\(\int \cos ^4(a+b x) \cot ^5(a+b x) \, dx\) [171]

Optimal result

Integrand size = 17, antiderivative size = 69 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\frac {2 \csc ^2(a+b x)}{b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {6 \log (\sin (a+b x))}{b}-\frac {2 \sin ^2(a+b x)}{b}+\frac {\sin ^4(a+b x)}{4 b} \]

[Out]

2*csc(b*x+a)^2/b-1/4*csc(b*x+a)^4/b+6*ln(sin(b*x+a))/b-2*sin(b*x+a)^2/b+1/4*sin(b*x+a)^4/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2670, 272, 45} \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\frac {\sin ^4(a+b x)}{4 b}-\frac {2 \sin ^2(a+b x)}{b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {2 \csc ^2(a+b x)}{b}+\frac {6 \log (\sin (a+b x))}{b} \]

[In]

Int[Cos[a + b*x]^4*Cot[a + b*x]^5,x]

[Out]

(2*Csc[a + b*x]^2)/b - Csc[a + b*x]^4/(4*b) + (6*Log[Sin[a + b*x]])/b - (2*Sin[a + b*x]^2)/b + Sin[a + b*x]^4/
(4*b)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2670

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^4}{x^5} \, dx,x,-\sin (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {(1-x)^4}{x^3} \, dx,x,\sin ^2(a+b x)\right )}{2 b} \\ & = \frac {\text {Subst}\left (\int \left (-4+\frac {1}{x^3}-\frac {4}{x^2}+\frac {6}{x}+x\right ) \, dx,x,\sin ^2(a+b x)\right )}{2 b} \\ & = \frac {2 \csc ^2(a+b x)}{b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {6 \log (\sin (a+b x))}{b}-\frac {2 \sin ^2(a+b x)}{b}+\frac {\sin ^4(a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\frac {2 \csc ^2(a+b x)}{b}-\frac {\csc ^4(a+b x)}{4 b}+\frac {6 \log (\sin (a+b x))}{b}-\frac {2 \sin ^2(a+b x)}{b}+\frac {\sin ^4(a+b x)}{4 b} \]

[In]

Integrate[Cos[a + b*x]^4*Cot[a + b*x]^5,x]

[Out]

(2*Csc[a + b*x]^2)/b - Csc[a + b*x]^4/(4*b) + (6*Log[Sin[a + b*x]])/b - (2*Sin[a + b*x]^2)/b + Sin[a + b*x]^4/
(4*b)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.29

method	result	size
derivativedivides	\(\frac {-\frac {\cos ^{10}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}+\frac {3 \left (\cos ^{10}\left (b x +a \right )\right )}{4 \sin \left (b x +a \right )^{2}}+\frac {3 \left (\cos ^{8}\left (b x +a \right )\right )}{4}+\cos ^{6}\left (b x +a \right )+\frac {3 \left (\cos ^{4}\left (b x +a \right )\right )}{2}+3 \left (\cos ^{2}\left (b x +a \right )\right )+6 \ln \left (\sin \left (b x +a \right )\right )}{b}\)	\(89\)
default	\(\frac {-\frac {\cos ^{10}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4}}+\frac {3 \left (\cos ^{10}\left (b x +a \right )\right )}{4 \sin \left (b x +a \right )^{2}}+\frac {3 \left (\cos ^{8}\left (b x +a \right )\right )}{4}+\cos ^{6}\left (b x +a \right )+\frac {3 \left (\cos ^{4}\left (b x +a \right )\right )}{2}+3 \left (\cos ^{2}\left (b x +a \right )\right )+6 \ln \left (\sin \left (b x +a \right )\right )}{b}\)	\(89\)
risch	\(-6 i x +\frac {{\mathrm e}^{4 i \left (b x +a \right )}}{64 b}+\frac {7 \,{\mathrm e}^{2 i \left (b x +a \right )}}{16 b}+\frac {7 \,{\mathrm e}^{-2 i \left (b x +a \right )}}{16 b}+\frac {{\mathrm e}^{-4 i \left (b x +a \right )}}{64 b}-\frac {12 i a}{b}-\frac {4 \left (2 \,{\mathrm e}^{6 i \left (b x +a \right )}-3 \,{\mathrm e}^{4 i \left (b x +a \right )}+2 \,{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4}}+\frac {6 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}\)	\(138\)
parallelrisch	\(-\frac {3 \left (\left (\frac {3}{4}-\cos \left (2 b x +2 a \right )+\frac {\cos \left (4 b x +4 a \right )}{4}\right ) \ln \left (\sec ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\left (\cos \left (2 b x +2 a \right )-\frac {\cos \left (4 b x +4 a \right )}{4}-\frac {3}{4}\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\frac {71 \cos \left (2 b x +2 a \right )}{192}+\frac {41 \cos \left (4 b x +4 a \right )}{192}-\frac {\cos \left (6 b x +6 a \right )}{64}-\frac {\cos \left (8 b x +8 a \right )}{1536}+\frac {131}{512}\right ) \left (\sec ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\csc ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{16 b}\)	\(144\)
norman	\(\frac {-\frac {1}{64 b}+\frac {3 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}+\frac {3 \left (\tan ^{14}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {\tan ^{16}\left (\frac {b x}{2}+\frac {a}{2}\right )}{64 b}-\frac {93 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {93 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {591 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{32 b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{4} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}+\frac {6 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {6 \ln \left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}\)	\(165\)

