Integrand size = 15, antiderivative size = 40 \[ \int \csc ^5(a+b x) \sec (a+b x) \, dx=-\frac {\cot ^2(a+b x)}{b}-\frac {\cot ^4(a+b x)}{4 b}+\frac {\log (\tan (a+b x))}{b} \] Output:
-cot(b*x+a)^2/b-1/4*cot(b*x+a)^4/b+ln(tan(b*x+a))/b
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.35 \[ \int \csc ^5(a+b x) \sec (a+b x) \, dx=-\frac {\csc ^2(a+b x)}{2 b}-\frac {\csc ^4(a+b x)}{4 b}-\frac {\log (\cos (a+b x))}{b}+\frac {\log (\sin (a+b x))}{b} \] Input:
Integrate[Csc[a + b*x]^5*Sec[a + b*x],x]
Output:
-1/2*Csc[a + b*x]^2/b - Csc[a + b*x]^4/(4*b) - Log[Cos[a + b*x]]/b + Log[S in[a + b*x]]/b
Time = 0.37 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3100, 243, 49, 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.
\(\displaystyle \int \csc ^5(a+b x) \sec (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (a+b x)^5 \sec (a+b x)dx\) |
\(\Big \downarrow \) 3100 |
\(\displaystyle \frac {\int \cot ^5(a+b x) \left (\tan ^2(a+b x)+1\right )^2d\tan (a+b x)}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \cot ^3(a+b x) \left (\tan ^2(a+b x)+1\right )^2d\tan ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\cot ^3(a+b x)+2 \cot ^2(a+b x)+\cot (a+b x)\right )d\tan ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{2} \cot ^2(a+b x)-2 \cot (a+b x)+\log \left (\tan ^2(a+b x)\right )}{2 b}\) |
Input:
Int[Csc[a + b*x]^5*Sec[a + b*x],x]
Output:
(-2*Cot[a + b*x] - Cot[a + b*x]^2/2 + Log[Tan[a + b*x]^2])/(2*b)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[1/f Subst[Int[(1 + x^2)^((m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]] , x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]
Time = 2.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {-\frac {1}{4 \sin \left (b x +a \right )^{4}}-\frac {1}{2 \sin \left (b x +a \right )^{2}}+\ln \left (\tan \left (b x +a \right )\right )}{b}\) | \(33\) |
default | \(\frac {-\frac {1}{4 \sin \left (b x +a \right )^{4}}-\frac {1}{2 \sin \left (b x +a \right )^{2}}+\ln \left (\tan \left (b x +a \right )\right )}{b}\) | \(33\) |
risch | \(\frac {2 \,{\mathrm e}^{6 i \left (b x +a \right )}-8 \,{\mathrm e}^{4 i \left (b x +a \right )}+2 \,{\mathrm e}^{2 i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}-\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}\) | \(83\) |
parallelrisch | \(\frac {-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\cot \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-12 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-12 \cot \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+64 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )-64 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )-64 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{64 b}\) | \(99\) |
norman | \(\frac {-\frac {1}{64 b}-\frac {3 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{16 b}-\frac {3 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}}{16 b}-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{8}}{64 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{b}\) | \(116\) |
Input:
int(csc(b*x+a)^5*sec(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/b*(-1/4/sin(b*x+a)^4-1/2/sin(b*x+a)^2+ln(tan(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (38) = 76\).
Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.62 \[ \int \csc ^5(a+b x) \sec (a+b x) \, dx=\frac {2 \, \cos \left (b x + a\right )^{2} - 2 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) + 2 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) - 3}{4 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \] Input:
integrate(csc(b*x+a)^5*sec(b*x+a),x, algorithm="fricas")
Output:
1/4*(2*cos(b*x + a)^2 - 2*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(cos( b*x + a)^2) + 2*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(-1/4*cos(b*x + a)^2 + 1/4) - 3)/(b*cos(b*x + a)^4 - 2*b*cos(b*x + a)^2 + b)
\[ \int \csc ^5(a+b x) \sec (a+b x) \, dx=\int \csc ^{5}{\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \] Input:
integrate(csc(b*x+a)**5*sec(b*x+a),x)
Output:
Integral(csc(a + b*x)**5*sec(a + b*x), x)
Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.28 \[ \int \csc ^5(a+b x) \sec (a+b x) \, dx=-\frac {\frac {2 \, \sin \left (b x + a\right )^{2} + 1}{\sin \left (b x + a\right )^{4}} + 2 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 2 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \] Input:
integrate(csc(b*x+a)^5*sec(b*x+a),x, algorithm="maxima")
Output:
-1/4*((2*sin(b*x + a)^2 + 1)/sin(b*x + a)^4 + 2*log(sin(b*x + a)^2 - 1) - 2*log(sin(b*x + a)^2))/b
Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.38 \[ \int \csc ^5(a+b x) \sec (a+b x) \, dx=-\frac {\log \left ({\left | \sin \left (b x + a\right )^{2} - 1 \right |}\right )}{2 \, b} + \frac {\log \left ({\left | \sin \left (b x + a\right ) \right |}\right )}{b} - \frac {2 \, \sin \left (b x + a\right )^{2} + 1}{4 \, b \sin \left (b x + a\right )^{4}} \] Input:
integrate(csc(b*x+a)^5*sec(b*x+a),x, algorithm="giac")
Output:
-1/2*log(abs(sin(b*x + a)^2 - 1))/b + log(abs(sin(b*x + a)))/b - 1/4*(2*si n(b*x + a)^2 + 1)/(b*sin(b*x + a)^4)
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.98 \[ \int \csc ^5(a+b x) \sec (a+b x) \, dx=\frac {\ln \left (\frac {\cos \left (2\,a+2\,b\,x\right )}{2}-\frac {1}{2}\right )}{2\,b}-\frac {\ln \left (\cos \left (a+b\,x\right )\right )}{b}-\frac {\frac {\cos \left (2\,a+2\,b\,x\right )}{4}-\frac {1}{2}}{b\,\left (\cos \left (2\,a+2\,b\,x\right )-{\left (\frac {\cos \left (2\,a+2\,b\,x\right )}{2}+\frac {1}{2}\right )}^2\right )} \] Input:
int(1/(cos(a + b*x)*sin(a + b*x)^5),x)
Output:
log(cos(2*a + 2*b*x)/2 - 1/2)/(2*b) - log(cos(a + b*x))/b - (cos(2*a + 2*b *x)/4 - 1/2)/(b*(cos(2*a + 2*b*x) - (cos(2*a + 2*b*x)/2 + 1/2)^2))
Time = 0.16 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.48 \[ \int \csc ^5(a+b x) \sec (a+b x) \, dx=\frac {-32 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{4}-32 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{4}+32 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{4}+11 \sin \left (b x +a \right )^{4}-16 \sin \left (b x +a \right )^{2}-8}{32 \sin \left (b x +a \right )^{4} b} \] Input:
int(csc(b*x+a)^5*sec(b*x+a),x)
Output:
( - 32*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**4 - 32*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**4 + 32*log(tan((a + b*x)/2))*sin(a + b*x)**4 + 11*sin( a + b*x)**4 - 16*sin(a + b*x)**2 - 8)/(32*sin(a + b*x)**4*b)