Integrand size = 17, antiderivative size = 60 \[ \int \cos ^3(a+b x) \cot ^3(a+b x) \, dx=\frac {5 \text {arctanh}(\cos (a+b x))}{2 b}-\frac {2 \cos (a+b x)}{b}-\frac {\cos ^3(a+b x)}{3 b}-\frac {\cot (a+b x) \csc (a+b x)}{2 b} \] Output:
5/2*arctanh(cos(b*x+a))/b-2*cos(b*x+a)/b-1/3*cos(b*x+a)^3/b-1/2*cot(b*x+a) *csc(b*x+a)/b
Time = 0.19 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.72 \[ \int \cos ^3(a+b x) \cot ^3(a+b x) \, dx=-\frac {9 \cos (a+b x)}{4 b}-\frac {\cos (3 (a+b x))}{12 b}-\frac {\csc ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}+\frac {5 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}-\frac {5 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}+\frac {\sec ^2\left (\frac {1}{2} (a+b x)\right )}{8 b} \] Input:
Integrate[Cos[a + b*x]^3*Cot[a + b*x]^3,x]
Output:
(-9*Cos[a + b*x])/(4*b) - Cos[3*(a + b*x)]/(12*b) - Csc[(a + b*x)/2]^2/(8* b) + (5*Log[Cos[(a + b*x)/2]])/(2*b) - (5*Log[Sin[(a + b*x)/2]])/(2*b) + S ec[(a + b*x)/2]^2/(8*b)
Time = 0.42 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3042, 25, 3072, 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.
\(\displaystyle \int \cos ^3(a+b x) \cot ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\sin \left (a+b x+\frac {\pi }{2}\right )^3 \tan \left (a+b x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sin \left (\frac {1}{2} (2 a+\pi )+b x\right )^3 \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )^3dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle -\frac {\int \frac {\cos ^6(a+b x)}{\left (1-\cos ^2(a+b x)\right )^2}d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {\cos ^5(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\cos ^4(a+b x)}{1-\cos ^2(a+b x)}d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle -\frac {\frac {\cos ^5(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\cos ^2(a+b x)+\frac {1}{1-\cos ^2(a+b x)}-1\right )d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {\cos ^5(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\cos (a+b x))-\frac {1}{3} \cos ^3(a+b x)-\cos (a+b x)\right )}{b}\) |
Input:
Int[Cos[a + b*x]^3*Cot[a + b*x]^3,x]
Output:
-((Cos[a + b*x]^5/(2*(1 - Cos[a + b*x]^2)) - (5*(ArcTanh[Cos[a + b*x]] - C os[a + b*x] - Cos[a + b*x]^3/3))/2)/b)
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 = 2.89 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17
method | result | size |
derivativedivides | \(\frac {-\frac {\cos \left (b x +a \right )^{7}}{2 \sin \left (b x +a \right )^{2}}-\frac {\cos \left (b x +a \right )^{5}}{2}-\frac {5 \cos \left (b x +a \right )^{3}}{6}-\frac {5 \cos \left (b x +a \right )}{2}-\frac {5 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\) | \(70\) |
default | \(\frac {-\frac {\cos \left (b x +a \right )^{7}}{2 \sin \left (b x +a \right )^{2}}-\frac {\cos \left (b x +a \right )^{5}}{2}-\frac {5 \cos \left (b x +a \right )^{3}}{6}-\frac {5 \cos \left (b x +a \right )}{2}-\frac {5 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\) | \(70\) |
risch | \(-\frac {{\mathrm e}^{3 i \left (b x +a \right )}}{24 b}-\frac {9 \,{\mathrm e}^{i \left (b x +a \right )}}{8 b}-\frac {9 \,{\mathrm e}^{-i \left (b x +a \right )}}{8 b}-\frac {{\mathrm e}^{-3 i \left (b x +a \right )}}{24 b}+\frac {{\mathrm e}^{3 i \left (b x +a \right )}+{\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{2 b}+\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{2 b}\) | \(128\) |
Input:
int(cos(b*x+a)^3*cot(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b*(-1/2/sin(b*x+a)^2*cos(b*x+a)^7-1/2*cos(b*x+a)^5-5/6*cos(b*x+a)^3-5/2* cos(b*x+a)-5/2*ln(csc(b*x+a)-cot(b*x+a)))
Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.55 \[ \int \cos ^3(a+b x) \cot ^3(a+b x) \, dx=-\frac {4 \, \cos \left (b x + a\right )^{5} + 20 \, \cos \left (b x + a\right )^{3} - 15 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 30 \, \cos \left (b x + a\right )}{12 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \] Input:
integrate(cos(b*x+a)^3*cot(b*x+a)^3,x, algorithm="fricas")
Output:
-1/12*(4*cos(b*x + a)^5 + 20*cos(b*x + a)^3 - 15*(cos(b*x + a)^2 - 1)*log( 1/2*cos(b*x + a) + 1/2) + 15*(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)^2 - b)
Timed out. \[ \int \cos ^3(a+b x) \cot ^3(a+b x) \, dx=\text {Timed out} \] Input:
integrate(cos(b*x+a)**3*cot(b*x+a)**3,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int \cos ^3(a+b x) \cot ^3(a+b x) \, dx=-\frac {4 \, \cos \left (b x + a\right )^{3} - \frac {6 \, \cos \left (b x + a\right )}{\cos \left (b x + a\right )^{2} - 1} + 24 \, \cos \left (b x + a\right ) - 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{12 \, b} \] Input:
integrate(cos(b*x+a)^3*cot(b*x+a)^3,x, algorithm="maxima")
Output:
-1/12*(4*cos(b*x + a)^3 - 6*cos(b*x + a)/(cos(b*x + a)^2 - 1) + 24*cos(b*x + a) - 15*log(cos(b*x + a) + 1) + 15*log(cos(b*x + a) - 1))/b
Time = 0.14 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.38 \[ \int \cos ^3(a+b x) \cot ^3(a+b x) \, dx=\frac {5 \, \log \left ({\left | \cos \left (b x + a\right ) + 1 \right |}\right )}{4 \, b} - \frac {5 \, \log \left ({\left | \cos \left (b x + a\right ) - 1 \right |}\right )}{4 \, b} + \frac {\cos \left (b x + a\right )}{2 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} b} - \frac {b^{2} \cos \left (b x + a\right )^{3} + 6 \, b^{2} \cos \left (b x + a\right )}{3 \, b^{3}} \] Input:
integrate(cos(b*x+a)^3*cot(b*x+a)^3,x, algorithm="giac")
Output:
5/4*log(abs(cos(b*x + a) + 1))/b - 5/4*log(abs(cos(b*x + a) - 1))/b + 1/2* cos(b*x + a)/((cos(b*x + a)^2 - 1)*b) - 1/3*(b^2*cos(b*x + a)^3 + 6*b^2*co s(b*x + a))/b^3
Time = 26.64 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.15 \[ \int \cos ^3(a+b x) \cot ^3(a+b x) \, dx=\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{8\,b}-\frac {5\,\ln \left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{2\,b}-\frac {\frac {49\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6}{8}+\frac {67\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4}{8}+\frac {121\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{24}+\frac {1}{8}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8+3\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2\right )} \] Input:
int(cos(a + b*x)^3*cot(a + b*x)^3,x)
Output:
tan(a/2 + (b*x)/2)^2/(8*b) - (5*log(tan(a/2 + (b*x)/2)))/(2*b) - ((121*tan (a/2 + (b*x)/2)^2)/24 + (67*tan(a/2 + (b*x)/2)^4)/8 + (49*tan(a/2 + (b*x)/ 2)^6)/8 + 1/8)/(b*(tan(a/2 + (b*x)/2)^2 + 3*tan(a/2 + (b*x)/2)^4 + 3*tan(a /2 + (b*x)/2)^6 + tan(a/2 + (b*x)/2)^8))
Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.40 \[ \int \cos ^3(a+b x) \cot ^3(a+b x) \, dx=\frac {8 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{4}-56 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}-12 \cos \left (b x +a \right )-60 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2}+65 \sin \left (b x +a \right )^{2}}{24 \sin \left (b x +a \right )^{2} b} \] Input:
int(cos(b*x+a)^3*cot(b*x+a)^3,x)
Output:
(8*cos(a + b*x)*sin(a + b*x)**4 - 56*cos(a + b*x)*sin(a + b*x)**2 - 12*cos (a + b*x) - 60*log(tan((a + b*x)/2))*sin(a + b*x)**2 + 65*sin(a + b*x)**2) /(24*sin(a + b*x)**2*b)