\(\int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx\) [397]

Optimal result

Integrand size = 21, antiderivative size = 85 \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\frac {2 b^7}{9 f (b \sec (e+f x))^{9/2}}-\frac {6 b^5}{5 f (b \sec (e+f x))^{5/2}}+\frac {6 b^3}{f \sqrt {b \sec (e+f x)}}+\frac {2 b (b \sec (e+f x))^{3/2}}{3 f} \]

[Out]

2/9*b^7/f/(b*sec(f*x+e))^(9/2)-6/5*b^5/f/(b*sec(f*x+e))^(5/2)+2/3*b*(b*sec(f*x+e))^(3/2)/f+6*b^3/f/(b*sec(f*x+
e))^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2702, 276} \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\frac {2 b^7}{9 f (b \sec (e+f x))^{9/2}}-\frac {6 b^5}{5 f (b \sec (e+f x))^{5/2}}+\frac {6 b^3}{f \sqrt {b \sec (e+f x)}}+\frac {2 b (b \sec (e+f x))^{3/2}}{3 f} \]

[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 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*} \text {integral}& = \frac {b^7 \text {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 \text {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] (verified)

Time = 0.65 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.61 \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\frac {b (2366+1803 \cos (2 (e+f x))-78 \cos (4 (e+f x))+5 \cos (6 (e+f x))) (b \sec (e+f x))^{3/2}}{720 f} \]

[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(445\) vs. \(2(71)=142\).

Time = 0.87 (sec) , antiderivative size = 446, normalized size of antiderivative = 5.25

\[\frac {\sqrt {b \sec \left (f x +e \right )}\, b^{2} \left (20 \left (\cos ^{5}\left (f x +e \right )\right )+135 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-135 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-108 \left (\cos ^{3}\left (f x +e \right )\right )+135 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-135 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+540 \cos \left (f x +e \right )+60 \sec \left (f x +e \right )\right )}{90 f}\]

[In]

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

[Out]

1/90/f*(b*sec(f*x+e))^(1/2)*b^2*(20*cos(f*x+e)^5+135*ln((2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(
-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x
+e)-135*ln(2*(2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x
+e)+1)/(cos(f*x+e)+1))*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)-108*cos(f*x+e)^3+135*ln((2*cos(f*x+e)*(
-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))*(-cos
(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-135*ln(2*(2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(co
s(f*x+e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1))*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+540*cos(f*x+e)+60*sec(
f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82 \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\frac {2 \, {\left (5 \, b^{2} \cos \left (f x + e\right )^{6} - 27 \, b^{2} \cos \left (f x + e\right )^{4} + 135 \, b^{2} \cos \left (f x + e\right )^{2} + 15 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{45 \, f \cos \left (f x + e\right )} \]

[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))

Sympy [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.78 \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\frac {2 \, {\left (15 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}} + \frac {5 \, b^{6} - \frac {27 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac {135 \, b^{6}}{\cos \left (f x + e\right )^{4}}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {9}{2}}}\right )} b}{45 \, f} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.18 \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\frac {2 \, {\left (5 \, \sqrt {b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{4} - 27 \, \sqrt {b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{2} + 135 \, \sqrt {b \cos \left (f x + e\right )} b^{4} + \frac {15 \, b^{5}}{\sqrt {b \cos \left (f x + e\right )} \cos \left (f x + e\right )}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{45 \, b^{2} f} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^7\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2} \,d x \]

[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)