Integrand size = 42, antiderivative size = 202 \[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \left (5 a^3 B-15 a b^2 B-15 a^2 b C-3 b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (9 a^2 b B+b^3 B+3 a^3 C+3 a b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {2 b \left (6 a^2 B-b^2 B-3 a b C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}-\frac {2 b^2 (5 a B-b C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]
-2/5*(5*B*a^3-15*B*a*b^2-15*C*a^2*b-3*C*b^3)*(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))/d+2/3*(9*B*a^2*b+ B*b^3+3*C*a^3+3*C*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E llipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d-2/5*b^2*(5*B*a-C*b)*cos(d*x+c)^(3/2 )*sin(d*x+c)/d+2*a*B*(a+b*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(1/2)-2/3* b*(6*B*a^2-B*b^2-3*C*a*b)*sin(d*x+c)*cos(d*x+c)^(1/2)/d
Time = 2.79 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\left (-30 a^3 B+90 a b^2 B+90 a^2 b C+18 b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (9 a^2 b B+b^3 B+3 a^3 C+3 a b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {\left (10 b^2 (b B+3 a C) \cos (c+d x)+3 \left (10 a^3 B+b^3 C+b^3 C \cos (2 (c+d x))\right )\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}}{15 d} \]
((-30*a^3*B + 90*a*b^2*B + 90*a^2*b*C + 18*b^3*C)*EllipticE[(c + d*x)/2, 2 ] + 10*(9*a^2*b*B + b^3*B + 3*a^3*C + 3*a*b^2*C)*EllipticF[(c + d*x)/2, 2] + ((10*b^2*(b*B + 3*a*C)*Cos[c + d*x] + 3*(10*a^3*B + b^3*C + b^3*C*Cos[2 *(c + d*x)]))*Sin[c + d*x])/Sqrt[Cos[c + d*x]])/(15*d)
Time = 1.26 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3508, 3042, 3468, 27, 3042, 3512, 27, 3042, 3502, 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.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^3 (B+C \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3468 |
| \(\displaystyle 2 \int \frac {(a+b \cos (c+d x)) \left (-b (5 a B-b C) \cos ^2(c+d x)-\left (B a^2-2 b C a-b^2 B\right ) \cos (c+d x)+a (5 b B+a C)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {(a+b \cos (c+d x)) \left (-b (5 a B-b C) \cos ^2(c+d x)-\left (B a^2-2 b C a-b^2 B\right ) \cos (c+d x)+a (5 b B+a C)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-b (5 a B-b C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (-B a^2+2 b C a+b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a (5 b B+a C)\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {2}{5} \int \frac {5 (5 b B+a C) a^2-5 b \left (6 B a^2-3 b C a-b^2 B\right ) \cos ^2(c+d x)-\left (5 B a^3-15 b C a^2-15 b^2 B a-3 b^3 C\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {5 (5 b B+a C) a^2-5 b \left (6 B a^2-3 b C a-b^2 B\right ) \cos ^2(c+d x)-\left (5 B a^3-15 b C a^2-15 b^2 B a-3 b^3 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {5 (5 b B+a C) a^2-5 b \left (6 B a^2-3 b C a-b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (-5 B a^3+15 b C a^2+15 b^2 B a+3 b^3 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {5 \left (3 C a^3+9 b B a^2+3 b^2 C a+b^3 B\right )-3 \left (5 B a^3-15 b C a^2-15 b^2 B a-3 b^3 C\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx-\frac {10 b \left (6 a^2 B-3 a b C-b^2 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {5 \left (3 C a^3+9 b B a^2+3 b^2 C a+b^3 B\right )-3 \left (5 B a^3-15 b C a^2-15 b^2 B a-3 b^3 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {10 b \left (6 a^2 B-3 a b C-b^2 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {5 \left (3 C a^3+9 b B a^2+3 b^2 C a+b^3 B\right )-3 \left (5 B a^3-15 b C a^2-15 b^2 B a-3 b^3 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 b \left (6 a^2 B-3 a b C-b^2 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^3 C+9 a^2 b B+3 a b^2 C+b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 \left (5 a^3 B-15 a^2 b C-15 a b^2 B-3 b^3 C\right ) \int \sqrt {\cos (c+d x)}dx\right )-\frac {10 b \left (6 a^2 B-3 a b C-b^2 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^3 C+9 a^2 b B+3 a b^2 C+b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 \left (5 a^3 B-15 a^2 b C-15 a b^2 B-3 b^3 C\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {10 b \left (6 a^2 B-3 a b C-b^2 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 \left (3 a^3 C+9 a^2 b B+3 a b^2 C+b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 \left (5 a^3 B-15 a^2 b C-15 a b^2 B-3 b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {10 b \left (6 a^2 B-3 a b C-b^2 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {10 \left (3 a^3 C+9 a^2 b B+3 a b^2 C+b^3 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 \left (5 a^3 B-15 a^2 b C-15 a b^2 B-3 b^3 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {10 b \left (6 a^2 B-3 a b C-b^2 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt {\cos (c+d x)}}\) |
(-2*b^2*(5*a*B - b*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*a*B*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) + (((-6*(5*a^3*B - 15*a*b^2*B - 15*a^2*b*C - 3*b^3*C)*EllipticE[(c + d*x)/2, 2])/d + (10*(9* a^2*b*B + b^3*B + 3*a^3*C + 3*a*b^2*C)*EllipticF[(c + d*x)/2, 2])/d)/3 - ( 10*b*(6*a^2*B - b^2*B - 3*a*b*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d))/5
3.9.71.3.1 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[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] :> Si mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((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 - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a *B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, 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] && GtQ[m, 1] && LtQ[n, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -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[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(640\) vs. \(2(240)=480\).
