3.477 \(\int \frac{\cot ^6(e+f x)}{\sqrt{a-a \sin ^2(e+f x)}} \, dx\)
Optimal. Leaf size=96 \[ -\frac{\cot (e+f x)}{f \sqrt{a \cos ^2(e+f x)}}-\frac{\cot (e+f x) \csc ^4(e+f x)}{5 f \sqrt{a \cos ^2(e+f x)}}+\frac{2 \cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt{a \cos ^2(e+f x)}} \]
[Out]
-(Cot[e + f*x]/(f*Sqrt[a*Cos[e + f*x]^2])) + (2*Cot[e + f*x]*Csc[e + f*x]^2)/(3*f*Sqrt[a*Cos[e + f*x]^2]) - (C
ot[e + f*x]*Csc[e + f*x]^4)/(5*f*Sqrt[a*Cos[e + f*x]^2])
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Rubi [A] time = 0.12132, antiderivative size = 96, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used =
{3176, 3207, 2606, 194} \[ -\frac{\cot (e+f x)}{f \sqrt{a \cos ^2(e+f x)}}-\frac{\cot (e+f x) \csc ^4(e+f x)}{5 f \sqrt{a \cos ^2(e+f x)}}+\frac{2 \cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt{a \cos ^2(e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[Cot[e + f*x]^6/Sqrt[a - a*Sin[e + f*x]^2],x]
[Out]
-(Cot[e + f*x]/(f*Sqrt[a*Cos[e + f*x]^2])) + (2*Cot[e + f*x]*Csc[e + f*x]^2)/(3*f*Sqrt[a*Cos[e + f*x]^2]) - (C
ot[e + f*x]*Csc[e + f*x]^4)/(5*f*Sqrt[a*Cos[e + f*x]^2])
Rule 3176
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]
Rule 3207
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])
Rule 2606
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Rule 194
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]
Rubi steps
\begin{align*} \int \frac{\cot ^6(e+f x)}{\sqrt{a-a \sin ^2(e+f x)}} \, dx &=\int \frac{\cot ^6(e+f x)}{\sqrt{a \cos ^2(e+f x)}} \, dx\\ &=\frac{\cos (e+f x) \int \cot ^5(e+f x) \csc (e+f x) \, dx}{\sqrt{a \cos ^2(e+f x)}}\\ &=-\frac{\cos (e+f x) \operatorname{Subst}\left (\int \left (-1+x^2\right )^2 \, dx,x,\csc (e+f x)\right )}{f \sqrt{a \cos ^2(e+f x)}}\\ &=-\frac{\cos (e+f x) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\csc (e+f x)\right )}{f \sqrt{a \cos ^2(e+f x)}}\\ &=-\frac{\cot (e+f x)}{f \sqrt{a \cos ^2(e+f x)}}+\frac{2 \cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt{a \cos ^2(e+f x)}}-\frac{\cot (e+f x) \csc ^4(e+f x)}{5 f \sqrt{a \cos ^2(e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.0745488, size = 49, normalized size = 0.51 \[ -\frac{\cot (e+f x) \left (3 \csc ^4(e+f x)-10 \csc ^2(e+f x)+15\right )}{15 f \sqrt{a \cos ^2(e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[e + f*x]^6/Sqrt[a - a*Sin[e + f*x]^2],x]
[Out]
-(Cot[e + f*x]*(15 - 10*Csc[e + f*x]^2 + 3*Csc[e + f*x]^4))/(15*f*Sqrt[a*Cos[e + f*x]^2])
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Maple [A] time = 0.714, size = 54, normalized size = 0.6 \begin{align*} -{\frac{\cos \left ( fx+e \right ) \left ( 15\, \left ( \sin \left ( fx+e \right ) \right ) ^{4}-10\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}+3 \right ) }{15\, \left ( \sin \left ( fx+e \right ) \right ) ^{5}f}{\frac{1}{\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(f*x+e)^6/(a-a*sin(f*x+e)^2)^(1/2),x)
[Out]
-1/15*cos(f*x+e)*(15*sin(f*x+e)^4-10*sin(f*x+e)^2+3)/sin(f*x+e)^5/(a*cos(f*x+e)^2)^(1/2)/f
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Maxima [B] time = 1.77746, size = 1669, normalized size = 17.39 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^6/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
2/15*((15*sin(9*f*x + 9*e) - 20*sin(7*f*x + 7*e) + 58*sin(5*f*x + 5*e) - 20*sin(3*f*x + 3*e) + 15*sin(f*x + e)
)*cos(10*f*x + 10*e) + 75*(sin(8*f*x + 8*e) - 2*sin(6*f*x + 6*e) + 2*sin(4*f*x + 4*e) - sin(2*f*x + 2*e))*cos(
9*f*x + 9*e) + 5*(20*sin(7*f*x + 7*e) - 58*sin(5*f*x + 5*e) + 20*sin(3*f*x + 3*e) - 15*sin(f*x + e))*cos(8*f*x
