Integrand size = 21, antiderivative size = 102 \[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=\frac {8 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{7 b d^4 \sqrt {d \cos (a+b x)}}-\frac {4 \sin (a+b x)}{7 b d^3 (d \cos (a+b x))^{3/2}}+\frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}} \]
-4/7*sin(b*x+a)/b/d^3/(d*cos(b*x+a))^(3/2)+2/7*sin(b*x+a)^3/b/d/(d*cos(b*x +a))^(7/2)+8/7*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticF(s in(1/2*a+1/2*b*x),2^(1/2))*cos(b*x+a)^(1/2)/b/d^4/(d*cos(b*x+a))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.64 \[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=\frac {\cos ^3(a+b x) \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {11}{4},\frac {7}{2},\sin ^2(a+b x)\right ) \sin ^5(a+b x)}{5 b (d \cos (a+b x))^{9/2}} \]
(Cos[a + b*x]^3*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[5/2, 11/4, 7/2, S in[a + b*x]^2]*Sin[a + b*x]^5)/(5*b*(d*Cos[a + b*x])^(9/2))
Time = 0.48 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3046, 3042, 3046, 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.
\(\displaystyle \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{9/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (a+b x)^4}{(d \cos (a+b x))^{9/2}}dx\) |
\(\Big \downarrow \) 3046 |
\(\displaystyle \frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}}-\frac {6 \int \frac {\sin ^2(a+b x)}{(d \cos (a+b x))^{5/2}}dx}{7 d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}}-\frac {6 \int \frac {\sin (a+b x)^2}{(d \cos (a+b x))^{5/2}}dx}{7 d^2}\) |
\(\Big \downarrow \) 3046 |
\(\displaystyle \frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}}-\frac {6 \left (\frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 \int \frac {1}{\sqrt {d \cos (a+b x)}}dx}{3 d^2}\right )}{7 d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}}-\frac {6 \left (\frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 \int \frac {1}{\sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 d^2}\right )}{7 d^2}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}}-\frac {6 \left (\frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)}}dx}{3 d^2 \sqrt {d \cos (a+b x)}}\right )}{7 d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}}-\frac {6 \left (\frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 d^2 \sqrt {d \cos (a+b x)}}\right )}{7 d^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 \sin ^3(a+b x)}{7 b d (d \cos (a+b x))^{7/2}}-\frac {6 \left (\frac {2 \sin (a+b x)}{3 b d (d \cos (a+b x))^{3/2}}-\frac {4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b d^2 \sqrt {d \cos (a+b x)}}\right )}{7 d^2}\) |
(2*Sin[a + b*x]^3)/(7*b*d*(d*Cos[a + b*x])^(7/2)) - (6*((-4*Sqrt[Cos[a + b *x]]*EllipticF[(a + b*x)/2, 2])/(3*b*d^2*Sqrt[d*Cos[a + b*x]]) + (2*Sin[a + b*x])/(3*b*d*(d*Cos[a + b*x])^(3/2))))/(7*d^2)
3.3.20.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(a*Sin[e + f*x])^(m - 1)*((b*Cos[e + f*x])^(n + 1)/(b*f*(n + 1))), x] + Simp[a^2*((m - 1)/(b^2*(n + 1))) Int[(a*Sin[e + f *x])^(m - 2)*(b*Cos[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[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]
Leaf count of result is larger than twice the leaf count of optimal. \(397\) vs. \(2(114)=228\).
Time = 0.53 (sec) , antiderivative size = 398, normalized size of antiderivative = 3.90
method | result | size |
default | \(\frac {8 \left (8 \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}\, F\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-6 \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-12 \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}\, F\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+6 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+6 \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}\, F\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\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}\, F\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right ) \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 )}}{7 d^{4} {\left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}^{3} \sqrt {-d \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \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}\) | \(398\) |
8/7*(8*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*Ellip ticF(cos(1/2*b*x+1/2*a),2^(1/2))*sin(1/2*b*x+1/2*a)^6-6*sin(1/2*b*x+1/2*a) ^6*cos(1/2*b*x+1/2*a)-12*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a )^2-1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))*sin(1/2*b*x+1/2*a)^4+6* cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)^4+6*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2* sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))*sin(1/ 2*b*x+1/2*a)^2-sin(1/2*b*x+1/2*a)^2*cos(1/2*b*x+1/2*a)-(sin(1/2*b*x+1/2*a) ^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^ (1/2)))/d^4*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/(2*c os(1/2*b*x+1/2*a)^2-1)^3/(-d*(2*sin(1/2*b*x+1/2*a)^4-sin(1/2*b*x+1/2*a)^2) )^(1/2)/sin(1/2*b*x+1/2*a)/(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 = 114, normalized size of antiderivative = 1.12 \[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=-\frac {2 \, {\left (2 i \, \sqrt {2} \sqrt {d} \cos \left (b x + a\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 2 i \, \sqrt {2} \sqrt {d} \cos \left (b x + a\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + \sqrt {d \cos \left (b x + a\right )} {\left (3 \, \cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )\right )}}{7 \, b d^{5} \cos \left (b x + a\right )^{4}} \]
-2/7*(2*I*sqrt(2)*sqrt(d)*cos(b*x + a)^4*weierstrassPInverse(-4, 0, cos(b* x + a) + I*sin(b*x + a)) - 2*I*sqrt(2)*sqrt(d)*cos(b*x + a)^4*weierstrassP Inverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)) + sqrt(d*cos(b*x + a))*(3*co s(b*x + a)^2 - 1)*sin(b*x + a))/(b*d^5*cos(b*x + a)^4)
Timed out. \[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=\int { \frac {\sin \left (b x + a\right )^{4}}{\left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}}} \,d x } \]
\[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=\int { \frac {\sin \left (b x + a\right )^{4}}{\left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sin ^4(a+b x)}{(d \cos (a+b x))^{9/2}} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^4}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{9/2}} \,d x \]