Integrand size = 17, antiderivative size = 43 \[ \int \csc ^3(a+b x) \sec ^3(a+b x) \, dx=-\frac {\cot ^2(a+b x)}{2 b}+\frac {2 \log (\tan (a+b x))}{b}+\frac {\tan ^2(a+b x)}{2 b} \] Output:
-1/2*cot(b*x+a)^2/b+2*ln(tan(b*x+a))/b+1/2*tan(b*x+a)^2/b
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.42 \[ \int \csc ^3(a+b x) \sec ^3(a+b x) \, dx=8 \left (-\frac {\csc ^2(a+b x)}{16 b}-\frac {\log (\cos (a+b x))}{4 b}+\frac {\log (\sin (a+b x))}{4 b}+\frac {\sec ^2(a+b x)}{16 b}\right ) \] Input:
Integrate[Csc[a + b*x]^3*Sec[a + b*x]^3,x]
Output:
8*(-1/16*Csc[a + b*x]^2/b - Log[Cos[a + b*x]]/(4*b) + Log[Sin[a + b*x]]/(4 *b) + Sec[a + b*x]^2/(16*b))
Time = 0.38 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, 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.
\(\displaystyle \int \csc ^3(a+b x) \sec ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (a+b x)^3 \sec (a+b x)^3dx\) |
\(\Big \downarrow \) 3100 |
\(\displaystyle \frac {\int \cot ^3(a+b x) \left (\tan ^2(a+b x)+1\right )^2d\tan (a+b x)}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \cot ^2(a+b x) \left (\tan ^2(a+b x)+1\right )^2d\tan ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\cot ^2(a+b x)+2 \cot (a+b x)+1\right )d\tan ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\tan ^2(a+b x)-\cot (a+b x)+2 \log \left (\tan ^2(a+b x)\right )}{2 b}\) |
Input:
Int[Csc[a + b*x]^3*Sec[a + b*x]^3,x]
Output:
(-Cot[a + b*x] + 2*Log[Tan[a + b*x]^2] + 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 = 2.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {\frac {1}{2 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )^{2}}-\frac {1}{\sin \left (b x +a \right )^{2}}+2 \ln \left (\tan \left (b x +a \right )\right )}{b}\) | \(43\) |
default | \(\frac {\frac {1}{2 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )^{2}}-\frac {1}{\sin \left (b x +a \right )^{2}}+2 \ln \left (\tan \left (b x +a \right )\right )}{b}\) | \(43\) |
risch | \(\frac {4 \,{\mathrm e}^{6 i \left (b x +a \right )}+4 \,{\mathrm e}^{2 i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}\) | \(86\) |
norman | \(\frac {-\frac {1}{8 b}-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{8}}{8 b}+\frac {9 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}{4 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )^{2}}+\frac {2 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {2 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{b}-\frac {2 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{b}\) | \(116\) |
parallelrisch | \(\frac {\left (-8 \cos \left (2 b x +2 a \right )-8\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (-8 \cos \left (2 b x +2 a \right )-8\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+\left (8 \cos \left (2 b x +2 a \right )+8\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\left (-8 \cos \left (b x +a \right )+2 \cos \left (2 b x +2 a \right )+2\right ) \cot \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+\left (-\sec \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+4\right ) \csc \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{4 b \left (1+\cos \left (2 b x +2 a \right )\right )}\) | \(156\) |
Input:
int(csc(b*x+a)^3*sec(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b*(1/2/sin(b*x+a)^2/cos(b*x+a)^2-1/sin(b*x+a)^2+2*ln(tan(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (39) = 78\).
Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.37 \[ \int \csc ^3(a+b x) \sec ^3(a+b x) \, dx=\frac {2 \, \cos \left (b x + a\right )^{2} - 2 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) + 2 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) - 1}{2 \, {\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )}} \] Input:
integrate(csc(b*x+a)^3*sec(b*x+a)^3,x, algorithm="fricas")
Output:
1/2*(2*cos(b*x + a)^2 - 2*(cos(b*x + a)^4 - cos(b*x + a)^2)*log(cos(b*x + a)^2) + 2*(cos(b*x + a)^4 - cos(b*x + a)^2)*log(-1/4*cos(b*x + a)^2 + 1/4) - 1)/(b*cos(b*x + a)^4 - b*cos(b*x + a)^2)
\[ \int \csc ^3(a+b x) \sec ^3(a+b x) \, dx=\int \csc ^{3}{\left (a + b x \right )} \sec ^{3}{\left (a + b x \right )}\, dx \] Input:
integrate(csc(b*x+a)**3*sec(b*x+a)**3,x)
Output:
Integral(csc(a + b*x)**3*sec(a + b*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.49 \[ \int \csc ^3(a+b x) \sec ^3(a+b x) \, dx=-\frac {\frac {2 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{4} - \sin \left (b x + a\right )^{2}} + 2 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 2 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{2 \, b} \] Input:
integrate(csc(b*x+a)^3*sec(b*x+a)^3,x, algorithm="maxima")
Output:
-1/2*((2*sin(b*x + a)^2 - 1)/(sin(b*x + a)^4 - sin(b*x + a)^2) + 2*log(sin (b*x + a)^2 - 1) - 2*log(sin(b*x + a)^2))/b
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.60 \[ \int \csc ^3(a+b x) \sec ^3(a+b x) \, dx=-\frac {\log \left ({\left | \sin \left (b x + a\right )^{2} - 1 \right |}\right )}{b} + \frac {2 \, \log \left ({\left | \sin \left (b x + a\right ) \right |}\right )}{b} - \frac {2 \, \sin \left (b x + a\right )^{2} - 1}{2 \, {\left (\sin \left (b x + a\right )^{4} - \sin \left (b x + a\right )^{2}\right )} b} \] Input:
integrate(csc(b*x+a)^3*sec(b*x+a)^3,x, algorithm="giac")
Output:
-log(abs(sin(b*x + a)^2 - 1))/b + 2*log(abs(sin(b*x + a)))/b - 1/2*(2*sin( b*x + a)^2 - 1)/((sin(b*x + a)^4 - sin(b*x + a)^2)*b)
Time = 25.37 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \csc ^3(a+b x) \sec ^3(a+b x) \, dx=\frac {{\mathrm {tan}\left (a+b\,x\right )}^2}{2\,b}-\frac {1}{2\,b\,{\mathrm {tan}\left (a+b\,x\right )}^2}+\frac {2\,\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b} \] Input:
int(1/(cos(a + b*x)^3*sin(a + b*x)^3),x)
Output:
tan(a + b*x)^2/(2*b) - 1/(2*b*tan(a + b*x)^2) + (2*log(tan(a + b*x)))/b
Time = 0.22 (sec) , antiderivative size = 175, normalized size of antiderivative = 4.07 \[ \int \csc ^3(a+b x) \sec ^3(a+b x) \, dx=\frac {-4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{4}+4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{4}+4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{4}-4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2}-\sin \left (b x +a \right )^{4}-\sin \left (b x +a \right )^{2}+1}{2 \sin \left (b x +a \right )^{2} b \left (\sin \left (b x +a \right )^{2}-1\right )} \] Input:
int(csc(b*x+a)^3*sec(b*x+a)^3,x)
Output:
( - 4*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**4 + 4*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**2 - 4*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**4 + 4*log( tan((a + b*x)/2) + 1)*sin(a + b*x)**2 + 4*log(tan((a + b*x)/2))*sin(a + b* x)**4 - 4*log(tan((a + b*x)/2))*sin(a + b*x)**2 - sin(a + b*x)**4 - sin(a + b*x)**2 + 1)/(2*sin(a + b*x)**2*b*(sin(a + b*x)**2 - 1))