Integrand size = 43, antiderivative size = 184 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a^{5/2} (5 A+2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {a^3 (15 A+70 B+64 C) \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {a^2 (15 A-10 B-16 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}-\frac {a (5 A-2 C) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac {A (a+a \cos (c+d x))^{5/2} \tan (c+d x)}{d} \] Output:
a^(5/2)*(5*A+2*B)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/d+1/1 5*a^3*(15*A+70*B+64*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-1/15*a^2*(15*A- 10*B-16*C)*(a+a*cos(d*x+c))^(1/2)*sin(d*x+c)/d-1/5*a*(5*A-2*C)*(a+a*cos(d* x+c))^(3/2)*sin(d*x+c)/d+A*(a+a*cos(d*x+c))^(5/2)*tan(d*x+c)/d
Time = 0.94 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.79 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (30 \sqrt {2} (5 A+2 B) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos (c+d x)+2 (30 A+10 B+28 C+(60 A+160 B+181 C) \cos (c+d x)+2 (5 B+14 C) \cos (2 (c+d x))+3 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{60 d} \] Input:
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]^2,x]
Output:
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]*(30*Sqrt[2]* (5*A + 2*B)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x] + 2*(30*A + 10* B + 28*C + (60*A + 160*B + 181*C)*Cos[c + d*x] + 2*(5*B + 14*C)*Cos[2*(c + d*x)] + 3*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(60*d)
Time = 1.33 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 3522, 27, 3042, 3455, 27, 3042, 3455, 27, 3042, 3460, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{5/2} (a (5 A+2 B)-a (5 A-2 C) \cos (c+d x)) \sec (c+d x)dx}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{5/2} (a (5 A+2 B)-a (5 A-2 C) \cos (c+d x)) \sec (c+d x)dx}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+2 B)-a (5 A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {2}{5} \int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} \left (5 a^2 (5 A+2 B)-a^2 (15 A-10 B-16 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 a^2 (5 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \int (\cos (c+d x) a+a)^{3/2} \left (5 a^2 (5 A+2 B)-a^2 (15 A-10 B-16 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 a^2 (5 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (5 a^2 (5 A+2 B)-a^2 (15 A-10 B-16 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^2 (5 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (15 (5 A+2 B) a^3+(15 A+70 B+64 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx-\frac {2 a^3 (15 A-10 B-16 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \sqrt {\cos (c+d x) a+a} \left (15 (5 A+2 B) a^3+(15 A+70 B+64 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx-\frac {2 a^3 (15 A-10 B-16 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (15 (5 A+2 B) a^3+(15 A+70 B+64 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^3 (15 A-10 B-16 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (15 a^3 (5 A+2 B) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {2 a^4 (15 A+70 B+64 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^3 (15 A-10 B-16 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (15 a^3 (5 A+2 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^4 (15 A+70 B+64 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^3 (15 A-10 B-16 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\frac {2 a^4 (15 A+70 B+64 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {30 a^4 (5 A+2 B) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )-\frac {2 a^3 (15 A-10 B-16 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\frac {30 a^{7/2} (5 A+2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^4 (15 A+70 B+64 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^3 (15 A-10 B-16 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )-\frac {2 a^2 (5 A-2 C) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d}}{2 a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^{5/2}}{d}\) |
Input:
Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec [c + d*x]^2,x]
Output:
((-2*a^2*(5*A - 2*C)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d) + ((-2 *a^3*(15*A - 10*B - 16*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ( (30*a^(7/2)*(5*A + 2*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a^4*(15*A + 70*B + 64*C)*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/3)/5)/(2*a) + (A*(a + a*Cos[c + d*x])^(5/2)*Tan[c + d*x])/d
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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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. \(755\) vs. \(2(164)=328\).
Time = 14.20 (sec) , antiderivative size = 756, normalized size of antiderivative = 4.11
| method | result | size |
| parts | \(\frac {A \,a^{\frac {3}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (-8 \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-10 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -10 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +6 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+5 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a +5 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a \right )}{\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {B \,a^{\frac {3}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \left (-8 \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 \sqrt {2}\, \ln \left (-\frac {2 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a +3 \sqrt {2}\, \ln \left (\frac {2 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}+4 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +36 \sqrt {a}\, \sqrt {\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a}\right ) \sqrt {2}}{6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {8 C \,a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8\right ) \sqrt {2}}{15 \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(756\) |
| default | \(\text {Expression too large to display}\) | \(857\) |
Input:
int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, method=_RETURNVERBOSE)
Output:
A*a^(3/2)*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(-8*2^(1/2)*(s in(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^2-10*ln(-4/(2*cos( 1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin (1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*sin(1/2*d*x+1/2*c)^2*a-10*ln(4/(2*cos(1/2 *d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/ 2*d*x+1/2*c)^2*a)^(1/2)+2*a))*sin(1/2*d*x+1/2*c)^2*a+6*a^(1/2)*2^(1/2)*(si n(1/2*d*x+1/2*c)^2*a)^(1/2)+5*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1 /2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a) )*a+5*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^ (1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a)/(2*cos(1/2*d*x+1/2*c )-2^(1/2))/(2*cos(1/2*d*x+1/2*c)+2^(1/2))/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d* x+1/2*c)^2)^(1/2)/d+1/6*B*a^(3/2)*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2 *a)^(1/2)*(-8*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^2+ 3*2^(1/2)*ln(-2/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2* c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a+3*2^(1/2)*ln(2/( 2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2 )*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a+36*a^(1/2)*(sin(1/2*d*x+1/2*c)^2* a)^(1/2))/sin(1/2*d*x+1/2*c)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+8/15 *C*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)*(3*cos(1/2*d*x+1/2*c)^4+4*cos (1/2*d*x+1/2*c)^2+8)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.13 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.22 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {15 \, {\left ({\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (5 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (6 \, C a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (5 \, B + 14 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, A + 40 \, B + 43 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \] Input:
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 )^2,x, algorithm="fricas")
Output:
1/60*(15*((5*A + 2*B)*a^2*cos(d*x + c)^2 + (5*A + 2*B)*a^2*cos(d*x + c))*s qrt(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*(6*C*a^2*cos(d*x + c)^3 + 2*(5*B + 14*C)*a^2*cos(d*x + c) ^2 + 2*(15*A + 40*B + 43*C)*a^2*cos(d*x + c) + 15*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Timed out} \] Input:
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)**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 8175 vs. \(2 (164) = 328\).
