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

Optimal. Leaf size=65 \[ -\frac {2 b^5}{13 f (b \sec (e+f x))^{13/2}}+\frac {4 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac {2 b}{5 f (b \sec (e+f x))^{5/2}} \]

[Out]

-2/13*b^5/f/(b*sec(f*x+e))^(13/2)+4/9*b^3/f/(b*sec(f*x+e))^(9/2)-2/5*b/f/(b*sec(f*x+e))^(5/2)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 65, 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 = {2622, 270} \[ -\frac {2 b^5}{13 f (b \sec (e+f x))^{13/2}}+\frac {4 b^3}{9 f (b \sec (e+f x))^{9/2}}-\frac {2 b}{5 f (b \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^5)/(13*f*(b*Sec[e + f*x])^(13/2)) + (4*b^3)/(9*f*(b*Sec[e + f*x])^(9/2)) - (2*b)/(5*f*(b*Sec[e + f*x])^(
5/2))

Rule 270

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 2622

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

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 42, normalized size = 0.65 \[ \frac {b (340 \cos (2 (e+f x))-45 \cos (4 (e+f x))-551)}{2340 f (b \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(-551 + 340*Cos[2*(e + f*x)] - 45*Cos[4*(e + f*x)]))/(2340*f*(b*Sec[e + f*x])^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 51, normalized size = 0.78 \[ -\frac {2 \, {\left (45 \, \cos \left (f x + e\right )^{7} - 130 \, \cos \left (f x + e\right )^{5} + 117 \, \cos \left (f x + e\right )^{3}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{585 \, b^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/585*(45*cos(f*x + e)^7 - 130*cos(f*x + e)^5 + 117*cos(f*x + e)^3)*sqrt(b/cos(f*x + e))/(b^2*f)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)2/b/f*2/585*(-12480*b*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^10-6240*b*sqrt(-b
)*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^9+48672*b^2*(-sqrt(-b)*tan((f*x+exp(1))/2
)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^8+9984*b^2*sqrt(-b)*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+ex
p(1))/2)^4+b))^7-64896*b^3*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^6-7488*b^3*sqrt(
-b)*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^5+29120*b^4*(-sqrt(-b)*tan((f*x+exp(1))
/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^4+416*b^5*sqrt(-b)*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+e
xp(1))/2)^4+b))+3328*b^4*sqrt(-b)*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^3-2496*b^
5*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^2+32*b^6)/(-sqrt(-b)*tan((f*x+exp(1))/2)^
2+sqrt(-b*tan((f*x+exp(1))/2)^4+b)-sqrt(-b))^13/sign(tan((f*x+exp(1))/2)^2-1)

________________________________________________________________________________________

maple [A]  time = 0.16, size = 46, normalized size = 0.71 \[ -\frac {2 \left (45 \left (\cos ^{4}\left (f x +e \right )\right )-130 \left (\cos ^{2}\left (f x +e \right )\right )+117\right ) \cos \left (f x +e \right )}{585 f \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/585/f*(45*cos(f*x+e)^4-130*cos(f*x+e)^2+117)*cos(f*x+e)/(b/cos(f*x+e))^(3/2)

________________________________________________________________________________________

maxima [A]  time = 0.48, size = 50, normalized size = 0.77 \[ -\frac {2 \, {\left (45 \, b^{4} - \frac {130 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac {117 \, b^{4}}{\cos \left (f x + e\right )^{4}}\right )} b}{585 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/585*(45*b^4 - 130*b^4/cos(f*x + e)^2 + 117*b^4/cos(f*x + e)^4)*b/(f*(b/cos(f*x + e))^(13/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________