Integrand size = 37, antiderivative size = 246 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\frac {(163 A+19 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(17 A+C) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {5 (19 A+3 C) \sin (c+d x)}{48 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(299 A+27 C) \sin (c+d x)}{48 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \] Output:
1/32*(163*A+19*C)*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/( a+a*cos(d*x+c))^(1/2))*2^(1/2)/a^(5/2)/d-1/4*(A+C)*sin(d*x+c)/d/cos(d*x+c) ^(3/2)/(a+a*cos(d*x+c))^(5/2)-1/16*(17*A+C)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2 )/(a+a*cos(d*x+c))^(3/2)+5/48*(19*A+3*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+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 = 8.55 (sec) , antiderivative size = 861, normalized size of antiderivative = 3.50 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
Integrate[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^ (5/2)),x]
Output:
-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]) - (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*Hypergeome tricPFQ[{2, 2, 2, 2, 7/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*S in[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 - 1280*Cos[(c + d*x)/2]^6*Hype rgeometricPFQ[{2, 2, 2, 7/2}, {1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*S in[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12*(-6 + 5*Sin[c/2 + (d*x)/2]^2) + 33*(1 - 2*Sin[c/2 + (d*x)/2]^2)^3*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[ c/2 + (d*x)/2]^2)]*(-105*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Cos[(c + d*x)/2]^4*(-10935 + 72902*Sin[c/2 + (d*x)/2]^2 - 188110*Sin[c/2 + (d*x)/2]^4 + 234156*Sin[c/2 + (d*x)/2]^6 - 140732*Sin[c/ 2 + (d*x)/2]^8 + 33208*Sin[c/2 + (d*x)/2]^10) + Sqrt[Sin[c/2 + (d*x)/2]^2/ (-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-1148175 + 10333785*Sin[c/2 + (d*x)/2]^2 - 38990350*Sin[c/2 + (d*x)/2]^4 + 79946462*Sin[c/2 + (d*x)/2]^6 - 96281836* Sin[c/2 + (d*x)/2]^8 + 68243596*Sin[c/2 + (d*x)/2]^10 - 26448512*Sin[c/2 + (d*x)/2]^12 + 4344400*Sin[c/2 + (d*x)/2]^14))))/(41580*d*(a*(1 + Cos[c + d*x]))^(5/2)*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2))
Time = 1.47 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {3042, 3521, 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.
| \(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3521 |
| \(\displaystyle \frac {\int \frac {a (11 A+3 C)-2 a (3 A-C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (11 A+3 C)-2 a (3 A-C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (11 A+3 C)-2 a (3 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {5 a^2 (19 A+3 C)-4 a^2 (17 A+C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {5 a^2 (19 A+3 C)-4 a^2 (17 A+C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {5 a^2 (19 A+3 C)-4 a^2 (17 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {\frac {2 \int -\frac {a^3 (299 A+27 C)-10 a^3 (19 A+3 C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (299 A+27 C)-10 a^3 (19 A+3 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a^3 (299 A+27 C)-10 a^3 (19 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {3 a^4 (163 A+19 C)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^3 (299 A+27 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (299 A+27 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a^3 (163 A+19 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (299 A+27 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a^3 (163 A+19 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {6 a^4 (163 A+19 C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a^3 (299 A+27 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {10 a^2 (19 A+3 C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a^3 (299 A+27 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {3 \sqrt {2} a^{5/2} (163 A+19 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {a (17 A+C) \sin (c+d x)}{2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}\) |
Input:
Int[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(5/2)) ,x]
Output:
-1/4*((A + 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 + C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((10*a^2*(19*A + 3*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 + 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 + 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)
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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(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)) I nt[(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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 = 4.16 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.16
