Integrand size = 15, antiderivative size = 70 \[ \int \cos (a+b x) \cot ^5(a+b x) \, dx=-\frac {15 \text {arctanh}(\cos (a+b x))}{8 b}+\frac {15 \cos (a+b x)}{8 b}+\frac {5 \cos (a+b x) \cot ^2(a+b x)}{8 b}-\frac {\cos (a+b x) \cot ^4(a+b x)}{4 b} \]
-15/8*arctanh(cos(b*x+a))/b+15/8*cos(b*x+a)/b+5/8*cos(b*x+a)*cot(b*x+a)^2/ b-1/4*cos(b*x+a)*cot(b*x+a)^4/b
Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.76 \[ \int \cos (a+b x) \cot ^5(a+b x) \, dx=\frac {\cos (a+b x)}{b}+\frac {9 \csc ^2\left (\frac {1}{2} (a+b x)\right )}{32 b}-\frac {\csc ^4\left (\frac {1}{2} (a+b x)\right )}{64 b}-\frac {15 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{8 b}+\frac {15 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{8 b}-\frac {9 \sec ^2\left (\frac {1}{2} (a+b x)\right )}{32 b}+\frac {\sec ^4\left (\frac {1}{2} (a+b x)\right )}{64 b} \]
Cos[a + b*x]/b + (9*Csc[(a + b*x)/2]^2)/(32*b) - Csc[(a + b*x)/2]^4/(64*b) - (15*Log[Cos[(a + b*x)/2]])/(8*b) + (15*Log[Sin[(a + b*x)/2]])/(8*b) - ( 9*Sec[(a + b*x)/2]^2)/(32*b) + Sec[(a + b*x)/2]^4/(64*b)
Time = 0.23 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.19, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 25, 3072, 252, 252, 262, 219}
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.
| \(\displaystyle \int \cos (a+b x) \cot ^5(a+b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\sin \left (a+b x+\frac {\pi }{2}\right ) \tan \left (a+b x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \sin \left (\frac {1}{2} (2 a+\pi )+b x\right ) \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )^5dx\) |
| \(\Big \downarrow \) 3072 |
| \(\displaystyle -\frac {\int \frac {\cos ^6(a+b x)}{\left (1-\cos ^2(a+b x)\right )^3}d\cos (a+b x)}{b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\frac {\cos ^5(a+b x)}{4 \left (1-\cos ^2(a+b x)\right )^2}-\frac {5}{4} \int \frac {\cos ^4(a+b x)}{\left (1-\cos ^2(a+b x)\right )^2}d\cos (a+b x)}{b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\frac {\cos ^5(a+b x)}{4 \left (1-\cos ^2(a+b x)\right )^2}-\frac {5}{4} \left (\frac {\cos ^3(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {3}{2} \int \frac {\cos ^2(a+b x)}{1-\cos ^2(a+b x)}d\cos (a+b x)\right )}{b}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle -\frac {\frac {\cos ^5(a+b x)}{4 \left (1-\cos ^2(a+b x)\right )^2}-\frac {5}{4} \left (\frac {\cos ^3(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\cos ^2(a+b x)}d\cos (a+b x)-\cos (a+b x)\right )\right )}{b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {\cos ^5(a+b x)}{4 \left (1-\cos ^2(a+b x)\right )^2}-\frac {5}{4} \left (\frac {\cos ^3(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {3}{2} (\text {arctanh}(\cos (a+b x))-\cos (a+b x))\right )}{b}\) |
-((Cos[a + b*x]^5/(4*(1 - Cos[a + b*x]^2)^2) - (5*((-3*(ArcTanh[Cos[a + b* x]] - Cos[a + b*x]))/2 + Cos[a + b*x]^3/(2*(1 - Cos[a + b*x]^2))))/4)/b)
3.2.74.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.26
| method | result | size |
| derivativedivides | \(\frac {-\frac {\cos ^{7}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}+\frac {3 \left (\cos ^{7}\left (b x +a \right )\right )}{8 \sin \left (b x +a \right )^{2}}+\frac {3 \left (\cos ^{5}\left (b x +a \right )\right )}{8}+\frac {5 \left (\cos ^{3}\left (b x +a \right )\right )}{8}+\frac {15 \cos \left (b x +a \right )}{8}+\frac {15 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{8}}{b}\) | \(88\) |
| default | \(\frac {-\frac {\cos ^{7}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}+\frac {3 \left (\cos ^{7}\left (b x +a \right )\right )}{8 \sin \left (b x +a \right )^{2}}+\frac {3 \left (\cos ^{5}\left (b x +a \right )\right )}{8}+\frac {5 \left (\cos ^{3}\left (b x +a \right )\right )}{8}+\frac {15 \cos \left (b x +a \right )}{8}+\frac {15 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{8}}{b}\) | \(88\) |
| parallelrisch | \(\frac {120 \left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )-\left (\cot ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-15 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+15 \left (\cot ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-160 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{64 b \left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\) | \(110\) |
| norman | \(\frac {-\frac {1}{64 b}+\frac {15 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{64 b}-\frac {15 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{64 b}+\frac {\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )}{64 b}+\frac {5 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}+\frac {15 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}\) | \(114\) |
