Integrand size = 17, antiderivative size = 69 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=-\frac {2 \cot ^2(a+b x)}{b}-\frac {\cot ^4(a+b x)}{4 b}+\frac {6 \log (\tan (a+b x))}{b}+\frac {2 \tan ^2(a+b x)}{b}+\frac {\tan ^4(a+b x)}{4 b} \]
Time = 0.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.32 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=32 \left (-\frac {3 \csc ^2(a+b x)}{64 b}-\frac {\csc ^4(a+b x)}{128 b}-\frac {3 \log (\cos (a+b x))}{16 b}+\frac {3 \log (\sin (a+b x))}{16 b}+\frac {3 \sec ^2(a+b x)}{64 b}+\frac {\sec ^4(a+b x)}{128 b}\right ) \]
32*((-3*Csc[a + b*x]^2)/(64*b) - Csc[a + b*x]^4/(128*b) - (3*Log[Cos[a + b *x]])/(16*b) + (3*Log[Sin[a + b*x]])/(16*b) + (3*Sec[a + b*x]^2)/(64*b) + Sec[a + b*x]^4/(128*b))
Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.88, 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.
\(\displaystyle \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (a+b x)^5 \sec (a+b x)^5dx\) |
\(\Big \downarrow \) 3100 |
\(\displaystyle \frac {\int \cot ^5(a+b x) \left (\tan ^2(a+b x)+1\right )^4d\tan (a+b x)}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \cot ^3(a+b x) \left (\tan ^2(a+b x)+1\right )^4d\tan ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\cot ^3(a+b x)+4 \cot ^2(a+b x)+6 \cot (a+b x)+\tan ^2(a+b x)+4\right )d\tan ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} \tan ^4(a+b x)+4 \tan ^2(a+b x)-\frac {1}{2} \cot ^2(a+b x)-4 \cot (a+b x)+6 \log \left (\tan ^2(a+b x)\right )}{2 b}\) |
(-4*Cot[a + b*x] - Cot[a + b*x]^2/2 + 6*Log[Tan[a + b*x]^2] + 4*Tan[a + b* x]^2 + Tan[a + b*x]^4/2)/(2*b)
3.2.84.3.1 Defintions of rubi rules used
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 = 0.47 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14
method | result | size |
derivativedivides | \(\frac {\frac {1}{4 \sin \left (b x +a \right )^{4} \cos \left (b x +a \right )^{4}}-\frac {1}{2 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{4}}+\frac {3}{2 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2}}-\frac {3}{\sin \left (b x +a \right )^{2}}+6 \ln \left (\tan \left (b x +a \right )\right )}{b}\) | \(79\) |
default | \(\frac {\frac {1}{4 \sin \left (b x +a \right )^{4} \cos \left (b x +a \right )^{4}}-\frac {1}{2 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{4}}+\frac {3}{2 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )^{2}}-\frac {3}{\sin \left (b x +a \right )^{2}}+6 \ln \left (\tan \left (b x +a \right )\right )}{b}\) | \(79\) |
risch | \(\frac {12 \,{\mathrm e}^{14 i \left (b x +a \right )}-44 \,{\mathrm e}^{10 i \left (b x +a \right )}-44 \,{\mathrm e}^{6 i \left (b x +a \right )}+12 \,{\mathrm e}^{2 i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4}}+\frac {6 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}-\frac {6 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}\) | \(112\) |
parallelrisch | \(\frac {\left (-384 \cos \left (2 b x +2 a \right )-96 \cos \left (4 b x +4 a \right )-288\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (-384 \cos \left (2 b x +2 a \right )-96 \cos \left (4 b x +4 a \right )-288\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+\left (384 \cos \left (2 b x +2 a \right )+96 \cos \left (4 b x +4 a \right )+288\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\left (3384 \cos \left (b x +a \right )-1284 \cos \left (2 b x +2 a \right )+328 \cos \left (3 b x +3 a \right )-41 \cos \left (4 b x +4 a \right )-3227\right ) \left (\cot ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+592 \left (\cot ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\csc ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-2 \left (\csc ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\sec ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )+16 \left (\sec ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-140\right )}{16 b \left (\cos \left (4 b x +4 a \right )+4 \cos \left (2 b x +2 a \right )+3\right )}\) | \(258\) |
1/b*(1/4/sin(b*x+a)^4/cos(b*x+a)^4-1/2/cos(b*x+a)^2/sin(b*x+a)^4+3/2/cos(b *x+a)^2/sin(b*x+a)^2-3/sin(b*x+a)^2+6*ln(tan(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (65) = 130\).
