Integrand size = 43, antiderivative size = 214 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {(20 A-35 B+56 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac {5 (A-2 B+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}-\frac {5 (A-2 B+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {(20 A-35 B+56 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a^2 d}-\frac {(A-2 B+3 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]
1/5*(20*A-35*B+56*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellip ticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d-5/3*(A-2*B+3*C)*(cos(1/2*d*x+1/2*c) ^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d+1 /15*(20*A-35*B+56*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a^2/d-(A-2*B+3*C)*cos(d*x +c)^(5/2)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A-B+C)*cos(d*x+c)^(7/2)*sin (d*x+c)/d/(a+a*cos(d*x+c))^2-5/3*(A-2*B+3*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/a ^2/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.40 (sec) , antiderivative size = 1433, normalized size of antiderivative = 6.70 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx =\text {Too large to display} \]
Integrate[(Cos[c + d*x]^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^2,x]
(10*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Si n[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan [Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x]) ^2*Sqrt[1 + Cot[c]^2]) - (20*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Hypergeometri cPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - Ar cTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2 ]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/ (3*d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2]) + (10*C*Cos[c/2 + (d*x)/2] ^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]] ^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]] *Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin [d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2]) + ( Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*((-4*(10*A - 15*B + 20*C + 10*A*Co s[c] - 20*B*Cos[c] + 36*C*Cos[c])*Csc[c])/(5*d) + (8*(B - 2*C)*Cos[d*x]*Si n[c])/(3*d) + (4*C*Cos[2*d*x]*Sin[2*c])/(5*d) + (2*Sec[c/2]*Sec[c/2 + (d*x )/2]^3*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) - (4*Sec[ c/2]*Sec[c/2 + (d*x)/2]*(2*A*Sin[(d*x)/2] - 3*B*Sin[(d*x)/2] + 4*C*Sin[(d* x)/2]))/d + (8*(B - 2*C)*Cos[c]*Sin[d*x])/(3*d) + (4*C*Cos[2*c]*Sin[2*d*x] )/(5*d) + (2*(A - B + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/(a + a*...
Time = 1.03 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.96, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 3520, 27, 3042, 3456, 3042, 3227, 3042, 3115, 3042, 3119, 3120}
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.
\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 3520 |
\(\displaystyle \frac {\int -\frac {\cos ^{\frac {5}{2}}(c+d x) (a (A-7 B+7 C)-a (5 A-5 B+11 C) \cos (c+d x))}{2 (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\cos ^{\frac {5}{2}}(c+d x) (a (A-7 B+7 C)-a (5 A-5 B+11 C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (A-7 B+7 C)-a (5 A-5 B+11 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3456 |
\(\displaystyle -\frac {\frac {\int \cos ^{\frac {3}{2}}(c+d x) \left (15 a^2 (A-2 B+3 C)-a^2 (20 A-35 B+56 C) \cos (c+d x)\right )dx}{a^2}+\frac {6 (A-2 B+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (15 a^2 (A-2 B+3 C)-a^2 (20 A-35 B+56 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {6 (A-2 B+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle -\frac {\frac {15 a^2 (A-2 B+3 C) \int \cos ^{\frac {3}{2}}(c+d x)dx-a^2 (20 A-35 B+56 C) \int \cos ^{\frac {5}{2}}(c+d x)dx}{a^2}+\frac {6 (A-2 B+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {15 a^2 (A-2 B+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-a^2 (20 A-35 B+56 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx}{a^2}+\frac {6 (A-2 B+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle -\frac {\frac {15 a^2 (A-2 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a^2 (20 A-35 B+56 C) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (A-2 B+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {15 a^2 (A-2 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a^2 (20 A-35 B+56 C) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (A-2 B+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {\frac {15 a^2 (A-2 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a^2 (20 A-35 B+56 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (A-2 B+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {\frac {15 a^2 (A-2 B+3 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a^2 (20 A-35 B+56 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (A-2 B+3 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*((A - B + C)*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]) ^2) - ((6*(A - 2*B + 3*C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + (15*a^2*(A - 2*B + 3*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + ( 2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)) - a^2*(20*A - 35*B + 56*C)*((6*E llipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d )))/a^2)/(6*a^2)
3.5.63.3.1 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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{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[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 0])
