Integrand size = 17, antiderivative size = 89 \[ \int \cos ^3(a+b x) \cot ^5(a+b x) \, dx=-\frac {35 \text {arctanh}(\cos (a+b x))}{8 b}+\frac {35 \cos (a+b x)}{8 b}+\frac {35 \cos ^3(a+b x)}{24 b}+\frac {7 \cos ^3(a+b x) \cot ^2(a+b x)}{8 b}-\frac {\cos ^3(a+b x) \cot ^4(a+b x)}{4 b} \]
-35/8*arctanh(cos(b*x+a))/b+35/8*cos(b*x+a)/b+35/24*cos(b*x+a)^3/b+7/8*cos (b*x+a)^3*cot(b*x+a)^2/b-1/4*cos(b*x+a)^3*cot(b*x+a)^4/b
Time = 0.13 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.58 \[ \int \cos ^3(a+b x) \cot ^5(a+b x) \, dx=\frac {13 \cos (a+b x)}{4 b}+\frac {\cos (3 (a+b x))}{12 b}+\frac {13 \csc ^2\left (\frac {1}{2} (a+b x)\right )}{32 b}-\frac {\csc ^4\left (\frac {1}{2} (a+b x)\right )}{64 b}-\frac {35 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{8 b}+\frac {35 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{8 b}-\frac {13 \sec ^2\left (\frac {1}{2} (a+b x)\right )}{32 b}+\frac {\sec ^4\left (\frac {1}{2} (a+b x)\right )}{64 b} \]
(13*Cos[a + b*x])/(4*b) + Cos[3*(a + b*x)]/(12*b) + (13*Csc[(a + b*x)/2]^2 )/(32*b) - Csc[(a + b*x)/2]^4/(64*b) - (35*Log[Cos[(a + b*x)/2]])/(8*b) + (35*Log[Sin[(a + b*x)/2]])/(8*b) - (13*Sec[(a + b*x)/2]^2)/(32*b) + Sec[(a + b*x)/2]^4/(64*b)
Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 25, 3072, 252, 252, 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.
| \(\displaystyle \int \cos ^3(a+b x) \cot ^5(a+b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\sin \left (a+b x+\frac {\pi }{2}\right )^3 \tan \left (a+b x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \sin \left (\frac {1}{2} (2 a+\pi )+b x\right )^3 \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )^5dx\) |
| \(\Big \downarrow \) 3072 |
| \(\displaystyle -\frac {\int \frac {\cos ^8(a+b x)}{\left (1-\cos ^2(a+b x)\right )^3}d\cos (a+b x)}{b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\frac {\cos ^7(a+b x)}{4 \left (1-\cos ^2(a+b x)\right )^2}-\frac {7}{4} \int \frac {\cos ^6(a+b x)}{\left (1-\cos ^2(a+b x)\right )^2}d\cos (a+b x)}{b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\frac {\cos ^7(a+b x)}{4 \left (1-\cos ^2(a+b x)\right )^2}-\frac {7}{4} \left (\frac {\cos ^5(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\cos ^4(a+b x)}{1-\cos ^2(a+b x)}d\cos (a+b x)\right )}{b}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle -\frac {\frac {\cos ^7(a+b x)}{4 \left (1-\cos ^2(a+b x)\right )^2}-\frac {7}{4} \left (\frac {\cos ^5(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\cos ^2(a+b x)+\frac {1}{1-\cos ^2(a+b x)}-1\right )d\cos (a+b x)\right )}{b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {\cos ^7(a+b x)}{4 \left (1-\cos ^2(a+b x)\right )^2}-\frac {7}{4} \left (\frac {\cos ^5(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\cos (a+b x))-\frac {1}{3} \cos ^3(a+b x)-\cos (a+b x)\right )\right )}{b}\) |
-((Cos[a + b*x]^7/(4*(1 - Cos[a + b*x]^2)^2) - (7*(Cos[a + b*x]^5/(2*(1 - Cos[a + b*x]^2)) - (5*(ArcTanh[Cos[a + b*x]] - Cos[a + b*x] - Cos[a + b*x] ^3/3))/2))/4)/b)
3.2.72.3.1 Defintions of rubi rules used
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[(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[((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 = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.10
| method | result | size |
| derivativedivides | \(\frac {-\frac {\cos ^{9}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}+\frac {5 \left (\cos ^{9}\left (b x +a \right )\right )}{8 \sin \left (b x +a \right )^{2}}+\frac {5 \left (\cos ^{7}\left (b x +a \right )\right )}{8}+\frac {7 \left (\cos ^{5}\left (b x +a \right )\right )}{8}+\frac {35 \left (\cos ^{3}\left (b x +a \right )\right )}{24}+\frac {35 \cos \left (b x +a \right )}{8}+\frac {35 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{8}}{b}\) | \(98\) |
