3.640 \(\int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=73 \[ -\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}-\frac {x}{a^2} \]

[Out]

-x/a^2-arctanh(cos(d*x+c))/a^2/d-cot(d*x+c)/a^2/d-1/3*cot(d*x+c)^3/a^2/d+cot(d*x+c)*csc(d*x+c)/a^2/d

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Rubi [A]  time = 0.32, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2875, 2873, 3473, 8, 2611, 3770, 2607, 30} \[ -\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}-\frac {x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

-(x/a^2) - ArcTanh[Cos[c + d*x]]/(a^2*d) - Cot[c + d*x]/(a^2*d) - Cot[c + d*x]^3/(3*a^2*d) + (Cot[c + d*x]*Csc
[c + d*x])/(a^2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2875

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[((g*Cos[e + f*x])^(2*m + p)*(d*Sin[e + f*x])^n)/(a - b*Sin[e +
f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \cot ^2(c+d x) \csc ^2(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \cot ^2(c+d x)-2 a^2 \cot ^2(c+d x) \csc (c+d x)+a^2 \cot ^2(c+d x) \csc ^2(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \cot ^2(c+d x) \, dx}{a^2}+\frac {\int \cot ^2(c+d x) \csc ^2(c+d x) \, dx}{a^2}-\frac {2 \int \cot ^2(c+d x) \csc (c+d x) \, dx}{a^2}\\ &=-\frac {\cot (c+d x)}{a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}-\frac {\int 1 \, dx}{a^2}+\frac {\int \csc (c+d x) \, dx}{a^2}+\frac {\operatorname {Subst}\left (\int x^2 \, dx,x,-\cot (c+d x)\right )}{a^2 d}\\ &=-\frac {x}{a^2}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a^2 d}-\frac {\cot ^3(c+d x)}{3 a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.28, size = 124, normalized size = 1.70 \[ -\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \left (\cot \left (\frac {1}{2} (c+d x)\right )+1\right )^4 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-6 \sin (2 (c+d x))+6 \cos (c+d x)-2 \cos (3 (c+d x))+12 \sin ^3(c+d x) \left (-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+c+d x\right )\right )}{96 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]

[Out]

-1/96*((1 + Cot[(c + d*x)/2])^4*Sec[(c + d*x)/2]^2*(6*Cos[c + d*x] - 2*Cos[3*(c + d*x)] + 12*(c + d*x + Log[Co
s[(c + d*x)/2]] - Log[Sin[(c + d*x)/2]])*Sin[c + d*x]^3 - 6*Sin[2*(c + d*x)])*Tan[(c + d*x)/2])/(a^2*d*(1 + Si
n[c + d*x])^2)

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fricas [A]  time = 0.49, size = 139, normalized size = 1.90 \[ -\frac {4 \, \cos \left (d x + c\right )^{3} + 3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 6 \, {\left (d x \cos \left (d x + c\right )^{2} - d x + \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 6 \, \cos \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/6*(4*cos(d*x + c)^3 + 3*(cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*(cos(d*x + c)^2 -
1)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 6*(d*x*cos(d*x + c)^2 - d*x + cos(d*x + c))*sin(d*x + c) - 6*c
os(d*x + c))/((a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x + c))

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giac [A]  time = 0.24, size = 137, normalized size = 1.88 \[ -\frac {\frac {24 \, {\left (d x + c\right )}}{a^{2}} - \frac {24 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {44 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/24*(24*(d*x + c)/a^2 - 24*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + (44*tan(1/2*d*x + 1/2*c)^3 + 9*tan(1/2*d*x +
1/2*c)^2 - 6*tan(1/2*d*x + 1/2*c) + 1)/(a^2*tan(1/2*d*x + 1/2*c)^3) - (a^4*tan(1/2*d*x + 1/2*c)^3 - 6*a^4*tan
(1/2*d*x + 1/2*c)^2 + 9*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [B]  time = 0.57, size = 149, normalized size = 2.04 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{2}}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2} d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {1}{24 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {1}{4 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {3}{8 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x)

[Out]

1/24/d/a^2*tan(1/2*d*x+1/2*c)^3-1/4/d/a^2*tan(1/2*d*x+1/2*c)^2+3/8/d/a^2*tan(1/2*d*x+1/2*c)-2/d/a^2*arctan(tan
(1/2*d*x+1/2*c))-1/24/a^2/d/tan(1/2*d*x+1/2*c)^3+1/4/a^2/d/tan(1/2*d*x+1/2*c)^2-3/8/d/a^2/tan(1/2*d*x+1/2*c)+1
/d/a^2*ln(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.45, size = 176, normalized size = 2.41 \[ \frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {6 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {48 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {24 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {{\left (\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a^{2} \sin \left (d x + c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/24*((9*sin(d*x + c)/(cos(d*x + c) + 1) - 6*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^3/(cos(d*x + c
) + 1)^3)/a^2 - 48*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 24*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 +
(6*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)*(cos(d*x + c) + 1)^3/(a^2*sin
(d*x + c)^3))/d

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mupad [B]  time = 9.30, size = 261, normalized size = 3.58 \[ \frac {{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-6\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+9\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-9\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+48\,\mathrm {atan}\left (\frac {\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+24\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(sin(c + d*x)^4*(a + a*sin(c + d*x))^2),x)

[Out]

(sin(c/2 + (d*x)/2)^6 - cos(c/2 + (d*x)/2)^6 - 6*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^5 + 6*cos(c/2 + (d*x)/2
)^5*sin(c/2 + (d*x)/2) + 9*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^4 - 9*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/
2)^2 + 48*atan((cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2)))*cos(c/2 +
(d*x)/2)^3*sin(c/2 + (d*x)/2)^3 + 24*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^3*sin(c/2 +
(d*x)/2)^3)/(24*a^2*d*cos(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*csc(d*x+c)**4/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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