[In]

int(cos(b*x+a)^9/sin(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

1/b*(-1/4/sin(b*x+a)^4*cos(b*x+a)^10+3/4/sin(b*x+a)^2*cos(b*x+a)^10+3/4*cos(b*x+a)^8+cos(b*x+a)^6+3/2*cos(b*x+
a)^4+3*cos(b*x+a)^2+6*ln(sin(b*x+a)))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.45 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\frac {8 \, \cos \left (b x + a\right )^{8} + 32 \, \cos \left (b x + a\right )^{6} - 115 \, \cos \left (b x + a\right )^{4} + 38 \, \cos \left (b x + a\right )^{2} + 192 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (b x + a\right )\right ) + 29}{32 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \]

[In]

integrate(cos(b*x+a)^9/sin(b*x+a)^5,x, algorithm="fricas")

[Out]

1/32*(8*cos(b*x + a)^8 + 32*cos(b*x + a)^6 - 115*cos(b*x + a)^4 + 38*cos(b*x + a)^2 + 192*(cos(b*x + a)^4 - 2*
cos(b*x + a)^2 + 1)*log(1/2*sin(b*x + a)) + 29)/(b*cos(b*x + a)^4 - 2*b*cos(b*x + a)^2 + b)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1664 vs. \(2 (58) = 116\).

Time = 7.68 (sec) , antiderivative size = 1664, normalized size of antiderivative = 24.12 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\text {Too large to display} \]

[In]

integrate(cos(b*x+a)**9/sin(b*x+a)**5,x)

[Out]

Piecewise((-384*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**12/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 +
b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 1536*log(tan
(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**10/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan
(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 2304*log(tan(a/2 + b*x/2)**2 + 1)*t
an(a/2 + b*x/2)**8/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b
*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 1536*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**6/(64*b
*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 6
4*b*tan(a/2 + b*x/2)**4) - 384*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**4/(64*b*tan(a/2 + b*x/2)**12 + 2
56*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4)
+ 384*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**12/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384
*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 1536*log(tan(a/2 + b*x/2))*ta
n(a/2 + b*x/2)**10/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b
*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 2304*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**8/(64*b*tan(a/
2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan
(a/2 + b*x/2)**4) + 1536*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2
+ b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 384*log(tan
(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b
*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - tan(a/2 + b*x/2)**16/(64*b*tan(a/2 + b*x/2)
**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x
/2)**4) + 24*tan(a/2 + b*x/2)**14/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*
x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 744*tan(a/2 + b*x/2)**10/(64*b*tan(a/2 + b*x
/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 +
b*x/2)**4) - 1182*tan(a/2 + b*x/2)**8/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2
+ b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 744*tan(a/2 + b*x/2)**6/(64*b*tan(a/2 +
b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2
+ b*x/2)**4) + 24*tan(a/2 + b*x/2)**2/(64*b*tan(a/2 + b*x/2)**12 + 256*b*tan(a/2 + b*x/2)**10 + 384*b*tan(a/2
+ b*x/2)**8 + 256*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)**12 + 256*b*ta
n(a/2 + b*x/2)**10 + 384*b*tan(a/2 + b*x/2)**8 + 256*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4), Ne(b,
0)), (x*cos(a)**9/sin(a)**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.81 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\frac {\sin \left (b x + a\right )^{4} - 8 \, \sin \left (b x + a\right )^{2} + \frac {8 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{4}} + 12 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \]

[In]

integrate(cos(b*x+a)^9/sin(b*x+a)^5,x, algorithm="maxima")

[Out]

1/4*(sin(b*x + a)^4 - 8*sin(b*x + a)^2 + (8*sin(b*x + a)^2 - 1)/sin(b*x + a)^4 + 12*log(sin(b*x + a)^2))/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (65) = 130\).

Time = 0.32 (sec) , antiderivative size = 277, normalized size of antiderivative = 4.01 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=-\frac {\frac {{\left (\frac {28 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {288 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} + \frac {28 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {32 \, {\left (\frac {84 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {126 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {84 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - \frac {25 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} - 25\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{4}} - 192 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 384 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1 \right |}\right )}{64 \, b} \]

[In]

integrate(cos(b*x+a)^9/sin(b*x+a)^5,x, algorithm="giac")

[Out]

-1/64*((28*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 288*(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 + 28*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x +
a) + 1)^2 + 32*(84*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 126*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 84
*(cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 - 25*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 - 25)/((cos(b*x + a)
- 1)/(cos(b*x + a) + 1) - 1)^4 - 192*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 384*log(abs(-(cos(b*
x + a) - 1)/(cos(b*x + a) + 1) + 1)))/b

Mupad [B] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.33 \[ \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx=\frac {6\,\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b}-\frac {3\,\ln \left ({\mathrm {tan}\left (a+b\,x\right )}^2+1\right )}{b}+\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^6+\frac {9\,{\mathrm {tan}\left (a+b\,x\right )}^4}{2}+{\mathrm {tan}\left (a+b\,x\right )}^2-\frac {1}{4}}{b\,\left ({\mathrm {tan}\left (a+b\,x\right )}^8+2\,{\mathrm {tan}\left (a+b\,x\right )}^6+{\mathrm {tan}\left (a+b\,x\right )}^4\right )} \]

[In]

int(cos(a + b*x)^9/sin(a + b*x)^5,x)

[Out]

(6*log(tan(a + b*x)))/b - (3*log(tan(a + b*x)^2 + 1))/b + (tan(a + b*x)^2 + (9*tan(a + b*x)^4)/2 + 3*tan(a + b
*x)^6 - 1/4)/(b*(tan(a + b*x)^4 + 2*tan(a + b*x)^6 + tan(a + b*x)^8))