Integrand size = 35, antiderivative size = 122 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {b^2 (4 A+3 C) x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b^2 (4 A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b^2 C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d} \]
1/4*b^2*C*cos(d*x+c)^(5/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+1/8*b^2*(4*A+ 3*C)*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+1/8*b^2*(4*A+3*C)*sin(d*x+c)* cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)/d
Time = 0.72 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.55 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {(b \cos (c+d x))^{5/2} (4 (4 A+3 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+C \sin (4 (c+d x)))}{32 d \cos ^{\frac {5}{2}}(c+d x)} \]
((b*Cos[c + d*x])^(5/2)*(4*(4*A + 3*C)*(c + d*x) + 8*(A + C)*Sin[2*(c + d* x)] + C*Sin[4*(c + d*x)]))/(32*d*Cos[c + d*x]^(5/2))
Time = 0.34 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.70, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2031, 3042, 3493, 3042, 3115, 24}
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.
\(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \cos ^2(c+d x) \left (C \cos ^2(c+d x)+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3493 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} (4 A+3 C) \int \cos ^2(c+d x)dx+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} (4 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} (4 A+3 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} (4 A+3 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
(b^2*Sqrt[b*Cos[c + d*x]]*((C*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + ((4*A + 3*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/Sqrt[Cos[c + d*x]]
3.2.8.3.1 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[((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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f *(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2) Int[(b*Sin[e + f*x] )^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && !LtQ[m, -1]
Time = 7.87 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75
method | result | size |
default | \(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (2 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 C \cos \left (d x +c \right ) \sin \left (d x +c \right )+4 A \left (d x +c \right )+3 C \left (d x +c \right )\right )}{8 d \sqrt {\cos \left (d x +c \right )}}\) | \(91\) |
risch | \(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, x \left (8 A +6 C \right )}{16 \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, C \sin \left (4 d x +4 c \right )}{32 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (A +C \right ) \sin \left (2 d x +2 c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(107\) |
parts | \(\frac {A \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}+\frac {C \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right )}{8 d \sqrt {\cos \left (d x +c \right )}}\) | \(112\) |
1/8*b^2/d*(cos(d*x+c)*b)^(1/2)*(2*C*cos(d*x+c)^3*sin(d*x+c)+4*A*sin(d*x+c) *cos(d*x+c)+3*C*cos(d*x+c)*sin(d*x+c)+4*A*(d*x+c)+3*C*(d*x+c))/cos(d*x+c)^ (1/2)
Time = 0.29 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.80 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\left [\frac {{\left (4 \, A + 3 \, C\right )} \sqrt {-b} b^{2} \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 3 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, d}, \frac {{\left (4 \, A + 3 \, C\right )} b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) + {\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 3 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, d}\right ] \]
[1/16*((4*A + 3*C)*sqrt(-b)*b^2*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*b^2*cos(d*x + c)^2 + (4*A + 3*C)*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/d, 1/8*((4*A + 3*C)*b^(5/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/ (sqrt(b)*cos(d*x + c)^(3/2))) + (2*C*b^2*cos(d*x + c)^2 + (4*A + 3*C)*b^2) *sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/d]
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \]
Time = 0.41 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.75 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {8 \, {\left (2 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} A \sqrt {b} + {\left (12 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b^{2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} C \sqrt {b}}{32 \, d} \]
1/32*(8*(2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c))*A*sqrt(b) + (12*(d*x + c) *b^2 + b^2*sin(4*d*x + 4*c) + 8*b^2*sin(1/2*arctan2(sin(4*d*x + 4*c), cos( 4*d*x + 4*c))))*C*sqrt(b))/d
\[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
Time = 0.93 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.59 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (8\,A\,\sin \left (2\,c+2\,d\,x\right )+8\,C\,\sin \left (2\,c+2\,d\,x\right )+C\,\sin \left (4\,c+4\,d\,x\right )+16\,A\,d\,x+12\,C\,d\,x\right )}{32\,d\,\sqrt {\cos \left (c+d\,x\right )}} \]