Integrand size = 17, antiderivative size = 61 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=-\frac {15 x}{8}-\frac {15 \cot (a+b x)}{8 b}+\frac {5 \cos ^2(a+b x) \cot (a+b x)}{8 b}+\frac {\cos ^4(a+b x) \cot (a+b x)}{4 b} \]
Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.67 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=-\frac {60 a+60 b x+32 \cot (a+b x)+16 \sin (2 (a+b x))+\sin (4 (a+b x))}{32 b} \]
Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.30, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3042, 3071, 252, 252, 262, 216}
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.
\(\displaystyle \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (a+b x+\frac {\pi }{2}\right )^4 \tan \left (a+b x+\frac {\pi }{2}\right )^2dx\) |
\(\Big \downarrow \) 3071 |
\(\displaystyle -\frac {\int \frac {\cot ^6(a+b x)}{\left (\cot ^2(a+b x)+1\right )^3}d\cot (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {5}{4} \int \frac {\cot ^4(a+b x)}{\left (\cot ^2(a+b x)+1\right )^2}d\cot (a+b x)-\frac {\cot ^5(a+b x)}{4 \left (\cot ^2(a+b x)+1\right )^2}}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {5}{4} \left (\frac {3}{2} \int \frac {\cot ^2(a+b x)}{\cot ^2(a+b x)+1}d\cot (a+b x)-\frac {\cot ^3(a+b x)}{2 \left (\cot ^2(a+b x)+1\right )}\right )-\frac {\cot ^5(a+b x)}{4 \left (\cot ^2(a+b x)+1\right )^2}}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {\frac {5}{4} \left (\frac {3}{2} \left (\cot (a+b x)-\int \frac {1}{\cot ^2(a+b x)+1}d\cot (a+b x)\right )-\frac {\cot ^3(a+b x)}{2 \left (\cot ^2(a+b x)+1\right )}\right )-\frac {\cot ^5(a+b x)}{4 \left (\cot ^2(a+b x)+1\right )^2}}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {5}{4} \left (\frac {3}{2} (\cot (a+b x)-\arctan (\cot (a+b x)))-\frac {\cot ^3(a+b x)}{2 \left (\cot ^2(a+b x)+1\right )}\right )-\frac {\cot ^5(a+b x)}{4 \left (\cot ^2(a+b x)+1\right )^2}}{b}\) |
-((-1/4*Cot[a + b*x]^5/(1 + Cot[a + b*x]^2)^2 + (5*((3*(-ArcTan[Cot[a + b* x]] + Cot[a + b*x]))/2 - Cot[a + b*x]^3/(2*(1 + Cot[a + b*x]^2))))/4)/b)
3.2.34.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[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[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[I nt[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]
Time = 0.14 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(\frac {\left (-120 b x \sin \left (b x +a \right )-80 \cos \left (b x +a \right )+15 \cos \left (3 b x +3 a \right )+\cos \left (5 b x +5 a \right )\right ) \sec \left (\frac {b x}{2}+\frac {a}{2}\right ) \csc \left (\frac {b x}{2}+\frac {a}{2}\right )}{128 b}\) | \(63\) |
derivativedivides | \(\frac {-\frac {\cos ^{7}\left (b x +a \right )}{\sin \left (b x +a \right )}-\left (\cos ^{5}\left (b x +a \right )+\frac {5 \left (\cos ^{3}\left (b x +a \right )\right )}{4}+\frac {15 \cos \left (b x +a \right )}{8}\right ) \sin \left (b x +a \right )-\frac {15 b x}{8}-\frac {15 a}{8}}{b}\) | \(66\) |
default | \(\frac {-\frac {\cos ^{7}\left (b x +a \right )}{\sin \left (b x +a \right )}-\left (\cos ^{5}\left (b x +a \right )+\frac {5 \left (\cos ^{3}\left (b x +a \right )\right )}{4}+\frac {15 \cos \left (b x +a \right )}{8}\right ) \sin \left (b x +a \right )-\frac {15 b x}{8}-\frac {15 a}{8}}{b}\) | \(66\) |
risch | \(-\frac {15 x}{8}+\frac {i {\mathrm e}^{2 i \left (b x +a \right )}}{4 b}-\frac {i {\mathrm e}^{-2 i \left (b x +a \right )}}{4 b}-\frac {2 i}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {\sin \left (4 b x +4 a \right )}{32 b}\) | \(68\) |
