Integrand size = 15, antiderivative size = 45 \[ \int \cos (a+b x) \cot ^3(a+b x) \, dx=\frac {3 \text {arctanh}(\cos (a+b x))}{2 b}-\frac {\cos (a+b x)}{b}-\frac {\cot (a+b x) \csc (a+b x)}{2 b} \] Output:
3/2*arctanh(cos(b*x+a))/b-cos(b*x+a)/b-1/2*cot(b*x+a)*csc(b*x+a)/b
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.91 \[ \int \cos (a+b x) \cot ^3(a+b x) \, dx=-\frac {\cos (a+b x)}{b}-\frac {\csc ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}+\frac {3 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}-\frac {3 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}+\frac {\sec ^2\left (\frac {1}{2} (a+b x)\right )}{8 b} \] Input:
Integrate[Cos[a + b*x]*Cot[a + b*x]^3,x]
Output:
-(Cos[a + b*x]/b) - Csc[(a + b*x)/2]^2/(8*b) + (3*Log[Cos[(a + b*x)/2]])/( 2*b) - (3*Log[Sin[(a + b*x)/2]])/(2*b) + Sec[(a + b*x)/2]^2/(8*b)
Time = 0.35 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 25, 3072, 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.
\(\displaystyle \int \cos (a+b x) \cot ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\sin \left (a+b x+\frac {\pi }{2}\right ) \tan \left (a+b x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sin \left (\frac {1}{2} (2 a+\pi )+b x\right ) \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )^3dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle -\frac {\int \frac {\cos ^4(a+b x)}{\left (1-\cos ^2(a+b x)\right )^2}d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {\cos ^3(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {3}{2} \int \frac {\cos ^2(a+b x)}{1-\cos ^2(a+b x)}d\cos (a+b x)}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {\frac {\cos ^3(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\cos ^2(a+b x)}d\cos (a+b x)-\cos (a+b x)\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {\cos ^3(a+b x)}{2 \left (1-\cos ^2(a+b x)\right )}-\frac {3}{2} (\text {arctanh}(\cos (a+b x))-\cos (a+b x))}{b}\) |
Input:
Int[Cos[a + b*x]*Cot[a + b*x]^3,x]
Output:
-(((-3*(ArcTanh[Cos[a + b*x]] - Cos[a + b*x]))/2 + Cos[a + b*x]^3/(2*(1 - Cos[a + b*x]^2)))/b)
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 = 1.59 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.33
method | result | size |
derivativedivides | \(\frac {-\frac {\cos \left (b x +a \right )^{5}}{2 \sin \left (b x +a \right )^{2}}-\frac {\cos \left (b x +a \right )^{3}}{2}-\frac {3 \cos \left (b x +a \right )}{2}-\frac {3 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\) | \(60\) |
default | \(\frac {-\frac {\cos \left (b x +a \right )^{5}}{2 \sin \left (b x +a \right )^{2}}-\frac {\cos \left (b x +a \right )^{3}}{2}-\frac {3 \cos \left (b x +a \right )}{2}-\frac {3 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2}}{b}\) | \(60\) |
risch | \(-\frac {{\mathrm e}^{i \left (b x +a \right )}}{2 b}-\frac {{\mathrm e}^{-i \left (b x +a \right )}}{2 b}+\frac {{\mathrm e}^{3 i \left (b x +a \right )}+{\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{2 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{2 b}\) | \(100\) |
Input:
int(cos(b*x+a)*cot(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b*(-1/2/sin(b*x+a)^2*cos(b*x+a)^5-1/2*cos(b*x+a)^3-3/2*cos(b*x+a)-3/2*ln (csc(b*x+a)-cot(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (41) = 82\).
Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.84 \[ \int \cos (a+b x) \cot ^3(a+b x) \, dx=-\frac {4 \, \cos \left (b x + a\right )^{3} - 3 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 6 \, \cos \left (b x + a\right )}{4 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \] Input:
integrate(cos(b*x+a)*cot(b*x+a)^3,x, algorithm="fricas")
Output:
-1/4*(4*cos(b*x + a)^3 - 3*(cos(b*x + a)^2 - 1)*log(1/2*cos(b*x + a) + 1/2 ) + 3*(cos(b*x + a)^2 - 1)*log(-1/2*cos(b*x + a) + 1/2) - 6*cos(b*x + a))/ (b*cos(b*x + a)^2 - b)
\[ \int \cos (a+b x) \cot ^3(a+b x) \, dx=\int \cos {\left (a + b x \right )} \cot ^{3}{\left (a + b x \right )}\, dx \] Input:
integrate(cos(b*x+a)*cot(b*x+a)**3,x)
Output:
Integral(cos(a + b*x)*cot(a + b*x)**3, x)
Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.24 \[ \int \cos (a+b x) \cot ^3(a+b x) \, dx=\frac {\frac {2 \, \cos \left (b x + a\right )}{\cos \left (b x + a\right )^{2} - 1} - 4 \, \cos \left (b x + a\right ) + 3 \, \log \left (\cos \left (b x + a\right ) + 1\right ) - 3 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{4 \, b} \] Input:
integrate(cos(b*x+a)*cot(b*x+a)^3,x, algorithm="maxima")
Output:
1/4*(2*cos(b*x + a)/(cos(b*x + a)^2 - 1) - 4*cos(b*x + a) + 3*log(cos(b*x + a) + 1) - 3*log(cos(b*x + a) - 1))/b
Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.44 \[ \int \cos (a+b x) \cot ^3(a+b x) \, dx=-\frac {\cos \left (b x + a\right )}{b} + \frac {3 \, \log \left ({\left | \cos \left (b x + a\right ) + 1 \right |}\right )}{4 \, b} - \frac {3 \, \log \left ({\left | \cos \left (b x + a\right ) - 1 \right |}\right )}{4 \, b} + \frac {\cos \left (b x + a\right )}{2 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )} b} \] Input:
integrate(cos(b*x+a)*cot(b*x+a)^3,x, algorithm="giac")
Output:
-cos(b*x + a)/b + 3/4*log(abs(cos(b*x + a) + 1))/b - 3/4*log(abs(cos(b*x + a) - 1))/b + 1/2*cos(b*x + a)/((cos(b*x + a)^2 - 1)*b)
Time = 25.62 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.71 \[ \int \cos (a+b x) \cot ^3(a+b x) \, dx=\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{8\,b}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{2\,b}-\frac {\frac {17\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{8}+\frac {1}{8}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2\right )} \] Input:
int(cos(a + b*x)*cot(a + b*x)^3,x)
Output:
tan(a/2 + (b*x)/2)^2/(8*b) - (3*log(tan(a/2 + (b*x)/2)))/(2*b) - ((17*tan( a/2 + (b*x)/2)^2)/8 + 1/8)/(b*(tan(a/2 + (b*x)/2)^2 + tan(a/2 + (b*x)/2)^4 ))
Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.51 \[ \int \cos (a+b x) \cot ^3(a+b x) \, dx=\frac {-8 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}-4 \cos \left (b x +a \right )-12 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2}+9 \sin \left (b x +a \right )^{2}}{8 \sin \left (b x +a \right )^{2} b} \] Input:
int(cos(b*x+a)*cot(b*x+a)^3,x)
Output:
( - 8*cos(a + b*x)*sin(a + b*x)**2 - 4*cos(a + b*x) - 12*log(tan((a + b*x) /2))*sin(a + b*x)**2 + 9*sin(a + b*x)**2)/(8*sin(a + b*x)**2*b)