Integrand size = 43, antiderivative size = 199 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {a^{5/2} (19 A+20 B+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 d}-\frac {a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}-\frac {a^2 (21 A+12 B-8 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{12 d}+\frac {a (5 A+4 B) (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d} \]
1/4*a^(5/2)*(19*A+20*B+8*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1 /2))/d-1/12*a^3*(27*A-12*B-56*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-1/12* a^2*(21*A+12*B-8*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+1/4*a*(5*A+4*B)*(a +a*cos(d*x+c))^(3/2)*tan(d*x+c)/d+1/2*A*(a+a*cos(d*x+c))^(5/2)*sec(d*x+c)* tan(d*x+c)/d
Time = 1.14 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.77 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (6 \sqrt {2} (19 A+20 B+8 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2(c+d x)+4 (6 A+12 B+32 C+3 (11 A+4 B+2 C) \cos (c+d x)+4 (3 B+8 C) \cos (2 (c+d x))+2 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \]
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]^3,x]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^2*(6*Sqrt[2] *(19*A + 20*B + 8*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^2 + 4* (6*A + 12*B + 32*C + 3*(11*A + 4*B + 2*C)*Cos[c + d*x] + 4*(3*B + 8*C)*Cos [2*(c + d*x)] + 2*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)
Time = 1.35 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.07, 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, 3454, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{5/2} (a (5 A+4 B)-a (3 A-4 C) \cos (c+d x)) \sec ^2(c+d x)dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{5/2} (a (5 A+4 B)-a (3 A-4 C) \cos (c+d x)) \sec ^2(c+d x)dx}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+4 B)-a (3 A-4 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} \left (a^2 (19 A+20 B+8 C)-a^2 (21 A+12 B-8 C) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {a^2 (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \int (\cos (c+d x) a+a)^{3/2} \left (a^2 (19 A+20 B+8 C)-a^2 (21 A+12 B-8 C) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {a^2 (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a^2 (19 A+20 B+8 C)-a^2 (21 A+12 B-8 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {2}{3} \int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (3 a^3 (19 A+20 B+8 C)-a^3 (27 A-12 B-56 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 a^3 (21 A+12 B-8 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a^2 (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \int \sqrt {\cos (c+d x) a+a} \left (3 a^3 (19 A+20 B+8 C)-a^3 (27 A-12 B-56 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {2 a^3 (21 A+12 B-8 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a^2 (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a^3 (19 A+20 B+8 C)-a^3 (27 A-12 B-56 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^3 (21 A+12 B-8 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a^2 (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (3 a^3 (19 A+20 B+8 C) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-\frac {2 a^4 (27 A-12 B-56 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^3 (21 A+12 B-8 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a^2 (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (3 a^3 (19 A+20 B+8 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^4 (27 A-12 B-56 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^3 (21 A+12 B-8 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a^2 (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (-\frac {6 a^4 (19 A+20 B+8 C) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {2 a^4 (27 A-12 B-56 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^3 (21 A+12 B-8 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )+\frac {a^2 (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {a^2 (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d}+\frac {1}{2} \left (\frac {1}{3} \left (\frac {6 a^{7/2} (19 A+20 B+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 a^4 (27 A-12 B-56 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )-\frac {2 a^3 (21 A+12 B-8 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}\right )}{4 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d}\) |
(A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((-2*a^3 *(21*A + 12*B - 8*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ((6*a^ (7/2)*(19*A + 20*B + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (2*a^4*(27*A - 12*B - 56*C)*Sin[c + d*x])/(d*Sqrt[a + a*Cos[ c + d*x]]))/3)/2 + (a^2*(5*A + 4*B)*(a + a*Cos[c + d*x])^(3/2)*Tan[c + d*x ])/d)/(4*a)
3.4.96.3.1 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[((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. \(1243\) vs. \(2(175)=350\).
Time = 68.25 (sec) , antiderivative size = 1244, normalized size of antiderivative = 6.25
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1244\) |
| default | \(\text {Expression too large to display}\) | \(1530\) |
int((a+cos(d*x+c)*a)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, method=_RETURNVERBOSE)
1/2*A*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(76*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))+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)^4+(-44*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^ (1/2)*a^(1/2)-76*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-76*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+26*2^(1/2)* (a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+19*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+19*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*cos( 1/2*d*x+1/2*c)-2^(1/2))^2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^2/sin(1/2*d*x+1/2 *c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+1/2*B*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*s in(1/2*d*x+1/2*c)^2)^(1/2)*(-10*2^(1/2)*ln(2/(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)^2*a-10*2^(1/2)*ln(-2/(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)^2*a-16*(a*sin(1/2*d*x+1/2*c)^2)^(1/...
Time = 0.33 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.17 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {3 \, {\left ({\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (8 \, C a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (3 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (11 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right ) + 6 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]
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 )^3,x, algorithm="fricas")
1/48*(3*((19*A + 20*B + 8*C)*a^2*cos(d*x + c)^3 + (19*A + 20*B + 8*C)*a^2* cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqr t(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*(8*C*a^2*cos(d*x + c)^3 + 8*(3*B + 8*C)* a^2*cos(d*x + c)^2 + 3*(11*A + 4*B)*a^2*cos(d*x + c) + 6*A*a^2)*sqrt(a*cos (d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 11782 vs. \(2 (175) = 350\).
Time = 3.51 (sec) , antiderivative size = 11782, normalized size of antiderivative = 59.21 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Too large to display} \]
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 )^3,x, algorithm="maxima")
-1/1008*(63*(150*sqrt(2)*a^2*cos(7/2*d*x + 7/2*c)*sin(2*d*x + 2*c) + 154*s qrt(2)*a^2*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) - 28*sqrt(2)*a^2*sin(3/2* d*x + 3/2*c) + 44*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) - (3*sqrt(2)*a^2*sin(7/ 2*d*x + 7/2*c) + 5*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) - 17*sqrt(2)*a^2*sin(3 /2*d*x + 3/2*c) - 55*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) + 19*a^2*log(2*cos(1 /2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2 *c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 19*a^2*log(2*cos(1/2*d*x + 1/2 *c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt (2)*sin(1/2*d*x + 1/2*c) + 2) + 19*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*si n(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2* d*x + 1/2*c) + 2) - 19*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(4*d*x + 4*c)^2 + 4*(17*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 55*sq rt(2)*a^2*sin(1/2*d*x + 1/2*c) - 19*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*s in(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2 *d*x + 1/2*c) + 2) + 19*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c ) + 2) - 19*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 19* a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)...
Time = 1.90 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.54 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=-\frac {\sqrt {2} {\left (64 \, 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} - 96 \, 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 ) - 288 \, 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 \, \sqrt {2} {\left (19 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 20 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 8 \, 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 {12 \, {\left (22 \, 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} + 8 \, 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} - 13 \, 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 ) - 4 \, 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 )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}\right )} \sqrt {a}}{48 \, d} \]
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 )^3,x, algorithm="giac")
-1/48*sqrt(2)*(64*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 96*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 288*C*a^2*sgn(c os(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 3*sqrt(2)*(19*A*a^2*sgn(cos(1/ 2*d*x + 1/2*c)) + 20*B*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 8*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))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))) + 12*(22*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1 /2*d*x + 1/2*c)^3 + 8*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) ^3 - 13*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 4*B*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 )^2)*sqrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^3} \,d x \]