Integrand size = 43, antiderivative size = 199 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {b^2 (4 A+3 C) x \sqrt {b \cos (c+d x)}}{8 \sqrt {\cos (c+d x)}}+\frac {b^2 B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {b^2 (4 A+3 C) \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{8 d}+\frac {b^2 C \cos ^{\frac {5}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{4 d}-\frac {b^2 B \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}} \]
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)+b^2*B*sin(d*x+c)*(b*cos(d*x+c ))^(1/2)/d/cos(d*x+c)^(1/2)-1/3*b^2*B*sin(d*x+c)^3*(b*cos(d*x+c))^(1/2)/d/ 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 = 1.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.46 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {(b \cos (c+d x))^{5/2} (48 A c+36 c C+48 A d x+36 C d x+72 B \sin (c+d x)+24 (A+C) \sin (2 (c+d x))+8 B \sin (3 (c+d x))+3 C \sin (4 (c+d x)))}{96 d \cos ^{\frac {5}{2}}(c+d x)} \]
Integrate[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) /Sqrt[Cos[c + d*x]],x]
((b*Cos[c + d*x])^(5/2)*(48*A*c + 36*c*C + 48*A*d*x + 36*C*d*x + 72*B*Sin[ c + d*x] + 24*(A + C)*Sin[2*(c + d*x)] + 8*B*Sin[3*(c + d*x)] + 3*C*Sin[4* (c + d*x)]))/(96*d*Cos[c + d*x]^(5/2))
Time = 0.52 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.57, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2031, 3042, 3502, 3042, 3227, 3042, 3113, 2009, 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.
| \(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \cos ^2(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \int \cos ^2(c+d x) (4 A+3 C+4 B \cos (c+d x))dx+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 A+3 C+4 B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \int \cos ^2(c+d x)dx+4 B \int \cos ^3(c+d x)dx\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+4 B \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 B \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {4 B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{4} \left ((4 A+3 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {4 B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )}{\sqrt {\cos (c+d x)}}\) |
(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*B*(-Sin[c + d*x] + Si n[c + d*x]^3/3))/d)/4))/Sqrt[Cos[c + d*x]]
3.4.7.3.1 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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 8.34 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.59
| method | result | size |
| default | \(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (6 C \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+8 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+12 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+9 C \cos \left (d x +c \right ) \sin \left (d x +c \right )+12 A \left (d x +c \right )+16 B \sin \left (d x +c \right )+9 C \left (d x +c \right )\right )}{24 d \sqrt {\cos \left (d x +c \right )}}\) | \(117\) |
| parts | \(\frac {A \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}+\frac {B \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{3 d \sqrt {\cos \left (d x +c \right )}}+\frac {C \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 d x +3 c \right )}{8 d \sqrt {\cos \left (d x +c \right )}}\) | \(155\) |
| risch | \(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, x \left (8 A +6 C \right )}{16 \sqrt {\cos \left (d x +c \right )}}+\frac {3 b^{2} B \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{4 d \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, C \sin \left (4 d x +4 c \right )}{32 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, B \sin \left (3 d x +3 c \right )}{12 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (A +C \right ) \sin \left (2 d x +2 c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(176\) |
int((cos(d*x+c)*b)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2), x,method=_RETURNVERBOSE)
1/24*b^2/d*(cos(d*x+c)*b)^(1/2)*(6*C*cos(d*x+c)^3*sin(d*x+c)+8*B*sin(d*x+c )*cos(d*x+c)^2+12*A*sin(d*x+c)*cos(d*x+c)+9*C*cos(d*x+c)*sin(d*x+c)+12*A*( d*x+c)+16*B*sin(d*x+c)+9*C*(d*x+c))/cos(d*x+c)^(1/2)
Time = 0.33 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.52 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\left [\frac {3 \, {\left (4 \, A + 3 \, C\right )} \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (6 \, C b^{2} \cos \left (d x + c\right )^{3} + 8 \, B b^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} b^{2} \cos \left (d x + c\right ) + 16 \, B b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )}, \frac {3 \, {\left (4 \, A + 3 \, C\right )} b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (6 \, C b^{2} \cos \left (d x + c\right )^{3} + 8 \, B b^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} b^{2} \cos \left (d x + c\right ) + 16 \, B b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{24 \, d \cos \left (d x + c\right )}\right ] \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (1/2),x, algorithm="fricas")
[1/48*(3*(4*A + 3*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*(6* C*b^2*cos(d*x + c)^3 + 8*B*b^2*cos(d*x + c)^2 + 3*(4*A + 3*C)*b^2*cos(d*x + c) + 16*B*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d* cos(d*x + c)), 1/24*(3*(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)))*cos(d*x + c) + (6*C*b^2*cos(d*x + c)^3 + 8*B*b^2*cos(d*x + c)^2 + 3*(4*A + 3*C)*b^2*cos(d*x + c) + 16*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. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \]
Time = 0.50 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.70 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {24 \, {\left (2 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} A \sqrt {b} + 8 \, {\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} B \sqrt {b} + 3 \, {\left (12 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (4 \, d x + 4 \, c\right ) + 8 \, b^{2} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )\right )} C \sqrt {b}}{96 \, d} \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (1/2),x, algorithm="maxima")
1/96*(24*(2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c))*A*sqrt(b) + 8*(b^2*sin(3 *d*x + 3*c) + 9*b^2*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))* B*sqrt(b) + 3*(12*(d*x + c)*b^2 + b^2*sin(4*d*x + 4*c) + 8*b^2*sin(1/2*arc tan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))*C*sqrt(b))/d
\[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (1/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)/s qrt(cos(d*x + c)), x)
Time = 1.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.47 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (72\,B\,\sin \left (c+d\,x\right )+24\,A\,\sin \left (2\,c+2\,d\,x\right )+8\,B\,\sin \left (3\,c+3\,d\,x\right )+24\,C\,\sin \left (2\,c+2\,d\,x\right )+3\,C\,\sin \left (4\,c+4\,d\,x\right )+48\,A\,d\,x+36\,C\,d\,x\right )}{96\,d\,\sqrt {\cos \left (c+d\,x\right )}} \]