Integrand size = 43, antiderivative size = 135 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {A x \sqrt {\cos (c+d x)}}{b \sqrt {b \cos (c+d x)}}+\frac {C x \sqrt {\cos (c+d x)}}{2 b \sqrt {b \cos (c+d x)}}+\frac {B \sqrt {\cos (c+d x)} \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}+\frac {C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 b d \sqrt {b \cos (c+d x)}} \]
1/2*C*cos(d*x+c)^(3/2)*sin(d*x+c)/b/d/(b*cos(d*x+c))^(1/2)+A*x*cos(d*x+c)^ (1/2)/b/(b*cos(d*x+c))^(1/2)+1/2*C*x*cos(d*x+c)^(1/2)/b/(b*cos(d*x+c))^(1/ 2)+B*sin(d*x+c)*cos(d*x+c)^(1/2)/b/d/(b*cos(d*x+c))^(1/2)
Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.45 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) (2 (2 A+C) (c+d x)+4 B \sin (c+d x)+C \sin (2 (c+d x)))}{4 d (b \cos (c+d x))^{3/2}} \]
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b* Cos[c + d*x])^(3/2),x]
(Cos[c + d*x]^(3/2)*(2*(2*A + C)*(c + d*x) + 4*B*Sin[c + d*x] + C*Sin[2*(c + d*x)]))/(4*d*(b*Cos[c + d*x])^(3/2))
Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.50, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {2031, 2009}
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.
\(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{b \sqrt {b \cos (c+d x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {\cos (c+d x)} \left (A x+\frac {B \sin (c+d x)}{d}+\frac {C \sin (c+d x) \cos (c+d x)}{2 d}+\frac {C x}{2}\right )}{b \sqrt {b \cos (c+d x)}}\) |
(Sqrt[Cos[c + d*x]]*(A*x + (C*x)/2 + (B*Sin[c + d*x])/d + (C*Cos[c + d*x]* Sin[c + d*x])/(2*d)))/(b*Sqrt[b*Cos[c + d*x]])
3.4.25.3.1 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]
Time = 10.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.49
method | result | size |
default | \(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (C \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 A \left (d x +c \right )+2 B \sin \left (d x +c \right )+C \left (d x +c \right )\right )}{2 b d \sqrt {\cos \left (d x +c \right ) b}}\) | \(66\) |
risch | \(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) x \left (4 A +2 C \right )}{4 b \sqrt {\cos \left (d x +c \right ) b}}+\frac {B \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{b d \sqrt {\cos \left (d x +c \right ) b}}+\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) C \sin \left (2 d x +2 c \right )}{4 b \sqrt {\cos \left (d x +c \right ) b}\, d}\) | \(101\) |
parts | \(\frac {A \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right ) b}\, b}+\frac {B \sin \left (d x +c \right ) \left (\sqrt {\cos }\left (d x +c \right )\right )}{b d \sqrt {\cos \left (d x +c \right ) b}}+\frac {C \left (\sqrt {\cos }\left (d x +c \right )\right ) \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d b \sqrt {\cos \left (d x +c \right ) b}}\) | \(110\) |
int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(cos(d*x+c)*b)^(3/2), x,method=_RETURNVERBOSE)
1/2/b/d*cos(d*x+c)^(1/2)*(C*cos(d*x+c)*sin(d*x+c)+2*A*(d*x+c)+2*B*sin(d*x+ c)+C*(d*x+c))/(cos(d*x+c)*b)^(1/2)
Time = 0.31 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.61 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\left [-\frac {{\left (2 \, A + C\right )} \sqrt {-b} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, {\left (C \cos \left (d x + c\right ) + 2 \, B\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b^{2} d \cos \left (d x + c\right )}, \frac {{\left (2 \, A + C\right )} \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (C \cos \left (d x + c\right ) + 2 \, B\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{2} d \cos \left (d x + c\right )}\right ] \]
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^ (3/2),x, algorithm="fricas")
[-1/4*((2*A + C)*sqrt(-b)*cos(d*x + c)*log(2*b*cos(d*x + c)^2 + 2*sqrt(b*c os(d*x + c))*sqrt(-b)*sqrt(cos(d*x + c))*sin(d*x + c) - b) - 2*(C*cos(d*x + c) + 2*B)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^2*d*c os(d*x + c)), 1/2*((2*A + C)*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) + (C*cos(d*x + c) + 2*B)*sq rt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(b^2*d*cos(d*x + c))]
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Time = 0.48 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.47 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C}{b^{\frac {3}{2}}} + \frac {8 \, A \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{b^{\frac {3}{2}}} + \frac {4 \, B \sin \left (d x + c\right )}{b^{\frac {3}{2}}}}{4 \, d} \]
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^ (3/2),x, algorithm="maxima")
1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*C/b^(3/2) + 8*A*arctan(sin(d*x + c)/ (cos(d*x + c) + 1))/b^(3/2) + 4*B*sin(d*x + c)/b^(3/2))/d
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^ (3/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*co s(d*x + c))^(3/2), x)
Time = 0.90 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (C\,\sin \left (c+d\,x\right )+4\,B\,\sin \left (2\,c+2\,d\,x\right )+C\,\sin \left (3\,c+3\,d\,x\right )+8\,A\,d\,x\,\cos \left (c+d\,x\right )+4\,C\,d\,x\,\cos \left (c+d\,x\right )\right )}{4\,b^2\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]