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

Optimal. Leaf size=65 \[ -\frac{2 b^5}{15 f (b \sec (e+f x))^{15/2}}+\frac{4 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac{2 b}{7 f (b \sec (e+f x))^{7/2}} \]

[Out]

(-2*b^5)/(15*f*(b*Sec[e + f*x])^(15/2)) + (4*b^3)/(11*f*(b*Sec[e + f*x])^(11/2)) - (2*b)/(7*f*(b*Sec[e + f*x])
^(7/2))

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Rubi [A]  time = 0.0585754, antiderivative size = 65, 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, 270} \[ -\frac{2 b^5}{15 f (b \sec (e+f x))^{15/2}}+\frac{4 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac{2 b}{7 f (b \sec (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^5)/(15*f*(b*Sec[e + f*x])^(15/2)) + (4*b^3)/(11*f*(b*Sec[e + f*x])^(11/2)) - (2*b)/(7*f*(b*Sec[e + f*x])
^(7/2))

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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\sin ^5(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{\left (-1+\frac{x^2}{b^2}\right )^2}{x^{17/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{b^5 \operatorname{Subst}\left (\int \left (\frac{1}{x^{17/2}}-\frac{2}{b^2 x^{13/2}}+\frac{1}{b^4 x^{9/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=-\frac{2 b^5}{15 f (b \sec (e+f x))^{15/2}}+\frac{4 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac{2 b}{7 f (b \sec (e+f x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.219008, size = 52, normalized size = 0.8 \[ \frac{\cos ^4(e+f x) (532 \cos (2 (e+f x))-77 \cos (4 (e+f x))-711) \sqrt{b \sec (e+f x)}}{4620 b^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]^4*(-711 + 532*Cos[2*(e + f*x)] - 77*Cos[4*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(4620*b^3*f)

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Maple [A]  time = 0.165, size = 46, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 154\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-420\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+330 \right ) \cos \left ( fx+e \right ) }{1155\,f} \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/1155/f*(77*cos(f*x+e)^4-210*cos(f*x+e)^2+165)*cos(f*x+e)/(b/cos(f*x+e))^(5/2)

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Maxima [A]  time = 1.00631, size = 68, normalized size = 1.05 \begin{align*} -\frac{2 \,{\left (77 \, b^{4} - \frac{210 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac{165 \, b^{4}}{\cos \left (f x + e\right )^{4}}\right )} b}{1155 \, f \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{15}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/1155*(77*b^4 - 210*b^4/cos(f*x + e)^2 + 165*b^4/cos(f*x + e)^4)*b/(f*(b/cos(f*x + e))^(15/2))

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Fricas [A]  time = 2.67109, size = 135, normalized size = 2.08 \begin{align*} -\frac{2 \,{\left (77 \, \cos \left (f x + e\right )^{8} - 210 \, \cos \left (f x + e\right )^{6} + 165 \, \cos \left (f x + e\right )^{4}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{1155 \, b^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/1155*(77*cos(f*x + e)^8 - 210*cos(f*x + e)^6 + 165*cos(f*x + e)^4)*sqrt(b/cos(f*x + e))/(b^3*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)**5/(b*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{5}}{\left (b \sec \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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