Integrand size = 23, antiderivative size = 107 \[ \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {3 b^2 x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {3 b^2 \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b^2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d} \]
1/4*b^2*cos(d*x+c)^(5/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+3/8*b^2*x*(b*co s(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+3/8*b^2*sin(d*x+c)*cos(d*x+c)^(1/2)*(b*co s(d*x+c))^(1/2)/d
Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.51 \[ \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {(b \cos (c+d x))^{5/2} (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{32 d \cos ^{\frac {5}{2}}(c+d x)} \]
((b*Cos[c + d*x])^(5/2)*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d* x)]))/(32*d*Cos[c + d*x]^(5/2))
Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2031, 3042, 3115, 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.
\(\displaystyle \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \cos ^4(c+d x)dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )}{\sqrt {\cos (c+d x)}}\) |
(b^2*Sqrt[b*Cos[c + d*x]]*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/Sqrt[Cos[c + d*x]]
3.2.61.3.1 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]
Time = 2.88 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.61
method | result | size |
default | \(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right )}{8 d \sqrt {\cos \left (d x +c \right )}}\) | \(65\) |
risch | \(\frac {3 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{i \left (d x +c \right )} x}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{5 i \left (d x +c \right )}}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-i \left (d x +c \right )}}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {7 i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \cos \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {9 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \sin \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\) | \(243\) |
1/8/d*b^2*(cos(d*x+c)*b)^(1/2)*(2*sin(d*x+c)*cos(d*x+c)^3+3*cos(d*x+c)*sin (d*x+c)+3*d*x+3*c)/cos(d*x+c)^(1/2)
Time = 0.30 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.80 \[ \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx=\left [\frac {3 \, \sqrt {-b} b^{2} \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (2 \, b^{2} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{16 \, d}, \frac {3 \, b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) + {\left (2 \, b^{2} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{8 \, d}\right ] \]
[1/16*(3*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*b^2*cos(d*x + c)^2 + 3*b^ 2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/d, 1/8*(3*b^(5/2) *arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2))) + (2*b^2*cos(d*x + c)^2 + 3*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin (d*x + c))/d]
Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx=\text {Timed out} \]
Time = 0.40 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.55 \[ \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {{\left (12 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b^{2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} \sqrt {b}}{32 \, d} \]
1/32*(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))))*sqrt(b)/d
Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (89) = 178\).
Time = 2.48 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.90 \[ \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {3 \, b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 12 \, b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 10 \, b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 18 \, b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6 \, b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, b^{\frac {5}{2}} d x + 10 \, b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{8 \, {\left (d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 4 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 6 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + d\right )}} \]
1/8*(3*b^(5/2)*d*x*tan(1/2*d*x + 1/2*c)^8 + 12*b^(5/2)*d*x*tan(1/2*d*x + 1 /2*c)^6 - 10*b^(5/2)*tan(1/2*d*x + 1/2*c)^7 + 18*b^(5/2)*d*x*tan(1/2*d*x + 1/2*c)^4 + 6*b^(5/2)*tan(1/2*d*x + 1/2*c)^5 + 12*b^(5/2)*d*x*tan(1/2*d*x + 1/2*c)^2 - 6*b^(5/2)*tan(1/2*d*x + 1/2*c)^3 + 3*b^(5/2)*d*x + 10*b^(5/2) *tan(1/2*d*x + 1/2*c))/(d*tan(1/2*d*x + 1/2*c)^8 + 4*d*tan(1/2*d*x + 1/2*c )^6 + 6*d*tan(1/2*d*x + 1/2*c)^4 + 4*d*tan(1/2*d*x + 1/2*c)^2 + d)
Time = 15.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.73 \[ \int \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^{5/2} \, dx=\frac {b^2\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (8\,\sin \left (c+d\,x\right )+9\,\sin \left (3\,c+3\,d\,x\right )+\sin \left (5\,c+5\,d\,x\right )+24\,d\,x\,\cos \left (c+d\,x\right )\right )}{32\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]