3.5.37 \(\int \frac {\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\) [437]

Optimal. Leaf size=87 \[ \frac {2 b^7}{19 f (b \sec (e+f x))^{19/2}}-\frac {2 b^5}{5 f (b \sec (e+f x))^{15/2}}+\frac {6 b^3}{11 f (b \sec (e+f x))^{11/2}}-\frac {2 b}{7 f (b \sec (e+f x))^{7/2}} \]

[Out]

2/19*b^7/f/(b*sec(f*x+e))^(19/2)-2/5*b^5/f/(b*sec(f*x+e))^(15/2)+6/11*b^3/f/(b*sec(f*x+e))^(11/2)-2/7*b/f/(b*s
ec(f*x+e))^(7/2)

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Rubi [A]
time = 0.04, antiderivative size = 87, 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, 276} \begin {gather*} \frac {2 b^7}{19 f (b \sec (e+f x))^{19/2}}-\frac {2 b^5}{5 f (b \sec (e+f x))^{15/2}}+\frac {6 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]^7/(b*Sec[e + f*x])^(5/2),x]

[Out]

(2*b^7)/(19*f*(b*Sec[e + f*x])^(19/2)) - (2*b^5)/(5*f*(b*Sec[e + f*x])^(15/2)) + (6*b^3)/(11*f*(b*Sec[e + f*x]
)^(11/2)) - (2*b)/(7*f*(b*Sec[e + f*x])^(7/2))

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*} \int \frac {\sin ^7(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx &=\frac {b^7 \text {Subst}\left (\int \frac {\left (-1+\frac {x^2}{b^2}\right )^3}{x^{21/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {b^7 \text {Subst}\left (\int \left (-\frac {1}{x^{21/2}}+\frac {3}{b^2 x^{17/2}}-\frac {3}{b^4 x^{13/2}}+\frac {1}{b^6 x^{9/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {2 b^7}{19 f (b \sec (e+f x))^{19/2}}-\frac {2 b^5}{5 f (b \sec (e+f x))^{15/2}}+\frac {6 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 = 0.26, size = 62, normalized size = 0.71 \begin {gather*} \frac {\cos ^4(e+f x) (-15226+14287 \cos (2 (e+f x))-3542 \cos (4 (e+f x))+385 \cos (6 (e+f x))) \sqrt {b \sec (e+f x)}}{117040 b^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^7/(b*Sec[e + f*x])^(5/2),x]

[Out]

(Cos[e + f*x]^4*(-15226 + 14287*Cos[2*(e + f*x)] - 3542*Cos[4*(e + f*x)] + 385*Cos[6*(e + f*x)])*Sqrt[b*Sec[e
+ f*x]])/(117040*b^3*f)

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Maple [A]
time = 0.27, size = 56, normalized size = 0.64

method	result	size
default	\(\frac {2 \left (385 \left (\cos ^{6}\left (f x +e \right )\right )-1463 \left (\cos ^{4}\left (f x +e \right )\right )+1995 \left (\cos ^{2}\left (f x +e \right )\right )-1045\right ) \cos \left (f x +e \right )}{7315 f \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^7/(b*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/7315/f*(385*cos(f*x+e)^6-1463*cos(f*x+e)^4+1995*cos(f*x+e)^2-1045)*cos(f*x+e)/(b/cos(f*x+e))^(5/2)

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Maxima [A]
time = 0.29, size = 67, normalized size = 0.77 \begin {gather*} \frac {2 \, {\left (385 \, b^{6} - \frac {1463 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac {1995 \, b^{6}}{\cos \left (f x + e\right )^{4}} - \frac {1045 \, b^{6}}{\cos \left (f x + e\right )^{6}}\right )} b}{7315 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {19}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^7/(b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

2/7315*(385*b^6 - 1463*b^6/cos(f*x + e)^2 + 1995*b^6/cos(f*x + e)^4 - 1045*b^6/cos(f*x + e)^6)*b/(f*(b/cos(f*x
+ e))^(19/2))

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Fricas [A]
time = 0.41, size = 66, normalized size = 0.76 \begin {gather*} \frac {2 \, {\left (385 \, \cos \left (f x + e\right )^{10} - 1463 \, \cos \left (f x + e\right )^{8} + 1995 \, \cos \left (f x + e\right )^{6} - 1045 \, \cos \left (f x + e\right )^{4}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{7315 \, b^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^7/(b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

2/7315*(385*cos(f*x + e)^10 - 1463*cos(f*x + e)^8 + 1995*cos(f*x + e)^6 - 1045*cos(f*x + e)^4)*sqrt(b/cos(f*x
+ e))/(b^3*f)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**7/(b*sec(f*x+e))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5007 deep

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Giac [A]
time = 5.32, size = 119, normalized size = 1.37 \begin {gather*} \frac {2 \, {\left (385 \, \sqrt {b \cos \left (f x + e\right )} b^{9} \cos \left (f x + e\right )^{9} - 1463 \, \sqrt {b \cos \left (f x + e\right )} b^{9} \cos \left (f x + e\right )^{7} + 1995 \, \sqrt {b \cos \left (f x + e\right )} b^{9} \cos \left (f x + e\right )^{5} - 1045 \, \sqrt {b \cos \left (f x + e\right )} b^{9} \cos \left (f x + e\right )^{3}\right )}}{7315 \, b^{12} 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)^7/(b*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

2/7315*(385*sqrt(b*cos(f*x + e))*b^9*cos(f*x + e)^9 - 1463*sqrt(b*cos(f*x + e))*b^9*cos(f*x + e)^7 + 1995*sqrt
(b*cos(f*x + e))*b^9*cos(f*x + e)^5 - 1045*sqrt(b*cos(f*x + e))*b^9*cos(f*x + e)^3)/(b^12*f*sgn(cos(f*x + e)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (e+f\,x\right )}^7}{{\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)^7/(b/cos(e + f*x))^(5/2),x)

[Out]

int(sin(e + f*x)^7/(b/cos(e + f*x))^(5/2), x)

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