\(\int \frac {\sin ^7(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\) [410]

Optimal result

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

[Out]

2/15*b^7/f/(b*sec(f*x+e))^(15/2)-6/11*b^5/f/(b*sec(f*x+e))^(11/2)+6/7*b^3/f/(b*sec(f*x+e))^(7/2)-2/3*b/f/(b*se
c(f*x+e))^(3/2)

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

[In]

Int[Sin[e + f*x]^7/Sqrt[b*Sec[e + f*x]],x]

[Out]

(2*b^7)/(15*f*(b*Sec[e + f*x])^(15/2)) - (6*b^5)/(11*f*(b*Sec[e + f*x])^(11/2)) + (6*b^3)/(7*f*(b*Sec[e + f*x]
)^(7/2)) - (2*b)/(3*f*(b*Sec[e + f*x])^(3/2))

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^{17/2}} \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {b^7 \text {Subst}\left (\int \left (-\frac {1}{x^{17/2}}+\frac {3}{b^2 x^{13/2}}-\frac {3}{b^4 x^{9/2}}+\frac {1}{b^6 x^{5/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {2 b^7}{15 f (b \sec (e+f x))^{15/2}}-\frac {6 b^5}{11 f (b \sec (e+f x))^{11/2}}+\frac {6 b^3}{7 f (b \sec (e+f x))^{7/2}}-\frac {2 b}{3 f (b \sec (e+f x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int \frac {\sin ^7(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {b (-7410+4035 \cos (2 (e+f x))-798 \cos (4 (e+f x))+77 \cos (6 (e+f x)))}{18480 f (b \sec (e+f x))^{3/2}} \]

[In]

Integrate[Sin[e + f*x]^7/Sqrt[b*Sec[e + f*x]],x]

[Out]

(b*(-7410 + 4035*Cos[2*(e + f*x)] - 798*Cos[4*(e + f*x)] + 77*Cos[6*(e + f*x)]))/(18480*f*(b*Sec[e + f*x])^(3/
2))

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.63

method	result	size
default	\(\frac {\frac {2 \left (\cos ^{7}\left (f x +e \right )\right )}{15}-\frac {6 \left (\cos ^{5}\left (f x +e \right )\right )}{11}+\frac {6 \left (\cos ^{3}\left (f x +e \right )\right )}{7}-\frac {2 \cos \left (f x +e \right )}{3}}{f \sqrt {b \sec \left (f x +e \right )}}\)	\(55\)

[In]

int(sin(f*x+e)^7/(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/1155/f/(b*sec(f*x+e))^(1/2)*(77*cos(f*x+e)^7-315*cos(f*x+e)^5+495*cos(f*x+e)^3-385*cos(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.70 \[ \int \frac {\sin ^7(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {2 \, {\left (77 \, \cos \left (f x + e\right )^{8} - 315 \, \cos \left (f x + e\right )^{6} + 495 \, \cos \left (f x + e\right )^{4} - 385 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{1155 \, b f} \]

[In]

integrate(sin(f*x+e)^7/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/1155*(77*cos(f*x + e)^8 - 315*cos(f*x + e)^6 + 495*cos(f*x + e)^4 - 385*cos(f*x + e)^2)*sqrt(b/cos(f*x + e))
/(b*f)

Sympy [F(-1)]

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

[In]

integrate(sin(f*x+e)**7/(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \frac {\sin ^7(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {2 \, {\left (77 \, b^{6} - \frac {315 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac {495 \, b^{6}}{\cos \left (f x + e\right )^{4}} - \frac {385 \, b^{6}}{\cos \left (f x + e\right )^{6}}\right )} b}{1155 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {15}{2}}} \]

[In]

integrate(sin(f*x+e)^7/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

2/1155*(77*b^6 - 315*b^6/cos(f*x + e)^2 + 495*b^6/cos(f*x + e)^4 - 385*b^6/cos(f*x + e)^6)*b/(f*(b/cos(f*x + e
))^(15/2))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.24 \[ \int \frac {\sin ^7(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {2 \, {\left (77 \, \sqrt {b \cos \left (f x + e\right )} b^{7} \cos \left (f x + e\right )^{7} - 315 \, \sqrt {b \cos \left (f x + e\right )} b^{7} \cos \left (f x + e\right )^{5} + 495 \, \sqrt {b \cos \left (f x + e\right )} b^{7} \cos \left (f x + e\right )^{3} - 385 \, \sqrt {b \cos \left (f x + e\right )} b^{7} \cos \left (f x + e\right )\right )}}{1155 \, b^{8} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]

[In]

integrate(sin(f*x+e)^7/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

2/1155*(77*sqrt(b*cos(f*x + e))*b^7*cos(f*x + e)^7 - 315*sqrt(b*cos(f*x + e))*b^7*cos(f*x + e)^5 + 495*sqrt(b*
cos(f*x + e))*b^7*cos(f*x + e)^3 - 385*sqrt(b*cos(f*x + e))*b^7*cos(f*x + e))/(b^8*f*sgn(cos(f*x + e)))

Mupad [F(-1)]

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

[In]

int(sin(e + f*x)^7/(b/cos(e + f*x))^(1/2),x)

[Out]

int(sin(e + f*x)^7/(b/cos(e + f*x))^(1/2), x)