Integrand size = 33, antiderivative size = 187 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {3 b^2 B x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {3 b^2 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b^2 B \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}-\frac {A b^2 \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}} \]
1/4*b^2*B*cos(d*x+c)^(5/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+3/8*b^2*B*x*( b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+A*b^2*sin(d*x+c)*(b*cos(d*x+c))^(1/2) /d/cos(d*x+c)^(1/2)-1/3*A*b^2*sin(d*x+c)^3*(b*cos(d*x+c))^(1/2)/d/cos(d*x+ c)^(1/2)+3/8*b^2*B*sin(d*x+c)*cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)/d
Time = 0.40 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.43 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {(b \cos (c+d x))^{5/2} (36 B c+36 B d x+72 A \sin (c+d x)+24 B \sin (2 (c+d x))+8 A \sin (3 (c+d x))+3 B \sin (4 (c+d x)))}{96 d \cos ^{\frac {5}{2}}(c+d x)} \]
((b*Cos[c + d*x])^(5/2)*(36*B*c + 36*B*d*x + 72*A*Sin[c + d*x] + 24*B*Sin[ 2*(c + d*x)] + 8*A*Sin[3*(c + d*x)] + 3*B*Sin[4*(c + d*x)]))/(96*d*Cos[c + d*x]^(5/2))
Time = 0.46 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.57, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {2031, 3042, 3227, 3042, 3113, 2009, 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.
\(\displaystyle \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \cos ^3(c+d x) (A+B \cos (c+d x))dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (A \int \cos ^3(c+d x)dx+B \int \cos ^4(c+d x)dx\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (A \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx+B \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (B \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {A \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (B \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {A \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (B \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {A \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (B \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {A \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (B \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {A \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (B \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {A \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
(b^2*Sqrt[b*Cos[c + d*x]]*(-((A*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d) + B *((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]]
3.9.58.3.1 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[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
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_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Time = 5.10 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.50
method | result | size |
default | \(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (6 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+8 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+9 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+16 A \sin \left (d x +c \right )+9 B \left (d x +c \right )\right )}{24 d \sqrt {\cos \left (d x +c \right )}}\) | \(94\) |
parts | \(\frac {A \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}\, \sin \left (d x +c \right )}{3 d \sqrt {\cos \left (d x +c \right )}}+\frac {B \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right )}{8 d \sqrt {\cos \left (d x +c \right )}}\) | \(110\) |
risch | \(\frac {3 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{i \left (d x +c \right )} x B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{5 i \left (d x +c \right )} B}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{4 i \left (d x +c \right )} A}{12 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {3 i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{2 i \left (d x +c \right )} A}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {3 i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) A}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-i \left (d x +c \right )} B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-2 i \left (d x +c \right )} A}{12 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {7 i b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) B \cos \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {9 b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) B \sin \left (3 d x +3 c \right )}{32 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\) | \(439\) |
1/24*b^2/d*(cos(d*x+c)*b)^(1/2)*(6*B*sin(d*x+c)*cos(d*x+c)^3+8*A*sin(d*x+c )*cos(d*x+c)^2+9*B*sin(d*x+c)*cos(d*x+c)+16*A*sin(d*x+c)+9*B*(d*x+c))/cos( d*x+c)^(1/2)
Time = 0.36 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.49 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\left [\frac {9 \, B \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (6 \, B b^{2} \cos \left (d x + c\right )^{3} + 8 \, A b^{2} \cos \left (d x + c\right )^{2} + 9 \, B b^{2} \cos \left (d x + c\right ) + 16 \, A b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )}, \frac {9 \, B b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (6 \, B b^{2} \cos \left (d x + c\right )^{3} + 8 \, A b^{2} \cos \left (d x + c\right )^{2} + 9 \, B b^{2} \cos \left (d x + c\right ) + 16 \, A b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )}\right ] \]
[1/48*(9*B*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*(6*B*b^2*cos( d*x + c)^3 + 8*A*b^2*cos(d*x + c)^2 + 9*B*b^2*cos(d*x + c) + 16*A*b^2)*sqr t(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/24* (9*B*b^(5/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c )^(3/2)))*cos(d*x + c) + (6*B*b^2*cos(d*x + c)^3 + 8*A*b^2*cos(d*x + c)^2 + 9*B*b^2*cos(d*x + c) + 16*A*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c)) *sin(d*x + c))/(d*cos(d*x + c))]
Timed out. \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\text {Timed out} \]
Time = 0.51 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.59 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {8 \, {\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} A \sqrt {b} + 3 \, {\left (12 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b^{2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} B \sqrt {b}}{96 \, d} \]
1/96*(8*(b^2*sin(3*d*x + 3*c) + 9*b^2*sin(1/3*arctan2(sin(3*d*x + 3*c), co s(3*d*x + 3*c))))*A*sqrt(b) + 3*(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))))*B*sqrt(b))/d
Time = 3.70 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.49 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {9 \, B b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 36 \, B b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 48 \, A b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, B b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 54 \, B b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 80 \, A b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, B b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, B b^{\frac {5}{2}} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 80 \, A b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 18 \, B b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B b^{\frac {5}{2}} d x + 48 \, A b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 30 \, B b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{24 \, {\left (d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 4 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 6 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + d\right )}} \]
1/24*(9*B*b^(5/2)*d*x*tan(1/2*d*x + 1/2*c)^8 + 36*B*b^(5/2)*d*x*tan(1/2*d* x + 1/2*c)^6 + 48*A*b^(5/2)*tan(1/2*d*x + 1/2*c)^7 - 30*B*b^(5/2)*tan(1/2* d*x + 1/2*c)^7 + 54*B*b^(5/2)*d*x*tan(1/2*d*x + 1/2*c)^4 + 80*A*b^(5/2)*ta n(1/2*d*x + 1/2*c)^5 + 18*B*b^(5/2)*tan(1/2*d*x + 1/2*c)^5 + 36*B*b^(5/2)* d*x*tan(1/2*d*x + 1/2*c)^2 + 80*A*b^(5/2)*tan(1/2*d*x + 1/2*c)^3 - 18*B*b^ (5/2)*tan(1/2*d*x + 1/2*c)^3 + 9*B*b^(5/2)*d*x + 48*A*b^(5/2)*tan(1/2*d*x + 1/2*c) + 30*B*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 = 15.84 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.58 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx=\frac {b^2\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (24\,B\,\sin \left (c+d\,x\right )+80\,A\,\sin \left (2\,c+2\,d\,x\right )+8\,A\,\sin \left (4\,c+4\,d\,x\right )+27\,B\,\sin \left (3\,c+3\,d\,x\right )+3\,B\,\sin \left (5\,c+5\,d\,x\right )+72\,B\,d\,x\,\cos \left (c+d\,x\right )\right )}{96\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]