Integrand size = 21, antiderivative size = 128 \[ \int (b \cos (c+d x))^{5/2} \sec ^8(c+d x) \, dx=-\frac {14 b^2 \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {2 b^7 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {14 b^5 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {14 b^3 \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}} \]
2/9*b^7*sin(d*x+c)/d/(b*cos(d*x+c))^(9/2)+14/45*b^5*sin(d*x+c)/d/(b*cos(d* x+c))^(5/2)+14/15*b^3*sin(d*x+c)/d/(b*cos(d*x+c))^(1/2)-14/15*b^2*(cos(1/2 *d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/ 2))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)
Time = 0.85 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.62 \[ \int (b \cos (c+d x))^{5/2} \sec ^8(c+d x) \, dx=\frac {(b \cos (c+d x))^{5/2} \sec ^7(c+d x) \left (-336 \cos ^{\frac {9}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+150 \sin (c+d x)+91 \sin (3 (c+d x))+21 \sin (5 (c+d x))\right )}{360 d} \]
((b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^7*(-336*Cos[c + d*x]^(9/2)*EllipticE[ (c + d*x)/2, 2] + 150*Sin[c + d*x] + 91*Sin[3*(c + d*x)] + 21*Sin[5*(c + d *x)]))/(360*d)
Time = 0.60 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 2030, 3116, 3042, 3116, 3042, 3116, 3042, 3121, 3042, 3119}
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.
\(\displaystyle \int \sec ^8(c+d x) (b \cos (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )^8}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^8 \int \frac {1}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{11/2}}dx\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle b^8 \left (\frac {7 \int \frac {1}{(b \cos (c+d x))^{7/2}}dx}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^8 \left (\frac {7 \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle b^8 \left (\frac {7 \left (\frac {3 \int \frac {1}{(b \cos (c+d x))^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^8 \left (\frac {7 \left (\frac {3 \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle b^8 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \cos (c+d x)}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^8 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle b^8 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^8 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle b^8 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
b^8*((2*Sin[c + d*x])/(9*b*d*(b*Cos[c + d*x])^(9/2)) + (7*((2*Sin[c + d*x] )/(5*b*d*(b*Cos[c + d*x])^(5/2)) + (3*((-2*Sqrt[b*Cos[c + d*x]]*EllipticE[ (c + d*x)/2, 2])/(b^2*d*Sqrt[Cos[c + d*x]]) + (2*Sin[c + d*x])/(b*d*Sqrt[b *Cos[c + d*x]])))/(5*b^2)))/(9*b^2))
3.2.1.3.1 Defintions of rubi rules used
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Timed out.
hanged
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.08 \[ \int (b \cos (c+d x))^{5/2} \sec ^8(c+d x) \, dx=\frac {-21 i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} b^{\frac {5}{2}} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (21 \, b^{2} \cos \left (d x + c\right )^{4} + 7 \, b^{2} \cos \left (d x + c\right )^{2} + 5 \, b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{45 \, d \cos \left (d x + c\right )^{5}} \]
1/45*(-21*I*sqrt(2)*b^(5/2)*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstr assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*I*sqrt(2)*b^(5/2)* cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(21*b^2*cos(d*x + c)^4 + 7*b^2*cos(d*x + c)^2 + 5*b^2)*sqrt(b*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)
Timed out. \[ \int (b \cos (c+d x))^{5/2} \sec ^8(c+d x) \, dx=\text {Timed out} \]
\[ \int (b \cos (c+d x))^{5/2} \sec ^8(c+d x) \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{8} \,d x } \]
\[ \int (b \cos (c+d x))^{5/2} \sec ^8(c+d x) \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{8} \,d x } \]
Timed out. \[ \int (b \cos (c+d x))^{5/2} \sec ^8(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^8} \,d x \]