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[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x ] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c *(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c ^2 - d^2, 0] && LtQ[m, -2^(-1)]
Time = 11.30 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.29
method | result | size |
default | \(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-2 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (25 A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+60 A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-50 B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 B E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+75 C F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+168 C E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (25 A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+60 A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-50 B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 B E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+75 C F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+168 C E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+96 C \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-80 B -128 C \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-120 A +380 B -328 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (170 A -420 B +526 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-55 A +125 B -171 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{30 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(491\) |
int(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+cos(d*x+c)*a)^2,x, method=_RETURNVERBOSE)
1/30*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*(2*sin(1/ 2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(25*A*EllipticF(cos(1 /2*d*x+1/2*c),2^(1/2))+60*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-50*B*Ell ipticF(cos(1/2*d*x+1/2*c),2^(1/2))-105*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1 /2))+75*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+168*C*EllipticE(cos(1/2*d* x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*(2*sin(1/2*d* x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(25*A*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2))+60*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-50*B*Ellipti cF(cos(1/2*d*x+1/2*c),2^(1/2))-105*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)) +75*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+168*C*EllipticE(cos(1/2*d*x+1/ 2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)+96*C*sin(1/2*d*x+1/2*c)^10+(-80*B-128*C) *sin(1/2*d*x+1/2*c)^8+(-120*A+380*B-328*C)*sin(1/2*d*x+1/2*c)^6+(170*A-420 *B+526*C)*sin(1/2*d*x+1/2*c)^4+(-55*A+125*B-171*C)*sin(1/2*d*x+1/2*c)^2)/a ^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/ 2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 426, normalized size of antiderivative = 1.99 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\frac {2 \, {\left (6 \, C \cos \left (d x + c\right )^{3} + 2 \, {\left (5 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (30 \, A - 65 \, B + 94 \, C\right )} \cos \left (d x + c\right ) - 25 \, A + 50 \, B - 75 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 25 \, {\left (\sqrt {2} {\left (-i \, A + 2 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, A + 2 i \, B - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + 2 i \, B - 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 25 \, {\left (\sqrt {2} {\left (i \, A - 2 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, A - 2 i \, B + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - 2 i \, B + 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-20 i \, A + 35 i \, B - 56 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-20 i \, A + 35 i \, B - 56 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-20 i \, A + 35 i \, B - 56 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (20 i \, A - 35 i \, B + 56 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (20 i \, A - 35 i \, B + 56 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (20 i \, A - 35 i \, B + 56 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{30 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c) )^2,x, algorithm="fricas")
1/30*(2*(6*C*cos(d*x + c)^3 + 2*(5*B - 4*C)*cos(d*x + c)^2 - (30*A - 65*B + 94*C)*cos(d*x + c) - 25*A + 50*B - 75*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 25*(sqrt(2)*(-I*A + 2*I*B - 3*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-I*A + 2 *I*B - 3*I*C)*cos(d*x + c) + sqrt(2)*(-I*A + 2*I*B - 3*I*C))*weierstrassPI nverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 25*(sqrt(2)*(I*A - 2*I*B + 3*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(I*A - 2*I*B + 3*I*C)*cos(d*x + c) + sqr t(2)*(I*A - 2*I*B + 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*si n(d*x + c)) - 3*(sqrt(2)*(-20*I*A + 35*I*B - 56*I*C)*cos(d*x + c)^2 + 2*sq rt(2)*(-20*I*A + 35*I*B - 56*I*C)*cos(d*x + c) + sqrt(2)*(-20*I*A + 35*I*B - 56*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(20*I*A - 35*I*B + 56*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(20*I*A - 35*I*B + 56*I*C)*cos(d*x + c) + sqrt(2)*(20*I*A - 3 5*I*B + 56*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a ^2*d)
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c) )^2,x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(a*co s(d*x + c) + a)^2, x)
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
integrate(cos(d*x+c)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c) )^2,x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(5/2)/(a*co s(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]