Time = 0.57 (sec) , antiderivative size = 8175, normalized size of antiderivative = 44.43 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\text {Too large to display} \] Input:
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 )^2,x, algorithm="maxima")
Output:
1/1260*(42*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d* x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*C*sqrt(a) - 5*(1449*sqr t(2)*a^2*cos(5/2*d*x + 5/2*c)^3*sin(2*d*x + 2*c) - 1260*sqrt(2)*a^2*sin(1/ 2*d*x + 1/2*c)^3 - 1449*(sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*sin(5 /2*d*x + 5/2*c)^3 + 21*(25*sqrt(2)*a^2*cos(2*d*x + 2*c)^2*sin(3/2*d*x + 3/ 2*c) + 25*sqrt(2)*a^2*sin(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) - 60*sqrt(2) *a^2*sin(1/2*d*x + 1/2*c) + 5*(5*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) - 12*sqr t(2)*a^2*sin(1/2*d*x + 1/2*c))*cos(2*d*x + 2*c) + (25*sqrt(2)*a^2*cos(3/2* d*x + 3/2*c) + 198*sqrt(2)*a^2*cos(1/2*d*x + 1/2*c))*sin(2*d*x + 2*c))*cos (5/2*d*x + 5/2*c)^2 - 21*(12*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) - 25*(sqrt(2 )*a^2*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*a^2*sin(1/2*d*x + 1/2*c)^2)*sin(3/2 *d*x + 3/2*c))*cos(2*d*x + 2*c)^2 + 21*(25*sqrt(2)*a^2*cos(2*d*x + 2*c)^2* sin(3/2*d*x + 3/2*c) + 25*sqrt(2)*a^2*sin(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2 *c) + 69*sqrt(2)*a^2*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) - 198*sqrt(2)*a ^2*sin(1/2*d*x + 1/2*c) + (25*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) - 198*sqrt( 2)*a^2*sin(1/2*d*x + 1/2*c))*cos(2*d*x + 2*c) + 5*(5*sqrt(2)*a^2*cos(3/2*d *x + 3/2*c) + 12*sqrt(2)*a^2*cos(1/2*d*x + 1/2*c))*sin(2*d*x + 2*c))*sin(5 /2*d*x + 5/2*c)^2 - 21*(12*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) - 25*(sqrt(2)* a^2*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*a^2*sin(1/2*d*x + 1/2*c)^2)*sin(3/2*d *x + 3/2*c))*sin(2*d*x + 2*c)^2 - 35*(sqrt(2)*a^2*cos(5/2*d*x + 5/2*c)^...
Time = 1.59 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.57 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {\sqrt {2} {\left (96 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 320 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 480 \, C a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {60 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 15 \, \sqrt {2} {\left (5 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )\right )} \sqrt {a}}{60 \, d} \] Input:
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 )^2,x, algorithm="giac")
Output:
1/60*sqrt(2)*(96*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 80*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 320*C*a^2*sgn( cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 120*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 360*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/ 2*d*x + 1/2*c) + 480*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 60*A*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) - 15*sqrt(2)*(5*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 2*B*a^2*s gn(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))))*sqrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^2} \,d x \] Input:
int(((a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/c os(c + d*x)^2,x)
Output:
int(((a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/c os(c + d*x)^2, x)
\[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\sqrt {a}\, a^{2} \left (2 \left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}d x \right ) b +2 \left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) a +2 \left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:
int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x)
Output:
sqrt(a)*a**2*(2*int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**2,x) *a + int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**2,x)*b + int(sq rt(cos(c + d*x) + 1)*cos(c + d*x)**4*sec(c + d*x)**2,x)*c + int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**2,x)*b + 2*int(sqrt(cos(c + d*x ) + 1)*cos(c + d*x)**3*sec(c + d*x)**2,x)*c + int(sqrt(cos(c + d*x) + 1)*c os(c + d*x)**2*sec(c + d*x)**2,x)*a + 2*int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**2,x)*b + int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2 *sec(c + d*x)**2,x)*c + int(sqrt(cos(c + d*x) + 1)*sec(c + d*x)**2,x)*a)