\(\int \csc ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx\) [383]

Optimal result

Integrand size = 21, antiderivative size = 95 \[ \int \csc ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx=-\frac {5 b \csc (e+f x)}{6 f \sqrt {b \sec (e+f x)}}-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{6 f} \]

[Out]

-5/6*b*csc(f*x+e)/f/(b*sec(f*x+e))^(1/2)-1/3*b*csc(f*x+e)^3/f/(b*sec(f*x+e))^(1/2)+5/6*(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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2705, 3856, 2720} \[ \int \csc ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx=-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}-\frac {5 b \csc (e+f x)}{6 f \sqrt {b \sec (e+f x)}}+\frac {5 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{6 f} \]

[In]

Int[Csc[e + f*x]^4*Sqrt[b*Sec[e + f*x]],x]

[Out]

(-5*b*Csc[e + f*x])/(6*f*Sqrt[b*Sec[e + f*x]]) - (b*Csc[e + f*x]^3)/(3*f*Sqrt[b*Sec[e + f*x]]) + (5*Sqrt[Cos[e
+ f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(6*f)

Rule 2705

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(-a)*b*(a*Cs
c[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n - 1)/(f*(m - 1))), x] + Dist[a^2*((m + n - 2)/(m - 1)), Int[(a*Csc[e
+ f*x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && IntegersQ[2*m, 2*n] &&
!GtQ[n, m]

Rule 2720

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]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5}{6} \int \csc ^2(e+f x) \sqrt {b \sec (e+f x)} \, dx \\ & = -\frac {5 b \csc (e+f x)}{6 f \sqrt {b \sec (e+f x)}}-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5}{12} \int \sqrt {b \sec (e+f x)} \, dx \\ & = -\frac {5 b \csc (e+f x)}{6 f \sqrt {b \sec (e+f x)}}-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {1}{12} \left (5 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx \\ & = -\frac {5 b \csc (e+f x)}{6 f \sqrt {b \sec (e+f x)}}-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{6 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.66 \[ \int \csc ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx=\frac {\left (-\cot (e+f x) \left (5+2 \csc ^2(e+f x)\right )+5 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )\right ) \sqrt {b \sec (e+f x)}}{6 f} \]

[In]

Integrate[Csc[e + f*x]^4*Sqrt[b*Sec[e + f*x]],x]

[Out]

((-(Cot[e + f*x]*(5 + 2*Csc[e + f*x]^2)) + 5*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])*Sqrt[b*Sec[e + f*x]
])/(6*f)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.96

method	result	size
default	\(\frac {i \sqrt {b \sec \left (f x +e \right )}\, \left (-5 \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \left (\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )-5 \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \left (\sin ^{2}\left (f x +e \right )\right )+5 i \left (\cos ^{2}\left (f x +e \right )\right ) \cot \left (f x +e \right )-7 i \cot \left (f x +e \right )\right )}{6 f \left (\cos ^{2}\left (f x +e \right )-1\right )}\)	\(186\)

[In]

int(csc(f*x+e)^4*(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*I/f*(b*sec(f*x+e))^(1/2)/(cos(f*x+e)^2-1)*(-5*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*E
llipticF(I*(cot(f*x+e)-csc(f*x+e)),I)*sin(f*x+e)^2*cos(f*x+e)-5*(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)*sin(f*x+e)^2+5*I*cos(f*x+e)^2*cot(f*x+e)-7*I*cot(f*x+e))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.56 \[ \int \csc ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, \cos \left (f x + e\right )^{2} - i\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 \, \sqrt {2} {\left (-i \, \cos \left (f x + e\right )^{2} + i\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (5 \, \cos \left (f x + e\right )^{3} - 7 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{12 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )} \sin \left (f x + e\right )} \]

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/12*(5*sqrt(2)*(I*cos(f*x + e)^2 - I)*sqrt(b)*sin(f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f
*x + e)) + 5*sqrt(2)*(-I*cos(f*x + e)^2 + I)*sqrt(b)*sin(f*x + e)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*
sin(f*x + e)) + 2*(5*cos(f*x + e)^3 - 7*cos(f*x + e))*sqrt(b/cos(f*x + e)))/((f*cos(f*x + e)^2 - f)*sin(f*x +
e))

Sympy [F]

\[ \int \csc ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx=\int \sqrt {b \sec {\left (e + f x \right )}} \csc ^{4}{\left (e + f x \right )}\, dx \]

[In]

integrate(csc(f*x+e)**4*(b*sec(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(b*sec(e + f*x))*csc(e + f*x)**4, x)

Maxima [F]

\[ \int \csc ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )} \csc \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(f*x + e))*csc(f*x + e)^4, x)

Giac [F]

\[ \int \csc ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )} \csc \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(f*x + e))*csc(f*x + e)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \csc ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx=\int \frac {\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{{\sin \left (e+f\,x\right )}^4} \,d x \]

[In]

int((b/cos(e + f*x))^(1/2)/sin(e + f*x)^4,x)

[Out]

int((b/cos(e + f*x))^(1/2)/sin(e + f*x)^4, x)