Integrand size = 33, antiderivative size = 221 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 \left (3 a^2 A+5 b (A b+2 a B)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (2 a A b+a^2 B+3 b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 \left (3 a^2 A+5 b (A b+2 a B)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (7 A b+5 a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 a A \sec ^{\frac {3}{2}}(c+d x) (b+a \sec (c+d x)) \sin (c+d x)}{5 d} \] Output:
-2/5*(3*A*a^2+5*b*(A*b+2*B*a))*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*(2*A*a*b+B*a^2+3*B*b^2)*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^2+5 *b*(A*b+2*B*a))*sec(d*x+c)^(1/2)*sin(d*x+c)/d+2/15*a*(7*A*b+5*B*a)*sec(d*x +c)^(3/2)*sin(d*x+c)/d+2/5*a*A*sec(d*x+c)^(3/2)*(b+a*sec(d*x+c))*sin(d*x+c )/d
Time = 4.38 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.77 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {\sec ^{\frac {5}{2}}(c+d x) \left (-12 \left (3 a^2 A+5 A b^2+10 a b B\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (2 a A b+a^2 B+3 b^2 B\right ) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 \left (15 \left (a^2 A+A b^2+2 a b B\right )+10 a (2 A b+a B) \cos (c+d x)+3 \left (3 a^2 A+5 A b^2+10 a b B\right ) \cos (2 (c+d x))\right ) \sin (c+d x)\right )}{30 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2),x ]
Output:
(Sec[c + d*x]^(5/2)*(-12*(3*a^2*A + 5*A*b^2 + 10*a*b*B)*Cos[c + d*x]^(5/2) *EllipticE[(c + d*x)/2, 2] + 20*(2*a*A*b + a^2*B + 3*b^2*B)*Cos[c + d*x]^( 5/2)*EllipticF[(c + d*x)/2, 2] + 2*(15*(a^2*A + A*b^2 + 2*a*b*B) + 10*a*(2 *A*b + a*B)*Cos[c + d*x] + 3*(3*a^2*A + 5*A*b^2 + 10*a*b*B)*Cos[2*(c + d*x )])*Sin[c + d*x]))/(30*d)
Time = 1.42 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.93, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 3439, 3042, 4514, 27, 3042, 4535, 3042, 4255, 3042, 4258, 3042, 3119, 4534, 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.
| \(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3439 |
| \(\displaystyle \int \sqrt {\sec (c+d x)} (a \sec (c+d x)+b)^2 (A \sec (c+d x)+B)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+b\right )^2 \left (A \csc \left (c+d x+\frac {\pi }{2}\right )+B\right )dx\) |
| \(\Big \downarrow \) 4514 |
| \(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {\sec (c+d x)} \left (a (7 A b+5 a B) \sec ^2(c+d x)+\left (3 A a^2+5 b (A b+2 a B)\right ) \sec (c+d x)+b (a A+5 b B)\right )dx+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\sec (c+d x)} \left (a (7 A b+5 a B) \sec ^2(c+d x)+\left (3 A a^2+5 b (A b+2 a B)\right ) \sec (c+d x)+b (a A+5 b B)\right )dx+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (7 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 A a^2+5 b (A b+2 a B)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+b (a A+5 b B)\right )dx+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (\left (3 a^2 A+5 b (2 a B+A b)\right ) \int \sec ^{\frac {3}{2}}(c+d x)dx+\int \sqrt {\sec (c+d x)} \left (a (7 A b+5 a B) \sec ^2(c+d x)+b (a A+5 b B)\right )dx\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\left (3 a^2 A+5 b (2 a B+A b)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (7 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (a A+5 b B)\right )dx\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\left (3 a^2 A+5 b (2 a B+A b)\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (7 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (a A+5 b B)\right )dx\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\left (3 a^2 A+5 b (2 a B+A b)\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (7 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (a A+5 b B)\right )dx\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\left (3 a^2 A+5 b (2 a B+A b)\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (7 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (a A+5 b B)\right )dx\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\left (3 a^2 A+5 b (2 a B+A b)\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (7 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (a A+5 b B)\right )dx\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (7 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (a A+5 b B)\right )dx+\left (3 a^2 A+5 b (2 a B+A b)\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (a^2 B+2 a A b+3 b^2 B\right ) \int \sqrt {\sec (c+d x)}dx+\left (3 a^2 A+5 b (2 a B+A b)\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a (5 a B+7 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (a^2 B+2 a A b+3 b^2 B\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\left (3 a^2 A+5 b (2 a B+A b)\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a (5 a B+7 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (a^2 B+2 a A b+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\left (3 a^2 A+5 b (2 a B+A b)\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a (5 a B+7 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{3} \left (a^2 B+2 a A b+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (3 a^2 A+5 b (2 a B+A b)\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a (5 a B+7 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {10 \left (a^2 B+2 a A b+3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\left (3 a^2 A+5 b (2 a B+A b)\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a (5 a B+7 A b) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+b)}{5 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2),x]