| default | \(\frac {-\frac {\cos ^{9}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}+\frac {5 \left (\cos ^{9}\left (b x +a \right )\right )}{8 \sin \left (b x +a \right )^{2}}+\frac {5 \left (\cos ^{7}\left (b x +a \right )\right )}{8}+\frac {7 \left (\cos ^{5}\left (b x +a \right )\right )}{8}+\frac {35 \left (\cos ^{3}\left (b x +a \right )\right )}{24}+\frac {35 \cos \left (b x +a \right )}{8}+\frac {35 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{8}}{b}\) | \(98\) |
| parallelrisch | \(\frac {\left (\csc ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\sec ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (-3360 \left (\cos \left (2 b x +2 a \right )-\frac {\cos \left (4 b x +4 a \right )}{4}-\frac {3}{4}\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )+840 \cos \left (b x +a \right )-3388 \cos \left (2 b x +2 a \right )-1512 \cos \left (3 b x +3 a \right )+847 \cos \left (4 b x +4 a \right )+280 \cos \left (5 b x +5 a \right )+8 \cos \left (7 b x +7 a \right )+2541\right )}{24576 b}\) | \(127\) |
| norman | \(\frac {-\frac {1}{64 b}+\frac {21 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{64 b}-\frac {21 \left (\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{64 b}+\frac {\tan ^{14}\left (\frac {b x}{2}+\frac {a}{2}\right )}{64 b}+\frac {21 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2 b}+\frac {511 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{32 b}+\frac {847 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{96 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{3} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}+\frac {35 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}\) | \(146\) |
| risch | \(\frac {{\mathrm e}^{3 i \left (b x +a \right )}}{24 b}+\frac {13 \,{\mathrm e}^{i \left (b x +a \right )}}{8 b}+\frac {13 \,{\mathrm e}^{-i \left (b x +a \right )}}{8 b}+\frac {{\mathrm e}^{-3 i \left (b x +a \right )}}{24 b}-\frac {13 \,{\mathrm e}^{7 i \left (b x +a \right )}-5 \,{\mathrm e}^{5 i \left (b x +a \right )}-5 \,{\mathrm e}^{3 i \left (b x +a \right )}+13 \,{\mathrm e}^{i \left (b x +a \right )}}{4 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4}}-\frac {35 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{8 b}+\frac {35 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{8 b}\) | \(155\) |
1/b*(-1/4*cos(b*x+a)^9/sin(b*x+a)^4+5/8/sin(b*x+a)^2*cos(b*x+a)^9+5/8*cos( b*x+a)^7+7/8*cos(b*x+a)^5+35/24*cos(b*x+a)^3+35/8*cos(b*x+a)+35/8*ln(csc(b *x+a)-cot(b*x+a)))
Time = 0.30 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.48 \[ \int \cos ^3(a+b x) \cot ^5(a+b x) \, dx=\frac {16 \, \cos \left (b x + a\right )^{7} + 112 \, \cos \left (b x + a\right )^{5} - 350 \, \cos \left (b x + a\right )^{3} - 105 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 105 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 210 \, \cos \left (b x + a\right )}{48 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \]
1/48*(16*cos(b*x + a)^7 + 112*cos(b*x + a)^5 - 350*cos(b*x + a)^3 - 105*(c os(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(1/2*cos(b*x + a) + 1/2) + 105*(c os(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(-1/2*cos(b*x + a) + 1/2) + 210*c os(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. 869 vs. \(2 (80) = 160\).
