Integrand size = 35, antiderivative size = 180 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {(A-9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {2 (2 A-3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(A-9 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
1/10*(A-9*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/6*(A+3*C)*Inver seJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d-1/5*(A+C)*cos(d*x+c)^(3/2)*sin(d*x +c)/d/(a+a*cos(d*x+c))^3+2/15*(2*A-3*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a +a*cos(d*x+c))^2-1/10*(A-9*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a^3+a^3*cos(d *x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.37 (sec) , antiderivative size = 1023, normalized size of antiderivative = 5.68 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:
Integrate[(Sqrt[Cos[c + d*x]]*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x]) ^3,x]
Output:
(-2*A*Cos[c/2 + (d*x)/2]^6*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]) ^3*Sqrt[1 + Cot[c]^2]) - (2*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Hypergeometric PFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - Arc Tan[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])^3*Sqrt[1 + Cot[c]^2]) + (Cos[c/2 + (d*x)/2]^6*Sqrt[ Cos[c + d*x]]*((-4*(A - 9*C)*Csc[c])/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2 ]*(A*Sin[(d*x)/2] - 9*C*Sin[(d*x)/2]))/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x) /2]^3*(A*Sin[(d*x)/2] - 9*C*Sin[(d*x)/2]))/(15*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*(A - 9*C)*Sec[c/2 + (d*x)/2]^2*Tan[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 - (A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2 ]*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Si n[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[ 1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[ 1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/S qrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[...
Time = 1.13 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3521, 27, 3042, 3456, 3042, 3457, 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.
\(\displaystyle \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3521 |
\(\displaystyle \frac {\int \frac {\sqrt {\cos (c+d x)} (a (7 A-3 C)-a (A-9 C) \cos (c+d x))}{2 (\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {\cos (c+d x)} (a (7 A-3 C)-a (A-9 C) \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a (7 A-3 C)-a (A-9 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3456 |
\(\displaystyle \frac {\frac {\int \frac {2 (2 A-3 C) a^2+(A+21 C) \cos (c+d x) a^2}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{3 a^2}+\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {2 (2 A-3 C) a^2+(A+21 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}+\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\frac {\int \frac {5 (A+3 C) a^3+3 (A-9 C) \cos (c+d x) a^3}{2 \sqrt {\cos (c+d x)}}dx}{a^2}-\frac {3 a^2 (A-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {\int \frac {5 (A+3 C) a^3+3 (A-9 C) \cos (c+d x) a^3}{\sqrt {\cos (c+d x)}}dx}{2 a^2}-\frac {3 a^2 (A-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \frac {5 (A+3 C) a^3+3 (A-9 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {3 a^2 (A-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^3 (A-9 C) \int \sqrt {\cos (c+d x)}dx}{2 a^2}-\frac {3 a^2 (A-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^3 (A-9 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}-\frac {3 a^2 (A-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^3 (A-9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {3 a^2 (A-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {\frac {10 a^3 (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^3 (A-9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {3 a^2 (A-9 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[(Sqrt[Cos[c + d*x]]*(A + C*Cos[c + d*x]^2))/(a + a*Cos[c + d*x])^3,x]
Output:
-1/5*((A + C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + ((4*a*(2*A - 3*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Cos[c + d *x])^2) + (((6*a^3*(A - 9*C)*EllipticE[(c + d*x)/2, 2])/d + (10*a^3*(A + 3 *C)*EllipticF[(c + d*x)/2, 2])/d)/(2*a^2) - (3*a^2*(A - 9*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_)*((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)]
Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(167)=334\).
Time = 3.48 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.51
method | result | size |
default | \(\frac {\sqrt {\left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\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 \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 ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+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 )-108 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-30 C \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 ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-54 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 )-22 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+198 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+6 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-114 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+27 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 {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(451\) |
Input:
int(cos(d*x+c)^(1/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x,method=_RETUR NVERBOSE)
Output:
1/60*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/ 2*d*x+1/2*c)^8-10*A*(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))*cos(1/2*d*x+1/2*c)^5+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))-108*C*cos(1/2*d*x+1/2*c)^8-3 0*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Ellipti cF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5-54*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)*Ellipti cE(cos(1/2*d*x+1/2*c),2^(1/2))-22*A*cos(1/2*d*x+1/2*c)^6+198*C*cos(1/2*d*x +1/2*c)^6+6*A*cos(1/2*d*x+1/2*c)^4-114*C*cos(1/2*d*x+1/2*c)^4+7*A*cos(1/2* d*x+1/2*c)^2+27*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)/(-1 +2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.57 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^(1/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algori thm="fricas")
Output:
-1/60*(2*(3*(A - 9*C)*cos(d*x + c)^2 + 4*(A - 9*C)*cos(d*x + c) - 5*A - 15 *C)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*(I*A + 3*I*C)*cos(d*x + c )^3 + 3*sqrt(2)*(I*A + 3*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A + 3*I*C)*cos (d*x + c) + sqrt(2)*(I*A + 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c)^3 + 3*sqrt(2)* (-I*A - 3*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c) + sq rt(2)*(-I*A - 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-I*A + 9*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A + 9*I*C )*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A + 9*I*C)*cos(d*x + c) + sqrt(2)*(-I*A + 9*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(I*A - 9*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I* A - 9*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A - 9*I*C)*cos(d*x + c) + sqrt(2) *(I*A - 9*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(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. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(1/2)*(A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**3,x)
Output:
Timed out
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:
integrate(cos(d*x+c)^(1/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algori thm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*sqrt(cos(d*x + c))/(a*cos(d*x + c) + a)^3 , x)
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:
integrate(cos(d*x+c)^(1/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x, algori thm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*sqrt(cos(d*x + c))/(a*cos(d*x + c) + a)^3 , x)
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:
int((cos(c + d*x)^(1/2)*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^3,x)
Output:
int((cos(c + d*x)^(1/2)*(A + C*cos(c + d*x)^2))/(a + a*cos(c + d*x))^3, x)
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx=\frac {\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) c}{a^{3}} \] Input:
int(cos(d*x+c)^(1/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^3,x)
Output:
(int(sqrt(cos(c + d*x))/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d *x) + 1),x)*a + int((sqrt(cos(c + d*x))*cos(c + d*x)**2)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*c)/a**3