Integrand size = 35, antiderivative size = 200 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a^{3/2} (75 A+112 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {a^2 (75 A+112 C) \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (13 A+16 C) \sec (c+d x) \tan (c+d x)}{32 d \sqrt {a+a \cos (c+d x)}}+\frac {a A \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{8 d}+\frac {A (a+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
1/64*a^(3/2)*(75*A+112*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2 ))/d+1/4*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^3*tan(d*x+c)/d+1/64*a^2*(75*A +112*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/32*a^2*(13*A+16*C)*sec(d*x+c )*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/8*a*A*sec(d*x+c)^2*(a+a*cos(d*x+c) )^(1/2)*tan(d*x+c)/d
Time = 1.04 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.76 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (2 \sqrt {2} (75 A+112 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4(c+d x)+(164 A+64 C+7 (55 A+48 C) \cos (c+d x)+4 (25 A+16 C) \cos (2 (c+d x))+75 A \cos (3 (c+d x))+112 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{256 d} \]
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^4*(2*Sqrt[2]*( 75*A + 112*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^4 + (164*A + 64*C + 7*(55*A + 48*C)*Cos[c + d*x] + 4*(25*A + 16*C)*Cos[2*(c + d*x)] + 7 5*A*Cos[3*(c + d*x)] + 112*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(256*d)
Time = 1.20 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3523, 27, 3042, 3454, 27, 3042, 3459, 3042, 3251, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a \cos (c+d x)+a)^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (3 a A+a (3 A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{3/2} (3 a A+a (3 A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a A+a (3 A+8 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {3}{2} \sqrt {\cos (c+d x) a+a} \left ((13 A+16 C) a^2+(9 A+16 C) \cos (c+d x) a^2\right ) \sec ^3(c+d x)dx+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \left ((13 A+16 C) a^2+(9 A+16 C) \cos (c+d x) a^2\right ) \sec ^3(c+d x)dx+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((13 A+16 C) a^2+(9 A+16 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \int \sqrt {\cos (c+d x) a+a} \sec ^2(c+d x)dx+\frac {a^3 (13 A+16 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^3 (13 A+16 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (13 A+16 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (13 A+16 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{4} a^2 (75 A+112 C) \left (\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (13 A+16 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 A \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {a^2 A \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{d}+\frac {1}{2} \left (\frac {a^3 (13 A+16 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}+\frac {1}{4} a^2 (75 A+112 C) \left (\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )\right )}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
(A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((a^2*A *Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/d + ((a^3*(13*A + 1 6*C)*Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(75* A + 112*C)*((Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x ]]])/d + (a*Tan[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])))/4)/2)/(8*a)
3.1.90.3.1 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_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))) 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}, x ] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
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. \(1446\) vs. \(2(176)=352\).
Time = 10.11 (sec) , antiderivative size = 1447, normalized size of antiderivative = 7.24
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1447\) |
| default | \(\text {Expression too large to display}\) | \(1650\) |
1/8*A*a^(1/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(1200*a*(l n(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))+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)^8-1200*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^ 2)^(1/2)*a^(1/2)+2*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+2*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*si n(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+200*(11*2^ (1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+9*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+9*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)^4+(-1460*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-60 0*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-600*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)^2+362*2^(1/2)*(a*sin(1/2*d* x+1/2*c)^2)^(1/2)*a^(1/2)+75*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...
Time = 0.32 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.06 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {{\left ({\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right )^{5} + {\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left ({\left (75 \, A + 112 \, C\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (25 \, A + 16 \, C\right )} a \cos \left (d x + c\right )^{2} + 40 \, A a \cos \left (d x + c\right ) + 16 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{256 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]
1/256*(((75*A + 112*C)*a*cos(d*x + c)^5 + (75*A + 112*C)*a*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 + co s(d*x + c)^2)) + 4*((75*A + 112*C)*a*cos(d*x + c)^3 + 2*(25*A + 16*C)*a*co s(d*x + c)^2 + 40*A*a*cos(d*x + c) + 16*A*a)*sqrt(a*cos(d*x + c) + a)*sin( d*x + c))/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)
Timed out. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 6985 vs. \(2 (176) = 352\).
