3.418 \(\int \frac{\sin ^4(e+f x)}{\sqrt{b \sec (e+f x)}} \, dx\)

Optimal. Leaf size=95 \[ -\frac{2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}-\frac{4 b \sin (e+f x)}{15 f (b \sec (e+f x))^{3/2}}+\frac{8 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{15 f \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}} \]

[Out]

(8*EllipticE[(e + f*x)/2, 2])/(15*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (4*b*Sin[e + f*x])/(15*f*(b*Sec
[e + f*x])^(3/2)) - (2*b*Sin[e + f*x]^3)/(9*f*(b*Sec[e + f*x])^(3/2))

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Rubi [A]  time = 0.0998004, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2627, 3771, 2639} \[ -\frac{2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}-\frac{4 b \sin (e+f x)}{15 f (b \sec (e+f x))^{3/2}}+\frac{8 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{15 f \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*EllipticE[(e + f*x)/2, 2])/(15*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (4*b*Sin[e + f*x])/(15*f*(b*Sec
[e + f*x])^(3/2)) - (2*b*Sin[e + f*x]^3)/(9*f*(b*Sec[e + f*x])^(3/2))

Rule 2627

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 3771

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]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sin ^4(e+f x)}{\sqrt{b \sec (e+f x)}} \, dx &=-\frac{2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}+\frac{2}{3} \int \frac{\sin ^2(e+f x)}{\sqrt{b \sec (e+f x)}} \, dx\\ &=-\frac{4 b \sin (e+f x)}{15 f (b \sec (e+f x))^{3/2}}-\frac{2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}+\frac{4}{15} \int \frac{1}{\sqrt{b \sec (e+f x)}} \, dx\\ &=-\frac{4 b \sin (e+f x)}{15 f (b \sec (e+f x))^{3/2}}-\frac{2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}+\frac{4 \int \sqrt{\cos (e+f x)} \, dx}{15 \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=\frac{8 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{15 f \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{4 b \sin (e+f x)}{15 f (b \sec (e+f x))^{3/2}}-\frac{2 b \sin ^3(e+f x)}{9 f (b \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.33787, size = 63, normalized size = 0.66 \[ \frac{-68 \sin (2 (e+f x))+10 \sin (4 (e+f x))+\frac{192 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{\sqrt{\cos (e+f x)}}}{360 f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((192*EllipticE[(e + f*x)/2, 2])/Sqrt[Cos[e + f*x]] - 68*Sin[2*(e + f*x)] + 10*Sin[4*(e + f*x)])/(360*f*Sqrt[b
*Sec[e + f*x]])

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Maple [C]  time = 0.15, size = 328, normalized size = 3.5 \begin{align*} -{\frac{2}{45\,f\sin \left ( fx+e \right ) b} \left ( 5\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}+12\,i{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},i \right ) \cos \left ( fx+e \right ) \sin \left ( fx+e \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}-12\,i\cos \left ( fx+e \right ){\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sin \left ( fx+e \right ) +12\,i{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sin \left ( fx+e \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}-12\,i{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sin \left ( fx+e \right ) -16\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+23\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-12\,\cos \left ( fx+e \right ) \right ) \sqrt{{\frac{b}{\cos \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45/f*(5*cos(f*x+e)^6+12*I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*(1/(cos(f*x+e)+1)
)^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-12*I*cos(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(cos(f*
x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+12*I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin
(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-12*I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e)
,I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-16*cos(f*x+e)^4+23*cos(f*x+e)^2-12*c
os(f*x+e))*(b/cos(f*x+e))^(1/2)/sin(f*x+e)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{4}}{\sqrt{b \sec \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt{b \sec \left (f x + e\right )}}{b \sec \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(b*sec(f*x + e))/(b*sec(f*x + e)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{4}{\left (e + f x \right )}}{\sqrt{b \sec{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{4}}{\sqrt{b \sec \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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