Time = 6.39 (sec) , antiderivative size = 641, normalized size of antiderivative = 3.17
| method | result | size |
| default | \(-\frac {2 \left (-24 C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{3}+20 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+60 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}+24 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}-30 B \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3}-10 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+45 B \,a^{2} b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 B \,b^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+15 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{3}-45 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2}-30 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}-6 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+15 C \,a^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+15 C a \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-45 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2} b -9 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{3}\right )}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(641\) |
| parts | \(\text {Expression too large to display}\) | \(760\) |
int((a+cos(d*x+c)*b)^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x,me thod=_RETURNVERBOSE)
-2/15*(-24*C*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)*b^3+20*B*cos(1/2*d*x+ 1/2*c)*sin(1/2*d*x+1/2*c)^4*b^3+60*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c) ^4*a*b^2+24*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^3-30*B*sin(1/2*d*x +1/2*c)^2*cos(1/2*d*x+1/2*c)*a^3-10*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c )^2*b^3+45*B*a^2*b*(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))+5*B*b^3*(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))+15*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El lipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-45*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)* (2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b ^2-30*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b^2-6*C*cos(1/2*d*x+1/2* c)*sin(1/2*d*x+1/2*c)^2*b^3+15*C*a^3*(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))+15*C*a*b^2* (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(co s(1/2*d*x+1/2*c),2^(1/2))-45*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-9*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)*EllipticE(cos(1/2*d *x+1/2*c),2^(1/2))*b^3)/sin(1/2*d*x+1/2*c)/(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 = 0.13 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.50 \[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, C a^{3} + 9 i \, B a^{2} b + 3 i \, C a b^{2} + i \, B b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, C a^{3} - 9 i \, B a^{2} b - 3 i \, C a b^{2} - i \, B b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, B a^{3} - 15 i \, C a^{2} b - 15 i \, B a b^{2} - 3 i \, C b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, B a^{3} + 15 i \, C a^{2} b + 15 i \, B a b^{2} + 3 i \, C b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, C b^{3} \cos \left (d x + c\right )^{2} + 15 \, B a^{3} + 5 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )} \]
integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2 ),x, algorithm="fricas")
-1/15*(5*sqrt(2)*(3*I*C*a^3 + 9*I*B*a^2*b + 3*I*C*a*b^2 + I*B*b^3)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2 )*(-3*I*C*a^3 - 9*I*B*a^2*b - 3*I*C*a*b^2 - I*B*b^3)*cos(d*x + c)*weierstr assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(5*I*B*a^3 - 15*I*C*a^2*b - 15*I*B*a*b^2 - 3*I*C*b^3)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2) *(-5*I*B*a^3 + 15*I*C*a^2*b + 15*I*B*a*b^2 + 3*I*C*b^3)*cos(d*x + c)*weier strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) )) - 2*(3*C*b^3*cos(d*x + c)^2 + 15*B*a^3 + 5*(3*C*a*b^2 + B*b^3)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))
Timed out. \[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2 ),x, algorithm="maxima")
\[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2 ),x, algorithm="giac")
Time = 2.87 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {B\,b^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,B\,a\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,B\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,C\,a^2\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {3\,C\,a\,b^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(B*b^3*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2 , 2))/3))/d + (2*C*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (6*B*a*b^2*ellipti cE(c/2 + (d*x)/2, 2))/d + (6*B*a^2*b*ellipticF(c/2 + (d*x)/2, 2))/d + (6*C *a^2*b*ellipticE(c/2 + (d*x)/2, 2))/d + (3*C*a*b^2*((2*cos(c + d*x)^(1/2)* sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d + (2*B*a^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*( sin(c + d*x)^2)^(1/2)) - (2*C*b^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeo m([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))