\(\int \frac {\sin ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\) [411]

Optimal result

Integrand size = 21, antiderivative size = 65 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=-\frac {2 b^5}{11 f (b \sec (e+f x))^{11/2}}+\frac {4 b^3}{7 f (b \sec (e+f x))^{7/2}}-\frac {2 b}{3 f (b \sec (e+f x))^{3/2}} \]

[Out]

-2/11*b^5/f/(b*sec(f*x+e))^(11/2)+4/7*b^3/f/(b*sec(f*x+e))^(7/2)-2/3*b/f/(b*sec(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, 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} \[ \int \frac {\sin ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=-\frac {2 b^5}{11 f (b \sec (e+f x))^{11/2}}+\frac {4 b^3}{7 f (b \sec (e+f x))^{7/2}}-\frac {2 b}{3 f (b \sec (e+f x))^{3/2}} \]

[In]

Int[Sin[e + f*x]^5/Sqrt[b*Sec[e + f*x]],x]

[Out]

(-2*b^5)/(11*f*(b*Sec[e + f*x])^(11/2)) + (4*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^5 \text {Subst}\left (\int \frac {\left (-1+\frac {x^2}{b^2}\right )^2}{x^{13/2}} \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {b^5 \text {Subst}\left (\int \left (\frac {1}{x^{13/2}}-\frac {2}{b^2 x^{9/2}}+\frac {1}{b^4 x^{5/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f} \\ & = -\frac {2 b^5}{11 f (b \sec (e+f x))^{11/2}}+\frac {4 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 = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {b (-415+180 \cos (2 (e+f x))-21 \cos (4 (e+f x)))}{924 f (b \sec (e+f x))^{3/2}} \]

[In]

Integrate[Sin[e + f*x]^5/Sqrt[b*Sec[e + f*x]],x]

[Out]

(b*(-415 + 180*Cos[2*(e + f*x)] - 21*Cos[4*(e + f*x)]))/(924*f*(b*Sec[e + f*x])^(3/2))

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69

method	result	size
default	\(-\frac {2 \left (21 \left (\cos ^{5}\left (f x +e \right )\right )-66 \left (\cos ^{3}\left (f x +e \right )\right )+77 \cos \left (f x +e \right )\right )}{231 f \sqrt {b \sec \left (f x +e \right )}}\)	\(45\)

[In]

int(sin(f*x+e)^5/(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/231/f/(b*sec(f*x+e))^(1/2)*(21*cos(f*x+e)^5-66*cos(f*x+e)^3+77*cos(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=-\frac {2 \, {\left (21 \, \cos \left (f x + e\right )^{6} - 66 \, \cos \left (f x + e\right )^{4} + 77 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{231 \, b f} \]

[In]

integrate(sin(f*x+e)^5/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-2/231*(21*cos(f*x + e)^6 - 66*cos(f*x + e)^4 + 77*cos(f*x + e)^2)*sqrt(b/cos(f*x + e))/(b*f)

Sympy [F(-1)]

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

[In]

integrate(sin(f*x+e)**5/(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=-\frac {2 \, {\left (21 \, b^{4} - \frac {66 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac {77 \, b^{4}}{\cos \left (f x + e\right )^{4}}\right )} b}{231 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {11}{2}}} \]

[In]

integrate(sin(f*x+e)^5/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-2/231*(21*b^4 - 66*b^4/cos(f*x + e)^2 + 77*b^4/cos(f*x + e)^4)*b/(f*(b/cos(f*x + e))^(11/2))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.31 \[ \int \frac {\sin ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=-\frac {2 \, {\left (21 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )^{5} - 66 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )^{3} + 77 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )\right )}}{231 \, b^{6} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]

[In]

integrate(sin(f*x+e)^5/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-2/231*(21*sqrt(b*cos(f*x + e))*b^5*cos(f*x + e)^5 - 66*sqrt(b*cos(f*x + e))*b^5*cos(f*x + e)^3 + 77*sqrt(b*co
s(f*x + e))*b^5*cos(f*x + e))/(b^6*f*sgn(cos(f*x + e)))

Mupad [F(-1)]

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

[In]

int(sin(e + f*x)^5/(b/cos(e + f*x))^(1/2),x)

[Out]

int(sin(e + f*x)^5/(b/cos(e + f*x))^(1/2), x)