\(\int \frac {\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\) [437]

Optimal result

Integrand size = 21, antiderivative size = 87 \[ \int \frac {\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 b^7}{19 f (b \sec (e+f x))^{19/2}}-\frac {2 b^5}{5 f (b \sec (e+f x))^{15/2}}+\frac {6 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac {2 b}{7 f (b \sec (e+f x))^{7/2}} \]

[Out]

2/19*b^7/f/(b*sec(f*x+e))^(19/2)-2/5*b^5/f/(b*sec(f*x+e))^(15/2)+6/11*b^3/f/(b*sec(f*x+e))^(11/2)-2/7*b/f/(b*s
ec(f*x+e))^(7/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2702, 276} \[ \int \frac {\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 b^7}{19 f (b \sec (e+f x))^{19/2}}-\frac {2 b^5}{5 f (b \sec (e+f x))^{15/2}}+\frac {6 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac {2 b}{7 f (b \sec (e+f x))^{7/2}} \]

[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 276

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]

Rule 2702

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])

Rubi steps \begin{align*} \text {integral}& = \frac {b^7 \text {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 \text {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] (verified)

Time = 0.54 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.71 \[ \int \frac {\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\cos ^4(e+f x) (-15226+14287 \cos (2 (e+f x))-3542 \cos (4 (e+f x))+385 \cos (6 (e+f x))) \sqrt {b \sec (e+f x)}}{117040 b^3 f} \]

[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)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.69

method	result	size
default	\(\frac {\frac {2 \left (\cos ^{9}\left (f x +e \right )\right )}{19}-\frac {2 \left (\cos ^{7}\left (f x +e \right )\right )}{5}+\frac {6 \left (\cos ^{5}\left (f x +e \right )\right )}{11}-\frac {2 \left (\cos ^{3}\left (f x +e \right )\right )}{7}}{f \,b^{2} \sqrt {b \sec \left (f x +e \right )}}\)	\(60\)

[In]

int(sin(f*x+e)^7/(b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/7315/f/b^2/(b*sec(f*x+e))^(1/2)*(385*cos(f*x+e)^9-1463*cos(f*x+e)^7+1995*cos(f*x+e)^5-1045*cos(f*x+e)^3)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.70 \[ \int \frac {\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (385 \, \cos \left (f x + e\right )^{10} - 1463 \, \cos \left (f x + e\right )^{8} + 1995 \, \cos \left (f x + e\right )^{6} - 1045 \, \cos \left (f x + e\right )^{4}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{7315 \, b^{3} f} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \frac {\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (385 \, b^{6} - \frac {1463 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac {1995 \, b^{6}}{\cos \left (f x + e\right )^{4}} - \frac {1045 \, b^{6}}{\cos \left (f x + e\right )^{6}}\right )} b}{7315 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {19}{2}}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.26 \[ \int \frac {\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (385 \, \sqrt {b \cos \left (f x + e\right )} b^{9} \cos \left (f x + e\right )^{9} - 1463 \, \sqrt {b \cos \left (f x + e\right )} b^{9} \cos \left (f x + e\right )^{7} + 1995 \, \sqrt {b \cos \left (f x + e\right )} b^{9} \cos \left (f x + e\right )^{5} - 1045 \, \sqrt {b \cos \left (f x + e\right )} b^{9} \cos \left (f x + e\right )^{3}\right )}}{7315 \, b^{12} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]

[In]

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

[Out]

2/7315*(385*sqrt(b*cos(f*x + e))*b^9*cos(f*x + e)^9 - 1463*sqrt(b*cos(f*x + e))*b^9*cos(f*x + e)^7 + 1995*sqrt
(b*cos(f*x + e))*b^9*cos(f*x + e)^5 - 1045*sqrt(b*cos(f*x + e))*b^9*cos(f*x + e)^3)/(b^12*f*sgn(cos(f*x + e)))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^7}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int(sin(e + f*x)^7/(b/cos(e + f*x))^(5/2),x)

[Out]

int(sin(e + f*x)^7/(b/cos(e + f*x))^(5/2), x)