} + 10 i \, A a b + 3 i \, B b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, B b^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(1/2),x, algorith m="fricas")
Output:
-1/15*(5*sqrt(2)*(3*I*A*a^2 + 2*I*B*a*b + I*A*b^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-3*I*A*a^2 - 2*I*B*a*b - I *A*b^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt (2)*(-5*I*B*a^2 - 10*I*A*a*b - 3*I*B*b^2)*weierstrassZeta(-4, 0, weierstra ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(5*I*B*a^2 + 10*I*A*a*b + 3*I*B*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*B*b^2*cos(d*x + c)^2 + 5*(2*B*a*b + A*b^2)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d
\[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sqrt {\sec (c+d x)} \, dx=\int \left (A + B \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{2} \sqrt {\sec {\left (c + d x \right )}}\, dx \] Input:
integrate((a+b*cos(d*x+c))**2*(A+B*cos(d*x+c))*sec(d*x+c)**(1/2),x)
Output:
Integral((A + B*cos(c + d*x))*(a + b*cos(c + d*x))**2*sqrt(sec(c + d*x)), x)
\[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sqrt {\sec (c+d x)} \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(1/2),x, algorith m="maxima")
Output:
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*sqrt(sec(d*x + c)), x)
\[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sqrt {\sec (c+d x)} \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(1/2),x, algorith m="giac")
Output:
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*sqrt(sec(d*x + c)), x)
Timed out. \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sqrt {\sec (c+d x)} \, dx=\int \left (A+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,d x \] Input:
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^2,x)
Output:
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(1/2)*(a + b*cos(c + d*x))^2, x)
\[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sqrt {\sec (c+d x)} \, dx=\left (\int \sqrt {\sec \left (d x +c \right )}d x \right ) a^{3}+3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) a^{2} b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}d x \right ) b^{3}+3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) a \,b^{2} \] Input:
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(1/2),x)
Output:
int(sqrt(sec(c + d*x)),x)*a**3 + 3*int(sqrt(sec(c + d*x))*cos(c + d*x),x)* a**2*b + int(sqrt(sec(c + d*x))*cos(c + d*x)**3,x)*b**3 + 3*int(sqrt(sec(c + d*x))*cos(c + d*x)**2,x)*a*b**2