3.397 \(\int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0623046, 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} \[ \frac{2 b^7}{9 f (b \sec (e+f x))^{9/2}}-\frac{6 b^5}{5 f (b \sec (e+f x))^{5/2}}+\frac{6 b^3}{f \sqrt{b \sec (e+f x)}}+\frac{2 b (b \sec (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx &=\frac{b^7 \operatorname{Subst}\left (\int \frac{\left (-1+\frac{x^2}{b^2}\right )^3}{x^{11/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{b^7 \operatorname{Subst}\left (\int \left (-\frac{1}{x^{11/2}}+\frac{3}{b^2 x^{7/2}}-\frac{3}{b^4 x^{3/2}}+\frac{\sqrt{x}}{b^6}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac{2 b^7}{9 f (b \sec (e+f x))^{9/2}}-\frac{6 b^5}{5 f (b \sec (e+f x))^{5/2}}+\frac{6 b^3}{f \sqrt{b \sec (e+f x)}}+\frac{2 b (b \sec (e+f x))^{3/2}}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.425678, size = 52, normalized size = 0.61 \[ \frac{b (1803 \cos (2 (e+f x))-78 \cos (4 (e+f x))+5 \cos (6 (e+f x))+2366) (b \sec (e+f x))^{3/2}}{720 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(2366 + 1803*Cos[2*(e + f*x)] - 78*Cos[4*(e + f*x)] + 5*Cos[6*(e + f*x)])*(b*Sec[e + f*x])^(3/2))/(720*f)

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Maple [B]  time = 0.152, size = 532, normalized size = 6.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/90/f*(-1+cos(f*x+e))^2*(20*cos(f*x+e)^6-108*cos(f*x+e)^4-135*cos(f*x+e)^2*(-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)+135*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*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)-135*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*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)+135*cos(f*x+e)*(-cos
(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*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)+540*cos(f*x+e)^2+60)*cos(f*x+e)*(cos(f*x+e)+1)^2*
(b/cos(f*x+e))^(5/2)/sin(f*x+e)^4

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Maxima [A]  time = 1.05393, size = 89, normalized size = 1.05 \begin{align*} \frac{2 \,{\left (15 \, \left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{3}{2}} + \frac{5 \, b^{6} - \frac{27 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac{135 \, b^{6}}{\cos \left (f x + e\right )^{4}}}{\left (\frac{b}{\cos \left (f x + e\right )}\right )^{\frac{9}{2}}}\right )} b}{45 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(15*(b/cos(f*x + e))^(3/2) + (5*b^6 - 27*b^6/cos(f*x + e)^2 + 135*b^6/cos(f*x + e)^4)/(b/cos(f*x + e))^(9
/2))*b/f

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Fricas [A]  time = 2.33944, size = 169, normalized size = 1.99 \begin{align*} \frac{2 \,{\left (5 \, b^{2} \cos \left (f x + e\right )^{6} - 27 \, b^{2} \cos \left (f x + e\right )^{4} + 135 \, b^{2} \cos \left (f x + e\right )^{2} + 15 \, b^{2}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}}}{45 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5*b^2*cos(f*x + e)^6 - 27*b^2*cos(f*x + e)^4 + 135*b^2*cos(f*x + e)^2 + 15*b^2)*sqrt(b/cos(f*x + e))/(f*
cos(f*x + e))

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

[Out]

Timed out

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Giac [A]  time = 1.20923, size = 146, normalized size = 1.72 \begin{align*} \frac{2 \,{\left (5 \, \sqrt{b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{4} - 27 \, \sqrt{b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{2} + 135 \, \sqrt{b \cos \left (f x + e\right )} b^{4} + \frac{15 \, b^{5}}{\sqrt{b \cos \left (f x + e\right )} \cos \left (f x + e\right )}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{45 \, b^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5*sqrt(b*cos(f*x + e))*b^4*cos(f*x + e)^4 - 27*sqrt(b*cos(f*x + e))*b^4*cos(f*x + e)^2 + 135*sqrt(b*cos(
f*x + e))*b^4 + 15*b^5/(sqrt(b*cos(f*x + e))*cos(f*x + e)))*sgn(cos(f*x + e))/(b^2*f)