Integrand size = 19, antiderivative size = 80 \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=-\frac {b^5 (b \sec (e+f x))^{-5+n}}{f (5-n)}+\frac {2 b^3 (b \sec (e+f x))^{-3+n}}{f (3-n)}-\frac {b (b \sec (e+f x))^{-1+n}}{f (1-n)} \] Output:
-b^5*(b*sec(f*x+e))^(-5+n)/f/(5-n)+2*b^3*(b*sec(f*x+e))^(-3+n)/f/(3-n)-b*( b*sec(f*x+e))^(-1+n)/f/(1-n)
Time = 0.74 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\frac {b \left (89-28 n+3 n^2-4 \left (7-8 n+n^2\right ) \cos (2 (e+f x))+\left (3-4 n+n^2\right ) \cos (4 (e+f x))\right ) (b \sec (e+f x))^{-1+n}}{8 f (-5+n) (-3+n) (-1+n)} \] Input:
Integrate[(b*Sec[e + f*x])^n*Sin[e + f*x]^5,x]
Output:
(b*(89 - 28*n + 3*n^2 - 4*(7 - 8*n + n^2)*Cos[2*(e + f*x)] + (3 - 4*n + n^ 2)*Cos[4*(e + f*x)])*(b*Sec[e + f*x])^(-1 + n))/(8*f*(-5 + n)*(-3 + n)*(-1 + n))
Time = 0.44 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 3102, 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.
\(\displaystyle \int \sin ^5(e+f x) (b \sec (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(b \sec (e+f x))^n}{\csc (e+f x)^5}dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {b^5 \int \frac {(b \sec (e+f x))^{n-6} \left (b^2-b^2 \sec ^2(e+f x)\right )^2}{b^4}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \int (b \sec (e+f x))^{n-6} \left (b^2-b^2 \sec ^2(e+f x)\right )^2d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {b \int \left (b^4 (b \sec (e+f x))^{n-6}-2 b^2 (b \sec (e+f x))^{n-4}+(b \sec (e+f x))^{n-2}\right )d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (-\frac {b^4 (b \sec (e+f x))^{n-5}}{5-n}+\frac {2 b^2 (b \sec (e+f x))^{n-3}}{3-n}-\frac {(b \sec (e+f x))^{n-1}}{1-n}\right )}{f}\) |
Input:
Int[(b*Sec[e + f*x])^n*Sin[e + f*x]^5,x]
Output:
(b*(-((b^4*(b*Sec[e + f*x])^(-5 + n))/(5 - n)) + (2*b^2*(b*Sec[e + f*x])^( -3 + n))/(3 - n) - (b*Sec[e + f*x])^(-1 + n)/(1 - n)))/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])
Time = 2.42 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \(\frac {\left (b \sec \left (f x +e \right )\right )^{n} \left (\left (-\frac {3}{2} n^{2}+14 n -\frac {25}{2}\right ) \cos \left (3 f x +3 e \right )+\left (\frac {1}{2} n^{2}-2 n +\frac {3}{2}\right ) \cos \left (5 f x +5 e \right )+\cos \left (f x +e \right ) \left (n^{2}-12 n +75\right )\right )}{8 \left (n^{3}-9 n^{2}+23 n -15\right ) f}\) | \(87\) |
default | \(\frac {\cos \left (f x +e \right ) {\mathrm e}^{n \ln \left (\frac {b}{\cos \left (f x +e \right )}\right )}}{f \left (-1+n \right )}-\frac {2 \cos \left (f x +e \right )^{3} {\mathrm e}^{n \ln \left (\frac {b}{\cos \left (f x +e \right )}\right )}}{f \left (-3+n \right )}+\frac {\cos \left (f x +e \right )^{5} {\mathrm e}^{n \ln \left (\frac {b}{\cos \left (f x +e \right )}\right )}}{f \left (-5+n \right )}\) | \(94\) |
Input:
int((b*sec(f*x+e))^n*sin(f*x+e)^5,x,method=_RETURNVERBOSE)
Output:
