Integrand size = 43, antiderivative size = 155 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {b^2 B x \sqrt {b \cos (c+d x)}}{2 \sqrt {\cos (c+d x)}}+\frac {b^2 (3 A+2 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {b^2 B \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)} \sin (c+d x)}{2 d}+\frac {b^2 C \cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d} \]
1/3*b^2*C*cos(d*x+c)^(3/2)*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+1/2*b^2*B*x*( b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)+1/3*b^2*(3*A+2*C)*sin(d*x+c)*(b*cos(d *x+c))^(1/2)/d/cos(d*x+c)^(1/2)+1/2*b^2*B*sin(d*x+c)*cos(d*x+c)^(1/2)*(b*c os(d*x+c))^(1/2)/d
Time = 1.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.48 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(b \cos (c+d x))^{5/2} (6 B c+6 B d x+3 (4 A+3 C) \sin (c+d x)+3 B \sin (2 (c+d x))+C \sin (3 (c+d x)))}{12 d \cos ^{\frac {5}{2}}(c+d x)} \]
Integrate[((b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)) /Cos[c + d*x]^(3/2),x]
((b*Cos[c + d*x])^(5/2)*(6*B*c + 6*B*d*x + 3*(4*A + 3*C)*Sin[c + d*x] + 3* B*Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)]))/(12*d*Cos[c + d*x]^(5/2))
Time = 0.36 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.63, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {2031, 3042, 3502, 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.
| \(\displaystyle \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 2031 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \cos (c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{3} \int \cos (c+d x) (3 A+2 C+3 B \cos (c+d x))dx+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{3} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (3 A+2 C+3 B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )}{\sqrt {\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3213 |
| \(\displaystyle \frac {b^2 \sqrt {b \cos (c+d x)} \left (\frac {1}{3} \left (\frac {(3 A+2 C) \sin (c+d x)}{d}+\frac {3 B \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 B x}{2}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )}{\sqrt {\cos (c+d x)}}\) |
(b^2*Sqrt[b*Cos[c + d*x]]*((C*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + ((3*B*x )/2 + ((3*A + 2*C)*Sin[c + d*x])/d + (3*B*Cos[c + d*x]*Sin[c + d*x])/(2*d) )/3))/Sqrt[Cos[c + d*x]]
3.4.8.3.1 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]
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 = 9.72 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.55
| method | result | size |
| default | \(\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (2 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+6 A \sin \left (d x +c \right )+3 B \left (d x +c \right )+4 \sin \left (d x +c \right ) C \right )}{6 d \sqrt {\cos \left (d x +c \right )}}\) | \(86\) |
| parts | \(\frac {A \,b^{2} \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{d \sqrt {\cos \left (d x +c \right )}}+\frac {B \,b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )+d x +c \right )}{2 d \sqrt {\cos \left (d x +c \right )}}+\frac {C \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\cos \left (d x +c \right ) b}}{3 d \sqrt {\cos \left (d x +c \right )}}\) | \(122\) |
| risch | \(\frac {b^{2} B x \sqrt {\cos \left (d x +c \right ) b}}{2 \sqrt {\cos \left (d x +c \right )}}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, \left (4 A +3 C \right ) \sin \left (d x +c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, C \sin \left (3 d x +3 c \right )}{12 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {b^{2} \sqrt {\cos \left (d x +c \right ) b}\, B \sin \left (2 d x +2 c \right )}{4 \sqrt {\cos \left (d x +c \right )}\, d}\) | \(138\) |
int((cos(d*x+c)*b)^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2), x,method=_RETURNVERBOSE)
1/6*b^2/d*(cos(d*x+c)*b)^(1/2)*(2*C*cos(d*x+c)^2*sin(d*x+c)+3*B*sin(d*x+c) *cos(d*x+c)+6*A*sin(d*x+c)+3*B*(d*x+c)+4*sin(d*x+c)*C)/cos(d*x+c)^(1/2)
Time = 0.30 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.70 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {3 \, B \sqrt {-b} b^{2} \cos \left (d x + c\right ) \log \left (2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) + 2 \, {\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + 3 \, B b^{2} \cos \left (d x + c\right ) + 2 \, {\left (3 \, A + 2 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )}, \frac {3 \, B b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right ) + {\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + 3 \, B b^{2} \cos \left (d x + c\right ) + 2 \, {\left (3 \, A + 2 \, C\right )} b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )}\right ] \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (3/2),x, algorithm="fricas")
[1/12*(3*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*(2*C*b^2*cos( d*x + c)^2 + 3*B*b^2*cos(d*x + c) + 2*(3*A + 2*C)*b^2)*sqrt(b*cos(d*x + c) )*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)), 1/6*(3*B*b^(5/2)*arct an(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c) + (2*C*b^2*cos(d*x + c)^2 + 3*B*b^2*cos(d*x + c) + 2*(3*A + 2*C)*b^2 )*sqrt(b*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))]
Timed out. \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Time = 0.53 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.61 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {12 \, A b^{\frac {5}{2}} \sin \left (d x + c\right ) + 3 \, {\left (2 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} B \sqrt {b} + {\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )\right )} C \sqrt {b}}{12 \, d} \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (3/2),x, algorithm="maxima")
1/12*(12*A*b^(5/2)*sin(d*x + c) + 3*(2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c ))*B*sqrt(b) + (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))))*C*sqrt(b))/d
\[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ (3/2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)/c os(d*x + c)^(3/2), x)
Time = 0.79 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.47 \[ \int \frac {(b \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {b^2\,\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (12\,A\,\sin \left (c+d\,x\right )+9\,C\,\sin \left (c+d\,x\right )+3\,B\,\sin \left (2\,c+2\,d\,x\right )+C\,\sin \left (3\,c+3\,d\,x\right )+6\,B\,d\,x\right )}{12\,d\,\sqrt {\cos \left (c+d\,x\right )}} \]