Integrand size = 43, antiderivative size = 207 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a^{5/2} (25 A+38 B+40 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}-\frac {a^3 (49 A+54 B-24 C) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (31 A+42 B+24 C) \sqrt {a+a \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {a (5 A+6 B) (a+a \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d} \]
1/8*a^(5/2)*(25*A+38*B+40*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^( 1/2))/d-1/24*a^3*(49*A+54*B-24*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/12 *a*(5*A+6*B)*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)*tan(d*x+c)/d+1/3*A*(a+a*cos (d*x+c))^(5/2)*sec(d*x+c)^2*tan(d*x+c)/d+1/24*a^2*(31*A+42*B+24*C)*(a+a*co s(d*x+c))^(1/2)*tan(d*x+c)/d
Time = 1.62 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.75 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (3 \sqrt {2} (25 A+38 B+40 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3(c+d x)+(91 A+66 B+24 C+4 (17 A+6 (B+3 C)) \cos (c+d x)+3 (25 A+22 B+8 C) \cos (2 (c+d x))+24 C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \]
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]^4,x]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^3*(3*Sqrt[2] *(25*A + 38*B + 40*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^3 + ( 91*A + 66*B + 24*C + 4*(17*A + 6*(B + 3*C))*Cos[c + d*x] + 3*(25*A + 22*B + 8*C)*Cos[2*(c + d*x)] + 24*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)
Time = 1.44 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.06, 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, 3454, 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.
| \(\displaystyle \int \sec ^4(c+d x) (a \cos (c+d x)+a)^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int \frac {1}{2} (\cos (c+d x) a+a)^{5/2} (a (5 A+6 B)-a (A-6 C) \cos (c+d x)) \sec ^3(c+d x)dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^{5/2} (a (5 A+6 B)-a (A-6 C) \cos (c+d x)) \sec ^3(c+d x)dx}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (5 A+6 B)-a (A-6 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {1}{2} (\cos (c+d x) a+a)^{3/2} \left (a^2 (31 A+42 B+24 C)-3 a^2 (3 A+2 B-8 C) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{4} \int (\cos (c+d x) a+a)^{3/2} \left (a^2 (31 A+42 B+24 C)-3 a^2 (3 A+2 B-8 C) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a^2 (31 A+42 B+24 C)-3 a^2 (3 A+2 B-8 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{4} \left (\int \frac {1}{2} \sqrt {\cos (c+d x) a+a} \left (3 a^3 (25 A+38 B+40 C)-a^3 (49 A+54 B-24 C) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {a^3 (31 A+42 B+24 C) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \sqrt {\cos (c+d x) a+a} \left (3 a^3 (25 A+38 B+40 C)-a^3 (49 A+54 B-24 C) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {a^3 (31 A+42 B+24 C) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a^3 (25 A+38 B+40 C)-a^3 (49 A+54 B-24 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^3 (31 A+42 B+24 C) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 a^3 (25 A+38 B+40 C) \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-\frac {2 a^4 (49 A+54 B-24 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (31 A+42 B+24 C) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (3 a^3 (25 A+38 B+40 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 a^4 (49 A+54 B-24 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (31 A+42 B+24 C) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (-\frac {6 a^4 (25 A+38 B+40 C) \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {2 a^4 (49 A+54 B-24 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^3 (31 A+42 B+24 C) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\right )+\frac {a^2 (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {a^2 (5 A+6 B) \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{3/2}}{2 d}+\frac {1}{4} \left (\frac {a^3 (31 A+42 B+24 C) \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}+\frac {1}{2} \left (\frac {6 a^{7/2} (25 A+38 B+40 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 a^4 (49 A+54 B-24 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )\right )}{6 a}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2}}{3 d}\) |
(A*(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a^2*( 5*A + 6*B)*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ( ((6*a^(7/2)*(25*A + 38*B + 40*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a *Cos[c + d*x]]])/d - (2*a^4*(49*A + 54*B - 24*C)*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/2 + (a^3*(31*A + 42*B + 24*C)*Sqrt[a + a*Cos[c + d*x]]* Tan[c + d*x])/d)/4)/(6*a)
3.4.97.3.1 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[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. \(1723\) vs. \(2(183)=366\).
Time = 193.76 (sec) , antiderivative size = 1724, normalized size of antiderivative = 8.33
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1724\) |
| default | \(\text {Expression too large to display}\) | \(1949\) |
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)^4,x, method=_RETURNVERBOSE)
1/6*A*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-600*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)^6+300*(2*2^(1/2)*(a*sin(1/2*d*x+1/2*c) ^2)^(1/2)*a^(1/2)+3*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+3*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*s in(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+(-736*2^( 1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-450*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-450*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*c os(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+234*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))*a+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))*a)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^3/(2*cos(1/2*d*x+ 1/2*c)+2^(1/2))^3/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+1/4* B*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(76*2^(1/2)...
