Integrand size = 43, antiderivative size = 175 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx=-\frac {(4 A-B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(5 A-2 B-C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {(4 A-B) \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {(5 A-2 B-C) \sin (c+d x)}{3 a^2 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \]
-(4*A-B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2 *d*x+1/2*c),2^(1/2))/a^2/d-1/3*(5*A-2*B-C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/co s(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d+(4*A-B)*sin(d *x+c)/a^2/d/cos(d*x+c)^(1/2)-1/3*(5*A-2*B-C)*sin(d*x+c)/a^2/d/(1+cos(d*x+c ))/cos(d*x+c)^(1/2)-1/3*(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2/cos(d*x+c) ^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.25 (sec) , antiderivative size = 1142, normalized size of antiderivative = 6.53 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx =\text {Too large to display} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2),x]
(10*A*Cos[c/2 + (d*x)/2]^4*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]) ^2*Sqrt[1 + Cot[c]^2]) - (4*B*Cos[c/2 + (d*x)/2]^4*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]]]])/( 3*d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2]) - (2*C*Cos[c/2 + (d*x)/2]^4 *Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2 ]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*S qrt[-(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])^2*Sqrt[1 + Cot[c]^2]) + ( Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*((2*(2*A + 2*A*Cos[c] - B*Cos[c])* Csc[c/2]*Sec[c/2]*Sec[c])/d + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(2*A*Sin[(d*x )/2] - B*Sin[(d*x)/2]))/d + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/ 2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) + (8*A*Sec[c]*Sec[c + d*x]*Si n[d*x])/d + (2*(A - B + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/(a + a*C os[c + d*x])^2 + (4*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*((Hypergeomet ricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcT...
Time = 1.00 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 3520, 27, 3042, 3457, 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.
\(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 3520 |
\(\displaystyle \frac {\int \frac {a (7 A-B+C)-3 a (A-B-C) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a (7 A-B+C)-3 a (A-B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a (7 A-B+C)-3 a (A-B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\int \frac {3 a^2 (4 A-B)-a^2 (5 A-2 B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {2 (5 A-2 B-C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {3 a^2 (4 A-B)-a^2 (5 A-2 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{a^2}-\frac {2 (5 A-2 B-C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {3 a^2 (4 A-B) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx-a^2 (5 A-2 B-C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}-\frac {2 (5 A-2 B-C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 a^2 (4 A-B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx-a^2 (5 A-2 B-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {2 (5 A-2 B-C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {\frac {3 a^2 (4 A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )-a^2 (5 A-2 B-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {2 (5 A-2 B-C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 a^2 (4 A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-a^2 (5 A-2 B-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {2 (5 A-2 B-C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {3 a^2 (4 A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-a^2 (5 A-2 B-C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {2 (5 A-2 B-C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {3 a^2 (4 A-B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^2 (5 A-2 B-C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}-\frac {2 (5 A-2 B-C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\) |
-1/3*((A - B + C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x]) ^2) + ((-2*(5*A - 2*B - C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])) + ((-2*a^2*(5*A - 2*B - C)*EllipticF[(c + d*x)/2, 2])/d + 3*a^2*( 4*A - B)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[ c + d*x]])))/a^2)/(6*a^2)
3.5.67.3.1 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[((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_.) + (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. \(562\) vs. \(2(215)=430\).
Time = 8.54 (sec) , antiderivative size = 563, normalized size of antiderivative = 3.22
method | result | size |
default | \(-\frac {\sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (5 A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 B E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-C F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (5 A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 B E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-C F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (4 A -B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (43 A -10 B +C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (37 A -7 B +C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{6 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-1+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(563\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+cos(d*x+c)*a)^2,x, method=_RETURNVERBOSE)
-1/6*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*(2*(-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)^2-1 )^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^( 1/2))-12*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-2*B*EllipticF(cos(1/2*d*x +1/2*c),2^(1/2))+3*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-C*EllipticF(cos (1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2*(-2*si n(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)^2-1)^ (1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/ 2))-12*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-2*B*EllipticF(cos(1/2*d*x+1 /2*c),2^(1/2))+3*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-C*EllipticF(cos(1 /2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*(-2*sin(1/2*d*x+1/2*c)^4+sin (1/2*d*x+1/2*c)^2)^(1/2)*(4*A-B)*sin(1/2*d*x+1/2*c)^6+2*(-2*sin(1/2*d*x+1/ 2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(43*A-10*B+C)*sin(1/2*d*x+1/2*c)^4-(-2* sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(37*A-7*B+C)*sin(1/2*d*x+ 1/2*c)^2)/sin(1/2*d*x+1/2*c)^3/(2*sin(1/2*d*x+1/2*c)^2-1)/cos(1/2*d*x+1/2* c)/(-1+sin(1/2*d*x+1/2*c)^2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.43 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx=\frac {2 \, {\left (3 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (19 \, A - 4 \, B + C\right )} \cos \left (d x + c\right ) + 6 \, A\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (5 i \, A - 2 i \, B - i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (-5 i \, A + 2 i \, B + i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (5 i \, A - 2 i \, B - 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 ) + {\left (\sqrt {2} {\left (-5 i \, A + 2 i \, B + i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (5 i \, A - 2 i \, B - i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-5 i \, A + 2 i \, B + 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 (4 i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (4 i \, A - i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (4 i \, A - i \, B\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 (-4 i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (-4 i \, A + i \, B\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-4 i \, A + i \, B\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 )}{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 )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c) )^2,x, algorithm="fricas")
1/6*(2*(3*(4*A - B)*cos(d*x + c)^2 + (19*A - 4*B + C)*cos(d*x + c) + 6*A)* sqrt(cos(d*x + c))*sin(d*x + c) + (sqrt(2)*(5*I*A - 2*I*B - I*C)*cos(d*x + c)^3 - 2*sqrt(2)*(-5*I*A + 2*I*B + I*C)*cos(d*x + c)^2 + sqrt(2)*(5*I*A - 2*I*B - I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*si n(d*x + c)) + (sqrt(2)*(-5*I*A + 2*I*B + I*C)*cos(d*x + c)^3 - 2*sqrt(2)*( 5*I*A - 2*I*B - I*C)*cos(d*x + c)^2 + sqrt(2)*(-5*I*A + 2*I*B + I*C)*cos(d *x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(sq rt(2)*(4*I*A - I*B)*cos(d*x + c)^3 + 2*sqrt(2)*(4*I*A - I*B)*cos(d*x + c)^ 2 + sqrt(2)*(4*I*A - I*B)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrass PInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-4*I*A + I*B )*cos(d*x + c)^3 + 2*sqrt(2)*(-4*I*A + I*B)*cos(d*x + c)^2 + sqrt(2)*(-4*I *A + I*B)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(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. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c) )^2,x, algorithm="maxima")
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c) )^2,x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^2* cos(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]