3.5.39 \(\int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\) [439]

Optimal. Leaf size=43 \[ \frac {2 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac {2 b}{7 f (b \sec (e+f x))^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2702, 14} \begin {gather*} \frac {2 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac {2 b}{7 f (b \sec (e+f x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^3/(b*Sec[e + f*x])^(5/2),x]

[Out]

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

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*} \int \frac {\sin ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx &=\frac {b^3 \text {Subst}\left (\int \frac {-1+\frac {x^2}{b^2}}{x^{13/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {b^3 \text {Subst}\left (\int \left (-\frac {1}{x^{13/2}}+\frac {1}{b^2 x^{9/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {2 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*}

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Mathematica [A]
time = 0.11, size = 42, normalized size = 0.98 \begin {gather*} \frac {\cos ^4(e+f x) (-15+7 \cos (2 (e+f x))) \sqrt {b \sec (e+f x)}}{77 b^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^3/(b*Sec[e + f*x])^(5/2),x]

[Out]

(Cos[e + f*x]^4*(-15 + 7*Cos[2*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(77*b^3*f)

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Maple [A]
time = 0.18, size = 36, normalized size = 0.84

method	result	size
default	\(\frac {2 \left (7 \left (\cos ^{2}\left (f x +e \right )\right )-11\right ) \cos \left (f x +e \right )}{77 f \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^3/(b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/77/f*(7*cos(f*x+e)^2-11)*cos(f*x+e)/(b/cos(f*x+e))^(5/2)

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Maxima [A]
time = 0.29, size = 39, normalized size = 0.91 \begin {gather*} \frac {2 \, {\left (7 \, b^{2} - \frac {11 \, b^{2}}{\cos \left (f x + e\right )^{2}}\right )} b}{77 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3/(b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

2/77*(7*b^2 - 11*b^2/cos(f*x + e)^2)*b/(f*(b/cos(f*x + e))^(11/2))

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Fricas [A]
time = 0.40, size = 44, normalized size = 1.02 \begin {gather*} \frac {2 \, {\left (7 \, \cos \left (f x + e\right )^{6} - 11 \, \cos \left (f x + e\right )^{4}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{77 \, b^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3/(b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

2/77*(7*cos(f*x + e)^6 - 11*cos(f*x + e)^4)*sqrt(b/cos(f*x + e))/(b^3*f)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**3/(b*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 4.73, size = 69, normalized size = 1.60 \begin {gather*} \frac {2 \, {\left (7 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )^{5} - 11 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )^{3}\right )}}{77 \, b^{8} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3/(b*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

2/77*(7*sqrt(b*cos(f*x + e))*b^5*cos(f*x + e)^5 - 11*sqrt(b*cos(f*x + e))*b^5*cos(f*x + e)^3)/(b^8*f*sgn(cos(f
*x + e)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\sin \left (e+f\,x\right )}^3}{{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^3/(b/cos(e + f*x))^(5/2),x)

[Out]

int(sin(e + f*x)^3/(b/cos(e + f*x))^(5/2), x)

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