Integrand size = 21, antiderivative size = 96 \[ \int (d \cos (a+b x))^{9/2} \csc ^2(a+b x) \, dx=-\frac {d (d \cos (a+b x))^{7/2} \csc (a+b x)}{b}-\frac {21 d^4 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b \sqrt {\cos (a+b x)}}-\frac {7 d^3 (d \cos (a+b x))^{3/2} \sin (a+b x)}{5 b} \]
-d*(d*cos(b*x+a))^(7/2)*csc(b*x+a)/b-7/5*d^3*(d*cos(b*x+a))^(3/2)*sin(b*x+ a)/b-21/5*d^4*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticE(si n(1/2*a+1/2*b*x),2^(1/2))*(d*cos(b*x+a))^(1/2)/b/cos(b*x+a)^(1/2)
Time = 0.48 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77 \[ \int (d \cos (a+b x))^{9/2} \csc ^2(a+b x) \, dx=-\frac {d^4 \sqrt {d \cos (a+b x)} \left (21 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )+\sqrt {\cos (a+b x)} (5 \cot (a+b x)+\sin (2 (a+b x)))\right )}{5 b \sqrt {\cos (a+b x)}} \]
-1/5*(d^4*Sqrt[d*Cos[a + b*x]]*(21*EllipticE[(a + b*x)/2, 2] + Sqrt[Cos[a + b*x]]*(5*Cot[a + b*x] + Sin[2*(a + b*x)])))/(b*Sqrt[Cos[a + b*x]])
Time = 0.46 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3047, 3042, 3115, 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.
\(\displaystyle \int \csc ^2(a+b x) (d \cos (a+b x))^{9/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \cos (a+b x))^{9/2}}{\sin (a+b x)^2}dx\) |
\(\Big \downarrow \) 3047 |
\(\displaystyle -\frac {7}{2} d^2 \int (d \cos (a+b x))^{5/2}dx-\frac {d \csc (a+b x) (d \cos (a+b x))^{7/2}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {7}{2} d^2 \int \left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{5/2}dx-\frac {d \csc (a+b x) (d \cos (a+b x))^{7/2}}{b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle -\frac {7}{2} d^2 \left (\frac {3}{5} d^2 \int \sqrt {d \cos (a+b x)}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{7/2}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {7}{2} d^2 \left (\frac {3}{5} d^2 \int \sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{7/2}}{b}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {7}{2} d^2 \left (\frac {3 d^2 \sqrt {d \cos (a+b x)} \int \sqrt {\cos (a+b x)}dx}{5 \sqrt {\cos (a+b x)}}+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{7/2}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {7}{2} d^2 \left (\frac {3 d^2 \sqrt {d \cos (a+b x)} \int \sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (a+b x)}}+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{7/2}}{b}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {7}{2} d^2 \left (\frac {6 d^2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{5 b \sqrt {\cos (a+b x)}}+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{7/2}}{b}\) |
-((d*(d*Cos[a + b*x])^(7/2)*Csc[a + b*x])/b) - (7*d^2*((6*d^2*Sqrt[d*Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2])/(5*b*Sqrt[Cos[a + b*x]]) + (2*d*(d*Cos [a + b*x])^(3/2)*Sin[a + b*x])/(5*b)))/2
3.3.33.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[a*(a*Cos[e + f*x])^(m - 1)*((b*Sin[e + f*x])^(n + 1)/ (b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1))) Int[(a*Cos[e + f*x] )^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ [m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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. \(228\) vs. \(2(110)=220\).
Time = 2.85 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.39
method | result | size |
default | \(\frac {\sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, d^{6} \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (-64 \left (\sin ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+160 \left (\sin ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+42 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) {\left (2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}^{\frac {3}{2}} E\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}-104 \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-4 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+22 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-5\right )}{10 {\left (-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}^{\frac {3}{2}} \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b}\) | \(229\) |
1/10*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^6/(-2*sin (1/2*b*x+1/2*a)^4*d+d*sin(1/2*b*x+1/2*a)^2)^(3/2)/cos(1/2*b*x+1/2*a)*sin(1 /2*b*x+1/2*a)*(-64*sin(1/2*b*x+1/2*a)^10+160*sin(1/2*b*x+1/2*a)^8+42*cos(1 /2*b*x+1/2*a)*(2*sin(1/2*b*x+1/2*a)^2-1)^(3/2)*EllipticE(cos(1/2*b*x+1/2*a ),2^(1/2))*(sin(1/2*b*x+1/2*a)^2)^(1/2)-104*sin(1/2*b*x+1/2*a)^6-4*sin(1/2 *b*x+1/2*a)^4+22*sin(1/2*b*x+1/2*a)^2-5)/(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1 /2)/b
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.26 \[ \int (d \cos (a+b x))^{9/2} \csc ^2(a+b x) \, dx=\frac {-21 i \, \sqrt {2} d^{\frac {9}{2}} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 21 i \, \sqrt {2} d^{\frac {9}{2}} \sin \left (b x + a\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + 2 \, {\left (2 \, d^{4} \cos \left (b x + a\right )^{3} - 7 \, d^{4} \cos \left (b x + a\right )\right )} \sqrt {d \cos \left (b x + a\right )}}{10 \, b \sin \left (b x + a\right )} \]
1/10*(-21*I*sqrt(2)*d^(9/2)*sin(b*x + a)*weierstrassZeta(-4, 0, weierstras sPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a))) + 21*I*sqrt(2)*d^(9/2)*si n(b*x + a)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a))) + 2*(2*d^4*cos(b*x + a)^3 - 7*d^4*cos(b*x + a))*sqrt(d* cos(b*x + a)))/(b*sin(b*x + a))
Timed out. \[ \int (d \cos (a+b x))^{9/2} \csc ^2(a+b x) \, dx=\text {Timed out} \]
\[ \int (d \cos (a+b x))^{9/2} \csc ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} \csc \left (b x + a\right )^{2} \,d x } \]
\[ \int (d \cos (a+b x))^{9/2} \csc ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} \csc \left (b x + a\right )^{2} \,d x } \]
Timed out. \[ \int (d \cos (a+b x))^{9/2} \csc ^2(a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{9/2}}{{\sin \left (a+b\,x\right )}^2} \,d x \]