Integrand size = 15, antiderivative size = 39 \[ \int \csc (a+b x) \sec ^5(a+b x) \, dx=\frac {\log (\tan (a+b x))}{b}+\frac {\tan ^2(a+b x)}{b}+\frac {\tan ^4(a+b x)}{4 b} \] Output:
ln(tan(b*x+a))/b+tan(b*x+a)^2/b+1/4*tan(b*x+a)^4/b
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.38 \[ \int \csc (a+b x) \sec ^5(a+b x) \, dx=-\frac {\log (\cos (a+b x))}{b}+\frac {\log (\sin (a+b x))}{b}+\frac {\sec ^2(a+b x)}{2 b}+\frac {\sec ^4(a+b x)}{4 b} \] Input:
Integrate[Csc[a + b*x]*Sec[a + b*x]^5,x]
Output:
-(Log[Cos[a + b*x]]/b) + Log[Sin[a + b*x]]/b + Sec[a + b*x]^2/(2*b) + Sec[ a + b*x]^4/(4*b)
Time = 0.34 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, 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.
\(\displaystyle \int \csc (a+b x) \sec ^5(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (a+b x) \sec (a+b x)^5dx\) |
\(\Big \downarrow \) 3100 |
\(\displaystyle \frac {\int \cot (a+b x) \left (\tan ^2(a+b x)+1\right )^2d\tan (a+b x)}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \cot (a+b x) \left (\tan ^2(a+b x)+1\right )^2d\tan ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\tan ^2(a+b x)+\cot (a+b x)+2\right )d\tan ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} \tan ^4(a+b x)+2 \tan ^2(a+b x)+\log \left (\tan ^2(a+b x)\right )}{2 b}\) |
Input:
Int[Csc[a + b*x]*Sec[a + b*x]^5,x]
Output:
(Log[Tan[a + b*x]^2] + 2*Tan[a + b*x]^2 + Tan[a + b*x]^4/2)/(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 = 2.43 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(\frac {\frac {1}{4 \cos \left (b x +a \right )^{4}}+\frac {1}{2 \cos \left (b x +a \right )^{2}}+\ln \left (\tan \left (b x +a \right )\right )}{b}\) | \(33\) |
default | \(\frac {\frac {1}{4 \cos \left (b x +a \right )^{4}}+\frac {1}{2 \cos \left (b x +a \right )^{2}}+\ln \left (\tan \left (b x +a \right )\right )}{b}\) | \(33\) |
risch | \(\frac {2 \,{\mathrm e}^{6 i \left (b x +a \right )}+8 \,{\mathrm e}^{4 i \left (b x +a \right )}+2 \,{\mathrm e}^{2 i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}-\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}\) | \(83\) |
norman | \(\frac {\frac {2}{3 b}+\frac {4 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{3 b}+\frac {4 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}}{3 b}+\frac {2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{8}}{3 b}}{\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )^{4}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{b}\) | \(120\) |
parallelrisch | \(\frac {\left (-4 \cos \left (4 b x +4 a \right )-16 \cos \left (2 b x +2 a \right )-12\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (-4 \cos \left (4 b x +4 a \right )-16 \cos \left (2 b x +2 a \right )-12\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+\left (4 \cos \left (4 b x +4 a \right )+16 \cos \left (2 b x +2 a \right )+12\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )-4 \cos \left (2 b x +2 a \right )-3 \cos \left (4 b x +4 a \right )+7}{4 b \left (\cos \left (4 b x +4 a \right )+4 \cos \left (2 b x +2 a \right )+3\right )}\) | \(163\) |
Input:
int(csc(b*x+a)*sec(b*x+a)^5,x,method=_RETURNVERBOSE)
Output:
1/b*(1/4/cos(b*x+a)^4+1/2/cos(b*x+a)^2+ln(tan(b*x+a)))
Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.72 \[ \int \csc (a+b x) \sec ^5(a+b x) \, dx=-\frac {2 \, \cos \left (b x + a\right )^{4} \log \left (\cos \left (b x + a\right )^{2}\right ) - 2 \, \cos \left (b x + a\right )^{4} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) - 2 \, \cos \left (b x + a\right )^{2} - 1}{4 \, b \cos \left (b x + a\right )^{4}} \] Input:
integrate(csc(b*x+a)*sec(b*x+a)^5,x, algorithm="fricas")
Output:
-1/4*(2*cos(b*x + a)^4*log(cos(b*x + a)^2) - 2*cos(b*x + a)^4*log(-1/4*cos (b*x + a)^2 + 1/4) - 2*cos(b*x + a)^2 - 1)/(b*cos(b*x + a)^4)
\[ \int \csc (a+b x) \sec ^5(a+b x) \, dx=\int \csc {\left (a + b x \right )} \sec ^{5}{\left (a + b x \right )}\, dx \] Input:
integrate(csc(b*x+a)*sec(b*x+a)**5,x)
Output:
Integral(csc(a + b*x)*sec(a + b*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.67 \[ \int \csc (a+b x) \sec ^5(a+b x) \, dx=-\frac {\frac {2 \, \sin \left (b x + a\right )^{2} - 3}{\sin \left (b x + a\right )^{4} - 2 \, \sin \left (b x + a\right )^{2} + 1} + 2 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 2 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \] Input:
integrate(csc(b*x+a)*sec(b*x+a)^5,x, algorithm="maxima")
Output:
-1/4*((2*sin(b*x + a)^2 - 3)/(sin(b*x + a)^4 - 2*sin(b*x + a)^2 + 1) + 2*l og(sin(b*x + a)^2 - 1) - 2*log(sin(b*x + a)^2))/b
Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.44 \[ \int \csc (a+b x) \sec ^5(a+b x) \, dx=\frac {\log \left ({\left | \cos \left (b x + a\right )^{2} - 1 \right |}\right )}{2 \, b} - \frac {\log \left ({\left | \cos \left (b x + a\right ) \right |}\right )}{b} + \frac {2 \, \cos \left (b x + a\right )^{2} + 1}{4 \, b \cos \left (b x + a\right )^{4}} \] Input:
integrate(csc(b*x+a)*sec(b*x+a)^5,x, algorithm="giac")
Output:
1/2*log(abs(cos(b*x + a)^2 - 1))/b - log(abs(cos(b*x + a)))/b + 1/4*(2*cos (b*x + a)^2 + 1)/(b*cos(b*x + a)^4)
Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.18 \[ \int \csc (a+b x) \sec ^5(a+b x) \, dx=\frac {\frac {\ln \left ({\sin \left (a+b\,x\right )}^2\right )}{2}-\ln \left (\cos \left (a+b\,x\right )\right )+\frac {\frac {{\cos \left (a+b\,x\right )}^2}{2}+\frac {1}{4}}{{\cos \left (a+b\,x\right )}^4}}{b} \] Input:
int(1/(cos(a + b*x)^5*sin(a + b*x)),x)
Output:
(log(sin(a + b*x)^2)/2 - log(cos(a + b*x)) + (cos(a + b*x)^2/2 + 1/4)/cos( a + b*x)^4)/b
Time = 0.17 (sec) , antiderivative size = 216, normalized size of antiderivative = 5.54 \[ \int \csc (a+b x) \sec ^5(a+b x) \, dx=\frac {-4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{4}+8 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )-4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{4}+8 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{4}-8 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )-3 \sin \left (b x +a \right )^{4}+4 \sin \left (b x +a \right )^{2}}{4 b \left (\sin \left (b x +a \right )^{4}-2 \sin \left (b x +a \right )^{2}+1\right )} \] Input:
int(csc(b*x+a)*sec(b*x+a)^5,x)
Output:
( - 4*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**4 + 8*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**2 - 4*log(tan((a + b*x)/2) - 1) - 4*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**4 + 8*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**2 - 4*lo g(tan((a + b*x)/2) + 1) + 4*log(tan((a + b*x)/2))*sin(a + b*x)**4 - 8*log( tan((a + b*x)/2))*sin(a + b*x)**2 + 4*log(tan((a + b*x)/2)) - 3*sin(a + b* x)**4 + 4*sin(a + b*x)**2)/(4*b*(sin(a + b*x)**4 - 2*sin(a + b*x)**2 + 1))