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

Optimal. Leaf size=43 \[ \frac{2 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac{2 b}{5 f (b \sec (e+f x))^{5/2}} \]

[Out]

(2*b^3)/(9*f*(b*Sec[e + f*x])^(9/2)) - (2*b)/(5*f*(b*Sec[e + f*x])^(5/2))

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Rubi [A]  time = 0.0502984, antiderivative size = 43, 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} \[ \frac{2 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac{2 b}{5 f (b \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^3)/(9*f*(b*Sec[e + f*x])^(9/2)) - (2*b)/(5*f*(b*Sec[e + f*x])^(5/2))

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 \frac{\sin ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{-1+\frac{x^2}{b^2}}{x^{11/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{b^3 \operatorname{Subst}\left (\int \left (-\frac{1}{x^{11/2}}+\frac{1}{b^2 x^{7/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{2 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac{2 b}{5 f (b \sec (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.171556, size = 32, normalized size = 0.74 \[ \frac{b (5 \cos (2 (e+f x))-13)}{45 f (b \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(-13 + 5*Cos[2*(e + f*x)]))/(45*f*(b*Sec[e + f*x])^(5/2))

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Maple [A]  time = 0.102, size = 36, normalized size = 0.8 \begin{align*}{\frac{ \left ( 10\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-18 \right ) \cos \left ( fx+e \right ) }{45\,f} \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)^3/(b*sec(f*x+e))^(3/2),x)

[Out]

2/45/f*(5*cos(f*x+e)^2-9)*cos(f*x+e)/(b/cos(f*x+e))^(3/2)

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Maxima [A]  time = 1.001, size = 50, normalized size = 1.16 \begin{align*} \frac{2 \,{\left (5 \, b^{2} - \frac{9 \, b^{2}}{\cos \left (f x + e\right )^{2}}\right )} b}{45 \, f \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5*b^2 - 9*b^2/cos(f*x + e)^2)*b/(f*(b/cos(f*x + e))^(9/2))

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Fricas [A]  time = 2.67593, size = 99, normalized size = 2.3 \begin{align*} \frac{2 \,{\left (5 \, \cos \left (f x + e\right )^{5} - 9 \, \cos \left (f x + e\right )^{3}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{45 \, b^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5*cos(f*x + e)^5 - 9*cos(f*x + e)^3)*sqrt(b/cos(f*x + e))/(b^2*f)

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

[Out]

Timed out

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Giac [A]  time = 1.47836, size = 68, normalized size = 1.58 \begin{align*} \frac{2 \,{\left (5 \, b^{2} - \frac{9 \, b^{2}}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{4}}{45 \, b^{3} f \sqrt{\frac{b}{\cos \left (f x + e\right )}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5*b^2 - 9*b^2/cos(f*x + e)^2)*cos(f*x + e)^4/(b^3*f*sqrt(b/cos(f*x + e)))