Time = 4.65 (sec) , antiderivative size = 869, normalized size of antiderivative = 9.76 \[ \int \cos ^3(a+b x) \cot ^5(a+b x) \, dx=\text {Too large to display} \]
Piecewise((840*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**10/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b*x/2)**6 + 192* b*tan(a/2 + b*x/2)**4) + 2520*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**8/(1 92*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b* x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) + 2520*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576* b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) + 840*log(tan(a/2 + b*x /2))*tan(a/2 + b*x/2)**4/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x /2)**8 + 576*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) + 3*tan(a/ 2 + b*x/2)**14/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 5 76*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) - 63*tan(a/2 + b*x/2 )**12/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan( a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) + 2016*tan(a/2 + b*x/2)**8/(1 92*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b* x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) + 3066*tan(a/2 + b*x/2)**6/(192*b*tan (a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2 + b*x/2)**4) + 1694*tan(a/2 + b*x/2)**4/(192*b*tan(a/2 + b *x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b*x/2)**6 + 192*b* tan(a/2 + b*x/2)**4) + 63*tan(a/2 + b*x/2)**2/(192*b*tan(a/2 + b*x/2)**10 + 576*b*tan(a/2 + b*x/2)**8 + 576*b*tan(a/2 + b*x/2)**6 + 192*b*tan(a/2...
Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \cos ^3(a+b x) \cot ^5(a+b x) \, dx=\frac {16 \, \cos \left (b x + a\right )^{3} - \frac {6 \, {\left (13 \, \cos \left (b x + a\right )^{3} - 11 \, \cos \left (b x + a\right )\right )}}{\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1} + 144 \, \cos \left (b x + a\right ) - 105 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 105 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{48 \, b} \]
1/48*(16*cos(b*x + a)^3 - 6*(13*cos(b*x + a)^3 - 11*cos(b*x + a))/(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1) + 144*cos(b*x + a) - 105*log(cos(b*x + a) + 1) + 105*log(cos(b*x + a) - 1))/b
Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (79) = 158\).
Time = 0.35 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.35 \[ \int \cos ^3(a+b x) \cot ^5(a+b x) \, dx=-\frac {\frac {3 \, {\left (\frac {24 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {210 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} - \frac {72 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {3 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - \frac {256 \, {\left (\frac {9 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {6 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 5\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{3}} - 420 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{192 \, b} \]
-1/192*(3*(24*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 210*(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 - 72*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 3*(cos(b*x + a) - 1)^2/(cos(b* x + a) + 1)^2 - 256*(9*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 6*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 - 5)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)^3 - 420*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b
Time = 0.31 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.76 \[ \int \cos ^3(a+b x) \cot ^5(a+b x) \, dx=\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4}{64\,b}-\frac {3\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{8\,b}+\frac {35\,\ln \left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{8\,b}+\frac {\frac {67\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8}{8}+\frac {839\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6}{64}+\frac {1487\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4}{192}+\frac {21\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{64}-\frac {1}{64}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{10}+3\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8+3\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4\right )} \]
tan(a/2 + (b*x)/2)^4/(64*b) - (3*tan(a/2 + (b*x)/2)^2)/(8*b) + (35*log(tan (a/2 + (b*x)/2)))/(8*b) + ((21*tan(a/2 + (b*x)/2)^2)/64 + (1487*tan(a/2 + (b*x)/2)^4)/192 + (839*tan(a/2 + (b*x)/2)^6)/64 + (67*tan(a/2 + (b*x)/2)^8 )/8 - 1/64)/(b*(tan(a/2 + (b*x)/2)^4 + 3*tan(a/2 + (b*x)/2)^6 + 3*tan(a/2 + (b*x)/2)^8 + tan(a/2 + (b*x)/2)^10))