\(\int \sqrt {b \sec (e+f x)} \sin ^5(e+f x) \, dx\) [372]

Optimal result

Integrand size = 21, antiderivative size = 63 \[ \int \sqrt {b \sec (e+f x)} \sin ^5(e+f x) \, dx=-\frac {2 b^5}{9 f (b \sec (e+f x))^{9/2}}+\frac {4 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac {2 b}{f \sqrt {b \sec (e+f x)}} \]

[Out]

-2/9*b^5/f/(b*sec(f*x+e))^(9/2)+4/5*b^3/f/(b*sec(f*x+e))^(5/2)-2*b/f/(b*sec(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, 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} \[ \int \sqrt {b \sec (e+f x)} \sin ^5(e+f x) \, dx=-\frac {2 b^5}{9 f (b \sec (e+f x))^{9/2}}+\frac {4 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac {2 b}{f \sqrt {b \sec (e+f x)}} \]

[In]

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

[Out]

(-2*b^5)/(9*f*(b*Sec[e + f*x])^(9/2)) + (4*b^3)/(5*f*(b*Sec[e + f*x])^(5/2)) - (2*b)/(f*Sqrt[b*Sec[e + f*x]])

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.76 \[ \int \sqrt {b \sec (e+f x)} \sin ^5(e+f x) \, dx=-\frac {(554 \cos (e+f x)-47 \cos (3 (e+f x))+5 \cos (5 (e+f x))) \sqrt {b \sec (e+f x)}}{360 f} \]

[In]

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

[Out]

-1/360*((554*Cos[e + f*x] - 47*Cos[3*(e + f*x)] + 5*Cos[5*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/f

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(434\) vs. \(2(53)=106\).

Time = 0.21 (sec) , antiderivative size = 435, normalized size of antiderivative = 6.90

method	result	size
default	\(-\frac {\left (20 \left (\cos ^{5}\left (f x +e \right )\right )+45 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-45 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-72 \left (\cos ^{3}\left (f x +e \right )\right )+45 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-45 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+180 \cos \left (f x +e \right )\right ) \sqrt {b \sec \left (f x +e \right )}}{90 f}\)	\(435\)

[In]

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

[Out]

-1/90/f*(20*cos(f*x+e)^5+45*ln((2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1
)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)-45*ln(2*(2*cos(f*x+e)
*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))*(-c
os(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)-72*cos(f*x+e)^3+45*ln((2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2
)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1
/2)-45*ln(2*(2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+
e)+1)/(cos(f*x+e)+1))*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+180*cos(f*x+e))*(b*sec(f*x+e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.73 \[ \int \sqrt {b \sec (e+f x)} \sin ^5(e+f x) \, dx=-\frac {2 \, {\left (5 \, \cos \left (f x + e\right )^{5} - 18 \, \cos \left (f x + e\right )^{3} + 45 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{45 \, f} \]

[In]

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

[Out]

-2/45*(5*cos(f*x + e)^5 - 18*cos(f*x + e)^3 + 45*cos(f*x + e))*sqrt(b/cos(f*x + e))/f

Sympy [F(-1)]

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

[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 = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.79 \[ \int \sqrt {b \sec (e+f x)} \sin ^5(e+f x) \, dx=-\frac {2 \, {\left (5 \, b^{4} - \frac {18 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac {45 \, b^{4}}{\cos \left (f x + e\right )^{4}}\right )} b}{45 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {9}{2}}} \]

[In]

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

[Out]

-2/45*(5*b^4 - 18*b^4/cos(f*x + e)^2 + 45*b^4/cos(f*x + e)^4)*b/(f*(b/cos(f*x + e))^(9/2))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.22 \[ \int \sqrt {b \sec (e+f x)} \sin ^5(e+f x) \, dx=-\frac {2 \, {\left (5 \, \sqrt {b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{4} - 18 \, \sqrt {b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{2} + 45 \, \sqrt {b \cos \left (f x + e\right )} b^{4}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{45 \, b^{4} f} \]

[In]

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

[Out]

-2/45*(5*sqrt(b*cos(f*x + e))*b^4*cos(f*x + e)^4 - 18*sqrt(b*cos(f*x + e))*b^4*cos(f*x + e)^2 + 45*sqrt(b*cos(
f*x + e))*b^4)*sgn(cos(f*x + e))/(b^4*f)

Mupad [F(-1)]

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

[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)