| risch | \(\frac {{\mathrm e}^{i \left (b x +a \right )}}{2 b}+\frac {{\mathrm e}^{-i \left (b x +a \right )}}{2 b}-\frac {9 \,{\mathrm e}^{7 i \left (b x +a \right )}-{\mathrm e}^{5 i \left (b x +a \right )}-{\mathrm e}^{3 i \left (b x +a \right )}+9 \,{\mathrm e}^{i \left (b x +a \right )}}{4 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4}}-\frac {15 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{8 b}+\frac {15 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{8 b}\) | \(127\) |
1/b*(-1/4*cos(b*x+a)^7/sin(b*x+a)^4+3/8*cos(b*x+a)^7/sin(b*x+a)^2+3/8*cos( b*x+a)^5+5/8*cos(b*x+a)^3+15/8*cos(b*x+a)+15/8*ln(csc(b*x+a)-cot(b*x+a)))
Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.74 \[ \int \cos (a+b x) \cot ^5(a+b x) \, dx=\frac {16 \, \cos \left (b x + a\right )^{5} - 50 \, \cos \left (b x + a\right )^{3} - 15 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 30 \, \cos \left (b x + a\right )}{16 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \]
1/16*(16*cos(b*x + a)^5 - 50*cos(b*x + a)^3 - 15*(cos(b*x + a)^4 - 2*cos(b *x + a)^2 + 1)*log(1/2*cos(b*x + a) + 1/2) + 15*(cos(b*x + a)^4 - 2*cos(b* x + a)^2 + 1)*log(-1/2*cos(b*x + a) + 1/2) + 30*cos(b*x + a))/(b*cos(b*x + a)^4 - 2*b*cos(b*x + a)^2 + b)
Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (63) = 126\).
Time = 1.71 (sec) , antiderivative size = 330, normalized size of antiderivative = 4.71 \[ \int \cos (a+b x) \cot ^5(a+b x) \, dx=\begin {cases} \frac {120 \log {\left (\tan {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )} \tan ^{6}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} + \frac {120 \log {\left (\tan {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )} \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} + \frac {\tan ^{10}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} - \frac {15 \tan ^{8}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} + \frac {160 \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} + \frac {15 \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} - \frac {1}{64 b \tan ^{6}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{6}{\left (a \right )}}{\sin ^{5}{\left (a \right )}} & \text {otherwise} \end {cases} \]
Piecewise((120*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(64*b*tan(a/2 + b *x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 120*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(64*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + tan(a/2 + b*x/2)**10/(64*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 15*t an(a/2 + b*x/2)**8/(64*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 160*tan(a/2 + b*x/2)**4/(64*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2) **4) + 15*tan(a/2 + b*x/2)**2/(64*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b *x/2)**4) - 1/(64*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4), Ne(b, 0)), (x*cos(a)**6/sin(a)**5, True))
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.13 \[ \int \cos (a+b x) \cot ^5(a+b x) \, dx=-\frac {\frac {2 \, {\left (9 \, \cos \left (b x + a\right )^{3} - 7 \, \cos \left (b x + a\right )\right )}}{\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1} - 16 \, \cos \left (b x + a\right ) + 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) - 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{16 \, b} \]
-1/16*(2*(9*cos(b*x + a)^3 - 7*cos(b*x + a))/(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1) - 16*cos(b*x + a) + 15*log(cos(b*x + a) + 1) - 15*log(cos(b*x + a) - 1))/b
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (62) = 124\).
Time = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.34 \[ \int \cos (a+b x) \cot ^5(a+b x) \, dx=-\frac {\frac {{\left (\frac {16 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {90 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} - \frac {16 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {128}{\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1} - 60 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{64 \, b} \]
-1/64*((16*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 90*(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 - 16* (cos(b*x + a) - 1)/(cos(b*x + a) + 1) - (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 128/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1) - 60*log(abs(-co s(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b
Time = 0.35 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.50 \[ \int \cos (a+b x) \cot ^5(a+b x) \, dx=\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4}{64\,b}-\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{4\,b}+\frac {15\,\ln \left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{8\,b}+\frac {\frac {9\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4}{4}+\frac {15\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{64}-\frac {1}{64}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4\right )} \]