\(\int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx\) [379]

Optimal result

Integrand size = 21, antiderivative size = 95 \[ \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx=\frac {8 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{7 f}-\frac {4 b \sin (e+f x)}{7 f \sqrt {b \sec (e+f x)}}-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}} \]

[Out]

-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

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 = {2707, 3856, 2720} \[ \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx=-\frac {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}-\frac {4 b \sin (e+f x)}{7 f \sqrt {b \sec (e+f x)}}+\frac {8 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \sec (e+f x)}}{7 f} \]

[In]

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

[Out]

(8*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(7*f) - (4*b*Sin[e + f*x])/(7*f*Sqrt[b*S
ec[e + f*x]]) - (2*b*Sin[e + f*x]^3)/(7*f*Sqrt[b*Sec[e + f*x]])

Rule 2707

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] + Dist[(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]

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 {2 b \sin ^3(e+f x)}{7 f \sqrt {b \sec (e+f x)}}+\frac {6}{7} \int \sqrt {b \sec (e+f x)} \sin ^2(e+f x) \, dx \\ & = -\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)}}+\frac {4}{7} \int \sqrt {b \sec (e+f x)} \, dx \\ & = -\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)}}+\frac {1}{7} \left (4 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.64 \[ \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx=\frac {\sqrt {b \sec (e+f x)} \left (32 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )-10 \sin (2 (e+f x))+\sin (4 (e+f x))\right )}{28 f} \]

[In]

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

[Out]

(Sqrt[b*Sec[e + f*x]]*(32*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] - 10*Sin[2*(e + f*x)] + Sin[4*(e + f*x)
]))/(28*f)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.86 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.71

method	result	size
default	\(\frac {2 \left (4 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \cos \left (f x +e \right )+4 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )+\left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \sin \left (f x +e \right ) \cos \left (f x +e \right )\right ) \sqrt {b \sec \left (f x +e \right )}}{7 f}\)	\(162\)

[In]

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

[Out]

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)*c
os(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)

Fricas [C] (verification not implemented)

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 \[ \int \sqrt {b \sec (e+f x)} \sin ^4(e+f x) \, dx=\frac {2 \, {\left ({\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 2 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 2 i \, \sqrt {2} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}}{7 \, f} \]

[In]

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

[Out]

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)*weierstrassPInv
erse(-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*s
in(f*x + e)))/f

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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