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

Optimal. Leaf size=108 \[ -\frac {\cot (c+d x)}{a^4 d}+\frac {4 \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac {104 \cot (c+d x)}{15 a^4 d (\csc (c+d x)+1)}+\frac {31 \cot (c+d x)}{15 a^4 d (\csc (c+d x)+1)^2}-\frac {2 \cot (c+d x)}{5 a^4 d (\csc (c+d x)+1)^3} \]

[Out]

4*arctanh(cos(d*x+c))/a^4/d-94/15*cot(d*x+c)/a^4/d+2/5*cot(d*x+c)/a^4/d/(1+sin(d*x+c))^3+13/15*cot(d*x+c)/a^4/
d/(1+sin(d*x+c))^2+4*cot(d*x+c)/a^4/d/(1+sin(d*x+c))

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Rubi [A]  time = 0.32, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {2709, 3770, 3767, 8, 3777, 3922, 3919, 3794} \[ -\frac {\cot (c+d x)}{a^4 d}+\frac {4 \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac {104 \cot (c+d x)}{15 a^4 d (\csc (c+d x)+1)}+\frac {31 \cot (c+d x)}{15 a^4 d (\csc (c+d x)+1)^2}-\frac {2 \cot (c+d x)}{5 a^4 d (\csc (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*ArcTanh[Cos[c + d*x]])/(a^4*d) - Cot[c + d*x]/(a^4*d) - (2*Cot[c + d*x])/(5*a^4*d*(1 + Csc[c + d*x])^3) + (
31*Cot[c + d*x])/(15*a^4*d*(1 + Csc[c + d*x])^2) - (104*Cot[c + d*x])/(15*a^4*d*(1 + Csc[c + d*x]))

Rule 8

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

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

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

Rule 3777

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(Cot[c + d*x]*(a + b*Csc[c + d*x])^n)/(d*(
2*n + 1)), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3922

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[((b
*c - a*d)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(b*f*(2*m + 1)), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e
+ f*x])^(m + 1)*Simp[a*c*(2*m + 1) - (b*c - a*d)*(m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f},
x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\cot ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx &=\frac {\int \left (\frac {9}{a^2}-\frac {4 \csc (c+d x)}{a^2}+\frac {\csc ^2(c+d x)}{a^2}-\frac {2}{a^2 (1+\csc (c+d x))^3}+\frac {9}{a^2 (1+\csc (c+d x))^2}-\frac {16}{a^2 (1+\csc (c+d x))}\right ) \, dx}{a^2}\\ &=\frac {9 x}{a^4}+\frac {\int \csc ^2(c+d x) \, dx}{a^4}-\frac {2 \int \frac {1}{(1+\csc (c+d x))^3} \, dx}{a^4}-\frac {4 \int \csc (c+d x) \, dx}{a^4}+\frac {9 \int \frac {1}{(1+\csc (c+d x))^2} \, dx}{a^4}-\frac {16 \int \frac {1}{1+\csc (c+d x)} \, dx}{a^4}\\ &=\frac {9 x}{a^4}+\frac {4 \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac {2 \cot (c+d x)}{5 a^4 d (1+\csc (c+d x))^3}+\frac {3 \cot (c+d x)}{a^4 d (1+\csc (c+d x))^2}-\frac {16 \cot (c+d x)}{a^4 d (1+\csc (c+d x))}+\frac {2 \int \frac {-5+2 \csc (c+d x)}{(1+\csc (c+d x))^2} \, dx}{5 a^4}-\frac {3 \int \frac {-3+\csc (c+d x)}{1+\csc (c+d x)} \, dx}{a^4}+\frac {16 \int -1 \, dx}{a^4}-\frac {\operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^4 d}\\ &=\frac {2 x}{a^4}+\frac {4 \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^4 d}-\frac {2 \cot (c+d x)}{5 a^4 d (1+\csc (c+d x))^3}+\frac {31 \cot (c+d x)}{15 a^4 d (1+\csc (c+d x))^2}-\frac {16 \cot (c+d x)}{a^4 d (1+\csc (c+d x))}-\frac {2 \int \frac {15-7 \csc (c+d x)}{1+\csc (c+d x)} \, dx}{15 a^4}-\frac {12 \int \frac {\csc (c+d x)}{1+\csc (c+d x)} \, dx}{a^4}\\ &=\frac {4 \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^4 d}-\frac {2 \cot (c+d x)}{5 a^4 d (1+\csc (c+d x))^3}+\frac {31 \cot (c+d x)}{15 a^4 d (1+\csc (c+d x))^2}-\frac {4 \cot (c+d x)}{a^4 d (1+\csc (c+d x))}+\frac {44 \int \frac {\csc (c+d x)}{1+\csc (c+d x)} \, dx}{15 a^4}\\ &=\frac {4 \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^4 d}-\frac {2 \cot (c+d x)}{5 a^4 d (1+\csc (c+d x))^3}+\frac {31 \cot (c+d x)}{15 a^4 d (1+\csc (c+d x))^2}-\frac {104 \cot (c+d x)}{15 a^4 d (1+\csc (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 0.42, size = 315, normalized size = 2.92 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (24 \sin \left (\frac {1}{2} (c+d x)\right )+316 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4-38 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3+76 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2-12 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+120 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5-120 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5+15 \tan \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5-15 \cot \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5\right )}{30 d (a \sin (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(24*Sin[(c + d*x)/2] - 12*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 76*
Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 38*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + 316*Si
n[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - 15*Cot[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)
/2])^5 + 120*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5 - 120*Log[Sin[(c + d*x)/2]]*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])^5 + 15*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5*Tan[(c + d*x)/2]))/(30*d*(a + a*
Sin[c + d*x])^4)

