Integrand size = 43, antiderivative size = 261 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a^{5/2} (283 A+326 B+400 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{128 d}+\frac {a^3 (283 A+326 B+400 C) \tan (c+d x)}{128 d \sqrt {a+a \cos (c+d x)}}+\frac {a^3 (787 A+950 B+1040 C) \sec (c+d x) \tan (c+d x)}{960 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (79 A+110 B+80 C) \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{240 d}+\frac {a (A+2 B) (a+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{8 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^4(c+d x) \tan (c+d x)}{5 d} \]
1/128*a^(5/2)*(283*A+326*B+400*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+ c))^(1/2))/d+1/8*a*(A+2*B)*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^3*tan(d*x+c)/ d+1/5*A*(a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^4*tan(d*x+c)/d+1/128*a^3*(283*A+ 326*B+400*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/960*a^3*(787*A+950*B+10 40*C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/240*a^2*(79*A+110*B +80*C)*sec(d*x+c)^2*(a+a*cos(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 3.01 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.80 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (60 \sqrt {2} (283 A+326 B+400 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5(c+d x)+(24863 A+22030 B+20560 C+12 (2343 A+1950 B+1360 C) \cos (c+d x)+4 (6509 A+6730 B+6640 C) \cos (2 (c+d x))+5660 A \cos (3 (c+d x))+6520 B \cos (3 (c+d x))+5440 C \cos (3 (c+d x))+4245 A \cos (4 (c+d x))+4890 B \cos (4 (c+d x))+6000 C \cos (4 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{15360 d} \]
Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^ 2)*Sec[c + d*x]^6,x]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^5*(60*Sqrt[2 ]*(283*A + 326*B + 400*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^5 + (24863*A + 22030*B + 20560*C + 12*(2343*A + 1950*B + 1360*C)*Cos[c + d* x] + 4*(6509*A + 6730*B + 6640*C)*Cos[2*(c + d*x)] + 5660*A*Cos[3*(c + d*x )] + 6520*B*Cos[3*(c + d*x)] + 5440*C*Cos[3*(c + d*x)] + 4245*A*Cos[4*(c + d*x)] + 4890*B*Cos[4*(c + d*x)] + 6000*C*Cos[4*(c + d*x)])*Sin[(c + d*x)/ 2]))/(15360*d)
Time = 1.62 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.372, Rules used = {3042, 3522, 27, 3042, 3454, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{5/2} (5 a (A+2 B)+a (3 A+10 C) \cos (c+d x)) \sec ^5(c+d x)dx}{5 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{5/2} (5 a (A+2 B)+a (3 A+10 C) \cos (c+d x)) \sec ^5(c+d x)dx}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (5 a (A+2 B)+a (3 A+10 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{4} \int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} \left ((79 A+110 B+80 C) a^2+(39 A+30 B+80 C) \cos (c+d x) a^2\right ) \sec ^4(c+d x)dx+\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{8} \int (\cos (c+d x) a+a)^{3/2} \left ((79 A+110 B+80 C) a^2+(39 A+30 B+80 C) \cos (c+d x) a^2\right ) \sec ^4(c+d x)dx+\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((79 A+110 B+80 C) a^2+(39 A+30 B+80 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{3} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left ((787 A+950 B+1040 C) a^3+3 (157 A+170 B+240 C) \cos (c+d x) a^3\right ) \sec ^3(c+d x)dx+\frac {a^3 (79 A+110 B+80 C) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \int \sqrt {\cos (c+d x) a+a} \left ((787 A+950 B+1040 C) a^3+3 (157 A+170 B+240 C) \cos (c+d x) a^3\right ) \sec ^3(c+d x)dx+\frac {a^3 (79 A+110 B+80 C) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((787 A+950 B+1040 C) a^3+3 (157 A+170 B+240 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^3 (79 A+110 B+80 C) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (283 A+326 B+400 C) \int \sqrt {\cos (c+d x) a+a} \sec ^2(c+d x)dx+\frac {a^4 (787 A+950 B+1040 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (79 A+110 B+80 C) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (283 A+326 B+400 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^4 (787 A+950 B+1040 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (79 A+110 B+80 C) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (283 A+326 B+400 C) \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (787 A+950 B+1040 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (79 A+110 B+80 C) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (283 A+326 B+400 C) \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^4 (787 A+950 B+1040 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (79 A+110 B+80 C) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {15}{4} a^3 (283 A+326 B+400 C) \left (\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a^4 (787 A+950 B+1040 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (79 A+110 B+80 C) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {5 a^2 (A+2 B) \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}+\frac {1}{8} \left (\frac {a^3 (79 A+110 B+80 C) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}+\frac {1}{6} \left (\frac {a^4 (787 A+950 B+1040 C) \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}+\frac {15}{4} a^3 (283 A+326 B+400 C) \left (\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )\right )\right )}{10 a}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d}\) |
