3.431 \(\int \frac{\sin ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=126 \[ \frac{8 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{77 b^2 f}-\frac{2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}+\frac{8 \sin (e+f x)}{77 b f \sqrt{b \sec (e+f x)}}-\frac{12 b \sin (e+f x)}{77 f (b \sec (e+f x))^{5/2}} \]

[Out]

(8*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(77*b^2*f) - (12*b*Sin[e + f*x])/(77*f*(
b*Sec[e + f*x])^(5/2)) + (8*Sin[e + f*x])/(77*b*f*Sqrt[b*Sec[e + f*x]]) - (2*b*Sin[e + f*x]^3)/(11*f*(b*Sec[e
+ f*x])^(5/2))

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Rubi [A]  time = 0.133504, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2627, 3769, 3771, 2641} \[ \frac{8 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{77 b^2 f}-\frac{2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}+\frac{8 \sin (e+f x)}{77 b f \sqrt{b \sec (e+f x)}}-\frac{12 b \sin (e+f x)}{77 f (b \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4/(b*Sec[e + f*x])^(3/2),x]

[Out]

(8*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(77*b^2*f) - (12*b*Sin[e + f*x])/(77*f*(
b*Sec[e + f*x])^(5/2)) + (8*Sin[e + f*x])/(77*b*f*Sqrt[b*Sec[e + f*x]]) - (2*b*Sin[e + f*x]^3)/(11*f*(b*Sec[e
+ f*x])^(5/2))

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 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(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)}{(b \sec (e+f x))^{3/2}} \, dx &=-\frac{2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}+\frac{6}{11} \int \frac{\sin ^2(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\\ &=-\frac{12 b \sin (e+f x)}{77 f (b \sec (e+f x))^{5/2}}-\frac{2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}+\frac{12}{77} \int \frac{1}{(b \sec (e+f x))^{3/2}} \, dx\\ &=-\frac{12 b \sin (e+f x)}{77 f (b \sec (e+f x))^{5/2}}+\frac{8 \sin (e+f x)}{77 b f \sqrt{b \sec (e+f x)}}-\frac{2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}+\frac{4 \int \sqrt{b \sec (e+f x)} \, dx}{77 b^2}\\ &=-\frac{12 b \sin (e+f x)}{77 f (b \sec (e+f x))^{5/2}}+\frac{8 \sin (e+f x)}{77 b f \sqrt{b \sec (e+f x)}}-\frac{2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}+\frac{\left (4 \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx}{77 b^2}\\ &=\frac{8 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{77 b^2 f}-\frac{12 b \sin (e+f x)}{77 f (b \sec (e+f x))^{5/2}}+\frac{8 \sin (e+f x)}{77 b f \sqrt{b \sec (e+f x)}}-\frac{2 b \sin ^3(e+f x)}{11 f (b \sec (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.163047, size = 81, normalized size = 0.64 \[ \frac{\sec ^2(e+f x) \left (-5 \sin (2 (e+f x))-24 \sin (4 (e+f x))+7 \sin (6 (e+f x))+128 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )\right )}{1232 f (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4/(b*Sec[e + f*x])^(3/2),x]

[Out]

(Sec[e + f*x]^2*(128*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] - 5*Sin[2*(e + f*x)] - 24*Sin[4*(e + f*x)] +
7*Sin[6*(e + f*x)]))/(1232*f*(b*Sec[e + f*x])^(3/2))

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Maple [C]  time = 0.167, size = 173, normalized size = 1.4 \begin{align*} -{\frac{2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) }{77\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}} \left ( -7\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}+4\,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 ) +7\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}+13\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-13\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+4\,\cos \left ( fx+e \right ) \right ) \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4/(b*sec(f*x+e))^(3/2),x)

[Out]

-2/77/f*(cos(f*x+e)+1)^2*(-1+cos(f*x+e))*(-7*cos(f*x+e)^6+4*I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(co
s(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+7*cos(f*x+e)^5+13*cos(f*x+e)^4-13*cos(f*x+e)^3
-4*cos(f*x+e)^2+4*cos(f*x+e))/cos(f*x+e)^2/sin(f*x+e)^3/(b/cos(f*x+e))^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{4}}{\left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^4/(b*sec(f*x + e))^(3/2), 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^{2} \sec \left (f x + e\right )^{2}}, 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))^(3/2),x, algorithm="fricas")

[Out]

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(b*sec(f*x + e))/(b^2*sec(f*x + e)^2), 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 )}}{\left (b \sec{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, 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))**(3/2),x)

[Out]

Integral(sin(e + f*x)**4/(b*sec(e + f*x))**(3/2), 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}}{\left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^4/(b*sec(f*x + e))^(3/2), x)