Integrand size = 33, antiderivative size = 140 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {A b x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b B \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {A b \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d}-\frac {b B \sqrt {b \cos (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\cos (c+d x)}} \]
1/2*A*b*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+b*B*sin(d*x+c)*(b*cos(d*x+ c))^(1/2)/d/cos(d*x+c)^(1/2)-1/3*b*B*sin(d*x+c)^3*(b*cos(d*x+c))^(1/2)/d/c os(d*x+c)^(1/2)+1/2*A*b*sin(d*x+c)*cos(d*x+c)^(1/2)*(b*cos(d*x+c))^(1/2)/d
Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.49 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {(b \cos (c+d x))^{3/2} (6 A c+6 A d x+9 B \sin (c+d x)+3 A \sin (2 (c+d x))+B \sin (3 (c+d x)))}{12 d \cos ^{\frac {3}{2}}(c+d x)} \]
((b*Cos[c + d*x])^(3/2)*(6*A*c + 6*A*d*x + 9*B*Sin[c + d*x] + 3*A*Sin[2*(c + d*x)] + B*Sin[3*(c + d*x)]))/(12*d*Cos[c + d*x]^(3/2))
Time = 0.38 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.56, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {2031, 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.
| \(\displaystyle \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \cos ^2(c+d x) (A+B \cos (c+d x))dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (A \int \cos ^2(c+d x)dx+B \int \cos ^3(c+d x)dx\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (A \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+B \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (A \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {B \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (A \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (A \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (A \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {B \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{\sqrt {\cos (c+d x)}}\) |
(b*Sqrt[b*Cos[c + d*x]]*(A*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) - (B* (-Sin[c + d*x] + Sin[c + d*x]^3/3))/d))/Sqrt[Cos[c + d*x]]
3.9.51.3.1 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 = 5.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.54
| method | result | size |
| default | \(\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (2 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+3 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 A \left (d x +c \right )+4 B \sin \left (d x +c \right )\right )}{6 d \sqrt {\cos \left (d x +c \right )}}\) | \(75\) |
| parts | \(\frac {A b \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}+\frac {B b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{3 d \sqrt {\cos \left (d x +c \right )}}\) | \(86\) |
| risch | \(\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{i \left (d x +c \right )} x A}{{\mathrm e}^{2 i \left (d x +c \right )}+1}-\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{4 i \left (d x +c \right )} B}{12 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{3 i \left (d x +c \right )} A}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {3 i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{2 i \left (d x +c \right )} B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {3 i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) B}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-i \left (d x +c \right )} A}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i b \sqrt {\cos \left (d x +c \right ) b}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) {\mathrm e}^{-2 i \left (d x +c \right )} B}{12 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\) | \(325\) |
1/6*b/d*(cos(d*x+c)*b)^(1/2)*(2*B*sin(d*x+c)*cos(d*x+c)^2+3*A*sin(d*x+c)*c os(d*x+c)+3*A*(d*x+c)+4*B*sin(d*x+c))/cos(d*x+c)^(1/2)
Time = 0.35 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.69 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\left [\frac {3 \, A \sqrt {-b} b \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (2 \, B b \cos \left (d x + c\right )^{2} + 3 \, A b \cos \left (d x + c\right ) + 4 \, B b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )}, \frac {3 \, A b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (2 \, B b \cos \left (d x + c\right )^{2} + 3 \, A b \cos \left (d x + c\right ) + 4 \, B b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )}\right ] \]
[1/12*(3*A*sqrt(-b)*b*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*(2*B*b*cos(d*x + c)^2 + 3*A*b*cos(d*x + c) + 4*B*b)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c ))*sin(d*x + c))/(d*cos(d*x + c)), 1/6*(3*A*b^(3/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + (2*B*b*cos (d*x + c)^2 + 3*A*b*cos(d*x + c) + 4*B*b)*sqrt(b*cos(d*x + c))*sqrt(cos(d* x + c))*sin(d*x + c))/(d*cos(d*x + c))]
Timed out. \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\text {Timed out} \]
Time = 0.45 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.53 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (2 \, {\left (d x + c\right )} b + b \sin \left (2 \, d x + 2 \, c\right )\right )} A \sqrt {b} + {\left (b \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} B \sqrt {b}}{12 \, d} \]
1/12*(3*(2*(d*x + c)*b + b*sin(2*d*x + 2*c))*A*sqrt(b) + (b*sin(3*d*x + 3* c) + 9*b*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))*B*sqrt(b))/ d
Time = 2.59 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.39 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {{\left (3 \, A \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 9 \, A \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A \sqrt {b} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, B \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A \sqrt {b} d x + 6 \, A \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B \sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} b}{6 \, {\left (d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + d\right )}} \]
1/6*(3*A*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^6 + 9*A*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^4 - 6*A*sqrt(b)*tan(1/2*d*x + 1/2*c)^5 + 12*B*sqrt(b)*tan(1/2*d*x + 1/2*c)^5 + 9*A*sqrt(b)*d*x*tan(1/2*d*x + 1/2*c)^2 + 8*B*sqrt(b)*tan(1/2 *d*x + 1/2*c)^3 + 3*A*sqrt(b)*d*x + 6*A*sqrt(b)*tan(1/2*d*x + 1/2*c) + 12* B*sqrt(b)*tan(1/2*d*x + 1/2*c))*b/(d*tan(1/2*d*x + 1/2*c)^6 + 3*d*tan(1/2* d*x + 1/2*c)^4 + 3*d*tan(1/2*d*x + 1/2*c)^2 + d)
Time = 1.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.66 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx=\frac {b\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (3\,A\,\sin \left (c+d\,x\right )+3\,A\,\sin \left (3\,c+3\,d\,x\right )+10\,B\,\sin \left (2\,c+2\,d\,x\right )+B\,\sin \left (4\,c+4\,d\,x\right )+12\,A\,d\,x\,\cos \left (c+d\,x\right )\right )}{12\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]