Integrand size = 43, antiderivative size = 135 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {A b^2 x \sqrt {b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}+\frac {b^2 C x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b^2 B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {b^2 C \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d} \]
A*b^2*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+1/2*b^2*C*x*(b*cos(d*x+c))^( 1/2)/cos(d*x+c)^(1/2)+b^2*B*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^( 1/2)+1/2*b^2*C*sin(d*x+c)*cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)/d
Time = 0.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.45 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{5/2} (2 (2 A+C) (c+d x)+4 B \sin (c+d x)+C \sin (2 (c+d x)))}{4 d \cos ^{\frac {5}{2}}(c+d x)} \]
Integrate[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) /Cos[c + d*x]^(5/2),x]
((b*Cos[c + d*x])^(5/2)*(2*(2*A + C)*(c + d*x) + 4*B*Sin[c + d*x] + C*Sin[ 2*(c + d*x)]))/(4*d*Cos[c + d*x]^(5/2))
Time = 0.22 (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.
\(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (A x+\frac {B \sin (c+d x)}{d}+\frac {C \sin (c+d x) \cos (c+d x)}{2 d}+\frac {C x}{2}\right )}{\sqrt {\cos (c+d x)}}\) |
(b^2*Sqrt[b*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)))/Sqrt[Cos[c + d*x]]
3.4.9.3.1 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 = 10.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.49
method | result | size |
default | \(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (C \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 A \left (d x +c \right )+2 B \sin \left (d x +c \right )+C \left (d x +c \right )\right )}{2 d \sqrt {\cos \left (d x +c \right )}}\) | \(66\) |
risch | \(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, x \left (4 A +2 C \right )}{4 \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} B \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, C \sin \left (2 d x +2 c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(101\) |
parts | \(\frac {A \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} B \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {C \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}\) | \(110\) |
int((cos(d*x+c)*b)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2), x,method=_RETURNVERBOSE)
1/2*b^2/d*(cos(d*x+c)*b)^(1/2)*(C*cos(d*x+c)*sin(d*x+c)+2*A*(d*x+c)+2*B*si n(d*x+c)+C*(d*x+c))/cos(d*x+c)^(1/2)
Time = 0.31 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.68 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\left [\frac {{\left (2 \, A + C\right )} \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (C b^{2} \cos \left (d x + c\right ) + 2 \, B b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )}, \frac {{\left (2 \, A + C\right )} b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (C b^{2} \cos \left (d x + c\right ) + 2 \, B b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}\right ] \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (5/2),x, algorithm="fricas")
[1/4*((2*A + C)*sqrt(-b)*b^2*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*(C*b^2*c os(d*x + c) + 2*B*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c ))/(d*cos(d*x + c)), 1/2*((2*A + C)*b^(5/2)*arctan(sqrt(b*cos(d*x + c))*si n(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + (C*b^2*cos(d*x + c ) + 2*B*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos( d*x + c))]
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]
Time = 0.49 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.53 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {8 \, A b^{\frac {5}{2}} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + 4 \, B b^{\frac {5}{2}} \sin \left (d x + c\right ) + {\left (2 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} C \sqrt {b}}{4 \, d} \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (5/2),x, algorithm="maxima")
1/4*(8*A*b^(5/2)*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) + 4*B*b^(5/2)*sin (d*x + c) + (2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c))*C*sqrt(b))/d
\[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (5/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)/c os(d*x + c)^(5/2), x)
Time = 1.39 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.42 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (4\,B\,\sin \left (c+d\,x\right )+C\,\sin \left (2\,c+2\,d\,x\right )+4\,A\,d\,x+2\,C\,d\,x\right )}{4\,d\,\sqrt {\cos \left (c+d\,x\right )}} \]