Integrand size = 45, antiderivative size = 261 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\frac {(163 A-75 B+19 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(17 A-9 B+C) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(95 A-39 B+15 C) \sin (c+d x)}{48 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(299 A-147 B+27 C) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]
-1/4*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(5/2)-1/16*(17 *A-9*B+C)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2)+1/32*(163 *A-75*B+19*C)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a* cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+1/48*(95*A-39*B+15*C)*sin(d*x+c)/a^2/ d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)-1/48*(299*A-147*B+27*C)*sin(d*x+ c)/a^2/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 9.18 (sec) , antiderivative size = 1368, normalized size of antiderivative = 5.24 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx =\text {Too large to display} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(5/2)),x]
-1/4*(C*Cos[c/2 + (d*x)/2]^5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*(11 - 3 1*Sin[c/2 + (d*x)/2]^2 + 18*Sin[c/2 + (d*x)/2]^4 - (19*ArcTanh[Sqrt[-(Sin[ c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]]*Cos[(c + d*x)/2]^4)/Sqrt[ -(Sin[c/2 + (d*x)/2]^2/(1 - 2*Sin[c/2 + (d*x)/2]^2))]))/(d*(a*(1 + Cos[c + d*x]))^(5/2)*Sqrt[1 - 2*Sin[c/2 + (d*x)/2]^2]) + (2*B*Cos[c/2 + (d*x)/2]^ 5*Sec[(c + d*x)/2]^4*Sin[c/2 + (d*x)/2]*((8*Cos[(c + d*x)/2]^6*Hypergeomet ricPFQ[{2, 2, 2, 5/2}, {1, 1, 11/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^2)/(315*(-1 + 2*Sin[c/2 + (d*x)/2]^2)) + (Csc[c/2 + (d*x)/2]^8*(1 - 2*Sin[c/2 + (d*x)/2]^2)^2*Sqrt[Sin[c/2 + (d*x) /2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-15*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^ 2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Cos[(c + d*x)/2]^4*(-343 + 1465*Sin[c/2 + (d*x)/2]^2 - 2021*Sin[c/2 + (d*x)/2]^4 + 824*Sin[c/2 + (d*x)/2]^6) + Sqr t[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-5145 + 33980*Sin[c /2 + (d*x)/2]^2 - 87764*Sin[c/2 + (d*x)/2]^4 + 109737*Sin[c/2 + (d*x)/2]^6 - 66122*Sin[c/2 + (d*x)/2]^8 + 15344*Sin[c/2 + (d*x)/2]^10)))/120))/(d*(a *(1 + Cos[c + d*x]))^(5/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(3/2)) - (A*Cot[c/ 2 + (d*x)/2]^5*Csc[c/2 + (d*x)/2]^4*Sec[(c + d*x)/2]^4*(640*Cos[(c + d*x)/ 2]^8*HypergeometricPFQ[{2, 2, 2, 2, 7/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x) /2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 - 1280*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 7/2}, {1, 1, 13/2}, Sin[c/2 + (d...
Time = 1.50 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3520, 27, 3042, 3457, 27, 3042, 3463, 27, 3042, 3463, 27, 3042, 3261, 218}
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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {a (11 A-3 B+3 C)-2 a (3 A-3 B-C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (11 A-3 B+3 C)-2 a (3 A-3 B-C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (11 A-3 B+3 C)-2 a (3 A-3 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (95 A-39 B+15 C)-4 a^2 (17 A-9 B+C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a (17 A-9 B+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (95 A-39 B+15 C)-4 a^2 (17 A-9 B+C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a (17 A-9 B+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (95 A-39 B+15 C)-4 a^2 (17 A-9 B+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (17 A-9 B+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {\frac {2 \int -\frac {a^3 (299 A-147 B+27 C)-2 a^3 (95 A-39 B+15 C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 a^2 (95 A-39 B+15 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {a (17 A-9 B+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B+15 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (299 A-147 B+27 C)-2 a^3 (95 A-39 B+15 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A-9 B+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B+15 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (299 A-147 B+27 C)-2 a^3 (95 A-39 B+15 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A-9 B+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B+15 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {3 a^4 (163 A-75 B+19 C)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^3 (299 A-147 B+27 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{4 a^2}-\frac {a (17 A-9 B+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B+15 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (299 A-147 B+27 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a^3 (163 A-75 B+19 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A-9 B+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B+15 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (299 A-147 B+27 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a^3 (163 A-75 B+19 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A-9 B+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B+15 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {6 a^4 (163 A-75 B+19 C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^3 (299 A-147 B+27 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{4 a^2}-\frac {a (17 A-9 B+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B+15 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (299 A-147 B+27 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {3 \sqrt {2} a^{5/2} (163 A-75 B+19 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {a (17 A-9 B+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Cos [c + d*x])^(5/2)),x]
-1/4*((A - B + C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x]) ^(5/2)) + (-1/2*(a*(17*A - 9*B + C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((2*a^2*(95*A - 39*B + 15*C)*Sin[c + d*x])/(3* d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((-3*Sqrt[2]*a^(5/2)*(163 *A - 75*B + 19*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x] ]*Sqrt[a + a*Cos[c + d*x]])])/d + (2*a^3*(299*A - 147*B + 27*C)*Sin[c + d* x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]))/(3*a))/(4*a^2))/(8*a^ 2)
3.6.21.3.1 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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*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] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && 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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 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)]
Leaf count of result is larger than twice the leaf count of optimal. \(762\) vs. \(2(224)=448\).
