Integrand size = 15, antiderivative size = 53 \[ \int \csc (a+b x) \sec ^6(a+b x) \, dx=-\frac {\text {arctanh}(\cos (a+b x))}{b}+\frac {\sec (a+b x)}{b}+\frac {\sec ^3(a+b x)}{3 b}+\frac {\sec ^5(a+b x)}{5 b} \] Output:
-arctanh(cos(b*x+a))/b+sec(b*x+a)/b+1/3*sec(b*x+a)^3/b+1/5*sec(b*x+a)^5/b
Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.36 \[ \int \csc (a+b x) \sec ^6(a+b x) \, dx=-\frac {\log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{b}+\frac {\log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{b}+\frac {\sec (a+b x)}{b}+\frac {\sec ^3(a+b x)}{3 b}+\frac {\sec ^5(a+b x)}{5 b} \] Input:
Integrate[Csc[a + b*x]*Sec[a + b*x]^6,x]
Output:
-(Log[Cos[(a + b*x)/2]]/b) + Log[Sin[(a + b*x)/2]]/b + Sec[a + b*x]/b + Se c[a + b*x]^3/(3*b) + Sec[a + b*x]^5/(5*b)
Time = 0.35 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3102, 25, 254, 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.
\(\displaystyle \int \csc (a+b x) \sec ^6(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (a+b x) \sec (a+b x)^6dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int -\frac {\sec ^6(a+b x)}{1-\sec ^2(a+b x)}d\sec (a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\sec ^6(a+b x)}{1-\sec ^2(a+b x)}d\sec (a+b x)}{b}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle -\frac {\int \left (-\sec ^4(a+b x)-\sec ^2(a+b x)+\frac {1}{1-\sec ^2(a+b x)}-1\right )d\sec (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\text {arctanh}(\sec (a+b x))+\frac {1}{5} \sec ^5(a+b x)+\frac {1}{3} \sec ^3(a+b x)+\sec (a+b x)}{b}\) |
Input:
Int[Csc[a + b*x]*Sec[a + b*x]^6,x]
Output:
(-ArcTanh[Sec[a + b*x]] + Sec[a + b*x] + Sec[a + b*x]^3/3 + Sec[a + b*x]^5 /5)/b
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
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 = 2.23 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {\frac {1}{5 \cos \left (b x +a \right )^{5}}+\frac {1}{3 \cos \left (b x +a \right )^{3}}+\frac {1}{\cos \left (b x +a \right )}+\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{b}\) | \(50\) |
default | \(\frac {\frac {1}{5 \cos \left (b x +a \right )^{5}}+\frac {1}{3 \cos \left (b x +a \right )^{3}}+\frac {1}{\cos \left (b x +a \right )}+\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{b}\) | \(50\) |
norman | \(\frac {-\frac {46}{15 b}+\frac {12 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}}{b}-\frac {6 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{8}}{b}+\frac {28 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{3 b}-\frac {56 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}{3 b}}{\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )^{5}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}\) | \(102\) |
risch | \(\frac {2 \,{\mathrm e}^{9 i \left (b x +a \right )}+\frac {32 \,{\mathrm e}^{7 i \left (b x +a \right )}}{3}+\frac {356 \,{\mathrm e}^{5 i \left (b x +a \right )}}{15}+\frac {32 \,{\mathrm e}^{3 i \left (b x +a \right )}}{3}+2 \,{\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{5}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}\) | \(109\) |
parallelrisch | \(\frac {15 \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{5} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{5} \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )-90 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{8}+180 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}-280 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}+140 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-46}{15 b \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{5} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{5}}\) | \(124\) |
Input:
