Integrand size = 21, antiderivative size = 95 \[ \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx=\frac {8 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{7 f}-\frac {4 b \sin (e+f x)}{7 f \sqrt {b \sec (e+f x)}}-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}} \]
-4/7*b*sin(f*x+e)/f/(b*sec(f*x+e))^(1/2)-2/7*b*sin(f*x+e)^3/f/(b*sec(f*x+e ))^(1/2)+8/7*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin (1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(b*sec(f*x+e))^(1/2)/f
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.64 \[ \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx=\frac {\sqrt {b \sec (e+f x)} \left (32 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )-10 \sin (2 (e+f x))+\sin (4 (e+f x))\right )}{28 f} \]
(Sqrt[b*Sec[e + f*x]]*(32*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] - 1 0*Sin[2*(e + f*x)] + Sin[4*(e + f*x)]))/(28*f)
Time = 0.45 (sec) , antiderivative size = 100, 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, 3107, 3042, 3107, 3042, 4258, 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.
\(\displaystyle \int \sin ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {b \sec (e+f x)}}{\csc (e+f x)^4}dx\) |
\(\Big \downarrow \) 3107 |
\(\displaystyle \frac {6}{7} \int \sqrt {b \sec (e+f x)} \sin ^2(e+f x)dx-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6}{7} \int \frac {\sqrt {b \sec (e+f x)}}{\csc (e+f x)^2}dx-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3107 |
\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \int \sqrt {b \sec (e+f x)}dx-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \int \sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}dx-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {6}{7} \left (\frac {2}{3} \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {6}{7} \left (\frac {4 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{3 f}-\frac {2 b \sin (e+f x)}{3 f \sqrt {b \sec (e+f x)}}\right )-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}\) |
(-2*b*Sin[e + f*x]^3)/(7*f*Sqrt[b*Sec[e + f*x]]) + (6*((4*Sqrt[Cos[e + f*x ]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(3*f) - (2*b*Sin[e + f* x])/(3*f*Sqrt[b*Sec[e + f*x]])))/7
3.4.79.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n _.), x_Symbol] :> Simp[b*(a*Csc[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n - 1) /(a*f*(m + n))), x] + Simp[(m + 1)/(a^2*(m + n)) Int[(a*Csc[e + f*x])^(m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Result contains complex when optimal does not.
Time = 0.86 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.71
method | result | size |
default | \(\frac {2 \left (4 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \cos \left (f x +e \right )+4 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )+\left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \sin \left (f x +e \right ) \cos \left (f x +e \right )\right ) \sqrt {b \sec \left (f x +e \right )}}{7 f}\) | \(162\) |
2/7/f*(4*I*EllipticF(I*(cot(f*x+e)-csc(f*x+e)),I)*(1/(cos(f*x+e)+1))^(1/2) *(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)+4*I*(1/(cos(f*x+e)+1))^(1/2) *(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cot(f*x+e)-csc(f*x+e)),I)+ cos(f*x+e)^3*sin(f*x+e)-3*sin(f*x+e)*cos(f*x+e))*(b*sec(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99 \[ \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx=\frac {2 \, {\left ({\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 2 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 2 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}}{7 \, f} \]
2/7*((cos(f*x + e)^3 - 3*cos(f*x + e))*sqrt(b/cos(f*x + e))*sin(f*x + e) - 2*I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + 2*I*sqrt(2)*sqrt(b)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin (f*x + e)))/f
\[ \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx=\int \sqrt {b \sec {\left (e + f x \right )}} \sin ^{4}{\left (e + f x \right )}\, dx \]
\[ \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx=\int { \sqrt {b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{4} \,d x } \]
\[ \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx=\int { \sqrt {b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^4\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}} \,d x \]