Integrand size = 21, antiderivative size = 83 \[ \int (b \sec (e+f x))^{3/2} \sin ^7(e+f x) \, dx=\frac {2 b^7}{11 f (b \sec (e+f x))^{11/2}}-\frac {6 b^5}{7 f (b \sec (e+f x))^{7/2}}+\frac {2 b^3}{f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \] Output:
2/11*b^7/f/(b*sec(f*x+e))^(11/2)-6/7*b^5/f/(b*sec(f*x+e))^(7/2)+2*b^3/f/(b *sec(f*x+e))^(3/2)+2*b*(b*sec(f*x+e))^(1/2)/f
Time = 0.65 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.63 \[ \int (b \sec (e+f x))^{3/2} \sin ^7(e+f x) \, dx=\frac {b (3370+809 \cos (2 (e+f x))-90 \cos (4 (e+f x))+7 \cos (6 (e+f x))) \sqrt {b \sec (e+f x)}}{1232 f} \] Input:
Integrate[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^7,x]
Output:
(b*(3370 + 809*Cos[2*(e + f*x)] - 90*Cos[4*(e + f*x)] + 7*Cos[6*(e + f*x)] )*Sqrt[b*Sec[e + f*x]])/(1232*f)
Time = 0.42 (sec) , antiderivative size = 76, 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.
\(\displaystyle \int \sin ^7(e+f x) (b \sec (e+f x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(b \sec (e+f x))^{3/2}}{\csc (e+f x)^7}dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {b^7 \int -\frac {\left (b^2-b^2 \sec ^2(e+f x)\right )^3}{b^6 (b \sec (e+f x))^{13/2}}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {b^7 \int \frac {\left (b^2-b^2 \sec ^2(e+f x)\right )^3}{b^6 (b \sec (e+f x))^{13/2}}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b \int \frac {\left (b^2-b^2 \sec ^2(e+f x)\right )^3}{(b \sec (e+f x))^{13/2}}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {b \int \left (\frac {b^6}{(b \sec (e+f x))^{13/2}}-\frac {3 b^4}{(b \sec (e+f x))^{9/2}}+\frac {3 b^2}{(b \sec (e+f x))^{5/2}}-\frac {1}{\sqrt {b \sec (e+f x)}}\right )d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b \left (-\frac {2 b^6}{11 (b \sec (e+f x))^{11/2}}+\frac {6 b^4}{7 (b \sec (e+f x))^{7/2}}-\frac {2 b^2}{(b \sec (e+f x))^{3/2}}-2 \sqrt {b \sec (e+f x)}\right )}{f}\) |
Input:
Int[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^7,x]
Output:
-((b*((-2*b^6)/(11*(b*Sec[e + f*x])^(11/2)) + (6*b^4)/(7*(b*Sec[e + f*x])^ (7/2)) - (2*b^2)/(b*Sec[e + f*x])^(3/2) - 2*Sqrt[b*Sec[e + f*x]]))/f)
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. \(254\) vs. \(2(71)=142\).
Time = 6.64 (sec) , antiderivative size = 255, normalized size of antiderivative = 3.07
method | result | size |
default | \(\frac {b \left (2+\frac {\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right )}{2}+\frac {\left (-\cos \left (f x +e \right )-1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right )}{2}+\frac {2 \cos \left (f x +e \right )^{6}}{11}-\frac {6 \cos \left (f x +e \right )^{4}}{7}+2 \cos \left (f x +e \right )^{2}\right ) \sqrt {b \sec \left (f x +e \right )}}{f}\) | \(255\) |
Input:
int((b*sec(f*x+e))^(3/2)*sin(f*x+e)^7,x,method=_RETURNVERBOSE)
Output:
1/f*b*(2+1/2*(cos(f*x+e)+1)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*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))+1/2*(-cos(f*x+e)-1)*(-cos(f*x+e)/( cos(f*x+e)+1)^2)^(1/2)*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))+ 2/11*cos(f*x+e)^6-6/7*cos(f*x+e)^4+2*cos(f*x+e)^2)*(b*sec(f*x+e))^(1/2)
Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.65 \[ \int (b \sec (e+f x))^{3/2} \sin ^7(e+f x) \, dx=\frac {2 \, {\left (7 \, b \cos \left (f x + e\right )^{6} - 33 \, b \cos \left (f x + e\right )^{4} + 77 \, b \cos \left (f x + e\right )^{2} + 77 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{77 \, f} \] Input:
integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^7,x, algorithm="fricas")
Output:
2/77*(7*b*cos(f*x + e)^6 - 33*b*cos(f*x + e)^4 + 77*b*cos(f*x + e)^2 + 77* b)*sqrt(b/cos(f*x + e))/f
Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^7(e+f x) \, dx=\text {Timed out} \] Input:
integrate((b*sec(f*x+e))**(3/2)*sin(f*x+e)**7,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.87 \[ \int (b \sec (e+f x))^{3/2} \sin ^7(e+f x) \, dx=\frac {2 \, b {\left (\frac {7 \, b^{6}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {11}{2}}} - \frac {33 \, b^{4}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {7}{2}}} + \frac {77 \, b^{2}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}} + 77 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}\right )}}{77 \, f} \] Input:
integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^7,x, algorithm="maxima")
Output:
2/77*b*(7*b^6/(b/cos(f*x + e))^(11/2) - 33*b^4/(b/cos(f*x + e))^(7/2) + 77 *b^2/(b/cos(f*x + e))^(3/2) + 77*sqrt(b/cos(f*x + e)))/f
Time = 0.14 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.18 \[ \int (b \sec (e+f x))^{3/2} \sin ^7(e+f x) \, dx=\frac {2 \, {\left (7 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )^{5} - 33 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )^{3} + 77 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right ) + \frac {77 \, b^{6}}{\sqrt {b \cos \left (f x + e\right )}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{77 \, b^{4} f} \] Input:
integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^7,x, algorithm="giac")
Output:
2/77*(7*sqrt(b*cos(f*x + e))*b^5*cos(f*x + e)^5 - 33*sqrt(b*cos(f*x + e))* b^5*cos(f*x + e)^3 + 77*sqrt(b*cos(f*x + e))*b^5*cos(f*x + e) + 77*b^6/sqr t(b*cos(f*x + e)))*sgn(cos(f*x + e))/(b^4*f)
Timed out. \[ \int (b \sec (e+f x))^{3/2} \sin ^7(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^7\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \] Input:
int(sin(e + f*x)^7*(b/cos(e + f*x))^(3/2),x)
Output:
int(sin(e + f*x)^7*(b/cos(e + f*x))^(3/2), x)
\[ \int (b \sec (e+f x))^{3/2} \sin ^7(e+f x) \, dx=\sqrt {b}\, \left (\int \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right ) \sin \left (f x +e \right )^{7}d x \right ) b \] Input:
int((b*sec(f*x+e))^(3/2)*sin(f*x+e)^7,x)
Output:
sqrt(b)*int(sqrt(sec(e + f*x))*sec(e + f*x)*sin(e + f*x)**7,x)*b