Integrand size = 35, antiderivative size = 200 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a^{5/2} (163 A+304 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {a^3 (299 A+432 C) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (17 A+16 C) \sqrt {a+a \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{32 d}+\frac {5 a A (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
1/64*a^(5/2)*(163*A+304*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/ 2))/d+5/24*a*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^2*tan(d*x+c)/d+1/4*A*(a+a *cos(d*x+c))^(5/2)*sec(d*x+c)^3*tan(d*x+c)/d+1/192*a^3*(299*A+432*C)*tan(d *x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/32*a^2*(17*A+16*C)*sec(d*x+c)*(a+a*cos(d* x+c))^(1/2)*tan(d*x+c)/d
Time = 1.13 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.76 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (6 \sqrt {2} (163 A+304 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4(c+d x)+(844 A+192 C+(2203 A+1584 C) \cos (c+d x)+4 (163 A+48 C) \cos (2 (c+d x))+489 A \cos (3 (c+d x))+528 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{768 d} \]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^4*(6*Sqrt[2] *(163*A + 304*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^4 + (844*A + 192*C + (2203*A + 1584*C)*Cos[c + d*x] + 4*(163*A + 48*C)*Cos[2*(c + d* x)] + 489*A*Cos[3*(c + d*x)] + 528*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/ (768*d)
Time = 1.37 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3523, 27, 3042, 3454, 27, 3042, 3454, 27, 3042, 3459, 3042, 3252, 219}
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.
| \(\displaystyle \int \sec ^5(c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{5/2} (5 a A+a (A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{5/2} (5 a A+a (A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (5 a A+a (A+8 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} \left (3 (17 A+16 C) a^2+(11 A+48 C) \cos (c+d x) a^2\right ) \sec ^3(c+d x)dx+\frac {5 a^2 A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \int (\cos (c+d x) a+a)^{3/2} \left (3 (17 A+16 C) a^2+(11 A+48 C) \cos (c+d x) a^2\right ) \sec ^3(c+d x)dx+\frac {5 a^2 A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 (17 A+16 C) a^2+(11 A+48 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {5 a^2 A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left ((299 A+432 C) a^3+5 (19 A+48 C) \cos (c+d x) a^3\right ) \sec ^2(c+d x)dx+\frac {3 a^3 (17 A+16 C) \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \int \sqrt {\cos (c+d x) a+a} \left ((299 A+432 C) a^3+5 (19 A+48 C) \cos (c+d x) a^3\right ) \sec ^2(c+d x)dx+\frac {3 a^3 (17 A+16 C) \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((299 A+432 C) a^3+5 (19 A+48 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {3 a^3 (17 A+16 C) \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (163 A+304 C) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a^4 (299 A+432 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^3 (17 A+16 C) \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (163 A+304 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^4 (299 A+432 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {3 a^3 (17 A+16 C) \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {a^4 (299 A+432 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {3 a^4 (163 A+304 C) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {3 a^3 (17 A+16 C) \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}\right )+\frac {5 a^2 A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {5 a^2 A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d}+\frac {1}{6} \left (\frac {3 a^3 (17 A+16 C) \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {3 a^{7/2} (163 A+304 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a^4 (299 A+432 C) \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )\right )}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d}\) |
(A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((5*a^2 *A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*a^3 *(17*A + 16*C)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((3*a^(7/2)*(163*A + 304*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos [c + d*x]]])/d + (a^4*(299*A + 432*C)*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/4)/6)/(8*a)
3.1.99.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) *(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d)) Int[Sqrt[a + b*Sin[e + f*x] ]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x ] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a *d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* (c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(1649\) vs. \(2(176)=352\).
Time = 3.95 (sec) , antiderivative size = 1650, normalized size of antiderivative = 8.25
\[\text {Expression too large to display}\]
1/24*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(48*a*(163* A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/ 2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))+163*A*ln(4/(2*cos(1/2*d*x+ 1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c )^2)^(1/2)*a^(1/2)+2*a))+304*C*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/ 2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a ))+304*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c) +2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a)))*sin(1/2*d*x+1/2*c)^ 8-48*(163*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+176*C*2^(1/2)*( a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+326*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^ (1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2 )*a^(1/2)-2*a))*a+326*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos (1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a+608 *C*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1 /2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a+608*C*ln(4/(2*cos(1/2*d *x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/ 2*c)^2)^(1/2)*a^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^6+8*(1793*A*a^(1/2)*2^(1 /2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+1680*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2) ^(1/2)*a^(1/2)+1467*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos( 1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a+1...
Time = 0.31 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.13 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left ({\left (163 \, A + 304 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (163 \, A + 304 \, C\right )} a^{2} \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (163 \, A + 176 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (163 \, A + 48 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 184 \, A a^{2} \cos \left (d x + c\right ) + 48 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]
1/768*(3*((163*A + 304*C)*a^2*cos(d*x + c)^5 + (163*A + 304*C)*a^2*cos(d*x + c)^4)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos (d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c )^3 + cos(d*x + c)^2)) + 4*(3*(163*A + 176*C)*a^2*cos(d*x + c)^3 + 2*(163* A + 48*C)*a^2*cos(d*x + c)^2 + 184*A*a^2*cos(d*x + c) + 48*A*a^2)*sqrt(a*c os(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \]
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \]
Time = 1.32 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.61 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=-\frac {\sqrt {2} {\left (3 \, \sqrt {2} {\left (163 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 304 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) + \frac {4 \, {\left (3912 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4224 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 7172 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6720 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4606 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3552 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1047 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 624 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}\right )} \sqrt {a}}{768 \, d} \]
-1/768*sqrt(2)*(3*sqrt(2)*(163*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 304*C*a^2 *sgn(cos(1/2*d*x + 1/2*c)))*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/a bs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))) + 4*(3912*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 + 4224*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin( 1/2*d*x + 1/2*c)^7 - 7172*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/ 2*c)^5 - 6720*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 460 6*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 3552*C*a^2*sgn( cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 1047*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 624*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1 /2*d*x + 1/2*c))/(2*sin(1/2*d*x + 1/2*c)^2 - 1)^4)*sqrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^5} \,d x \]