| method | result | size |
| default | \(-\frac {\sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (489 \cos \left (d x +c \right )^{4}+1467 \cos \left (d x +c \right )^{3}+1467 \cos \left (d x +c \right )^{2}+489 \cos \left (d x +c \right )\right )+C \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (57 \cos \left (d x +c \right )^{4}+171 \cos \left (d x +c \right )^{3}+171 \cos \left (d x +c \right )^{2}+57 \cos \left (d x +c \right )\right )+\sin \left (d x +c \right ) \left (299 \cos \left (d x +c \right )^{3}+503 \cos \left (d x +c \right )^{2}+160 \cos \left (d x +c \right )-32\right ) \sqrt {2}\, A +\sin \left (d x +c \right ) \cos \left (d x +c \right )^{2} \left (27 \cos \left (d x +c \right )+39\right ) \sqrt {2}\, C \right )}{48 d \,a^{3} \cos \left (d x +c \right )^{\frac {3}{2}} \left (1+\cos \left (d x +c \right )\right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\) | \(285\) |
| parts | \(-\frac {A \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\sin \left (d x +c \right ) \left (299 \cos \left (d x +c \right )^{3}+503 \cos \left (d x +c \right )^{2}+160 \cos \left (d x +c \right )-32\right ) \sqrt {2}+\arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (489 \cos \left (d x +c \right )^{4}+1467 \cos \left (d x +c \right )^{3}+1467 \cos \left (d x +c \right )^{2}+489 \cos \left (d x +c \right )\right )\right )}{48 d \,a^{3} \cos \left (d x +c \right )^{\frac {3}{2}} \left (1+\cos \left (d x +c \right )\right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}-\frac {C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (2 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+11 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+19 \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )\right )}{32 d \,a^{3} \sqrt {\cos \left (d x +c \right )}}\) | \(333\) |
Input:
int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x,method=_R ETURNVERBOSE)
Output:
-1/48/d/a^3*(a*cos(1/2*d*x+1/2*c)^2)^(1/2)*(A*arcsin(-csc(d*x+c)+cot(d*x+c ))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(489*cos(d*x+c)^4+1467*cos(d*x+c)^3+1 467*cos(d*x+c)^2+489*cos(d*x+c))+C*arcsin(-csc(d*x+c)+cot(d*x+c))*(cos(d*x +c)/(1+cos(d*x+c)))^(1/2)*(57*cos(d*x+c)^4+171*cos(d*x+c)^3+171*cos(d*x+c) ^2+57*cos(d*x+c))+sin(d*x+c)*(299*cos(d*x+c)^3+503*cos(d*x+c)^2+160*cos(d* x+c)-32)*2^(1/2)*A+sin(d*x+c)*cos(d*x+c)^2*(27*cos(d*x+c)+39)*2^(1/2)*C)/c os(d*x+c)^(3/2)/(1+cos(d*x+c))/(cos(d*x+c)^2+2*cos(d*x+c)+1)
Time = 0.12 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.07 \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\frac {3 \, \sqrt {2} {\left ({\left (163 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (163 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (163 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (163 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (299 \, A + 27 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (503 \, A + 39 \, C\right )} \cos \left (d x + c\right )^{2} + 160 \, A \cos \left (d x + c\right ) - 32 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{96 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, al gorithm="fricas")
Output:
1/96*(3*sqrt(2)*((163*A + 19*C)*cos(d*x + c)^5 + 3*(163*A + 19*C)*cos(d*x + c)^4 + 3*(163*A + 19*C)*cos(d*x + c)^3 + (163*A + 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 + 27*C) *cos(d*x + c)^3 + (503*A + 39*C)*cos(d*x + c)^2 + 160*A*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*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(5/2),x)
Output:
Timed out
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, al gorithm="maxima")
Output:
Timed out
\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x, al gorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^ (5/2)), x)
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a*cos(c + d*x))^(5/2)) ,x)
Output:
int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a*cos(c + d*x))^(5/2)) , x)
\[ \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{6}+3 \cos \left (d x +c \right )^{5}+3 \cos \left (d x +c \right )^{4}+\cos \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}+3 \cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )}d x \right ) c \right )}{a^{3}} \] Input:
int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(5/2),x)
Output:
(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x)**6 + 3*cos(c + d*x)**5 + 3*cos(c + d*x)**4 + cos(c + d*x)**3),x)*a + int((sq rt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x)**4 + 3*cos(c + d*x) **3 + 3*cos(c + d*x)**2 + cos(c + d*x)),x)*c))/a**3