Integrand size = 25, antiderivative size = 119 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {64 a^3 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{5 d} \]
2/5*a*cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+64/15*a^3*sin(d *x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/15*a^2*sin(d*x+c)*cos(d *x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)/d
Time = 0.18 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.54 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {a^2 \sqrt {\cos (c+d x)} (89+28 \cos (c+d x)+3 \cos (2 (c+d x))) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{15 d} \]
(a^2*Sqrt[Cos[c + d*x]]*(89 + 28*Cos[c + d*x] + 3*Cos[2*(c + d*x)])*Sqrt[a *(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(15*d)
Time = 0.69 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4752, 3042, 4296, 3042, 4296, 3042, 4291}
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.
\(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}dx\) |
\(\Big \downarrow \) 4752 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\sec (c+d x) a+a)^{5/2}}{\sec ^{\frac {5}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4296 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {8}{5} a \int \frac {(\sec (c+d x) a+a)^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {8}{5} a \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4296 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {8}{5} a \left (\frac {4}{3} a \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {8}{5} a \left (\frac {4}{3} a \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4291 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {8}{5} a \left (\frac {8 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a*(a + a*Sec[c + d*x])^(3/2)*Sin [c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (8*a*((8*a^2*Sqrt[Sec[c + d*x]]*Sin[ c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[a + a*Sec[c + d*x]]*S in[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])))/5)
3.5.14.3.1 Defintions of rubi rules used
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] *(d_.)], x_Symbol] :> Simp[-2*a*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*S qrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-a)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1) *((d*Csc[e + f*x])^n/(f*m)), x] + Simp[b*((2*m - 1)/(d*m)) Int[(a + b*Csc [e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f , m, n}, x] && EqQ[a^2 - b^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2] && Integer Q[2*m]
Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Time = 1.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.55
\[-\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\cos \left (d x +c \right )}\, \left (3 \cos \left (d x +c \right )^{3}+11 \cos \left (d x +c \right )^{2}+29 \cos \left (d x +c \right )-43\right ) \csc \left (d x +c \right )}{15 d}\]
-2/15/d*a^2*(a*(1+sec(d*x+c)))^(1/2)*cos(d*x+c)^(1/2)*(3*cos(d*x+c)^3+11*c os(d*x+c)^2+29*cos(d*x+c)-43)*csc(d*x+c)
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.66 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{2} + 14 \, a^{2} \cos \left (d x + c\right ) + 43 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
2/15*(3*a^2*cos(d*x + c)^2 + 14*a^2*cos(d*x + c) + 43*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c) + d)
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \]
Time = 0.37 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.50 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {{\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 25 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 150 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \]
1/30*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d*x + 3/ 2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c))*sqrt(a)/d
\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int {\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]