Integrand size = 10, antiderivative size = 65 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(a+b x)} \, dx=-\frac {6 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b}+\frac {2 \sin (a+b x)}{5 b \cos ^{\frac {5}{2}}(a+b x)}+\frac {6 \sin (a+b x)}{5 b \sqrt {\cos (a+b x)}} \]
-6/5*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticE(sin(1/2*a+1 /2*b*x),2^(1/2))/b+2/5*sin(b*x+a)/b/cos(b*x+a)^(5/2)+6/5*sin(b*x+a)/b/cos( b*x+a)^(1/2)
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(a+b x)} \, dx=\frac {-6 \cos ^{\frac {3}{2}}(a+b x) E\left (\left .\frac {1}{2} (a+b x)\right |2\right )+3 \sin (2 (a+b x))+2 \tan (a+b x)}{5 b \cos ^{\frac {3}{2}}(a+b x)} \]
(-6*Cos[a + b*x]^(3/2)*EllipticE[(a + b*x)/2, 2] + 3*Sin[2*(a + b*x)] + 2* Tan[a + b*x])/(5*b*Cos[a + b*x]^(3/2))
Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3116, 3042, 3116, 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.
\(\displaystyle \int \frac {1}{\cos ^{\frac {7}{2}}(a+b x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (a+b x+\frac {\pi }{2}\right )^{7/2}}dx\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {3}{5} \int \frac {1}{\cos ^{\frac {3}{2}}(a+b x)}dx+\frac {2 \sin (a+b x)}{5 b \cos ^{\frac {5}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{5} \int \frac {1}{\sin \left (a+b x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 \sin (a+b x)}{5 b \cos ^{\frac {5}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {3}{5} \left (\frac {2 \sin (a+b x)}{b \sqrt {\cos (a+b x)}}-\int \sqrt {\cos (a+b x)}dx\right )+\frac {2 \sin (a+b x)}{5 b \cos ^{\frac {5}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{5} \left (\frac {2 \sin (a+b x)}{b \sqrt {\cos (a+b x)}}-\int \sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}dx\right )+\frac {2 \sin (a+b x)}{5 b \cos ^{\frac {5}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 \sin (a+b x)}{5 b \cos ^{\frac {5}{2}}(a+b x)}+\frac {3}{5} \left (\frac {2 \sin (a+b x)}{b \sqrt {\cos (a+b x)}}-\frac {2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b}\right )\) |
(2*Sin[a + b*x])/(5*b*Cos[a + b*x]^(5/2)) + (3*((-2*EllipticE[(a + b*x)/2, 2])/b + (2*Sin[a + b*x])/(b*Sqrt[Cos[a + b*x]])))/5
3.1.16.3.1 Defintions of rubi rules used
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]
Leaf count of result is larger than twice the leaf count of optimal. \(357\) vs. \(2(81)=162\).
Time = 2.31 (sec) , antiderivative size = 358, normalized size of antiderivative = 5.51
method | result | size |
default | \(-\frac {2 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \left (24 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-12 \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, E\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-24 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+12 \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, E\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+8 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, E\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}}{5 \left (8 \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{3} \sqrt {-1+2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, b}\) | \(358\) |
-2/5*(-(-2*cos(1/2*b*x+1/2*a)^2+1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/(8*sin(1/2* b*x+1/2*a)^6-12*sin(1/2*b*x+1/2*a)^4+6*sin(1/2*b*x+1/2*a)^2-1)/sin(1/2*b*x +1/2*a)^3*(24*cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)^6-12*(2*sin(1/2*b*x+1/ 2*a)^2-1)^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))*(sin(1/2*b*x+1/2*a)^ 2)^(1/2)*sin(1/2*b*x+1/2*a)^4-24*sin(1/2*b*x+1/2*a)^4*cos(1/2*b*x+1/2*a)+1 2*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))*( sin(1/2*b*x+1/2*a)^2)^(1/2)*sin(1/2*b*x+1/2*a)^2+8*sin(1/2*b*x+1/2*a)^2*co s(1/2*b*x+1/2*a)-3*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1) ^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2)))*(-2*sin(1/2*b*x+1/2*a)^4+sin (1/2*b*x+1/2*a)^2)^(1/2)/(-1+2*cos(1/2*b*x+1/2*a)^2)^(1/2)/b
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.69 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(a+b x)} \, dx=\frac {-3 i \, \sqrt {2} \cos \left (b x + a\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 i \, \sqrt {2} \cos \left (b x + a\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + 2 \, {\left (3 \, \cos \left (b x + a\right )^{2} + 1\right )} \sqrt {\cos \left (b x + a\right )} \sin \left (b x + a\right )}{5 \, b \cos \left (b x + a\right )^{3}} \]
1/5*(-3*I*sqrt(2)*cos(b*x + a)^3*weierstrassZeta(-4, 0, weierstrassPInvers e(-4, 0, cos(b*x + a) + I*sin(b*x + a))) + 3*I*sqrt(2)*cos(b*x + a)^3*weie rstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a ))) + 2*(3*cos(b*x + a)^2 + 1)*sqrt(cos(b*x + a))*sin(b*x + a))/(b*cos(b*x + a)^3)
Timed out. \[ \int \frac {1}{\cos ^{\frac {7}{2}}(a+b x)} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\cos ^{\frac {7}{2}}(a+b x)} \, dx=\int { \frac {1}{\cos \left (b x + a\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {1}{\cos ^{\frac {7}{2}}(a+b x)} \, dx=\int { \frac {1}{\cos \left (b x + a\right )^{\frac {7}{2}}} \,d x } \]
Time = 14.85 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(a+b x)} \, dx=\frac {2\,\sin \left (a+b\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (a+b\,x\right )}^2\right )}{5\,b\,{\cos \left (a+b\,x\right )}^{5/2}\,\sqrt {{\sin \left (a+b\,x\right )}^2}} \]