Time = 13.63 (sec) , antiderivative size = 763, normalized size of antiderivative = 2.92
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(763\) |
| parts | \(\text {Expression too large to display}\) | \(778\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*a)^(5/2 ),x,method=_RETURNVERBOSE)
-1/96/a^3/d*(489*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot(d* x+c)-csc(d*x+c))*cos(d*x+c)^4-225*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1 /2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)^4+57*2^(1/2)*(cos(d*x+c)/(1+c os(d*x+c)))^(1/2)*C*cos(d*x+c)^4*arcsin(cot(d*x+c)-csc(d*x+c))+1467*A*cos( d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*arcsin(cot(d*x+c)-csc(d *x+c))-675*B*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*arcsin (cot(d*x+c)-csc(d*x+c))+171*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/ 2)*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+1467*A*(cos(d*x+c)/(1+cos(d*x+c)) )^(1/2)*2^(1/2)*cos(d*x+c)^2*arcsin(cot(d*x+c)-csc(d*x+c))-675*B*(cos(d*x+ c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*cos(d*x+c)^2*arcsin(cot(d*x+c)-csc(d*x+c) )+171*C*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcsin(cot( d*x+c)-csc(d*x+c))+489*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*arcs in(cot(d*x+c)-csc(d*x+c))*2^(1/2)+598*A*cos(d*x+c)^3*sin(d*x+c)-225*B*(cos (d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*arcsin(cot(d*x+c)-csc(d*x+c))*2^( 1/2)-294*B*sin(d*x+c)*cos(d*x+c)^3+57*C*cos(d*x+c)*2^(1/2)*(cos(d*x+c)/(1+ cos(d*x+c)))^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+54*C*cos(d*x+c)^3*sin(d*x +c)+1006*A*sin(d*x+c)*cos(d*x+c)^2-510*B*sin(d*x+c)*cos(d*x+c)^2+78*C*cos( d*x+c)^2*sin(d*x+c)+320*A*sin(d*x+c)*cos(d*x+c)-192*B*sin(d*x+c)*cos(d*x+c )-64*A*sin(d*x+c))*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))^3/cos(d*x+c)^(3 /2)
Time = 0.33 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.10 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\frac {3 \, \sqrt {2} {\left ({\left (163 \, A - 75 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (163 \, A - 75 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (163 \, A - 75 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (163 \, A - 75 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (299 \, A - 147 \, B + 27 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (503 \, A - 255 \, B + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 32 \, {\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 32 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{96 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c) )^(5/2),x, algorithm="fricas")
1/96*(3*sqrt(2)*((163*A - 75*B + 19*C)*cos(d*x + c)^5 + 3*(163*A - 75*B + 19*C)*cos(d*x + c)^4 + 3*(163*A - 75*B + 19*C)*cos(d*x + c)^3 + (163*A - 7 5*B + 19*C)*cos(d*x + c)^2)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d* x + c))) - 2*((299*A - 147*B + 27*C)*cos(d*x + c)^3 + (503*A - 255*B + 39* C)*cos(d*x + c)^2 + 32*(5*A - 3*B)*cos(d*x + c) - 32*A)*sqrt(a*cos(d*x + c ) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*co s(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c) )^(5/2),x, algorithm="maxima")
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c) )^(5/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^(5 /2)*cos(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a*cos (c + d*x))^(5/2)),x)