Integrand size = 21, antiderivative size = 85 \[ \int \sqrt {b \sec (e+f x)} \sin ^7(e+f x) \, dx=\frac {2 b^7}{13 f (b \sec (e+f x))^{13/2}}-\frac {2 b^5}{3 f (b \sec (e+f x))^{9/2}}+\frac {6 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac {2 b}{f \sqrt {b \sec (e+f x)}} \]
2/13*b^7/f/(b*sec(f*x+e))^(13/2)-2/3*b^5/f/(b*sec(f*x+e))^(9/2)+6/5*b^3/f/ (b*sec(f*x+e))^(5/2)-2*b/f/(b*sec(f*x+e))^(1/2)
Time = 0.58 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68 \[ \int \sqrt {b \sec (e+f x)} \sin ^7(e+f x) \, dx=\frac {(-8939 \cos (e+f x)+887 \cos (3 (e+f x))-155 \cos (5 (e+f x))+15 \cos (7 (e+f x))) \sqrt {b \sec (e+f x)}}{6240 f} \]
((-8939*Cos[e + f*x] + 887*Cos[3*(e + f*x)] - 155*Cos[5*(e + f*x)] + 15*Co s[7*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(6240*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.
\(\displaystyle \int \sin ^7(e+f x) \sqrt {b \sec (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {b \sec (e+f x)}}{\csc (e+f x)^7}dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {b^7 \int -\frac {\left (b^2-b^2 \sec ^2(e+f x)\right )^3}{b^6 (b \sec (e+f x))^{15/2}}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {b^7 \int \frac {\left (b^2-b^2 \sec ^2(e+f x)\right )^3}{b^6 (b \sec (e+f x))^{15/2}}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b \int \frac {\left (b^2-b^2 \sec ^2(e+f x)\right )^3}{(b \sec (e+f x))^{15/2}}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {b \int \left (\frac {b^6}{(b \sec (e+f x))^{15/2}}-\frac {3 b^4}{(b \sec (e+f x))^{11/2}}+\frac {3 b^2}{(b \sec (e+f x))^{7/2}}-\frac {1}{(b \sec (e+f x))^{3/2}}\right )d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b \left (-\frac {2 b^6}{13 (b \sec (e+f x))^{13/2}}+\frac {2 b^4}{3 (b \sec (e+f x))^{9/2}}-\frac {6 b^2}{5 (b \sec (e+f x))^{5/2}}+\frac {2}{\sqrt {b \sec (e+f x)}}\right )}{f}\) |
-((b*((-2*b^6)/(13*(b*Sec[e + f*x])^(13/2)) + (2*b^4)/(3*(b*Sec[e + f*x])^ (9/2)) - (6*b^2)/(5*(b*Sec[e + f*x])^(5/2)) + 2/Sqrt[b*Sec[e + f*x]]))/f)
3.4.71.3.1 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. \(444\) vs. \(2(71)=142\).
Time = 1.01 (sec) , antiderivative size = 445, normalized size of antiderivative = 5.24
method | result | size |
default | \(\frac {\left (60 \left (\cos ^{7}\left (f x +e \right )\right )-260 \left (\cos ^{5}\left (f x +e \right )\right )+468 \left (\cos ^{3}\left (f x +e \right )\right )-195 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )+195 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-195 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+195 \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-780 \cos \left (f x +e \right )\right ) \sqrt {b \sec \left (f x +e \right )}}{390 f}\) | \(445\) |
1/390/f*(60*cos(f*x+e)^7-260*cos(f*x+e)^5+468*cos(f*x+e)^3-195*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)+195*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)-195*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)+195*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) -780*cos(f*x+e))*(b*sec(f*x+e))^(1/2)
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66 \[ \int \sqrt {b \sec (e+f x)} \sin ^7(e+f x) \, dx=\frac {2 \, {\left (15 \, \cos \left (f x + e\right )^{7} - 65 \, \cos \left (f x + e\right )^{5} + 117 \, \cos \left (f x + e\right )^{3} - 195 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{195 \, f} \]
2/195*(15*cos(f*x + e)^7 - 65*cos(f*x + e)^5 + 117*cos(f*x + e)^3 - 195*co s(f*x + e))*sqrt(b/cos(f*x + e))/f
Timed out. \[ \int \sqrt {b \sec (e+f x)} \sin ^7(e+f x) \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.74 \[ \int \sqrt {b \sec (e+f x)} \sin ^7(e+f x) \, dx=\frac {2 \, {\left (15 \, b^{6} - \frac {65 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac {117 \, b^{6}}{\cos \left (f x + e\right )^{4}} - \frac {195 \, b^{6}}{\cos \left (f x + e\right )^{6}}\right )} b}{195 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {13}{2}}} \]
2/195*(15*b^6 - 65*b^6/cos(f*x + e)^2 + 117*b^6/cos(f*x + e)^4 - 195*b^6/c os(f*x + e)^6)*b/(f*(b/cos(f*x + e))^(13/2))
Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.18 \[ \int \sqrt {b \sec (e+f x)} \sin ^7(e+f x) \, dx=\frac {2 \, {\left (15 \, \sqrt {b \cos \left (f x + e\right )} b^{6} \cos \left (f x + e\right )^{6} - 65 \, \sqrt {b \cos \left (f x + e\right )} b^{6} \cos \left (f x + e\right )^{4} + 117 \, \sqrt {b \cos \left (f x + e\right )} b^{6} \cos \left (f x + e\right )^{2} - 195 \, \sqrt {b \cos \left (f x + e\right )} b^{6}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{195 \, b^{6} f} \]
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*cos(f*x + e))*b^6*cos(f*x + e)^2 - 195*s qrt(b*cos(f*x + e))*b^6)*sgn(cos(f*x + e))/(b^6*f)
Timed out. \[ \int \sqrt {b \sec (e+f x)} \sin ^7(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^7\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}} \,d x \]