Time = 1.25 (sec) , antiderivative size = 6985, normalized size of antiderivative = 34.92 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Too large to display} \]
-1/256*((140*a*cos(8*d*x + 8*c)^2*sin(3/2*d*x + 3/2*c) + 2240*a*cos(6*d*x + 6*c)^2*sin(3/2*d*x + 3/2*c) + 5040*a*cos(4*d*x + 4*c)^2*sin(3/2*d*x + 3/ 2*c) + 2240*a*cos(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) + 140*a*sin(8*d*x + 8*c)^2*sin(3/2*d*x + 3/2*c) + 2240*a*sin(6*d*x + 6*c)^2*sin(3/2*d*x + 3/2* c) + 5040*a*sin(4*d*x + 4*c)^2*sin(3/2*d*x + 3/2*c) + 2240*a*sin(2*d*x + 2 *c)^2*sin(3/2*d*x + 3/2*c) + 4064*a*cos(7/2*d*x + 7/2*c)*sin(2*d*x + 2*c) + 336*a*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) - 240*a*cos(3/2*d*x + 3/2*c) *sin(2*d*x + 2*c) + 1360*a*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) - 36*(a*s in(8*d*x + 8*c) + 4*a*sin(6*d*x + 6*c) + 6*a*sin(4*d*x + 4*c) + 4*a*sin(2* d*x + 2*c))*cos(21/2*d*x + 21/2*c) + 140*(a*sin(8*d*x + 8*c) + 4*a*sin(6*d *x + 6*c) + 6*a*sin(4*d*x + 4*c) + 4*a*sin(2*d*x + 2*c))*cos(19/2*d*x + 19 /2*c) + 456*(a*sin(8*d*x + 8*c) + 4*a*sin(6*d*x + 6*c) + 6*a*sin(4*d*x + 4 *c) + 4*a*sin(2*d*x + 2*c))*cos(17/2*d*x + 17/2*c) + 4*(280*a*cos(6*d*x + 6*c)*sin(3/2*d*x + 3/2*c) + 420*a*cos(4*d*x + 4*c)*sin(3/2*d*x + 3/2*c) + 280*a*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) - 290*a*sin(15/2*d*x + 15/2*c) - 596*a*sin(13/2*d*x + 13/2*c) - 780*a*sin(11/2*d*x + 11/2*c) - 750*a*sin (9/2*d*x + 9/2*c) - 254*a*sin(7/2*d*x + 7/2*c) - 21*a*sin(5/2*d*x + 5/2*c) + 85*a*sin(3/2*d*x + 3/2*c))*cos(8*d*x + 8*c) + 2320*(2*a*sin(6*d*x + 6*c ) + 3*a*sin(4*d*x + 4*c) + 2*a*sin(2*d*x + 2*c))*cos(15/2*d*x + 15/2*c) + 4768*(2*a*sin(6*d*x + 6*c) + 3*a*sin(4*d*x + 4*c) + 2*a*sin(2*d*x + 2*c...
Time = 0.56 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.50 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=-\frac {\sqrt {2} {\left (\sqrt {2} {\left (75 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 112 \, C a \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 (600 \, A a \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} + 896 \, C a \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} - 1100 \, A a \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} - 1472 \, C a \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} + 730 \, A a \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} + 800 \, C a \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} - 181 \, A a \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 ) - 144 \, C a \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}}{256 \, d} \]
-1/256*sqrt(2)*(sqrt(2)*(75*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 112*C*a*sgn(co s(1/2*d*x + 1/2*c)))*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sq rt(2) + 4*sin(1/2*d*x + 1/2*c))) + 4*(600*A*a*sgn(cos(1/2*d*x + 1/2*c))*si n(1/2*d*x + 1/2*c)^7 + 896*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2 *c)^7 - 1100*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 1472*C *a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 730*A*a*sgn(cos(1/2* d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 800*C*a*sgn(cos(1/2*d*x + 1/2*c))*s in(1/2*d*x + 1/2*c)^3 - 181*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/ 2*c) - 144*C*a*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. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^5} \,d x \]