3.437 \(\int \frac{\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=87 \[ \frac{2 b^7}{19 f (b \sec (e+f x))^{19/2}}-\frac{2 b^5}{5 f (b \sec (e+f x))^{15/2}}+\frac{6 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac{2 b}{7 f (b \sec (e+f x))^{7/2}} \]

[Out]

(2*b^7)/(19*f*(b*Sec[e + f*x])^(19/2)) - (2*b^5)/(5*f*(b*Sec[e + f*x])^(15/2)) + (6*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 = 0.0630548, antiderivative size = 87, 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} \[ \frac{2 b^7}{19 f (b \sec (e+f x))^{19/2}}-\frac{2 b^5}{5 f (b \sec (e+f x))^{15/2}}+\frac{6 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac{2 b}{7 f (b \sec (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^7/(b*Sec[e + f*x])^(5/2),x]

[Out]

(2*b^7)/(19*f*(b*Sec[e + f*x])^(19/2)) - (2*b^5)/(5*f*(b*Sec[e + f*x])^(15/2)) + (6*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 ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx &=\frac{b^7 \operatorname{Subst}\left (\int \frac{\left (-1+\frac{x^2}{b^2}\right )^3}{x^{21/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{b^7 \operatorname{Subst}\left (\int \left (-\frac{1}{x^{21/2}}+\frac{3}{b^2 x^{17/2}}-\frac{3}{b^4 x^{13/2}}+\frac{1}{b^6 x^{9/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{2 b^7}{19 f (b \sec (e+f x))^{19/2}}-\frac{2 b^5}{5 f (b \sec (e+f x))^{15/2}}+\frac{6 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 = 0.361263, size = 62, normalized size = 0.71 \[ \frac{\cos ^4(e+f x) (14287 \cos (2 (e+f x))-3542 \cos (4 (e+f x))+385 \cos (6 (e+f x))-15226) \sqrt{b \sec (e+f x)}}{117040 b^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^7/(b*Sec[e + f*x])^(5/2),x]

[Out]

(Cos[e + f*x]^4*(-15226 + 14287*Cos[2*(e + f*x)] - 3542*Cos[4*(e + f*x)] + 385*Cos[6*(e + f*x)])*Sqrt[b*Sec[e
+ f*x]])/(117040*b^3*f)

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Maple [A]  time = 0.19, size = 56, normalized size = 0.6 \begin{align*}{\frac{ \left ( 770\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}-2926\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+3990\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2090 \right ) \cos \left ( fx+e \right ) }{7315\,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)^7/(b*sec(f*x+e))^(5/2),x)

[Out]

2/7315/f*(385*cos(f*x+e)^6-1463*cos(f*x+e)^4+1995*cos(f*x+e)^2-1045)*cos(f*x+e)/(b/cos(f*x+e))^(5/2)

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Maxima [A]  time = 1.01907, size = 85, normalized size = 0.98 \begin{align*} \frac{2 \,{\left (385 \, b^{6} - \frac{1463 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac{1995 \, b^{6}}{\cos \left (f x + e\right )^{4}} - \frac{1045 \, b^{6}}{\cos \left (f x + e\right )^{6}}\right )} b}{7315 \, f \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{19}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^7/(b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

2/7315*(385*b^6 - 1463*b^6/cos(f*x + e)^2 + 1995*b^6/cos(f*x + e)^4 - 1045*b^6/cos(f*x + e)^6)*b/(f*(b/cos(f*x
+ e))^(19/2))

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Fricas [A]  time = 2.92419, size = 169, normalized size = 1.94 \begin{align*} \frac{2 \,{\left (385 \, \cos \left (f x + e\right )^{10} - 1463 \, \cos \left (f x + e\right )^{8} + 1995 \, \cos \left (f x + e\right )^{6} - 1045 \, \cos \left (f x + e\right )^{4}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{7315 \, b^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^7/(b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

2/7315*(385*cos(f*x + e)^10 - 1463*cos(f*x + e)^8 + 1995*cos(f*x + e)^6 - 1045*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)**7/(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 )^{7}}{\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)^7/(b*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^7/(b*sec(f*x + e))^(5/2), x)