Integrand size = 35, antiderivative size = 229 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {(7 A+C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {2 (5 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(7 A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {2 (5 A+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac {(7 A+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:
(7*A+C)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^ (1/2)/a^2/d+2/3*(5*A+C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^( 1/2))*sec(d*x+c)^(1/2)/a^2/d-(7*A+C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a^2/d+2/3 *(5*A+C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d-1/3*(7*A+C)*sec(d*x+c)^(3/2)*si n(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A+C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a *cos(d*x+c))^2
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.51 (sec) , antiderivative size = 734, normalized size of antiderivative = 3.21 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:
Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/(a + a*Cos[c + d*x]) ^2,x]
Output:
(-7*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^( (2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4 , -E^((2*I)*(c + d*x))])*Sec[c/2])/(3*d*E^(I*d*x)*(a + a*Cos[c + d*x])^2) - (Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^(( 2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2])/(3*d*E^(I*d*x)*(a + a*Cos[c + d*x])^2) + (20*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x )/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(3*d*(a + a*Cos[c + d*x])^2) + (4*C*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x) /2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(3*d*(a + a*Cos[c + d*x])^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Sec[c + d*x]]*((-2*(7*A + C)*Cos[d*x]*Csc[c/2]* Sec[c/2])/d + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d* x)/2]))/(3*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(4*A*Sin[(d*x)/2] + C*Sin[( d*x)/2]))/(3*d) + (8*A*Sec[c]*Sec[c + d*x]*Sin[d*x])/(3*d) + (8*(A + 5*A*C os[c] + C*Cos[c])*Sec[c]*Tan[c/2])/(3*d) + (2*(A + C)*Sec[c/2 + (d*x)/2]^2 *Tan[c/2])/(3*d)))/(a + a*Cos[c + d*x])^2
Time = 1.19 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.92, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4709, 3042, 3521, 27, 3042, 3457, 27, 3042, 3227, 3042, 3116, 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.
\(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^{5/2} \left (A+C \cos (c+d x)^2\right )}{(a \cos (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 4709 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+A}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx\) |
\(\Big \downarrow \) 3521 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {3 a (3 A+C)-a (5 A-C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {3 a (3 A+C)-a (5 A-C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {3 a (3 A+C)-a (5 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {3 \left (2 a^2 (5 A+C)-a^2 (7 A+C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 \int \frac {2 a^2 (5 A+C)-a^2 (7 A+C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 \int \frac {2 a^2 (5 A+C)-a^2 (7 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 \left (2 a^2 (5 A+C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx-a^2 (7 A+C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 \left (2 a^2 (5 A+C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-a^2 (7 A+C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 \left (2 a^2 (5 A+C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^2 (7 A+C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 \left (2 a^2 (5 A+C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^2 (7 A+C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 \left (2 a^2 (5 A+C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^2 (7 A+C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3 \left (2 a^2 (5 A+C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )-a^2 (7 A+C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}\right )\) |
Input:
Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/(a + a*Cos[c + d*x])^2,x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/3*((A + C)*Sin[c + d*x])/(d*Cos[ c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2) + ((-2*(7*A + C)*Sin[c + d*x])/(d*C os[c + d*x]^(3/2)*(1 + Cos[c + d*x])) + (3*(2*a^2*(5*A + C)*((2*EllipticF[ (c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))) - a^2* (7*A + C)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos [c + d*x]]))))/a^2)/(6*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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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[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_)*((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)]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(710\) vs. \(2(208)=416\).
Time = 26.96 (sec) , antiderivative size = 711, normalized size of antiderivative = 3.10
Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^2,x,method=_RETUR NVERBOSE)
Output:
-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*(1/3*(A +C)*(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)*(2*El lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos(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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 /2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))- 3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x +1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2* c)/(-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 )^2-1)+4*A*(cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x +1/2*c)^2-1)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/ 2*d*x+1/2*c),2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)/cos(1/ 2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+4*A*(-1/ 6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/ (cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2* d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2) *EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-8*A/sin(1/2*d*x+1/2*c)^2/(2*sin(1/ 2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2* sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip ticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(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 complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.82 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (5 i \, A + i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (5 i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (5 i \, A + i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (\sqrt {2} {\left (-5 i \, A - i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (-5 i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-5 i \, A - i \, C\right )} \cos \left (d x + 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 (-7 i \, A - i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (-7 i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-7 i \, A - i \, C\right )} \cos \left (d x + 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 (7 i \, A + i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (7 i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (7 i \, A + i \, C\right )} \cos \left (d x + 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 ) + \frac {2 \, {\left (3 \, {\left (7 \, A + C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (8 \, A + C\right )} \cos \left (d x + c\right )^{2} + 8 \, A \cos \left (d x + c\right ) - 2 \, A\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^2,x, algori thm="fricas")
Output:
-1/6*(2*(sqrt(2)*(5*I*A + I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(5*I*A + I*C)*co s(d*x + c)^2 + sqrt(2)*(5*I*A + I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 2*(sqrt(2)*(-5*I*A - I*C)*cos(d*x + c )^3 + 2*sqrt(2)*(-5*I*A - I*C)*cos(d*x + c)^2 + sqrt(2)*(-5*I*A - I*C)*cos (d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*( sqrt(2)*(-7*I*A - I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(-7*I*A - I*C)*cos(d*x + c)^2 + sqrt(2)*(-7*I*A - I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weiers trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(7*I*A + I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(7*I*A + I*C)*cos(d*x + c)^2 + sqrt(2)*(7 *I*A + I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0 , cos(d*x + c) - I*sin(d*x + c))) + 2*(3*(7*A + C)*cos(d*x + c)^3 + 4*(8*A + C)*cos(d*x + c)^2 + 8*A*cos(d*x + c) - 2*A)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**(5/2)/(a+a*cos(d*x+c))**2,x)
Output:
Timed out
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^2,x, algori thm="maxima")
Output:
Timed out
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^2,x, algori thm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^2 , x)
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:
int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2))/(a + a*cos(c + d*x))^2 ,x)
Output:
int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(5/2))/(a + a*cos(c + d*x))^2 , x)
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1}d x \right ) a}{a^{2}} \] Input:
int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^2,x)
Output:
(int((sqrt(sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**2)/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x))*sec(c + d*x)**2)/(co s(c + d*x)**2 + 2*cos(c + d*x) + 1),x)*a)/a**2