Time = 0.33 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.14 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=\frac {12 \, \cos \left (b x + a\right )^{6} - 18 \, \cos \left (b x + a\right )^{4} + 4 \, \cos \left (b x + a\right )^{2} - 12 \, {\left (\cos \left (b x + a\right )^{8} - 2 \, \cos \left (b x + a\right )^{6} + \cos \left (b x + a\right )^{4}\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) + 12 \, {\left (\cos \left (b x + a\right )^{8} - 2 \, \cos \left (b x + a\right )^{6} + \cos \left (b x + a\right )^{4}\right )} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) + 1}{4 \, {\left (b \cos \left (b x + a\right )^{8} - 2 \, b \cos \left (b x + a\right )^{6} + b \cos \left (b x + a\right )^{4}\right )}} \]
1/4*(12*cos(b*x + a)^6 - 18*cos(b*x + a)^4 + 4*cos(b*x + a)^2 - 12*(cos(b* x + a)^8 - 2*cos(b*x + a)^6 + cos(b*x + a)^4)*log(cos(b*x + a)^2) + 12*(co s(b*x + a)^8 - 2*cos(b*x + a)^6 + cos(b*x + a)^4)*log(-1/4*cos(b*x + a)^2 + 1/4) + 1)/(b*cos(b*x + a)^8 - 2*b*cos(b*x + a)^6 + b*cos(b*x + a)^4)
\[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=\int \frac {\sec ^{5}{\left (a + b x \right )}}{\sin ^{5}{\left (a + b x \right )}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.33 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=-\frac {\frac {12 \, \sin \left (b x + a\right )^{6} - 18 \, \sin \left (b x + a\right )^{4} + 4 \, \sin \left (b x + a\right )^{2} + 1}{\sin \left (b x + a\right )^{8} - 2 \, \sin \left (b x + a\right )^{6} + \sin \left (b x + a\right )^{4}} + 12 \, \log \left (\sin \left (b x + a\right )^{2} - 1\right ) - 12 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \]
-1/4*((12*sin(b*x + a)^6 - 18*sin(b*x + a)^4 + 4*sin(b*x + a)^2 + 1)/(sin( b*x + a)^8 - 2*sin(b*x + a)^6 + sin(b*x + a)^4) + 12*log(sin(b*x + a)^2 - 1) - 12*log(sin(b*x + a)^2))/b
Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (65) = 130\).
Time = 0.32 (sec) , antiderivative size = 278, normalized size of antiderivative = 4.03 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=\frac {\frac {{\left (\frac {28 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {288 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} + \frac {28 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {32 \, {\left (\frac {84 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {126 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {84 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac {25 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 25\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{4}} + 192 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) - 384 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{64 \, b} \]
1/64*((28*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 288*(cos(b*x + a) - 1)^2 /(cos(b*x + a) + 1)^2 - 1)*(cos(b*x + a) + 1)^2/(cos(b*x + a) - 1)^2 + 28* (cos(b*x + a) - 1)/(cos(b*x + a) + 1) - (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 32*(84*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 126*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 84*(cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^ 3 + 25*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 25)/((cos(b*x + a) - 1) /(cos(b*x + a) + 1) + 1)^4 + 192*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) - 384*log(abs(-(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)))/b
Time = 0.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \csc ^5(a+b x) \sec ^5(a+b x) \, dx=\frac {2\,{\mathrm {tan}\left (a+b\,x\right )}^2}{b}+\frac {{\mathrm {tan}\left (a+b\,x\right )}^4}{4\,b}+\frac {6\,\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b}-\frac {{\mathrm {cot}\left (a+b\,x\right )}^4\,\left (2\,{\mathrm {tan}\left (a+b\,x\right )}^2+\frac {1}{4}\right )}{b} \]