\(\int \sqrt {b \sec (e+f x)} \sin ^3(e+f x) \, dx\) [373]

Optimal result

Integrand size = 21, antiderivative size = 41 \[ \int \sqrt {b \sec (e+f x)} \sin ^3(e+f x) \, dx=\frac {2 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac {2 b}{f \sqrt {b \sec (e+f x)}} \]

[Out]

2/5*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 = 41, 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, 14} \[ \int \sqrt {b \sec (e+f x)} \sin ^3(e+f x) \, dx=\frac {2 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac {2 b}{f \sqrt {b \sec (e+f x)}} \]

[In]

Int[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^3,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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^3 \text {Subst}\left (\int \frac {-1+\frac {x^2}{b^2}}{x^{7/2}} \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {b^3 \text {Subst}\left (\int \left (-\frac {1}{x^{7/2}}+\frac {1}{b^2 x^{3/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f} \\ & = \frac {2 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 = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \sqrt {b \sec (e+f x)} \sin ^3(e+f x) \, dx=\frac {(-17 \cos (e+f x)+\cos (3 (e+f x))) \sqrt {b \sec (e+f x)}}{10 f} \]

[In]

Integrate[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^3,x]

[Out]

((-17*Cos[e + f*x] + Cos[3*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(10*f)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(424\) vs. \(2(35)=70\).

Time = 0.26 (sec) , antiderivative size = 425, normalized size of antiderivative = 10.37

method	result	size
default	\(-\frac {\left (5 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-5 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-4 \left (\cos ^{3}\left (f x +e \right )\right )+5 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-5 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+20 \cos \left (f x +e \right )\right ) \sqrt {b \sec \left (f x +e \right )}}{10 f}\)	\(425\)

[In]

int(sin(f*x+e)^3*(b*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/10/f*(5*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)-5*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)*cos(f*x+e)-4*cos(f*x+e)^3+5*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)-5*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)+20*cos(f*x+e))*(b*sec(f*x+e))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \sqrt {b \sec (e+f x)} \sin ^3(e+f x) \, dx=\frac {2 \, {\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{5 \, f} \]

[In]

integrate(sin(f*x+e)^3*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \sqrt {b \sec (e+f x)} \sin ^3(e+f x) \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)**3*(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \sqrt {b \sec (e+f x)} \sin ^3(e+f x) \, dx=\frac {2 \, {\left (b^{2} - \frac {5 \, b^{2}}{\cos \left (f x + e\right )^{2}}\right )} b}{5 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {5}{2}}} \]

[In]

integrate(sin(f*x+e)^3*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

2/5*(b^2 - 5*b^2/cos(f*x + e)^2)*b/(f*(b/cos(f*x + e))^(5/2))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.29 \[ \int \sqrt {b \sec (e+f x)} \sin ^3(e+f x) \, dx=\frac {2 \, {\left (\sqrt {b \cos \left (f x + e\right )} b^{2} \cos \left (f x + e\right )^{2} - 5 \, \sqrt {b \cos \left (f x + e\right )} b^{2}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{5 \, b^{2} f} \]

[In]

integrate(sin(f*x+e)^3*(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {b \sec (e+f x)} \sin ^3(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^3\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}} \,d x \]

[In]

int(sin(e + f*x)^3*(b/cos(e + f*x))^(1/2),x)

[Out]

int(sin(e + f*x)^3*(b/cos(e + f*x))^(1/2), x)