Output:
(2*a*A*Sec[c + d*x]^(3/2)*(b + a*Sec[c + d*x])*Sin[c + d*x])/(5*d) + ((10* (2*a*A*b + a^2*B + 3*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*S qrt[Sec[c + d*x]])/(3*d) + (2*a*(7*A*b + 5*a*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d) + (3*a^2*A + 5*b*(A*b + 2*a*B))*((-2*Sqrt[Cos[c + d*x]]*Ellip ticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/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_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(m + n) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n* Simp[a^2*A*(m + n) + a*b*B*n + (a*(2*A*b + a*B)*(m + n) + b^2*B*(m + n - 1) )*Csc[e + f*x] + b*(A*b*(m + n) + a*B*(2*m + n - 1))*Csc[e + f*x]^2, x], x] , x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && !(IGtQ[n, 1] && !IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, 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. \(722\) vs. \(2(200)=400\).
Time = 516.69 (sec) , antiderivative size = 723, normalized size of antiderivative = 3.27
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(723\) |
| parts | \(\text {Expression too large to display}\) | \(914\) |
Input:
int((a+cos(d*x+c)*b)^2*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x,method=_RETURNV ERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B*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))+ 2*a*(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^2*A/(8*s in(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(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)^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)*EllipticE(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*(A*b+2*B*a)/sin(1/2*d*x+1/2*c)^2/(2*si n(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))))/si n(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 = 286, normalized size of antiderivative = 1.29 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {5 \, \sqrt {2} {\left (i \, B a^{2} + 2 i \, A a b + 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a^{2} - 2 i \, A a b - 3 i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (3 i \, A a^{2} + 10 i \, B a b + 5 i \, A b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-3 i \, A a^{2} - 10 i \, B a b - 5 i \, A b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, A a^{2} + 3 \, {\left (3 \, A a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorith m="fricas")
Output:
-1/15*(5*sqrt(2)*(I*B*a^2 + 2*I*A*a*b + 3*I*B*b^2)*cos(d*x + c)^2*weierstr assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*B*a^2 - 2*I*A*a*b - 3*I*B*b^2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(3*I*A*a^2 + 10*I*B*a*b + 5*I*A*b^2)*cos (d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-3*I*A*a^2 - 10*I*B*a*b - 5*I*A*b^2)*cos( d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*A*a^2 + 3*(3*A*a^2 + 10*B*a*b + 5*A*b^2)*cos(d*x + c)^2 + 5*(B*a^2 + 2*A*a*b)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c) ))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**2*(A+B*cos(d*x+c))*sec(d*x+c)**(7/2),x)
Output:
Timed out
\[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorith m="maxima")
Output:
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*sec(d*x + c)^(7/2), x)
\[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x, algorith m="giac")
Output:
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^2*sec(d*x + c)^(7/2), x)
Timed out. \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,d x \] Input:
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^2,x)
Output:
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^2, x)
\[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx=3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right ) a^{2} b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{3}d x \right ) b^{3}+3 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) a \,b^{2}+\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) a^{3} \] Input:
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^(7/2),x)
Output:
3*int(sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x)**3,x)*a**2*b + int(sqrt (sec(c + d*x))*cos(c + d*x)**3*sec(c + d*x)**3,x)*b**3 + 3*int(sqrt(sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**3,x)*a*b**2 + int(sqrt(sec(c + d*x)) *sec(c + d*x)**3,x)*a**3