Integrand size = 21, antiderivative size = 99 \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {8 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{7 b \sqrt {d \cos (a+b x)}}-\frac {4 \sqrt {d \cos (a+b x)} \sin (a+b x)}{7 b d}-\frac {2 \sqrt {d \cos (a+b x)} \sin ^3(a+b x)}{7 b d} \]
8/7*(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)-4/7*sin(b*x+a)*(d* cos(b*x+a))^(1/2)/b/d-2/7*sin(b*x+a)^3*(d*cos(b*x+a))^(1/2)/b/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59 \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {d \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{2},\sin ^2(a+b x)\right ) \sin ^5(a+b x)}{5 b (d \cos (a+b x))^{3/2}} \]
(d*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[3/4, 5/2, 7/2, Sin[a + b*x]^2] *Sin[a + b*x]^5)/(5*b*(d*Cos[a + b*x])^(3/2))
Time = 0.45 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3048, 3042, 3048, 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.
\(\displaystyle \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (a+b x)^4}{\sqrt {d \cos (a+b x)}}dx\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {6}{7} \int \frac {\sin ^2(a+b x)}{\sqrt {d \cos (a+b x)}}dx-\frac {2 \sin ^3(a+b x) \sqrt {d \cos (a+b x)}}{7 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6}{7} \int \frac {\sin (a+b x)^2}{\sqrt {d \cos (a+b x)}}dx-\frac {2 \sin ^3(a+b x) \sqrt {d \cos (a+b x)}}{7 b d}\) |
\(\Big \downarrow \) 3048 |
\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \int \frac {1}{\sqrt {d \cos (a+b x)}}dx-\frac {2 \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b d}\right )-\frac {2 \sin ^3(a+b x) \sqrt {d \cos (a+b x)}}{7 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \int \frac {1}{\sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}}dx-\frac {2 \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b d}\right )-\frac {2 \sin ^3(a+b x) \sqrt {d \cos (a+b x)}}{7 b d}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {6}{7} \left (\frac {2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\cos (a+b x)}}dx}{3 \sqrt {d \cos (a+b x)}}-\frac {2 \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b d}\right )-\frac {2 \sin ^3(a+b x) \sqrt {d \cos (a+b x)}}{7 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6}{7} \left (\frac {2 \sqrt {\cos (a+b x)} \int \frac {1}{\sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {d \cos (a+b x)}}-\frac {2 \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b d}\right )-\frac {2 \sin ^3(a+b x) \sqrt {d \cos (a+b x)}}{7 b d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {6}{7} \left (\frac {4 \sqrt {\cos (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{3 b \sqrt {d \cos (a+b x)}}-\frac {2 \sin (a+b x) \sqrt {d \cos (a+b x)}}{3 b d}\right )-\frac {2 \sin ^3(a+b x) \sqrt {d \cos (a+b x)}}{7 b d}\) |
(-2*Sqrt[d*Cos[a + b*x]]*Sin[a + b*x]^3)/(7*b*d) + (6*((4*Sqrt[Cos[a + b*x ]]*EllipticF[(a + b*x)/2, 2])/(3*b*Sqrt[d*Cos[a + b*x]]) - (2*Sqrt[d*Cos[a + b*x]]*Sin[a + b*x])/(3*b*d)))/7
3.3.16.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n)) Int[(b*Cos[e + f*x])^n *(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*m, 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 = 0.39 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.10
method | result | size |
default | \(-\frac {8 \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \left (4 \left (\sin ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-6 \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, F\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right )}{7 \sqrt {-d \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b}\) | \(208\) |
-8/7*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*(4*sin(1/2* b*x+1/2*a)^8*cos(1/2*b*x+1/2*a)-6*sin(1/2*b*x+1/2*a)^6*cos(1/2*b*x+1/2*a)+ 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*si n(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2)))/(-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 = 88, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {2 \, {\left (\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right )^{2} - 3\right )} \sin \left (b x + a\right ) - 2 i \, \sqrt {2} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 2 i \, \sqrt {2} \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}}{7 \, b d} \]
2/7*(sqrt(d*cos(b*x + a))*(cos(b*x + a)^2 - 3)*sin(b*x + a) - 2*I*sqrt(2)* sqrt(d)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) + 2*I*sq rt(2)*sqrt(d)*weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)))/( b*d)
Timed out. \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\int { \frac {\sin \left (b x + a\right )^{4}}{\sqrt {d \cos \left (b x + a\right )}} \,d x } \]
\[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\int { \frac {\sin \left (b x + a\right )^{4}}{\sqrt {d \cos \left (b x + a\right )}} \,d x } \]
Timed out. \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^4}{\sqrt {d\,\cos \left (a+b\,x\right )}} \,d x \]