Integrand size = 33, antiderivative size = 101 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {b B x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {A b \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {b B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d} \]
1/2*b*B*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+A*b*sin(d*x+c)*(b*cos(d*x+ c))^(1/2)/d/cos(d*x+c)^(1/2)+1/2*b*B*sin(d*x+c)*cos(d*x+c)^(1/2)*(b*cos(d* x+c))^(1/2)/d
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.57 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {b \sqrt {b \cos (c+d x)} (4 A \sin (c+d x)+B (2 (c+d x)+\sin (2 (c+d x))))}{4 d \sqrt {\cos (c+d x)}} \]
(b*Sqrt[b*Cos[c + d*x]]*(4*A*Sin[c + d*x] + B*(2*(c + d*x) + Sin[2*(c + d* x)])))/(4*d*Sqrt[Cos[c + d*x]])
Time = 0.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.61, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2031, 3042, 3213}
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.
\(\displaystyle \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \cos (c+d x) (A+B \cos (c+d x))dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {b \sqrt {b \cos (c+d x)} \left (\frac {A \sin (c+d x)}{d}+\frac {B \sin (c+d x) \cos (c+d x)}{2 d}+\frac {B x}{2}\right )}{\sqrt {\cos (c+d x)}}\) |
(b*Sqrt[b*Cos[c + d*x]]*((B*x)/2 + (A*Sin[c + d*x])/d + (B*Cos[c + d*x]*Si n[c + d*x])/(2*d)))/Sqrt[Cos[c + d*x]]
3.9.52.3.1 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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 4.90 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55
method | result | size |
default | \(\frac {b \sqrt {\cos \left (d x +c \right ) b}\, \left (B \sin \left (d x +c \right ) \cos \left (d x +c \right )+2 A \sin \left (d x +c \right )+B \left (d x +c \right )\right )}{2 d \sqrt {\cos \left (d x +c \right )}}\) | \(56\) |
parts | \(\frac {A b \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {B b \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}\) | \(75\) |
risch | \(\frac {b B x \sqrt {\cos \left (d x +c \right ) b}}{2 \sqrt {\cos \left (d x +c \right )}}+\frac {A b \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {b \sqrt {\cos \left (d x +c \right ) b}\, B \sin \left (2 d x +2 c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(89\) |
1/2*b/d*(cos(d*x+c)*b)^(1/2)*(B*sin(d*x+c)*cos(d*x+c)+2*A*sin(d*x+c)+B*(d* x+c))/cos(d*x+c)^(1/2)
Time = 0.32 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.07 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\left [\frac {B \sqrt {-b} b \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (B b \cos \left (d x + c\right ) + 2 \, A b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )}, \frac {B b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (B b \cos \left (d x + c\right ) + 2 \, A b\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}\right ] \]
[1/4*(B*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*(B*b*cos(d*x + c) + 2*A*b)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/2*(B*b^(3/2)*arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*co s(d*x + c)^(3/2)))*cos(d*x + c) + (B*b*cos(d*x + c) + 2*A*b)*sqrt(b*cos(d* x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]
Time = 31.58 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.50 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\begin {cases} \frac {A \left (b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}} \sin {\left (c + d x \right )}}{d \cos ^{\frac {3}{2}}{\left (c + d x \right )}} + \frac {B x \left (b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}} \sin ^{2}{\left (c + d x \right )}}{2 \cos ^{\frac {3}{2}}{\left (c + d x \right )}} + \frac {B x \left (b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}} \sqrt {\cos {\left (c + d x \right )}}}{2} + \frac {B \left (b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}} \sin {\left (c + d x \right )}}{2 d \sqrt {\cos {\left (c + d x \right )}}} & \text {for}\: d \neq 0 \\\frac {x \left (b \cos {\left (c \right )}\right )^{\frac {3}{2}} \left (A + B \cos {\left (c \right )}\right )}{\sqrt {\cos {\left (c \right )}}} & \text {otherwise} \end {cases} \]
Piecewise((A*(b*cos(c + d*x))**(3/2)*sin(c + d*x)/(d*cos(c + d*x)**(3/2)) + B*x*(b*cos(c + d*x))**(3/2)*sin(c + d*x)**2/(2*cos(c + d*x)**(3/2)) + B* x*(b*cos(c + d*x))**(3/2)*sqrt(cos(c + d*x))/2 + B*(b*cos(c + d*x))**(3/2) *sin(c + d*x)/(2*d*sqrt(cos(c + d*x))), Ne(d, 0)), (x*(b*cos(c))**(3/2)*(A + B*cos(c))/sqrt(cos(c)), True))
Time = 0.44 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.43 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {4 \, A b^{\frac {3}{2}} \sin \left (d x + c\right ) + {\left (2 \, {\left (d x + c\right )} b + b \sin \left (2 \, d x + 2 \, c\right )\right )} B \sqrt {b}}{4 \, d} \]
\[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
Time = 0.55 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.50 \[ \int \frac {(b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {b\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (4\,A\,\sin \left (c+d\,x\right )+B\,\sin \left (2\,c+2\,d\,x\right )+2\,B\,d\,x\right )}{4\,d\,\sqrt {\cos \left (c+d\,x\right )}} \]