Integrand size = 45, antiderivative size = 360 \[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (2 A b^2-5 a b B-3 a^2 (3 A+5 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a^3 d}-\frac {2 (a-b) \sqrt {a+b} (2 A b+a (9 A-5 B+15 C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a^2 d}+\frac {2 A \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (A b+5 a B) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 a d \cos ^{\frac {3}{2}}(c+d x)} \]
2/5*A*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)+2/15*(A*b+5*B*a )*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/a/d/cos(d*x+c)^(3/2)-2/15*(a-b)*(2*A*b ^2-5*B*a*b-3*a^2*(3*A+5*C))*cot(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1/2)/(a +b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*(a*(1-sec(d*x +c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/d-2/15*(a-b)*(2*A*b+a *(9*A-5*B+15*C))*cot(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/c os(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*(a*(1-sec(d*x+c))/(a+b)) ^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d
Result contains complex when optimal does not.
Time = 7.10 (sec) , antiderivative size = 1340, normalized size of antiderivative = 3.72 \[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx =\text {Too large to display} \]
Integrate[(Sqrt[a + b*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2 ))/Cos[c + d*x]^(7/2),x]
-1/15*((-4*a*(2*a^2*A*b - 2*A*b^3 - 5*a^3*B + 5*a*b^2*B)*Sqrt[((a + b)*Cot [(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2 )/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*Ellip ticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (- 2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b* Cos[c + d*x]]) - 4*a*(9*a^3*A - 2*a*A*b^2 + 5*a^2*b*B + 15*a^3*C)*((Sqrt[( (a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/S qrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]* Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*S qrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d *x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b), ArcSin[Sqrt[(( a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[ (c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])) + 2*(9*a^ 2*A*b - 2*A*b^3 + 5*a*b^2*B + 15*a^2*b*C)*((I*Cos[(c + d*x)/2]*Sqrt[a + b* Cos[c + d*x]]*EllipticE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x]]], (- 2*a)/(-a - b)]*Sec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt [((a + b*Cos[c + d*x])*Sec[c + d*x])/(a + b)]) + (2*a*((a*Sqrt[((a + b)*Co t[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/...
Time = 1.52 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 3526, 27, 3042, 3534, 27, 3042, 3477, 3042, 3295, 3473}
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.
\(\displaystyle \int \frac {\sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
\(\Big \downarrow \) 3526 |
\(\displaystyle \frac {2}{5} \int \frac {b (2 A+5 C) \cos ^2(c+d x)+(3 a A+5 b B+5 a C) \cos (c+d x)+A b+5 a B}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {b (2 A+5 C) \cos ^2(c+d x)+(3 a A+5 b B+5 a C) \cos (c+d x)+A b+5 a B}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {b (2 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(3 a A+5 b B+5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+A b+5 a B}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {1}{5} \left (\frac {2 \int -\frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \cos (c+d x) a+2 A b^2}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \cos (c+d x) a+2 A b^2}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+(a-b) (a (9 A-5 B+15 C)+2 A b) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+(a-b) (a (9 A-5 B+15 C)+2 A b) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) (a (9 A-5 B+15 C)+2 A b) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) (a (9 A-5 B+15 C)+2 A b) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
(2*A*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (-1 /3*((2*(a - b)*Sqrt[a + b]*(2*A*b^2 - 5*a*b*B - 3*a^2*(3*A + 5*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a* (1 + Sec[c + d*x]))/(a - b)])/(a^2*d) + (2*(a - b)*Sqrt[a + b]*(2*A*b + a* (9*A - 5*B + 15*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]] /(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d))/a + (2*(A* b + 5*a*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2 )))/5
3.12.20.3.1 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[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e , f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A, B]
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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d , 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
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*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Leaf count of result is larger than twice the leaf count of optimal. \(4059\) vs. \(2(328)=656\).
Time = 23.14 (sec) , antiderivative size = 4060, normalized size of antiderivative = 11.28
method | result | size |
parts | \(\text {Expression too large to display}\) | \(4060\) |
default | \(\text {Expression too large to display}\) | \(4476\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2)/cos(d*x+c)^(7/2 ),x,method=_RETURNVERBOSE)
-2/15*A/d/a^2*(-3*a^3*cos(d*x+c)*sin(d*x+c)+2*b^3*cos(d*x+c)^3*sin(d*x+c)- a*b^2*cos(d*x+c)^3*sin(d*x+c)-4*a^2*b*cos(d*x+c)^2*sin(d*x+c)+a*b^2*cos(d* x+c)^2*sin(d*x+c)-4*a^2*b*cos(d*x+c)*sin(d*x+c)-9*(cos(d*x+c)/(1+cos(d*x+c )))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(cot(d* x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*cos(d*x+c)^2+2*(cos(d*x+c)/(1+co s(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE (cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*b^3*cos(d*x+c)^2+7*(cos(d*x+c )/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*El lipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b*cos(d*x+c)^2-2*( cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c))) ^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^2*cos(d*x +c)^2-9*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos (d*x+c)))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2* b*cos(d*x+c)^2+2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c ))/(1+cos(d*x+c)))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1 /2))*a*b^2*cos(d*x+c)^2-3*a^3*sin(d*x+c)+2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/ 2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(cot(d*x+c)-cs c(d*x+c),(-(a-b)/(a+b))^(1/2))*b^3*cos(d*x+c)^4+18*(cos(d*x+c)/(1+cos(d*x+ c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(cot(d *x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*cos(d*x+c)^3+7*(cos(d*x+c)/(...
\[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2)/cos(d*x+c )^(7/2),x, algorithm="fricas")
integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)/ cos(d*x + c)^(7/2), x)
Timed out. \[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2)/cos(d*x+c )^(7/2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a) /cos(d*x + c)^(7/2), x)
\[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2)/cos(d*x+c )^(7/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a) /cos(d*x + c)^(7/2), x)
Timed out. \[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+b\,\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^{7/2}} \,d x \]
int(((a + b*cos(c + d*x))^(1/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/c os(c + d*x)^(7/2),x)