+ 8*e) + 100*(2*sin(6*f*x + 6*e) - 2*sin(4*f*x + 4*e) + sin(2*f*x + 2*e))*cos(7*f*x + 7*e) + 10*(58*sin(5*f*x
+ 5*e) - 20*sin(3*f*x + 3*e) + 15*sin(f*x + e))*cos(6*f*x + 6*e) + 290*(2*sin(4*f*x + 4*e) - sin(2*f*x + 2*e)
)*cos(5*f*x + 5*e) + 50*(4*sin(3*f*x + 3*e) - 3*sin(f*x + e))*cos(4*f*x + 4*e) - (15*cos(9*f*x + 9*e) - 20*cos
(7*f*x + 7*e) + 58*cos(5*f*x + 5*e) - 20*cos(3*f*x + 3*e) + 15*cos(f*x + e))*sin(10*f*x + 10*e) - 15*(5*cos(8*
f*x + 8*e) - 10*cos(6*f*x + 6*e) + 10*cos(4*f*x + 4*e) - 5*cos(2*f*x + 2*e) + 1)*sin(9*f*x + 9*e) - 5*(20*cos(
7*f*x + 7*e) - 58*cos(5*f*x + 5*e) + 20*cos(3*f*x + 3*e) - 15*cos(f*x + e))*sin(8*f*x + 8*e) - 20*(10*cos(6*f*
x + 6*e) - 10*cos(4*f*x + 4*e) + 5*cos(2*f*x + 2*e) - 1)*sin(7*f*x + 7*e) - 10*(58*cos(5*f*x + 5*e) - 20*cos(3
*f*x + 3*e) + 15*cos(f*x + e))*sin(6*f*x + 6*e) - 58*(10*cos(4*f*x + 4*e) - 5*cos(2*f*x + 2*e) + 1)*sin(5*f*x
+ 5*e) - 50*(4*cos(3*f*x + 3*e) - 3*cos(f*x + e))*sin(4*f*x + 4*e) - 20*(5*cos(2*f*x + 2*e) - 1)*sin(3*f*x + 3
*e) + 100*cos(3*f*x + 3*e)*sin(2*f*x + 2*e) - 75*cos(f*x + e)*sin(2*f*x + 2*e) + 75*cos(2*f*x + 2*e)*sin(f*x +
e) - 15*sin(f*x + e))*sqrt(a)/((a*cos(10*f*x + 10*e)^2 + 25*a*cos(8*f*x + 8*e)^2 + 100*a*cos(6*f*x + 6*e)^2 +
100*a*cos(4*f*x + 4*e)^2 + 25*a*cos(2*f*x + 2*e)^2 + a*sin(10*f*x + 10*e)^2 + 25*a*sin(8*f*x + 8*e)^2 + 100*a
*sin(6*f*x + 6*e)^2 + 100*a*sin(4*f*x + 4*e)^2 - 100*a*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 25*a*sin(2*f*x + 2*
e)^2 - 2*(5*a*cos(8*f*x + 8*e) - 10*a*cos(6*f*x + 6*e) + 10*a*cos(4*f*x + 4*e) - 5*a*cos(2*f*x + 2*e) + a)*cos
(10*f*x + 10*e) - 10*(10*a*cos(6*f*x + 6*e) - 10*a*cos(4*f*x + 4*e) + 5*a*cos(2*f*x + 2*e) - a)*cos(8*f*x + 8*
e) - 20*(10*a*cos(4*f*x + 4*e) - 5*a*cos(2*f*x + 2*e) + a)*cos(6*f*x + 6*e) - 20*(5*a*cos(2*f*x + 2*e) - a)*co
s(4*f*x + 4*e) - 10*a*cos(2*f*x + 2*e) - 10*(a*sin(8*f*x + 8*e) - 2*a*sin(6*f*x + 6*e) + 2*a*sin(4*f*x + 4*e)
- a*sin(2*f*x + 2*e))*sin(10*f*x + 10*e) - 50*(2*a*sin(6*f*x + 6*e) - 2*a*sin(4*f*x + 4*e) + a*sin(2*f*x + 2*e
))*sin(8*f*x + 8*e) - 100*(2*a*sin(4*f*x + 4*e) - a*sin(2*f*x + 2*e))*sin(6*f*x + 6*e) + a)*f)
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Fricas [A] time = 1.61589, size = 205, normalized size = 2.14 \begin{align*} -\frac{{\left (15 \, \cos \left (f x + e\right )^{4} - 20 \, \cos \left (f x + e\right )^{2} + 8\right )} \sqrt{a \cos \left (f x + e\right )^{2}}}{15 \,{\left (a f \cos \left (f x + e\right )^{5} - 2 \, a f \cos \left (f x + e\right )^{3} + a f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^6/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
-1/15*(15*cos(f*x + e)^4 - 20*cos(f*x + e)^2 + 8)*sqrt(a*cos(f*x + e)^2)/((a*f*cos(f*x + e)^5 - 2*a*f*cos(f*x
+ e)^3 + a*f*cos(f*x + e))*sin(f*x + e))
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)**6/(a-a*sin(f*x+e)**2)**(1/2),x)
[Out]
Timed out
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Giac [A] time = 1.40487, size = 173, normalized size = 1.8 \begin{align*} \frac{\frac{3 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 25 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 150 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )} + \frac{150 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 25 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5}}}{480 \, \sqrt{a} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^6/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
1/480*((3*tan(1/2*f*x + 1/2*e)^5 - 25*tan(1/2*f*x + 1/2*e)^3 + 150*tan(1/2*f*x + 1/2*e))/sgn(tan(1/2*f*x + 1/2
*e)^4 - 1) + (150*tan(1/2*f*x + 1/2*e)^4 - 25*tan(1/2*f*x + 1/2*e)^2 + 3)/(sgn(tan(1/2*f*x + 1/2*e)^4 - 1)*tan
(1/2*f*x + 1/2*e)^5))/(sqrt(a)*f)