(A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((5*a^2 *(A + 2*B)*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((a^3*(79*A + 110*B + 80*C)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a^4*(787*A + 950*B + 1040*C)*Sec[c + d*x]*Tan[c + d*x]) /(2*d*Sqrt[a + a*Cos[c + d*x]]) + (15*a^3*(283*A + 326*B + 400*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)/(10*a)
3.4.99.3.1 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_)])^(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. \(2878\) vs. \(2(233)=466\).
Time = 5.55 (sec) , antiderivative size = 2879, normalized size of antiderivative = 11.03
\[\text {output too large to display}\]
1/120*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-480*a*(2 83*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))+283*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))+326*B*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))+326*B*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))+400*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))+400*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)^10+240*(566*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d *x+1/2*c)^2)^(1/2)+652*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+80 0*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+1415*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+1415*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+1630*B*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+1630*B*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/...
Time = 0.40 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.02 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left ({\left (283 \, A + 326 \, B + 400 \, C\right )} a^{2} \cos \left (d x + c\right )^{6} + {\left (283 \, A + 326 \, B + 400 \, C\right )} a^{2} \cos \left (d x + c\right )^{5}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (15 \, {\left (283 \, A + 326 \, B + 400 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (283 \, A + 326 \, B + 272 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (283 \, A + 230 \, B + 80 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 48 \, {\left (29 \, A + 10 \, B\right )} a^{2} \cos \left (d x + c\right ) + 384 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{7680 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^6,x, algorithm="fricas")
1/7680*(15*((283*A + 326*B + 400*C)*a^2*cos(d*x + c)^6 + (283*A + 326*B + 400*C)*a^2*cos(d*x + c)^5)*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*(15*(283*A + 326*B + 400*C)*a ^2*cos(d*x + c)^4 + 10*(283*A + 326*B + 272*C)*a^2*cos(d*x + c)^3 + 8*(283 *A + 230*B + 80*C)*a^2*cos(d*x + c)^2 + 48*(29*A + 10*B)*a^2*cos(d*x + c) + 384*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^6 + d* cos(d*x + c)^5)
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^6,x, algorithm="maxima")
Leaf count of result is larger than twice the leaf count of optimal. 525 vs. \(2 (233) = 466\).
Time = 2.41 (sec) , antiderivative size = 525, normalized size of antiderivative = 2.01 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Too large to display} \]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c )^6,x, algorithm="giac")
-1/7680*sqrt(2)*(15*sqrt(2)*(283*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 326*B*a ^2*sgn(cos(1/2*d*x + 1/2*c)) + 400*C*a^2*sgn(cos(1/2*d*x + 1/2*c)))*log(ab s(-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*(67920*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^9 + 78240*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^9 + 96000*C*a^2 *sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^9 - 158480*A*a^2*sgn(cos(1 /2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^7 - 182560*B*a^2*sgn(cos(1/2*d*x + 1 /2*c))*sin(1/2*d*x + 1/2*c)^7 - 213760*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin (1/2*d*x + 1/2*c)^7 + 144896*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 163840*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 179200*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 62780*A* a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 67000*B*a^2*sgn(cos (1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 66880*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 11115*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin (1/2*d*x + 1/2*c) + 10470*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/ 2*c) + 9360*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)^5)*sqrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^6} \,d x \]