3.399 \(\int (b \sec (e+f x))^{5/2} \sin ^3(e+f x) \, dx\)

Optimal. Leaf size=41 \[ \frac{2 b^3}{f \sqrt{b \sec (e+f x)}}+\frac{2 b (b \sec (e+f x))^{3/2}}{3 f} \]

[Out]

(2*b^3)/(f*Sqrt[b*Sec[e + f*x]]) + (2*b*(b*Sec[e + f*x])^(3/2))/(3*f)

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Rubi [A]  time = 0.0492443, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2622, 14} \[ \frac{2 b^3}{f \sqrt{b \sec (e+f x)}}+\frac{2 b (b \sec (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^(5/2)*Sin[e + f*x]^3,x]

[Out]

(2*b^3)/(f*Sqrt[b*Sec[e + f*x]]) + (2*b*(b*Sec[e + f*x])^(3/2))/(3*f)

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.197709, size = 32, normalized size = 0.78 \[ \frac{b (3 \cos (2 (e+f x))+5) (b \sec (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^(5/2)*Sin[e + f*x]^3,x]

[Out]

(b*(5 + 3*Cos[2*(e + f*x)])*(b*Sec[e + f*x])^(3/2))/(3*f)

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Maple [B]  time = 0.151, size = 357, normalized size = 8.7 \begin{align*} -{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) }{6\,f \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 12\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -2\,{\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\cos \left ( fx+e \right ) -2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}-1 \right ) } \right ) -3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -{\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\cos \left ( fx+e \right ) -2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}-1 \right ) } \right ) +12\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}+4\,\cos \left ( fx+e \right ) \sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}+4\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}} \right ) \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}{\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^(5/2)*sin(f*x+e)^3,x)

[Out]

-1/6/f*(-1+cos(f*x+e))*(12*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+3*cos(f*x+e)^2*ln(-2*(2*cos(f*x+e
)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/s
in(f*x+e)^2)-3*cos(f*x+e)^2*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)
-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+12*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+
4*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*cos(f*x+e)*(b/cos(f*
x+e))^(5/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/sin(f*x+e)^2

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Maxima [A]  time = 1.03316, size = 49, normalized size = 1.2 \begin{align*} \frac{2 \,{\left (\frac{3 \, b^{2}}{\sqrt{\frac{b}{\cos \left (f x + e\right )}}} + \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{3}{2}}\right )} b}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^3,x, algorithm="maxima")

[Out]

2/3*(3*b^2/sqrt(b/cos(f*x + e)) + (b/cos(f*x + e))^(3/2))*b/f

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Fricas [A]  time = 2.08816, size = 97, normalized size = 2.37 \begin{align*} \frac{2 \,{\left (3 \, b^{2} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{3 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^3,x, algorithm="fricas")

[Out]

2/3*(3*b^2*cos(f*x + e)^2 + b^2)*sqrt(b/cos(f*x + e))/(f*cos(f*x + e))

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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))**(5/2)*sin(f*x+e)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.15711, size = 72, normalized size = 1.76 \begin{align*} \frac{2 \,{\left (3 \, \sqrt{b \cos \left (f x + e\right )} b + \frac{b^{2}}{\sqrt{b \cos \left (f x + e\right )} \cos \left (f x + e\right )}\right )} b \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^3,x, algorithm="giac")

[Out]

2/3*(3*sqrt(b*cos(f*x + e))*b + b^2/(sqrt(b*cos(f*x + e))*cos(f*x + e)))*b*sgn(cos(f*x + e))/f