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

Optimal. Leaf size=65 \[ -\frac {2 b^5}{15 f (b \sec (e+f x))^{15/2}}+\frac {4 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac {2 b}{7 f (b \sec (e+f x))^{7/2}} \]

[Out]

-2/15*b^5/f/(b*sec(f*x+e))^(15/2)+4/11*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 = 0.06, 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} \[ -\frac {2 b^5}{15 f (b \sec (e+f x))^{15/2}}+\frac {4 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac {2 b}{7 f (b \sec (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^5)/(15*f*(b*Sec[e + f*x])^(15/2)) + (4*b^3)/(11*f*(b*Sec[e + f*x])^(11/2)) - (2*b)/(7*f*(b*Sec[e + f*x])
^(7/2))

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))^{5/2}} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {\left (-1+\frac {x^2}{b^2}\right )^2}{x^{17/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {b^5 \operatorname {Subst}\left (\int \left (\frac {1}{x^{17/2}}-\frac {2}{b^2 x^{13/2}}+\frac {1}{b^4 x^{9/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=-\frac {2 b^5}{15 f (b \sec (e+f x))^{15/2}}+\frac {4 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 = 0.21, size = 52, normalized size = 0.80 \[ \frac {\cos ^4(e+f x) (532 \cos (2 (e+f x))-77 \cos (4 (e+f x))-711) \sqrt {b \sec (e+f x)}}{4620 b^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]^4*(-711 + 532*Cos[2*(e + f*x)] - 77*Cos[4*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(4620*b^3*f)

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fricas [A]  time = 0.75, size = 51, normalized size = 0.78 \[ -\frac {2 \, {\left (77 \, \cos \left (f x + e\right )^{8} - 210 \, \cos \left (f x + e\right )^{6} + 165 \, \cos \left (f x + e\right )^{4}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{1155 \, b^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/1155*(77*cos(f*x + e)^8 - 210*cos(f*x + e)^6 + 165*cos(f*x + e)^4)*sqrt(b/cos(f*x + e))/(b^3*f)

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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))^(5/2),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/f*2/b^2/1155*(24640*b*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^12+36960*b*sqrt
(-b)*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^11-170016*b^2*(-sqrt(-b)*tan((f*x+exp(
1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^10-110880*b^2*sqrt(-b)*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan
((f*x+exp(1))/2)^4+b))^9+343200*b^3*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^8+13728
0*b^3*sqrt(-b)*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^7-271040*b^4*(-sqrt(-b)*tan(
(f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^6-73920*b^4*sqrt(-b)*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt
(-b*tan((f*x+exp(1))/2)^4+b))^5+80640*b^5*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^4
+480*b^6*sqrt(-b)*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))+10080*b^5*sqrt(-b)*(-sqrt
(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b))^3-3360*b^6*(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(
-b*tan((f*x+exp(1))/2)^4+b))^2+32*b^7)/(-sqrt(-b)*tan((f*x+exp(1))/2)^2+sqrt(-b*tan((f*x+exp(1))/2)^4+b)-sqrt(
-b))^15/sign(tan((f*x+exp(1))/2)^2-1)

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maple [A]  time = 0.17, size = 46, normalized size = 0.71 \[ -\frac {2 \left (77 \left (\cos ^{4}\left (f x +e \right )\right )-210 \left (\cos ^{2}\left (f x +e \right )\right )+165\right ) \cos \left (f x +e \right )}{1155 f \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/1155/f*(77*cos(f*x+e)^4-210*cos(f*x+e)^2+165)*cos(f*x+e)/(b/cos(f*x+e))^(5/2)

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maxima [A]  time = 0.53, size = 50, normalized size = 0.77 \[ -\frac {2 \, {\left (77 \, b^{4} - \frac {210 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac {165 \, b^{4}}{\cos \left (f x + e\right )^{4}}\right )} b}{1155 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {15}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/1155*(77*b^4 - 210*b^4/cos(f*x + e)^2 + 165*b^4/cos(f*x + e)^4)*b/(f*(b/cos(f*x + e))^(15/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))**(5/2),x)

[Out]

Timed out

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