Integrand size = 21, antiderivative size = 98 \[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=-\frac {24 b^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}+\frac {12 b^3 \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)} \sin ^3(e+f x)}{f} \]
12/5*b^3*sin(f*x+e)/f/(b*sec(f*x+e))^(3/2)-24/5*b^2*(cos(1/2*f*x+1/2*e)^2) ^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/f/cos(f*x+ e)^(1/2)/(b*sec(f*x+e))^(1/2)+2*b*sin(f*x+e)^3*(b*sec(f*x+e))^(1/2)/f
Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.61 \[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=\frac {b \sqrt {b \sec (e+f x)} \left (-48 \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+21 \sin (e+f x)+\sin (3 (e+f x))\right )}{10 f} \]
(b*Sqrt[b*Sec[e + f*x]]*(-48*Sqrt[Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2] + 21*Sin[e + f*x] + Sin[3*(e + f*x)]))/(10*f)
Time = 0.47 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3104, 3042, 3107, 3042, 4258, 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.
\(\displaystyle \int \sin ^4(e+f x) (b \sec (e+f x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(b \sec (e+f x))^{3/2}}{\csc (e+f x)^4}dx\) |
\(\Big \downarrow \) 3104 |
\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \int \frac {\sin ^2(e+f x)}{\sqrt {b \sec (e+f x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \int \frac {1}{\csc (e+f x)^2 \sqrt {b \sec (e+f x)}}dx\) |
\(\Big \downarrow \) 3107 |
\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \left (\frac {2}{5} \int \frac {1}{\sqrt {b \sec (e+f x)}}dx-\frac {2 b \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \left (\frac {2}{5} \int \frac {1}{\sqrt {b \csc \left (e+f x+\frac {\pi }{2}\right )}}dx-\frac {2 b \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}\right )\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \left (\frac {2 \int \sqrt {\cos (e+f x)}dx}{5 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {2 b \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \left (\frac {2 \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {2 b \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 b \sin ^3(e+f x) \sqrt {b \sec (e+f x)}}{f}-6 b^2 \left (\frac {4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {2 b \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}\right )\) |
(2*b*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^3)/f - 6*b^2*((4*EllipticE[(e + f*x )/2, 2])/(5*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (2*b*Sin[e + f*x] )/(5*f*(b*Sec[e + f*x])^(3/2)))
3.4.92.3.1 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)/ (f*a*(n - 1))), x] + Simp[b^2*((m + 1)/(a^2*(n - 1))) Int[(a*Csc[e + f*x] )^(m + 2)*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ [n, 1] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
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[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[(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 = 1.42 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.98
method | result | size |
default | \(\frac {2 \sqrt {b \sec \left (f x +e \right )}\, b \left (i \left (12 \cos \left (f x +e \right )+12\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, E\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )+i \left (-12 \cos \left (f x +e \right )-12\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )-\left (\cos ^{3}\left (f x +e \right )\right ) \cot \left (f x +e \right )+8 \cos \left (f x +e \right ) \cot \left (f x +e \right )-12 \cot \left (f x +e \right )+5 \csc \left (f x +e \right )\right )}{5 f}\) | \(194\) |
2/5/f*(b*sec(f*x+e))^(1/2)*b*(I*(12*cos(f*x+e)+12)*(1/(cos(f*x+e)+1))^(1/2 )*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cot(f*x+e)-csc(f*x+e)),I) +I*(-12*cos(f*x+e)-12)*(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*cot(f*x+e)+8*c os(f*x+e)*cot(f*x+e)-12*cot(f*x+e)+5*csc(f*x+e))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=-\frac {2 \, {\left (6 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 6 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - {\left (b \cos \left (f x + e\right )^{2} + 5 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{5 \, f} \]
-2/5*(6*I*sqrt(2)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0 , cos(f*x + e) + I*sin(f*x + e))) - 6*I*sqrt(2)*b^(3/2)*weierstrassZeta(-4 , 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - (b*cos(f *x + e)^2 + 5*b)*sqrt(b/cos(f*x + e))*sin(f*x + e))/f
Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=\text {Timed out} \]
\[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sin \left (f x + e\right )^{4} \,d x } \]
\[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sin \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^4\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \]