Integrand size = 33, antiderivative size = 107 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {b^2 B x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {A b^2 \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {b^2 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d} \] Output:
1/2*b^2*B*x*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+A*b^2*(b*cos(d*x+c))^(1/ 2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+1/2*b^2*B*cos(d*x+c)^(1/2)*(b*cos(d*x+c)) ^(1/2)*sin(d*x+c)/d
Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.53 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{5/2} (4 A \sin (c+d x)+B (2 (c+d x)+\sin (2 (c+d x))))}{4 d \cos ^{\frac {5}{2}}(c+d x)} \] Input:
Integrate[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(3/2) ,x]
Output:
((b*Cos[c + d*x])^(5/2)*(4*A*Sin[c + d*x] + B*(2*(c + d*x) + Sin[2*(c + d* x)])))/(4*d*Cos[c + d*x]^(5/2))
Time = 0.25 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.60, 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.
\(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 2031 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \cos (c+d x) (A+B \cos (c+d x))dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{\sqrt {\cos (c+d x)}}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {A \sin (c+d x)}{d}+\frac {B \sin (c+d x) \cos (c+d x)}{2 d}+\frac {B x}{2}\right )}{\sqrt {\cos (c+d x)}}\) |
Input:
Int[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(3/2),x]
Output:
(b^2*Sqrt[b*Cos[c + d*x]]*((B*x)/2 + (A*Sin[c + d*x])/d + (B*Cos[c + d*x]* Sin[c + d*x])/(2*d)))/Sqrt[Cos[c + d*x]]
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 = 10.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.54
method | result | size |
default | \(\frac {b^{2} \left (B \cos \left (d x +c \right ) \sin \left (d x +c \right )+2 A \sin \left (d x +c \right )+B \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right ) b}}{2 d \sqrt {\cos \left (d x +c \right )}}\) | \(58\) |
parts | \(\frac {A \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \sin \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}+\frac {B \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right ) b^{2} \sqrt {\cos \left (d x +c \right ) b}}{2 d \sqrt {\cos \left (d x +c \right )}}\) | \(79\) |
risch | \(\frac {b^{2} B x \sqrt {\cos \left (d x +c \right ) b}}{2 \sqrt {\cos \left (d x +c \right )}}+\frac {A \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \sin \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, B \sin \left (2 d x +2 c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(95\) |
Input:
int((cos(d*x+c)*b)^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x,method=_RETUR NVERBOSE)
Output:
1/2*b^2/d*(B*cos(d*x+c)*sin(d*x+c)+2*A*sin(d*x+c)+B*(d*x+c))*(cos(d*x+c)*b )^(1/2)/cos(d*x+c)^(1/2)
Time = 0.11 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.05 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {B \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (B b^{2} \cos \left (d x + c\right ) + 2 \, A b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )}, \frac {B b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (B b^{2} \cos \left (d x + c\right ) + 2 \, A b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}\right ] \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algori thm="fricas")
Output:
[1/4*(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*(B*b^2*cos(d*x + c) + 2*A*b^2)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*co s(d*x + c)), 1/2*(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) + (B*b^2*cos(d*x + c) + 2*A*b^2)*sqr t(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(3/2),x)
Output:
Timed out
Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.44 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4 \, A b^{\frac {5}{2}} \sin \left (d x + c\right ) + {\left (2 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} B \sqrt {b}}{4 \, d} \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algori thm="maxima")
Output:
1/4*(4*A*b^(5/2)*sin(d*x + c) + (2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c))*B *sqrt(b))/d
Result contains complex when optimal does not.
Time = 0.36 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.34 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {{\left (-i \, B b^{2} \log \left (i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + i \, B b^{2} \log \left (-i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {2 \, {\left (2 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}\right )} \sqrt {b}}{2 \, d} \] Input:
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algori thm="giac")
Output:
-1/2*(-I*B*b^2*log(I*tan(1/2*d*x + 1/2*c) - 1) + I*B*b^2*log(-I*tan(1/2*d* x + 1/2*c) - 1) - 2*(2*A*b^2*tan(1/2*d*x + 1/2*c)^3 - B*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*A*b^2*tan(1/2*d*x + 1/2*c) + B*b^2*tan(1/2*d*x + 1/2*c))/(tan (1/2*d*x + 1/2*c)^4 + 2*tan(1/2*d*x + 1/2*c)^2 + 1))*sqrt(b)/d
Time = 41.61 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.49 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (4\,A\,\sin \left (c+d\,x\right )+B\,\sin \left (2\,c+2\,d\,x\right )+2\,B\,d\,x\right )}{4\,d\,\sqrt {\cos \left (c+d\,x\right )}} \] Input:
int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x)))/cos(c + d*x)^(3/2),x)
Output:
(b^2*(b*cos(c + d*x))^(1/2)*(4*A*sin(c + d*x) + B*sin(2*c + 2*d*x) + 2*B*d *x))/(4*d*cos(c + d*x)^(1/2))
Time = 0.17 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.36 \[ \int \frac {(b \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {b}\, b^{2} \left (\cos \left (d x +c \right ) \sin \left (d x +c \right ) b +2 \sin \left (d x +c \right ) a +b d x \right )}{2 d} \] Input:
int((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x)
Output:
(sqrt(b)*b**2*(cos(c + d*x)*sin(c + d*x)*b + 2*sin(c + d*x)*a + b*d*x))/(2 *d)