Integrand size = 21, antiderivative size = 125 \[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=-\frac {14 \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b^3 d \sqrt {\cos (c+d x)}}+\frac {2 b^2 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {14 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {14 \sin (c+d x)}{15 b^2 d \sqrt {b \cos (c+d x)}} \]
2/9*b^2*sin(d*x+c)/d/(b*cos(d*x+c))^(9/2)+14/45*sin(d*x+c)/d/(b*cos(d*x+c) )^(5/2)+14/15*sin(d*x+c)/b^2/d/(b*cos(d*x+c))^(1/2)-14/15*(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*c os(d*x+c))^(1/2)/b^3/d/cos(d*x+c)^(1/2)
Time = 0.39 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.64 \[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {-42 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+42 \sin (c+d x)+2 \sec (c+d x) \left (7+5 \sec ^2(c+d x)\right ) \tan (c+d x)}{45 b^2 d \sqrt {b \cos (c+d x)}} \]
(-42*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 42*Sin[c + d*x] + 2*Se c[c + d*x]*(7 + 5*Sec[c + d*x]^2)*Tan[c + d*x])/(45*b^2*d*Sqrt[b*Cos[c + d *x]])
Time = 0.59 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.15, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle b^3 \int \frac {1}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{11/2}}dx\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle b^3 \left (\frac {7 \int \frac {1}{(b \cos (c+d x))^{7/2}}dx}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^3 \left (\frac {7 \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle b^3 \left (\frac {7 \left (\frac {3 \int \frac {1}{(b \cos (c+d x))^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^3 \left (\frac {7 \left (\frac {3 \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle b^3 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \cos (c+d x)}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^3 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle b^3 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^3 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle b^3 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\) |
b^3*((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))
3.2.38.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(415\) vs. \(2(133)=266\).
Time = 3.17 (sec) , antiderivative size = 416, normalized size of antiderivative = 3.33
| method | result | size |
| default | \(-\frac {2 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}{144 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{5}}-\frac {7 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}{180 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{3}}-\frac {14 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}+\frac {7 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{15 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}-\frac {7 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}\right )}{b^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) | \(416\) |
-2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)/b^2*(-1/144 *cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^( 1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/ 2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^3-1 4/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1) *b*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1 /2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2) )^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-7/15*(sin(1/2*d*x+1/2*c)^2)^ (1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/ 2*d*x+1/2*c)^2))^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(co s(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/((2*cos(1/2*d*x+1/2*c)^2-1) *b)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\frac {-21 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (21 \, \cos \left (d x + c\right )^{4} + 7 \, \cos \left (d x + c\right )^{2} + 5\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{45 \, b^{3} d \cos \left (d x + c\right )^{5}} \]
1/45*(-21*I*sqrt(2)*sqrt(b)*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)*sqrt(b)* cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(21*cos(d*x + c)^4 + 7*cos(d*x + c)^2 + 5)*sqrt (b*cos(d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^5)
Timed out. \[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^3(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]