3.171 \(\int \cos ^4(a+b x) \cot ^5(a+b x) \, dx\)
Optimal. Leaf size=69 \[ \frac{\sin ^4(a+b x)}{4 b}-\frac{2 \sin ^2(a+b x)}{b}-\frac{\csc ^4(a+b x)}{4 b}+\frac{2 \csc ^2(a+b x)}{b}+\frac{6 \log (\sin (a+b x))}{b} \]
[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)
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Rubi [A] time = 0.0475192, antiderivative size = 69, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used =
{2590, 266, 43} \[ \frac{\sin ^4(a+b x)}{4 b}-\frac{2 \sin ^2(a+b x)}{b}-\frac{\csc ^4(a+b x)}{4 b}+\frac{2 \csc ^2(a+b x)}{b}+\frac{6 \log (\sin (a+b x))}{b} \]
Antiderivative was successfully verified.
[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 2590
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]
Rule 266
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 43
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])
Rubi steps
\begin{align*} \int \cos ^4(a+b x) \cot ^5(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^4}{x^5} \, dx,x,-\sin (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1-x)^4}{x^3} \, dx,x,\sin ^2(a+b x)\right )}{2 b}\\ &=\frac{\operatorname{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] time = 0.10923, size = 55, normalized size = 0.8 \[ \frac{\sin ^4(a+b x)-8 \sin ^2(a+b x)-\csc ^4(a+b x)+8 \csc ^2(a+b x)+24 \log (\sin (a+b x))}{4 b} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[a + b*x]^4*Cot[a + b*x]^5,x]
[Out]
(8*Csc[a + b*x]^2 - Csc[a + b*x]^4 + 24*Log[Sin[a + b*x]] - 8*Sin[a + b*x]^2 + Sin[a + b*x]^4)/(4*b)
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Maple [A] time = 0.012, size = 107, normalized size = 1.6 \begin{align*} -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{10}}{4\,b \left ( \sin \left ( bx+a \right ) \right ) ^{4}}}+{\frac{3\, \left ( \cos \left ( bx+a \right ) \right ) ^{10}}{4\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}}}+{\frac{3\, \left ( \cos \left ( bx+a \right ) \right ) ^{8}}{4\,b}}+{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{6}}{b}}+{\frac{3\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{2\,b}}+3\,{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{b}}+6\,{\frac{\ln \left ( \sin \left ( bx+a \right ) \right ) }{b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(b*x+a)^9/sin(b*x+a)^5,x)
[Out]
-1/4/b/sin(b*x+a)^4*cos(b*x+a)^10+3/4/b/sin(b*x+a)^2*cos(b*x+a)^10+3/4*cos(b*x+a)^8/b+cos(b*x+a)^6/b+3/2*cos(b
*x+a)^4/b+3*cos(b*x+a)^2/b+6*ln(sin(b*x+a))/b
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Maxima [A] time = 0.96805, size = 76, normalized size = 1.1 \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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
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Fricas [A] time = 2.42775, size = 274, normalized size = 3.97 \begin{align*} \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 )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)
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Sympy [A] time = 59.7554, size = 1664, normalized size = 24.12 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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))
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Giac [B] time = 1.21353, size = 374, normalized size = 5.42 \begin{align*} -\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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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