Integrand size = 21, antiderivative size = 85 \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\frac {2 b^7}{9 f (b \sec (e+f x))^{9/2}}-\frac {6 b^5}{5 f (b \sec (e+f x))^{5/2}}+\frac {6 b^3}{f \sqrt {b \sec (e+f x)}}+\frac {2 b (b \sec (e+f x))^{3/2}}{3 f} \]
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)
Time = 0.65 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.61 \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\frac {b (2366+1803 \cos (2 (e+f x))-78 \cos (4 (e+f x))+5 \cos (6 (e+f x))) (b \sec (e+f x))^{3/2}}{720 f} \]
(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)
Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3102, 25, 27, 244, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin ^7(e+f x) (b \sec (e+f x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(b \sec (e+f x))^{5/2}}{\csc (e+f x)^7}dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {b^7 \int -\frac {\left (b^2-b^2 \sec ^2(e+f x)\right )^3}{b^6 (b \sec (e+f x))^{11/2}}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {b^7 \int \frac {\left (b^2-b^2 \sec ^2(e+f x)\right )^3}{b^6 (b \sec (e+f x))^{11/2}}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b \int \frac {\left (b^2-b^2 \sec ^2(e+f x)\right )^3}{(b \sec (e+f x))^{11/2}}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {b \int \left (\frac {b^6}{(b \sec (e+f x))^{11/2}}-\frac {3 b^4}{(b \sec (e+f x))^{7/2}}+\frac {3 b^2}{(b \sec (e+f x))^{3/2}}-\sqrt {b \sec (e+f x)}\right )d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b \left (-\frac {2 b^6}{9 (b \sec (e+f x))^{9/2}}+\frac {6 b^4}{5 (b \sec (e+f x))^{5/2}}-\frac {6 b^2}{\sqrt {b \sec (e+f x)}}-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )}{f}\) |
-((b*((-2*b^6)/(9*(b*Sec[e + f*x])^(9/2)) + (6*b^4)/(5*(b*Sec[e + f*x])^(5 /2)) - (6*b^2)/Sqrt[b*Sec[e + f*x]] - (2*(b*Sec[e + f*x])^(3/2))/3))/f)
3.4.97.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[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])
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
\[\frac {\sqrt {b \sec \left (f x +e \right )}\, b^{2} \left (20 \left (\cos ^{5}\left (f x +e \right )\right )+135 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-135 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-108 \left (\cos ^{3}\left (f x +e \right )\right )+135 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-135 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+540 \cos \left (f x +e \right )+60 \sec \left (f x +e \right )\right )}{90 f}\]
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)*(-co s(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-co s(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)/(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)+540*cos(f*x+e)+60*sec(f*x+e))
Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82 \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\frac {2 \, {\left (5 \, b^{2} \cos \left (f x + e\right )^{6} - 27 \, b^{2} \cos \left (f x + e\right )^{4} + 135 \, b^{2} \cos \left (f x + e\right )^{2} + 15 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{45 \, f \cos \left (f x + e\right )} \]
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))
Timed out. \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.78 \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\frac {2 \, {\left (15 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}} + \frac {5 \, b^{6} - \frac {27 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac {135 \, b^{6}}{\cos \left (f x + e\right )^{4}}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {9}{2}}}\right )} b}{45 \, f} \]
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
Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.18 \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\frac {2 \, {\left (5 \, \sqrt {b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{4} - 27 \, \sqrt {b \cos \left (f x + e\right )} b^{4} \cos \left (f x + e\right )^{2} + 135 \, \sqrt {b \cos \left (f x + e\right )} b^{4} + \frac {15 \, b^{5}}{\sqrt {b \cos \left (f x + e\right )} \cos \left (f x + e\right )}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{45 \, b^{2} f} \]
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)
Timed out. \[ \int (b \sec (e+f x))^{5/2} \sin ^7(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^7\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2} \,d x \]