norman | \(\frac {-\frac {1}{2 b}-\frac {15 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}-\frac {5 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {5 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {15 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )}{2 b}-\frac {15 x \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{8}-\frac {15 x \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2}-\frac {45 x \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4}-\frac {15 x \left (\tan ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2}-\frac {15 x \left (\tan ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{4} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\) | \(182\) |
1/128*(-120*b*x*sin(b*x+a)-80*cos(b*x+a)+15*cos(3*b*x+3*a)+cos(5*b*x+5*a)) *sec(1/2*b*x+1/2*a)*csc(1/2*b*x+1/2*a)/b
Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.85 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=\frac {2 \, \cos \left (b x + a\right )^{5} + 5 \, \cos \left (b x + a\right )^{3} - 15 \, b x \sin \left (b x + a\right ) - 15 \, \cos \left (b x + a\right )}{8 \, b \sin \left (b x + a\right )} \]
1/8*(2*cos(b*x + a)^5 + 5*cos(b*x + a)^3 - 15*b*x*sin(b*x + a) - 15*cos(b* x + a))/(b*sin(b*x + a))
Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (54) = 108\).
Time = 0.65 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.95 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=\begin {cases} - \frac {15 x \sin ^{4}{\left (a + b x \right )}}{8} - \frac {15 x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} - \frac {15 x \cos ^{4}{\left (a + b x \right )}}{8} - \frac {15 \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{8 b} - \frac {25 \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} - \frac {\cos ^{5}{\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{6}{\left (a \right )}}{\sin ^{2}{\left (a \right )}} & \text {otherwise} \end {cases} \]
Piecewise((-15*x*sin(a + b*x)**4/8 - 15*x*sin(a + b*x)**2*cos(a + b*x)**2/ 4 - 15*x*cos(a + b*x)**4/8 - 15*sin(a + b*x)**3*cos(a + b*x)/(8*b) - 25*si n(a + b*x)*cos(a + b*x)**3/(8*b) - cos(a + b*x)**5/(b*sin(a + b*x)), Ne(b, 0)), (x*cos(a)**6/sin(a)**2, True))
Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=-\frac {15 \, b x + 15 \, a + \frac {15 \, \tan \left (b x + a\right )^{4} + 25 \, \tan \left (b x + a\right )^{2} + 8}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}}{8 \, b} \]
-1/8*(15*b*x + 15*a + (15*tan(b*x + a)^4 + 25*tan(b*x + a)^2 + 8)/(tan(b*x + a)^5 + 2*tan(b*x + a)^3 + tan(b*x + a)))/b
Time = 0.32 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=-\frac {15 \, b x + 15 \, a + \frac {7 \, \tan \left (b x + a\right )^{3} + 9 \, \tan \left (b x + a\right )}{{\left (\tan \left (b x + a\right )^{2} + 1\right )}^{2}} + \frac {8}{\tan \left (b x + a\right )}}{8 \, b} \]
-1/8*(15*b*x + 15*a + (7*tan(b*x + a)^3 + 9*tan(b*x + a))/(tan(b*x + a)^2 + 1)^2 + 8/tan(b*x + a))/b
Time = 0.33 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int \cos ^4(a+b x) \cot ^2(a+b x) \, dx=-\frac {15\,x}{8}-\frac {{\cos \left (a+b\,x\right )}^4\,\left (\frac {15\,{\mathrm {tan}\left (a+b\,x\right )}^4}{8}+\frac {25\,{\mathrm {tan}\left (a+b\,x\right )}^2}{8}+1\right )}{b\,\mathrm {tan}\left (a+b\,x\right )} \]