1/8*(b*sec(f*x+e))^n*((-3/2*n^2+14*n-25/2)*cos(3*f*x+3*e)+(1/2*n^2-2*n+3/2 )*cos(5*f*x+5*e)+cos(f*x+e)*(n^2-12*n+75))/(n^3-9*n^2+23*n-15)/f
Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06 \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\frac {{\left ({\left (n^{2} - 4 \, n + 3\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (n^{2} - 6 \, n + 5\right )} \cos \left (f x + e\right )^{3} + {\left (n^{2} - 8 \, n + 15\right )} \cos \left (f x + e\right )\right )} \left (\frac {b}{\cos \left (f x + e\right )}\right )^{n}}{f n^{3} - 9 \, f n^{2} + 23 \, f n - 15 \, f} \] Input:
integrate((b*sec(f*x+e))^n*sin(f*x+e)^5,x, algorithm="fricas")
Output:
((n^2 - 4*n + 3)*cos(f*x + e)^5 - 2*(n^2 - 6*n + 5)*cos(f*x + e)^3 + (n^2 - 8*n + 15)*cos(f*x + e))*(b/cos(f*x + e))^n/(f*n^3 - 9*f*n^2 + 23*f*n - 1 5*f)
Timed out. \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\text {Timed out} \] Input:
integrate((b*sec(f*x+e))**n*sin(f*x+e)**5,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06 \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\frac {\frac {b^{n} \cos \left (f x + e\right )^{-n} \cos \left (f x + e\right )^{5}}{n - 5} - \frac {2 \, b^{n} \cos \left (f x + e\right )^{-n} \cos \left (f x + e\right )^{3}}{n - 3} + \frac {b^{n} \cos \left (f x + e\right )^{-n} \cos \left (f x + e\right )}{n - 1}}{f} \] Input:
integrate((b*sec(f*x+e))^n*sin(f*x+e)^5,x, algorithm="maxima")
Output:
(b^n*cos(f*x + e)^(-n)*cos(f*x + e)^5/(n - 5) - 2*b^n*cos(f*x + e)^(-n)*co s(f*x + e)^3/(n - 3) + b^n*cos(f*x + e)^(-n)*cos(f*x + e)/(n - 1))/f
\[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\int { \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{5} \,d x } \] Input:
integrate((b*sec(f*x+e))^n*sin(f*x+e)^5,x, algorithm="giac")
Output:
integrate((b*sec(f*x + e))^n*sin(f*x + e)^5, x)
Time = 26.95 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.68 \[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=\frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n\,\left (150\,\cos \left (e+f\,x\right )-25\,\cos \left (3\,e+3\,f\,x\right )+3\,\cos \left (5\,e+5\,f\,x\right )-24\,n\,\cos \left (e+f\,x\right )+28\,n\,\cos \left (3\,e+3\,f\,x\right )-4\,n\,\cos \left (5\,e+5\,f\,x\right )+2\,n^2\,\cos \left (e+f\,x\right )-3\,n^2\,\cos \left (3\,e+3\,f\,x\right )+n^2\,\cos \left (5\,e+5\,f\,x\right )\right )}{16\,f\,\left (n^3-9\,n^2+23\,n-15\right )} \] Input:
int(sin(e + f*x)^5*(b/cos(e + f*x))^n,x)
Output:
((b/cos(e + f*x))^n*(150*cos(e + f*x) - 25*cos(3*e + 3*f*x) + 3*cos(5*e + 5*f*x) - 24*n*cos(e + f*x) + 28*n*cos(3*e + 3*f*x) - 4*n*cos(5*e + 5*f*x) + 2*n^2*cos(e + f*x) - 3*n^2*cos(3*e + 3*f*x) + n^2*cos(5*e + 5*f*x)))/(16 *f*(23*n - 9*n^2 + n^3 - 15))
\[ \int (b \sec (e+f x))^n \sin ^5(e+f x) \, dx=b^{n} \left (\int \sec \left (f x +e \right )^{n} \sin \left (f x +e \right )^{5}d x \right ) \] Input:
int((b*sec(f*x+e))^n*sin(f*x+e)^5,x)
Output:
b**n*int(sec(e + f*x)**n*sin(e + f*x)**5,x)