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

Optimal. Leaf size=98 \[ \frac{12 b^3 \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}-\frac{24 b^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 f \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{2 b \sin ^3(e+f x) \sqrt{b \sec (e+f x)}}{f} \]

[Out]

(-24*b^2*EllipticE[(e + f*x)/2, 2])/(5*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) + (12*b^3*Sin[e + f*x])/(5*f
*(b*Sec[e + f*x])^(3/2)) + (2*b*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^3)/f

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Rubi [A]  time = 0.107605, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2624, 2627, 3771, 2639} \[ \frac{12 b^3 \sin (e+f x)}{5 f (b \sec (e+f x))^{3/2}}-\frac{24 b^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 f \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{2 b \sin ^3(e+f x) \sqrt{b \sec (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*b^2*EllipticE[(e + f*x)/2, 2])/(5*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) + (12*b^3*Sin[e + f*x])/(5*f
*(b*Sec[e + f*x])^(3/2)) + (2*b*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^3)/f

Rule 2624

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] + Dist[(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] && Integer
sQ[2*m, 2*n]

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 (b \sec (e+f x))^{3/2} \sin ^4(e+f x) \, dx &=\frac{2 b \sqrt{b \sec (e+f x)} \sin ^3(e+f x)}{f}-\left (6 b^2\right ) \int \frac{\sin ^2(e+f x)}{\sqrt{b \sec (e+f x)}} \, dx\\ &=\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}-\frac{1}{5} \left (12 b^2\right ) \int \frac{1}{\sqrt{b \sec (e+f x)}} \, dx\\ &=\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}-\frac{\left (12 b^2\right ) \int \sqrt{\cos (e+f x)} \, dx}{5 \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=-\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}\\ \end{align*}

Mathematica [A]  time = 0.114691, size = 60, normalized size = 0.61 \[ \frac{b \sqrt{b \sec (e+f x)} \left (21 \sin (e+f x)+\sin (3 (e+f x))-48 \sqrt{\cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )\right )}{10 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

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Maple [C]  time = 0.179, size = 320, normalized size = 3.3 \begin{align*} -{\frac{2\,\cos \left ( fx+e \right ) }{5\,f\sin \left ( fx+e \right ) } \left ( -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\sin \left ( fx+e \right ){\it EllipticE} \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}}}+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 ) + \left ( \cos \left ( fx+e \right ) \right ) ^{4}-8\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+12\,\cos \left ( fx+e \right ) -5 \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((b*sec(f*x+e))^(3/2)*sin(f*x+e)^4,x)

[Out]

-2/5/f*(-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*sin(f*x+e)*EllipticE(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)+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)+cos(f*x+e)^4-8*cos(f*x+e)^2+12*cos(f*x+e)-5)*cos(f*
x+e)*(b/cos(f*x+e))^(3/2)/sin(f*x+e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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