3.371 \(\int \sqrt{b \sec (e+f x)} \sin ^7(e+f x) \, dx\)

Optimal. Leaf size=85 \[ \frac{2 b^7}{13 f (b \sec (e+f x))^{13/2}}-\frac{2 b^5}{3 f (b \sec (e+f x))^{9/2}}+\frac{6 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac{2 b}{f \sqrt{b \sec (e+f x)}} \]

[Out]

(2*b^7)/(13*f*(b*Sec[e + f*x])^(13/2)) - (2*b^5)/(3*f*(b*Sec[e + f*x])^(9/2)) + (6*b^3)/(5*f*(b*Sec[e + f*x])^
(5/2)) - (2*b)/(f*Sqrt[b*Sec[e + f*x]])

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Rubi [A]  time = 0.0592543, antiderivative size = 85, normalized size of antiderivative = 1., 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} \[ \frac{2 b^7}{13 f (b \sec (e+f x))^{13/2}}-\frac{2 b^5}{3 f (b \sec (e+f x))^{9/2}}+\frac{6 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac{2 b}{f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^7,x]

[Out]

(2*b^7)/(13*f*(b*Sec[e + f*x])^(13/2)) - (2*b^5)/(3*f*(b*Sec[e + f*x])^(9/2)) + (6*b^3)/(5*f*(b*Sec[e + f*x])^
(5/2)) - (2*b)/(f*Sqrt[b*Sec[e + f*x]])

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])

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]

Rubi steps

\begin{align*} \int \sqrt{b \sec (e+f x)} \sin ^7(e+f x) \, dx &=\frac{b^7 \operatorname{Subst}\left (\int \frac{\left (-1+\frac{x^2}{b^2}\right )^3}{x^{15/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{b^7 \operatorname{Subst}\left (\int \left (-\frac{1}{x^{15/2}}+\frac{3}{b^2 x^{11/2}}-\frac{3}{b^4 x^{7/2}}+\frac{1}{b^6 x^{3/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{2 b^7}{13 f (b \sec (e+f x))^{13/2}}-\frac{2 b^5}{3 f (b \sec (e+f x))^{9/2}}+\frac{6 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]  time = 0.340371, size = 58, normalized size = 0.68 \[ \frac{(-8939 \cos (e+f x)+887 \cos (3 (e+f x))-155 \cos (5 (e+f x))+15 \cos (7 (e+f x))) \sqrt{b \sec (e+f x)}}{6240 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^7,x]

[Out]

((-8939*Cos[e + f*x] + 887*Cos[3*(e + f*x)] - 155*Cos[5*(e + f*x)] + 15*Cos[7*(e + f*x)])*Sqrt[b*Sec[e + f*x]]
)/(6240*f)

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Maple [B]  time = 0.283, size = 517, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^7*(b*sec(f*x+e))^(1/2),x)

[Out]

1/390/f*(-1+cos(f*x+e))^2*(60*cos(f*x+e)^7-260*cos(f*x+e)^5+468*cos(f*x+e)^3+195*cos(f*x+e)*(-cos(f*x+e)/(cos(
f*x+e)+1)^2)^(1/2)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-co
s(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-195*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(-(2*c
os(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1
/2)-1)/sin(f*x+e)^2)+195*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1
)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-195*ln(-(2*cos(f*
x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1
)/sin(f*x+e)^2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-780*cos(f*x+e))*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(1/2)/sin
(f*x+e)^4

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Maxima [A]  time = 1.05537, size = 85, normalized size = 1. \begin{align*} \frac{2 \,{\left (15 \, b^{6} - \frac{65 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac{117 \, b^{6}}{\cos \left (f x + e\right )^{4}} - \frac{195 \, b^{6}}{\cos \left (f x + e\right )^{6}}\right )} b}{195 \, f \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{13}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^7*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

2/195*(15*b^6 - 65*b^6/cos(f*x + e)^2 + 117*b^6/cos(f*x + e)^4 - 195*b^6/cos(f*x + e)^6)*b/(f*(b/cos(f*x + e))
^(13/2))

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Fricas [A]  time = 2.24026, size = 149, normalized size = 1.75 \begin{align*} \frac{2 \,{\left (15 \, \cos \left (f x + e\right )^{7} - 65 \, \cos \left (f x + e\right )^{5} + 117 \, \cos \left (f x + e\right )^{3} - 195 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{195 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^7*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/195*(15*cos(f*x + e)^7 - 65*cos(f*x + e)^5 + 117*cos(f*x + e)^3 - 195*cos(f*x + e))*sqrt(b/cos(f*x + e))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**7*(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.17161, size = 146, normalized size = 1.72 \begin{align*} \frac{2 \,{\left (15 \, \sqrt{b \cos \left (f x + e\right )} b^{6} \cos \left (f x + e\right )^{6} - 65 \, \sqrt{b \cos \left (f x + e\right )} b^{6} \cos \left (f x + e\right )^{4} + 117 \, \sqrt{b \cos \left (f x + e\right )} b^{6} \cos \left (f x + e\right )^{2} - 195 \, \sqrt{b \cos \left (f x + e\right )} b^{6}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{195 \, b^{6} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^7*(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

2/195*(15*sqrt(b*cos(f*x + e))*b^6*cos(f*x + e)^6 - 65*sqrt(b*cos(f*x + e))*b^6*cos(f*x + e)^4 + 117*sqrt(b*co
s(f*x + e))*b^6*cos(f*x + e)^2 - 195*sqrt(b*cos(f*x + e))*b^6)*sgn(cos(f*x + e))/(b^6*f)