Time = 0.33 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.14 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left ({\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (25 \, A + 38 \, B + 40 \, C\right )} a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (48 \, C a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (25 \, A + 22 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \]
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 )^4,x, algorithm="fricas")
1/96*(3*((25*A + 38*B + 40*C)*a^2*cos(d*x + c)^4 + (25*A + 38*B + 40*C)*a^ 2*cos(d*x + c)^3)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*s qrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(co s(d*x + c)^3 + cos(d*x + c)^2)) + 4*(48*C*a^2*cos(d*x + c)^3 + 3*(25*A + 2 2*B + 8*C)*a^2*cos(d*x + c)^2 + 2*(17*A + 6*B)*a^2*cos(d*x + c) + 8*A*a^2) *sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^4 + d*cos(d*x + c) ^3)
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 16106 vs. \(2 (183) = 366\).
Time = 3.87 (sec) , antiderivative size = 16106, normalized size of antiderivative = 77.81 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Too large to display} \]
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 )^4,x, algorithm="maxima")
-1/2016*(21*(1530*a^2*cos(4*d*x + 4*c)^2*sin(3/2*d*x + 3/2*c) + 1530*a^2*c os(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) + 1530*a^2*sin(4*d*x + 4*c)^2*sin(3 /2*d*x + 3/2*c) + 1530*a^2*sin(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) + 4176* a^2*cos(7/2*d*x + 7/2*c)*sin(2*d*x + 2*c) + 2430*a^2*cos(5/2*d*x + 5/2*c)* sin(2*d*x + 2*c) + 678*a^2*cos(3/2*d*x + 3/2*c)*sin(2*d*x + 2*c) + 342*a^2 *cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) + 10*(a^2*sin(9/2*d*x + 9/2*c) + 17 *a^2*sin(3/2*d*x + 3/2*c))*cos(6*d*x + 6*c)^2 + 10*(a^2*sin(9/2*d*x + 9/2* c) + 17*a^2*sin(3/2*d*x + 3/2*c))*sin(6*d*x + 6*c)^2 - 56*a^2*sin(3/2*d*x + 3/2*c) + 10*(a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 3*a^2*sin(2 *d*x + 2*c))*cos(21/2*d*x + 21/2*c) - 30*(a^2*sin(6*d*x + 6*c) + 3*a^2*sin (4*d*x + 4*c) + 3*a^2*sin(2*d*x + 2*c))*cos(19/2*d*x + 19/2*c) - 48*(a^2*s in(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 3*a^2*sin(2*d*x + 2*c))*cos(17/ 2*d*x + 17/2*c) + 80*(a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 3*a^ 2*sin(2*d*x + 2*c))*cos(15/2*d*x + 15/2*c) + 396*(a^2*sin(6*d*x + 6*c) + 3 *a^2*sin(4*d*x + 4*c) + 3*a^2*sin(2*d*x + 2*c))*cos(13/2*d*x + 13/2*c) + 6 *(170*a^2*cos(4*d*x + 4*c)*sin(3/2*d*x + 3/2*c) + 170*a^2*cos(2*d*x + 2*c) *sin(3/2*d*x + 3/2*c) - 170*a^2*sin(11/2*d*x + 11/2*c) - 232*a^2*sin(7/2*d *x + 7/2*c) - 135*a^2*sin(5/2*d*x + 5/2*c) + 19*a^2*sin(3/2*d*x + 3/2*c) + 10*(a^2*cos(4*d*x + 4*c) + a^2*cos(2*d*x + 2*c) - 25*a^2)*sin(9/2*d*x + 9 /2*c))*cos(6*d*x + 6*c) + 3060*(a^2*sin(4*d*x + 4*c) + a^2*sin(2*d*x + ...
Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (183) = 366\).
Time = 2.32 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.87 \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\sqrt {2} {\left (192 \, 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 (25 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 38 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 40 \, 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 {4 \, {\left (300 \, 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 )^{5} + 264 \, 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 )^{5} + 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} - 368 \, 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} - 288 \, 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} - 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 )^{3} + 117 \, 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 ) + 78 \, 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 ) + 24 \, 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 )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )} \sqrt {a}}{96 \, 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 )^4,x, algorithm="giac")
1/96*sqrt(2)*(192*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) - 3 *sqrt(2)*(25*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 38*B*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 40*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))) - 4*(300*A*a^2*s gn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 264*B*a^2*sgn(cos(1/2*d* x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 + 96*C*a^2*sgn(cos(1/2*d*x + 1/2*c))*si n(1/2*d*x + 1/2*c)^5 - 368*A*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1 /2*c)^3 - 288*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 96* C*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 117*A*a^2*sgn(cos (1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 78*B*a^2*sgn(cos(1/2*d*x + 1/2*c ))*sin(1/2*d*x + 1/2*c) + 24*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)^3)*sqrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^4} \,d x \]