Integrand size = 31, antiderivative size = 180 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 (3 a A+5 b B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 (A b+a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 (3 a A+5 b B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 (A b+a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:
-2/5*(3*A*a+5*B*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)/d+2/3*(A*b+B*a)*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*(3*A*a+5*B*b)*sec(d*x+c)^(1/2)*sin(d *x+c)/d+2/3*(A*b+B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a*A*sec(d*x+c)^(5/ 2)*sin(d*x+c)/d
Time = 2.96 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.73 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {\sec ^{\frac {5}{2}}(c+d x) \left (-12 (3 a A+5 b B) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 (A b+a B) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 (15 (a A+b B)+10 (A b+a B) \cos (c+d x)+3 (3 a A+5 b B) \cos (2 (c+d x))) \sin (c+d x)\right )}{30 d} \] Input:
Integrate[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2),x]
Output:
(Sec[c + d*x]^(5/2)*(-12*(3*a*A + 5*b*B)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 20*(A*b + a*B)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 2*(15*(a*A + b*B) + 10*(A*b + a*B)*Cos[c + d*x] + 3*(3*a*A + 5*b*B)*Cos[ 2*(c + d*x)])*Sin[c + d*x]))/(30*d)
Time = 0.86 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.94, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 3439, 3042, 4485, 27, 3042, 4274, 3042, 4255, 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.
| \(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3439 |
| \(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b) (A \sec (c+d x)+B)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+b\right ) \left (A \csc \left (c+d x+\frac {\pi }{2}\right )+B\right )dx\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {2}{5} \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) (3 a A+5 b B+5 (A b+a B) \sec (c+d x))dx+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \sec ^{\frac {3}{2}}(c+d x) (3 a A+5 b B+5 (A b+a B) \sec (c+d x))dx+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (3 a A+5 b B+5 (A b+a B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {1}{5} \left ((3 a A+5 b B) \int \sec ^{\frac {3}{2}}(c+d x)dx+5 (a B+A b) \int \sec ^{\frac {5}{2}}(c+d x)dx\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left ((3 a A+5 b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+5 (a B+A b) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (5 (a B+A b) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+(3 a A+5 b B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (5 (a B+A b) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+(3 a A+5 b B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (5 (a B+A b) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+(3 a A+5 b B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (5 (a B+A b) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+(3 a A+5 b B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (5 (a B+A b) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+(3 a A+5 b B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (5 (a B+A b) \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )+(3 a A+5 b B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
Input:
Int[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2),x]
Output:
(2*a*A*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + ((3*a*A + 5*b*B)*((-2*Sqrt [Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[S ec[c + d*x]]*Sin[c + d*x])/d) + 5*(A*b + a*B)*((2*Sqrt[Cos[c + d*x]]*Ellip ticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*Sec[c + d*x]^(3/2)*Sin [c + d*x])/(3*d)))/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)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*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_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(635\) vs. \(2(159)=318\).
Time = 104.36 (sec) , antiderivative size = 636, normalized size of antiderivative = 3.53
| 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 \left (A b +B a \right ) \left (-\frac {\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}}}{6 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\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 )}{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}}}\right )+\frac {2 A a \left (24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {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}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {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}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8 \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 ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{5 \left (8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {2 B 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}}\, \left (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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\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}\) | \(636\) |
| parts | \(\text {Expression too large to display}\) | \(763\) |
Input:
int((a+cos(d*x+c)*b)*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x,method=_RETURNVER BOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*(A*b+B*a)*(- 1/6*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 )/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2/5*A*a/(8*sin(1/2*d*x+1/2*c)^6- 12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(24 *cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*EllipticE(cos(1/2*d*x+1/2*c),2 ^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin( 1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*EllipticE(c os(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2 *c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2 *c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt icE(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2* c)^2)^(1/2)+2*B*b/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin( 1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1 /2*d*x+1/2*c)-(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))))/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 = 235, normalized size of antiderivative = 1.31 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {5 \, \sqrt {2} {\left (i \, B a + i \, A b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a - i \, A b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (3 i \, A a + 5 i \, B b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-3 i \, A a - 5 i \, B b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, {\left (3 \, A a + 5 \, B b\right )} \cos \left (d x + c\right )^{2} + 3 \, A a + 5 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorithm= "fricas")
Output:
-1/15*(5*sqrt(2)*(I*B*a + I*A*b)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*B*a - I*A*b)*cos(d*x + c)^ 2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(3 *I*A*a + 5*I*B*b)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInvers e(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-3*I*A*a - 5*I*B*b)* cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*(3*A*a + 5*B*b)*cos(d*x + c)^2 + 3*A*a + 5*( B*a + A*b)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^ 2)
Timed out. \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)**(7/2),x)
Output:
Timed out
\[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorithm= "maxima")
Output:
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)*sec(d*x + c)^(7/2), x)
\[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorithm= "giac")
Output:
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)*sec(d*x + c)^(7/2), x)
Timed out. \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right ) \,d x \] Input:
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x)),x)
Output:
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x)), x)
\[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right ) a b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) b^{2}+\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) a^{2} \] Input:
int((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x)
Output:
2*int(sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x)**3,x)*a*b + int(sqrt(se c(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**3,x)*b**2 + int(sqrt(sec(c + d*x ))*sec(c + d*x)**3,x)*a**2