Integrand size = 21, antiderivative size = 96 \[ \int (d \cos (a+b x))^{7/2} \csc ^2(a+b x) \, dx=-\frac {d (d \cos (a+b x))^{5/2} \csc (a+b x)}{b}-\frac {5 d^4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b \sqrt {d \cos (a+b x)}}-\frac {5 d^3 \sqrt {d \cos (a+b x)} \sin (a+b x)}{3 b} \]
-d*(d*cos(b*x+a))^(5/2)*csc(b*x+a)/b-5/3*d^4*(cos(1/2*a+1/2*b*x)^2)^(1/2)/ cos(1/2*a+1/2*b*x)*EllipticF(sin(1/2*a+1/2*b*x),2^(1/2))*cos(b*x+a)^(1/2)/ b/(d*cos(b*x+a))^(1/2)-5/3*d^3*sin(b*x+a)*(d*cos(b*x+a))^(1/2)/b
Time = 0.38 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.76 \[ \int (d \cos (a+b x))^{7/2} \csc ^2(a+b x) \, dx=\frac {d^3 \sqrt {d \cos (a+b x)} \left (\sqrt {\cos (a+b x)} (-4+\cos (2 (a+b x))) \csc (a+b x)-5 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )\right )}{3 b \sqrt {\cos (a+b x)}} \]
(d^3*Sqrt[d*Cos[a + b*x]]*(Sqrt[Cos[a + b*x]]*(-4 + Cos[2*(a + b*x)])*Csc[ a + b*x] - 5*EllipticF[(a + b*x)/2, 2]))/(3*b*Sqrt[Cos[a + b*x]])
Time = 0.44 (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, 3120}
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.
\(\displaystyle \int \csc ^2(a+b x) (d \cos (a+b x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \cos (a+b x))^{7/2}}{\sin (a+b x)^2}dx\) |
\(\Big \downarrow \) 3047 |
\(\displaystyle -\frac {5}{2} d^2 \int (d \cos (a+b x))^{3/2}dx-\frac {d \csc (a+b x) (d \cos (a+b x))^{5/2}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {5}{2} d^2 \int \left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{3/2}dx-\frac {d \csc (a+b x) (d \cos (a+b x))^{5/2}}{b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle -\frac {5}{2} d^2 \left (\frac {1}{3} d^2 \int \frac {1}{\sqrt {d \cos (a+b x)}}dx+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{5/2}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {5}{2} d^2 \left (\frac {1}{3} d^2 \int \frac {1}{\sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}}dx+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{5/2}}{b}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {5}{2} d^2 \left (\frac {d^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)}}dx}{3 \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{5/2}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {5}{2} d^2 \left (\frac {d^2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{5/2}}{b}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {5}{2} d^2 \left (\frac {2 d^2 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b \sqrt {d \cos (a+b x)}}+\frac {2 d \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b}\right )-\frac {d \csc (a+b x) (d \cos (a+b x))^{5/2}}{b}\) |
-((d*(d*Cos[a + b*x])^(5/2)*Csc[a + b*x])/b) - (5*d^2*((2*d^2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(3*b*Sqrt[d*Cos[a + b*x]]) + (2*d*Sqrt[d *Cos[a + b*x]]*Sin[a + b*x])/(3*b)))/2
3.3.34.3.1 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[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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]
Time = 2.44 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.25
method | result | size |
default | \(-\frac {\sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, d^{5} \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (-32 \left (\sin ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+10 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) {\left (2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}^{\frac {3}{2}} F\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}+64 \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-28 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+3\right )}{6 {\left (-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}^{\frac {3}{2}} \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b}\) | \(216\) |
-1/6*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^5/(-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)*(-32*sin(1/2*b*x+1/2*a)^8+10*cos(1/2*b*x+1/2*a)*(2*sin(1/2*b *x+1/2*a)^2-1)^(3/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))*(sin(1/2*b*x+1/ 2*a)^2)^(1/2)+64*sin(1/2*b*x+1/2*a)^6-28*sin(1/2*b*x+1/2*a)^4-4*sin(1/2*b* x+1/2*a)^2+3)/(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 = 0.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.14 \[ \int (d \cos (a+b x))^{7/2} \csc ^2(a+b x) \, dx=\frac {5 i \, \sqrt {2} d^{\frac {7}{2}} \sin \left (b x + a\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 5 i \, \sqrt {2} d^{\frac {7}{2}} \sin \left (b x + a\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 2 \, {\left (2 \, d^{3} \cos \left (b x + a\right )^{2} - 5 \, d^{3}\right )} \sqrt {d \cos \left (b x + a\right )}}{6 \, b \sin \left (b x + a\right )} \]
1/6*(5*I*sqrt(2)*d^(7/2)*sin(b*x + a)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) - 5*I*sqrt(2)*d^(7/2)*sin(b*x + a)*weierstrassPInver se(-4, 0, cos(b*x + a) - I*sin(b*x + a)) + 2*(2*d^3*cos(b*x + a)^2 - 5*d^3 )*sqrt(d*cos(b*x + a)))/(b*sin(b*x + a))
Timed out. \[ \int (d \cos (a+b x))^{7/2} \csc ^2(a+b x) \, dx=\text {Timed out} \]
\[ \int (d \cos (a+b x))^{7/2} \csc ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}} \csc \left (b x + a\right )^{2} \,d x } \]
\[ \int (d \cos (a+b x))^{7/2} \csc ^2(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}} \csc \left (b x + a\right )^{2} \,d x } \]
Timed out. \[ \int (d \cos (a+b x))^{7/2} \csc ^2(a+b x) \, dx=\int \frac {{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2}}{{\sin \left (a+b\,x\right )}^2} \,d x \]