Integrand size = 35, antiderivative size = 218 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {(A-49 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {(A-13 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3 \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 (A-4 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)}+\frac {(A-13 C) \sin (c+d x)}{6 d \left (a^3+a^3 \cos (c+d x)\right ) \sqrt {\sec (c+d x)}} \] Output:
-1/10*(A-49*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^3/d+1/6*(A-13*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^3/d-1/5*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c ))^3/sec(d*x+c)^(5/2)+2/15*(A-4*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2/sec(d *x+c)^(3/2)+1/6*(A-13*C)*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))/sec(d*x+c)^(1/2 )
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.84 (sec) , antiderivative size = 813, normalized size of antiderivative = 3.73 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
Integrate[(A + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])^3*Sec[c + d*x]^(3/2 )),x]
Output:
(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]^6*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])/(15*d*E^(I*d*x)*(a + a*Cos[c + d*x])^3) - (49*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]^6*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])/(15*d*E^(I*d*x)*(a + a*Cos[c + d*x])^3) + (2*A*Cos[c/2 + (d*x)/2]^6*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])^3) - (26*C*Cos[c/2 + (d*x)/2]^6*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])^3) + (Cos[c/2 + (d*x)/2]^6*Sqrt[Sec[c + d*x]]*((-2*(-A + 39*C + 10*C*Cos[2*c] )*Cos[d*x]*Csc[c/2]*Sec[c/2])/(5*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A* Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*( 7*A*Sin[(d*x)/2] + 17*C*Sin[(d*x)/2]))/(15*d) + (4*Sec[c/2]*Sec[c/2 + (d*x )/2]*(A*Sin[(d*x)/2] + 23*C*Sin[(d*x)/2]))/(3*d) + (16*C*Cos[c]*Sin[d*x])/ d + (4*(A + 23*C)*Tan[c/2])/(3*d) - (4*(7*A + 17*C)*Sec[c/2 + (d*x)/2]^2*T an[c/2])/(15*d) + (2*(A + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(a + a *Cos[c + d*x])^3
Time = 1.34 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.98, 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, 3456, 3042, 3456, 27, 3042, 3227, 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.
| \(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \cos (c+d x)^2}{\sec (c+d x)^{3/2} (a \cos (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (C \cos ^2(c+d x)+A\right )}{(\cos (c+d x) a+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx\) |
| \(\Big \downarrow \) 3521 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) (5 a (A-C)+a (A+11 C) \cos (c+d x))}{2 (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) (5 a (A-C)+a (A+11 C) \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (5 a (A-C)+a (A+11 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {\cos (c+d x)} \left (6 (A-4 C) a^2+(A+41 C) \cos (c+d x) a^2\right )}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (6 (A-4 C) a^2+(A+41 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int \frac {5 a^3 (A-13 C)-3 a^3 (A-49 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx}{a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int \frac {5 a^3 (A-13 C)-3 a^3 (A-49 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int \frac {5 a^3 (A-13 C)-3 a^3 (A-49 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {5 a^3 (A-13 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^3 (A-49 C) \int \sqrt {\cos (c+d x)}dx}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {5 a^3 (A-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^3 (A-49 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {5 a^3 (A-13 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a^3 (A-49 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}+\frac {5 a^2 (A-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {5 a^2 (A-13 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}+\frac {\frac {10 a^3 (A-13 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^3 (A-49 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}}{3 a^2}+\frac {4 a (A-4 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )\) |
Input:
Int[(A + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])^3*Sec[c + d*x]^(3/2)),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/5*((A + C)*Cos[c + d*x]^(5/2)*Si n[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + ((4*a*(A - 4*C)*Cos[c + d*x]^(3/2 )*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (((-6*a^3*(A - 49*C)*Ellipt icE[(c + d*x)/2, 2])/d + (10*a^3*(A - 13*C)*EllipticF[(c + d*x)/2, 2])/d)/ (2*a^2) + (5*a^2*(A - 13*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos [c + d*x])))/(3*a^2))/(10*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[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_.) + (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. \(450\) vs. \(2(197)=394\).
Time = 5.72 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.07
| method | result | size |
| default | \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+10 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+6 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-348 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-130 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-294 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} A +578 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} A -264 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+17 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+37 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 A -3 C \right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(451\) |
Input:
int((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3/sec(d*x+c)^(3/2),x,method=_RETUR NVERBOSE)
Output:
-1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/ 2*d*x+1/2*c)^8+10*A*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2* cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*A*co s(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1 )^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-348*C*cos(1/2*d*x+1/2*c)^8-1 30*C*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2 *c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-294*C*cos(1/2*d*x+1/2 *c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Ellip ticE(cos(1/2*d*x+1/2*c),2^(1/2))-2*cos(1/2*d*x+1/2*c)^6*A+578*C*cos(1/2*d* x+1/2*c)^6-24*cos(1/2*d*x+1/2*c)^4*A-264*C*cos(1/2*d*x+1/2*c)^4+17*A*cos(1 /2*d*x+1/2*c)^2+37*C*cos(1/2*d*x+1/2*c)^2-3*A-3*C)/a^3/cos(1/2*d*x+1/2*c)^ 5/(-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 complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 474, normalized size of antiderivative = 2.17 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3/sec(d*x+c)^(3/2),x, algori thm="fricas")
Output:
-1/60*(5*(sqrt(2)*(I*A - 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A - 13*I*C) *cos(d*x + c)^2 + 3*sqrt(2)*(I*A - 13*I*C)*cos(d*x + c) + sqrt(2)*(I*A - 1 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqr t(2)*(-I*A + 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A + 13*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A + 13*I*C)*cos(d*x + c) + sqrt(2)*(-I*A + 13*I*C))*w eierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(I*A - 49*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A - 49*I*C)*cos(d*x + c)^2 + 3*sq rt(2)*(I*A - 49*I*C)*cos(d*x + c) + sqrt(2)*(I*A - 49*I*C))*weierstrassZet a(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(s qrt(2)*(-I*A + 49*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A + 49*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A + 49*I*C)*cos(d*x + c) + sqrt(2)*(-I*A + 49*I*C)) *weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* x + c))) - 2*(3*(A - 29*C)*cos(d*x + c)^3 + 2*(7*A - 73*C)*cos(d*x + c)^2 + 5*(A - 13*C)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d *x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**3/sec(d*x+c)**(3/2),x)
Output:
Timed out
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3/sec(d*x+c)^(3/2),x, algori thm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^3*sec(d*x + c)^(3/2 )), x)
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3/sec(d*x+c)^(3/2),x, algori thm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)/((a*cos(d*x + c) + a)^3*sec(d*x + c)^(3/2 )), x)
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:
int((A + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^3) ,x)
Output:
int((A + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))^3) , x)
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^3 \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}+3 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}+3 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )^{2}}d x \right ) c}{a^{3}} \] Input:
int((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3/sec(d*x+c)^(3/2),x)
Output:
(int(sqrt(sec(c + d*x))/(cos(c + d*x)**3*sec(c + d*x)**2 + 3*cos(c + d*x)* *2*sec(c + d*x)**2 + 3*cos(c + d*x)*sec(c + d*x)**2 + sec(c + d*x)**2),x)* a + int((sqrt(sec(c + d*x))*cos(c + d*x)**2)/(cos(c + d*x)**3*sec(c + d*x) **2 + 3*cos(c + d*x)**2*sec(c + d*x)**2 + 3*cos(c + d*x)*sec(c + d*x)**2 + sec(c + d*x)**2),x)*c)/a**3