Integrand size = 43, antiderivative size = 155 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {B x \sqrt {\cos (c+d x)}}{2 b^2 \sqrt {b \cos (c+d x)}}+\frac {(3 A+2 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 d \sqrt {b \cos (c+d x)}}+\frac {B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b^2 d \sqrt {b \cos (c+d x)}}+\frac {C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 b^2 d \sqrt {b \cos (c+d x)}} \]
1/2*B*cos(d*x+c)^(3/2)*sin(d*x+c)/b^2/d/(b*cos(d*x+c))^(1/2)+1/3*C*cos(d*x +c)^(5/2)*sin(d*x+c)/b^2/d/(b*cos(d*x+c))^(1/2)+1/2*B*x*cos(d*x+c)^(1/2)/b ^2/(b*cos(d*x+c))^(1/2)+1/3*(3*A+2*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/b^2/d/(b *cos(d*x+c))^(1/2)
Time = 0.80 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.50 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cos (c+d x)} (6 B c+6 B d x+3 (4 A+3 C) \sin (c+d x)+3 B \sin (2 (c+d x))+C \sin (3 (c+d x)))}{12 b^2 d \sqrt {b \cos (c+d x)}} \]
Integrate[(Cos[c + d*x]^(7/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b* Cos[c + d*x])^(5/2),x]
(Sqrt[Cos[c + d*x]]*(6*B*c + 6*B*d*x + 3*(4*A + 3*C)*Sin[c + d*x] + 3*B*Si n[2*(c + d*x)] + C*Sin[3*(c + d*x)]))/(12*b^2*d*Sqrt[b*Cos[c + d*x]])
Time = 0.35 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.63, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {2031, 3042, 3502, 3042, 3213}
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.
\(\displaystyle \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \cos (c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \int \cos (c+d x) (3 A+2 C+3 B \cos (c+d x))dx+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (3 A+2 C+3 B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{3} \left (\frac {(3 A+2 C) \sin (c+d x)}{d}+\frac {3 B \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 B x}{2}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )}{b^2 \sqrt {b \cos (c+d x)}}\) |
(Sqrt[Cos[c + d*x]]*((C*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + ((3*B*x)/2 + ((3*A + 2*C)*Sin[c + d*x])/d + (3*B*Cos[c + d*x]*Sin[c + d*x])/(2*d))/3))/ (b^2*Sqrt[b*Cos[c + d*x]])
3.4.32.3.1 Defintions of rubi rules used
Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v]) Int[v^(m + n)*Fx, x], x] /; FreeQ[{a , b, m}, x] && !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
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 = 9.40 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.55
method | result | size |
default | \(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (2 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+6 A \sin \left (d x +c \right )+3 B \left (d x +c \right )+4 \sin \left (d x +c \right ) C \right )}{6 b^{2} d \sqrt {\cos \left (d x +c \right ) b}}\) | \(86\) |
parts | \(\frac {A \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{b^{2} d \sqrt {\cos \left (d x +c \right ) b}}+\frac {B \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}+\frac {C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{3 d \,b^{2} \sqrt {\cos \left (d x +c \right ) b}}\) | \(122\) |
risch | \(\frac {B x \left (\sqrt {\cos }\left (d x +c \right )\right )}{2 b^{2} \sqrt {\cos \left (d x +c \right ) b}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (4 A +3 C \right ) \sin \left (d x +c \right )}{4 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) C \sin \left (3 d x +3 c \right )}{12 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) B \sin \left (2 d x +2 c \right )}{4 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, d}\) | \(138\) |
int(cos(d*x+c)^(7/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(cos(d*x+c)*b)^(5/2), x,method=_RETURNVERBOSE)
1/6/b^2/d*cos(d*x+c)^(1/2)*(2*C*cos(d*x+c)^2*sin(d*x+c)+3*B*sin(d*x+c)*cos (d*x+c)+6*A*sin(d*x+c)+3*B*(d*x+c)+4*sin(d*x+c)*C)/(cos(d*x+c)*b)^(1/2)
Time = 0.30 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.56 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\left [-\frac {3 \, B \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, {\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, b^{3} d \cos \left (d x + c\right )}, \frac {3 \, B \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, b^{3} d \cos \left (d x + c\right )}\right ] \]
integrate(cos(d*x+c)^(7/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^ (5/2),x, algorithm="fricas")
[-1/12*(3*B*sqrt(-b)*cos(d*x + c)*log(2*b*cos(d*x + c)^2 + 2*sqrt(b*cos(d* x + c))*sqrt(-b)*sqrt(cos(d*x + c))*sin(d*x + c) - b) - 2*(2*C*cos(d*x + c )^2 + 3*B*cos(d*x + c) + 6*A + 4*C)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c) )*sin(d*x + c))/(b^3*d*cos(d*x + c)), 1/6*(3*B*sqrt(b)*arctan(sqrt(b*cos(d *x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + (2*C*co s(d*x + c)^2 + 3*B*cos(d*x + c) + 6*A + 4*C)*sqrt(b*cos(d*x + c))*sqrt(cos (d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c))]
Timed out. \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Time = 0.50 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.52 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B}{b^{\frac {5}{2}}} + \frac {C {\left (\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )}}{b^{\frac {5}{2}}} + \frac {12 \, A \sin \left (d x + c\right )}{b^{\frac {5}{2}}}}{12 \, d} \]
integrate(cos(d*x+c)^(7/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^ (5/2),x, algorithm="maxima")
1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B/b^(5/2) + C*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))/b^(5/2) + 12*A*si n(d*x + c)/b^(5/2))/d
\[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {7}{2}}}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
integrate(cos(d*x+c)^(7/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^ (5/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(7/2)/(b*co s(d*x + c))^(5/2), x)
Time = 1.35 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (3\,B\,\sin \left (c+d\,x\right )+12\,A\,\sin \left (2\,c+2\,d\,x\right )+3\,B\,\sin \left (3\,c+3\,d\,x\right )+10\,C\,\sin \left (2\,c+2\,d\,x\right )+C\,\sin \left (4\,c+4\,d\,x\right )+12\,B\,d\,x\,\cos \left (c+d\,x\right )\right )}{12\,b^3\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]