3.424 \(\int \frac{\sin ^7(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=87 \[ \frac{2 b^7}{17 f (b \sec (e+f x))^{17/2}}-\frac{6 b^5}{13 f (b \sec (e+f x))^{13/2}}+\frac{2 b^3}{3 f (b \sec (e+f x))^{9/2}}-\frac{2 b}{5 f (b \sec (e+f x))^{5/2}} \]

[Out]

(2*b^7)/(17*f*(b*Sec[e + f*x])^(17/2)) - (6*b^5)/(13*f*(b*Sec[e + f*x])^(13/2)) + (2*b^3)/(3*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 = 0.0632463, 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} \[ \frac{2 b^7}{17 f (b \sec (e+f x))^{17/2}}-\frac{6 b^5}{13 f (b \sec (e+f x))^{13/2}}+\frac{2 b^3}{3 f (b \sec (e+f x))^{9/2}}-\frac{2 b}{5 f (b \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^7/(b*Sec[e + f*x])^(3/2),x]

[Out]

(2*b^7)/(17*f*(b*Sec[e + f*x])^(17/2)) - (6*b^5)/(13*f*(b*Sec[e + f*x])^(13/2)) + (2*b^3)/(3*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 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))^{3/2}} \, dx &=\frac{b^7 \operatorname{Subst}\left (\int \frac{\left (-1+\frac{x^2}{b^2}\right )^3}{x^{19/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{b^7 \operatorname{Subst}\left (\int \left (-\frac{1}{x^{19/2}}+\frac{3}{b^2 x^{15/2}}-\frac{3}{b^4 x^{11/2}}+\frac{1}{b^6 x^{7/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{2 b^7}{17 f (b \sec (e+f x))^{17/2}}-\frac{6 b^5}{13 f (b \sec (e+f x))^{13/2}}+\frac{2 b^3}{3 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 = 0.43607, size = 52, normalized size = 0.6 \[ \frac{b (8365 \cos (2 (e+f x))-1890 \cos (4 (e+f x))+195 \cos (6 (e+f x))-10766)}{53040 f (b \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^7/(b*Sec[e + f*x])^(3/2),x]

[Out]

(b*(-10766 + 8365*Cos[2*(e + f*x)] - 1890*Cos[4*(e + f*x)] + 195*Cos[6*(e + f*x)]))/(53040*f*(b*Sec[e + f*x])^
(5/2))

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Maple [A]  time = 0.146, size = 56, normalized size = 0.6 \begin{align*}{\frac{ \left ( 390\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}-1530\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+2210\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-1326 \right ) \cos \left ( fx+e \right ) }{3315\,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)^7/(b*sec(f*x+e))^(3/2),x)

[Out]

2/3315/f*(195*cos(f*x+e)^6-765*cos(f*x+e)^4+1105*cos(f*x+e)^2-663)*cos(f*x+e)/(b/cos(f*x+e))^(3/2)

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Maxima [A]  time = 1.00176, size = 85, normalized size = 0.98 \begin{align*} \frac{2 \,{\left (195 \, b^{6} - \frac{765 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac{1105 \, b^{6}}{\cos \left (f x + e\right )^{4}} - \frac{663 \, b^{6}}{\cos \left (f x + e\right )^{6}}\right )} b}{3315 \, f \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{17}{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))^(3/2),x, algorithm="maxima")

[Out]

2/3315*(195*b^6 - 765*b^6/cos(f*x + e)^2 + 1105*b^6/cos(f*x + e)^4 - 663*b^6/cos(f*x + e)^6)*b/(f*(b/cos(f*x +
e))^(17/2))

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Fricas [A]  time = 2.70848, size = 165, normalized size = 1.9 \begin{align*} \frac{2 \,{\left (195 \, \cos \left (f x + e\right )^{9} - 765 \, \cos \left (f x + e\right )^{7} + 1105 \, \cos \left (f x + e\right )^{5} - 663 \, \cos \left (f x + e\right )^{3}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{3315 \, b^{2} 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))^(3/2),x, algorithm="fricas")

[Out]

2/3315*(195*cos(f*x + e)^9 - 765*cos(f*x + e)^7 + 1105*cos(f*x + e)^5 - 663*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)**7/(b*sec(f*x+e))**(3/2),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{3}{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))^(3/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^7/(b*sec(f*x + e))^(3/2), x)