int(csc(b*x+a)*sec(b*x+a)^6,x,method=_RETURNVERBOSE)
Output:
1/b*(1/5/cos(b*x+a)^5+1/3/cos(b*x+a)^3+1/cos(b*x+a)+ln(csc(b*x+a)-cot(b*x+ a)))
Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.45 \[ \int \csc (a+b x) \sec ^6(a+b x) \, dx=-\frac {15 \, \cos \left (b x + a\right )^{5} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 15 \, \cos \left (b x + a\right )^{5} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 30 \, \cos \left (b x + a\right )^{4} - 10 \, \cos \left (b x + a\right )^{2} - 6}{30 \, b \cos \left (b x + a\right )^{5}} \] Input:
integrate(csc(b*x+a)*sec(b*x+a)^6,x, algorithm="fricas")
Output:
-1/30*(15*cos(b*x + a)^5*log(1/2*cos(b*x + a) + 1/2) - 15*cos(b*x + a)^5*l og(-1/2*cos(b*x + a) + 1/2) - 30*cos(b*x + a)^4 - 10*cos(b*x + a)^2 - 6)/( b*cos(b*x + a)^5)
\[ \int \csc (a+b x) \sec ^6(a+b x) \, dx=\int \csc {\left (a + b x \right )} \sec ^{6}{\left (a + b x \right )}\, dx \] Input:
integrate(csc(b*x+a)*sec(b*x+a)**6,x)
Output:
Integral(csc(a + b*x)*sec(a + b*x)**6, x)
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.13 \[ \int \csc (a+b x) \sec ^6(a+b x) \, dx=\frac {\frac {2 \, {\left (15 \, \cos \left (b x + a\right )^{4} + 5 \, \cos \left (b x + a\right )^{2} + 3\right )}}{\cos \left (b x + a\right )^{5}} - 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{30 \, b} \] Input:
integrate(csc(b*x+a)*sec(b*x+a)^6,x, algorithm="maxima")
Output:
1/30*(2*(15*cos(b*x + a)^4 + 5*cos(b*x + a)^2 + 3)/cos(b*x + a)^5 - 15*log (cos(b*x + a) + 1) + 15*log(cos(b*x + a) - 1))/b
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.25 \[ \int \csc (a+b x) \sec ^6(a+b x) \, dx=-\frac {\log \left ({\left | \cos \left (b x + a\right ) + 1 \right |}\right )}{2 \, b} + \frac {\log \left ({\left | \cos \left (b x + a\right ) - 1 \right |}\right )}{2 \, b} + \frac {15 \, \cos \left (b x + a\right )^{4} + 5 \, \cos \left (b x + a\right )^{2} + 3}{15 \, b \cos \left (b x + a\right )^{5}} \] Input:
integrate(csc(b*x+a)*sec(b*x+a)^6,x, algorithm="giac")
Output:
-1/2*log(abs(cos(b*x + a) + 1))/b + 1/2*log(abs(cos(b*x + a) - 1))/b + 1/1 5*(15*cos(b*x + a)^4 + 5*cos(b*x + a)^2 + 3)/(b*cos(b*x + a)^5)
Time = 25.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.85 \[ \int \csc (a+b x) \sec ^6(a+b x) \, dx=\frac {{\cos \left (a+b\,x\right )}^4+\frac {{\cos \left (a+b\,x\right )}^2}{3}+\frac {1}{5}}{b\,{\cos \left (a+b\,x\right )}^5}-\frac {\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{b} \] Input:
int(1/(cos(a + b*x)^6*sin(a + b*x)),x)
Output:
(cos(a + b*x)^2/3 + cos(a + b*x)^4 + 1/5)/(b*cos(a + b*x)^5) - atanh(cos(a + b*x))/b
Time = 0.20 (sec) , antiderivative size = 167, normalized size of antiderivative = 3.15 \[ \int \csc (a+b x) \sec ^6(a+b x) \, dx=\frac {15 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{4}-30 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2}+15 \cos \left (b x +a \right ) \mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )-23 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{4}+46 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}-23 \cos \left (b x +a \right )+15 \sin \left (b x +a \right )^{4}-35 \sin \left (b x +a \right )^{2}+23}{15 \cos \left (b x +a \right ) b \left (\sin \left (b x +a \right )^{4}-2 \sin \left (b x +a \right )^{2}+1\right )} \] Input:
int(csc(b*x+a)*sec(b*x+a)^6,x)
Output:
(15*cos(a + b*x)*log(tan((a + b*x)/2))*sin(a + b*x)**4 - 30*cos(a + b*x)*l og(tan((a + b*x)/2))*sin(a + b*x)**2 + 15*cos(a + b*x)*log(tan((a + b*x)/2 )) - 23*cos(a + b*x)*sin(a + b*x)**4 + 46*cos(a + b*x)*sin(a + b*x)**2 - 2 3*cos(a + b*x) + 15*sin(a + b*x)**4 - 35*sin(a + b*x)**2 + 23)/(15*cos(a + b*x)*b*(sin(a + b*x)**4 - 2*sin(a + b*x)**2 + 1))