Integrand size = 41, antiderivative size = 188 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {6 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)}}+\frac {2 (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 b d \sqrt {b \cos (c+d x)}}+\frac {2 (7 A+5 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b^2 d}+\frac {2 B (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b^3 d}+\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^4 d} \]
2/5*B*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/b^3/d+2/7*C*(b*cos(d*x+c))^(5/2)*sin (d*x+c)/b^4/d+2/21*(7*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2* c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/b/d/(b*cos(d*x+c ))^(1/2)+2/21*(7*A+5*C)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/b^2/d+6/5*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))*(b*cos(d*x+c))^(1/2)/b^2/d/cos(d*x+c)^(1/2)
Time = 1.36 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.57 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) \left (126 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 (7 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (70 A+65 C+42 B \cos (c+d x)+15 C \cos (2 (c+d x))) \sin (c+d x)\right )}{105 d (b \cos (c+d x))^{3/2}} \]
(Cos[c + d*x]^(3/2)*(126*B*EllipticE[(c + d*x)/2, 2] + 10*(7*A + 5*C)*Elli pticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(70*A + 65*C + 42*B*Cos[c + d*x ] + 15*C*Cos[2*(c + d*x)])*Sin[c + d*x]))/(105*d*(b*Cos[c + d*x])^(3/2))
Time = 0.80 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {2030, 3042, 3502, 27, 3042, 3227, 3042, 3115, 3042, 3121, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \cos (c+d x))^{3/2} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{b^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{b^3}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{2} (b \cos (c+d x))^{3/2} (b (7 A+5 C)+7 b B \cos (c+d x))dx}{7 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int (b \cos (c+d x))^{3/2} (b (7 A+5 C)+7 b B \cos (c+d x))dx}{7 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (b (7 A+5 C)+7 b B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^3}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {b (7 A+5 C) \int (b \cos (c+d x))^{3/2}dx+7 B \int (b \cos (c+d x))^{5/2}dx}{7 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {b (7 A+5 C) \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+7 B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx}{7 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^3}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\frac {b (7 A+5 C) \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+7 B \left (\frac {3}{5} b^2 \int \sqrt {b \cos (c+d x)}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{7 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {b (7 A+5 C) \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+7 B \left (\frac {3}{5} b^2 \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{7 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^3}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {\frac {b (7 A+5 C) \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+7 B \left (\frac {3 b^2 \sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{7 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {b (7 A+5 C) \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+7 B \left (\frac {3 b^2 \sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{7 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^3}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {b (7 A+5 C) \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+7 B \left (\frac {6 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{7 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^3}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {b (7 A+5 C) \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+7 B \left (\frac {6 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 d}\right )}{7 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^3}\) |
((2*C*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b*d) + (b*(7*A + 5*C)*((2*b^ 2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + (2*b*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)) + 7*B*((6*b^2*Sqrt[b*Cos [c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*b*(b*C os[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)))/(7*b))/b^3
3.3.72.3.1 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[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
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_.)*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]
Time = 15.41 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.88
method | result | size |
default | \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (240 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-168 B -360 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (140 A +168 B +280 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-70 A -42 B -80 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+35 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 b \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) | \(353\) |
parts | \(-\frac {2 A \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\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 )\right )}{3 b \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}-\frac {2 B \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-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}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 b \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}-\frac {2 C \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (48 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+128 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{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 )+16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 b \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) | \(615\) |
int(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(cos(d*x+c)*b)^(3/2),x,me thod=_RETURNVERBOSE)
-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)/b*(240*C* cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+(-168*B-360*C)*sin(1/2*d*x+1/2*c)^ 6*cos(1/2*d*x+1/2*c)+(140*A+168*B+280*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+ 1/2*c)+(-70*A-42*B-80*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+35*A*(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))-63*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))+25*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)*EllipticF(cos(1/2*d*x+1/2*c) ,2^(1/2)))/(-b*(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)*b)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=-\frac {5 \, \sqrt {2} {\left (7 i \, A + 5 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-7 i \, A - 5 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 63 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 63 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (15 \, C \cos \left (d x + c\right )^{2} + 21 \, B \cos \left (d x + c\right ) + 35 \, A + 25 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, b^{2} d} \]
integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2 ),x, algorithm="fricas")
-1/105*(5*sqrt(2)*(7*I*A + 5*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d *x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-7*I*A - 5*I*C)*sqrt(b)*weierstrass PInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 63*I*sqrt(2)*B*sqrt(b)*we ierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 63*I*sqrt(2)*B*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse( -4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(15*C*cos(d*x + c)^2 + 21*B*cos (d*x + c) + 35*A + 25*C)*sqrt(b*cos(d*x + c))*sin(d*x + c))/(b^2*d)
Timed out. \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2 ),x, algorithm="maxima")
\[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(3/2 ),x, algorithm="giac")
Timed out. \[ \int \frac {\cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]