Integrand size = 43, antiderivative size = 215 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a^{3/2} (75 A+88 B+112 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {a^2 (75 A+88 B+112 C) \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (39 A+56 B+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {a (3 A+8 B) \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
1/64*a^(3/2)*(75*A+88*B+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+88*B+112*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/96*a^2*(39*A+56*B+ 48*C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/24*a*(3*A+8*B)*sec( d*x+c)^2*(a+a*cos(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 3.66 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.81 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (6 \sqrt {2} (75 A+88 B+112 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4(c+d x)+(492 A+352 B+192 C+(1155 A+1048 B+1008 C) \cos (c+d x)+4 (75 A+88 B+48 C) \cos (2 (c+d x))+225 A \cos (3 (c+d x))+264 B \cos (3 (c+d x))+336 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{768 d} \]
Integrate[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 2)*Sec[c + d*x]^5,x]
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^4*(6*Sqrt[2]*( 75*A + 88*B + 112*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^4 + (4 92*A + 352*B + 192*C + (1155*A + 1048*B + 1008*C)*Cos[c + d*x] + 4*(75*A + 88*B + 48*C)*Cos[2*(c + d*x)] + 225*A*Cos[3*(c + d*x)] + 264*B*Cos[3*(c + d*x)] + 336*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(768*d)
Time = 1.25 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.302, Rules used = {3042, 3522, 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.
\(\displaystyle \int \sec ^5(c+d x) (a \cos (c+d x)+a)^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
\(\Big \downarrow \) 3522 |
\(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} (a (3 A+8 B)+a (3 A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (\cos (c+d x) a+a)^{3/2} (a (3 A+8 B)+a (3 A+8 C) \cos (c+d x)) \sec ^4(c+d x)dx}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (3 A+8 B)+a (3 A+8 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3454 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left ((39 A+56 B+48 C) a^2+3 (9 A+8 B+16 C) \cos (c+d x) a^2\right ) \sec ^3(c+d x)dx+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{6} \int \sqrt {\cos (c+d x) a+a} \left ((39 A+56 B+48 C) a^2+3 (9 A+8 B+16 C) \cos (c+d x) a^2\right ) \sec ^3(c+d x)dx+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{6} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((39 A+56 B+48 C) a^2+3 (9 A+8 B+16 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3459 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} a^2 (75 A+88 B+112 C) \int \sqrt {\cos (c+d x) a+a} \sec ^2(c+d x)dx+\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} a^2 (75 A+88 B+112 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3251 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} a^2 (75 A+88 B+112 C) \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} a^2 (75 A+88 B+112 C) \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 3252 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} a^2 (75 A+88 B+112 C) \left (\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {a^2 (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}+\frac {1}{6} \left (\frac {a^3 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}+\frac {3}{4} a^2 (75 A+88 B+112 C) \left (\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )\right )}{8 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}\) |
(A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((a^2*( 3*A + 8*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ( (a^3*(39*A + 56*B + 48*C)*Sec[c + d*x]*Tan[c + d*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*a^2*(75*A + 88*B + 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)/6)/(8*a)
3.4.89.3.1 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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + 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 - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(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, B, 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. \(2179\) vs. \(2(191)=382\).
Time = 17.17 (sec) , antiderivative size = 2180, normalized size of antiderivative = 10.14
method | result | size |
parts | \(\text {Expression too large to display}\) | \(2180\) |
default | \(\text {Expression too large to display}\) | \(2400\) |
int((a+cos(d*x+c)*a)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, method=_RETURNVERBOSE)
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 = 0.41 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.08 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left ({\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right )^{5} + {\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (75 \, A + 88 \, B + 48 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (15 \, A + 8 \, B\right )} a \cos \left (d x + c\right ) + 48 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^5,x, algorithm="fricas")
1/768*(3*((75*A + 88*B + 112*C)*a*cos(d*x + c)^5 + (75*A + 88*B + 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*sq rt(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*(75*A + 88*B + 112*C)*a*cos(d*x + c) ^3 + 2*(75*A + 88*B + 48*C)*a*cos(d*x + c)^2 + 8*(15*A + 8*B)*a*cos(d*x + c) + 48*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. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 12526 vs. \(2 (191) = 382\).
Time = 156.08 (sec) , antiderivative size = 12526, normalized size of antiderivative = 58.26 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Too large to display} \]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^5,x, algorithm="maxima")
-1/768*(3*(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 *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(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*s in(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...
Leaf count of result is larger than twice the leaf count of optimal. 414 vs. \(2 (191) = 382\).
Time = 0.83 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.93 \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=-\frac {\sqrt {2} {\left (3 \, \sqrt {2} {\left (75 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 88 \, B 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 (1800 \, 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} + 2112 \, B 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} + 2688 \, 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} - 3300 \, 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} - 3872 \, B 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} - 4416 \, 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} + 2190 \, 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} + 2416 \, B 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} + 2400 \, 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} - 543 \, 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 ) - 504 \, B 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 ) - 432 \, 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}}{768 \, d} \]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^5,x, algorithm="giac")
-1/768*sqrt(2)*(3*sqrt(2)*(75*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 88*B*a*sgn(c os(1/2*d*x + 1/2*c)) + 112*C*a*sgn(cos(1/2*d*x + 1/2*c)))*log(abs(-2*sqrt( 2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))) + 4* (1800*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 + 2112*B*a*sgn( cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 + 2688*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 - 3300*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/ 2*d*x + 1/2*c)^5 - 3872*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) ^5 - 4416*C*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 2190*A*a* sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 2416*B*a*sgn(cos(1/2*d* x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 2400*C*a*sgn(cos(1/2*d*x + 1/2*c))*si n(1/2*d*x + 1/2*c)^3 - 543*A*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2 *c) - 504*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 432*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. \[ \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^5} \,d x \]