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fricas [B]  time = 0.44, size = 369, normalized size = 3.42 \[ \frac {94 \, \cos \left (d x + c\right )^{4} + 222 \, \cos \left (d x + c\right )^{3} - 115 \, \cos \left (d x + c\right )^{2} + 30 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) + 4\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 30 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) + 4\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (94 \, \cos \left (d x + c\right )^{3} - 128 \, \cos \left (d x + c\right )^{2} - 243 \, \cos \left (d x + c\right ) - 6\right )} \sin \left (d x + c\right ) - 237 \, \cos \left (d x + c\right ) + 6}{15 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} - 2 \, a^{4} d \cos \left (d x + c\right )^{3} - 5 \, a^{4} d \cos \left (d x + c\right )^{2} + 2 \, a^{4} d \cos \left (d x + c\right ) + 4 \, a^{4} d - {\left (a^{4} d \cos \left (d x + c\right )^{3} + 3 \, a^{4} d \cos \left (d x + c\right )^{2} - 2 \, a^{4} d \cos \left (d x + c\right ) - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(94*cos(d*x + c)^4 + 222*cos(d*x + c)^3 - 115*cos(d*x + c)^2 + 30*(cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 5*
cos(d*x + c)^2 - (cos(d*x + c)^3 + 3*cos(d*x + c)^2 - 2*cos(d*x + c) - 4)*sin(d*x + c) + 2*cos(d*x + c) + 4)*l
og(1/2*cos(d*x + c) + 1/2) - 30*(cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 5*cos(d*x + c)^2 - (cos(d*x + c)^3 + 3*co
s(d*x + c)^2 - 2*cos(d*x + c) - 4)*sin(d*x + c) + 2*cos(d*x + c) + 4)*log(-1/2*cos(d*x + c) + 1/2) + (94*cos(d
*x + c)^3 - 128*cos(d*x + c)^2 - 243*cos(d*x + c) - 6)*sin(d*x + c) - 237*cos(d*x + c) + 6)/(a^4*d*cos(d*x + c
)^4 - 2*a^4*d*cos(d*x + c)^3 - 5*a^4*d*cos(d*x + c)^2 + 2*a^4*d*cos(d*x + c) + 4*a^4*d - (a^4*d*cos(d*x + c)^3
+ 3*a^4*d*cos(d*x + c)^2 - 2*a^4*d*cos(d*x + c) - 4*a^4*d)*sin(d*x + c))

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giac [A]  time = 0.70, size = 135, normalized size = 1.25 \[ -\frac {\frac {120 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{4}} - \frac {15 \, {\left (8 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {4 \, {\left (135 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 435 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 605 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 385 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 104\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(120*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 - 15*tan(1/2*d*x + 1/2*c)/a^4 - 15*(8*tan(1/2*d*x + 1/2*c) - 1)/
(a^4*tan(1/2*d*x + 1/2*c)) + 4*(135*tan(1/2*d*x + 1/2*c)^4 + 435*tan(1/2*d*x + 1/2*c)^3 + 605*tan(1/2*d*x + 1/
2*c)^2 + 385*tan(1/2*d*x + 1/2*c) + 104)/(a^4*(tan(1/2*d*x + 1/2*c) + 1)^5))/d

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maple [A]  time = 0.31, size = 161, normalized size = 1.49 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{4} d}-\frac {1}{2 a^{4} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4} d}-\frac {16}{5 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {8}{a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {44}{3 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {14}{a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {18}{a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a^4/d*tan(1/2*d*x+1/2*c)-1/2/a^4/d/tan(1/2*d*x+1/2*c)-4/a^4/d*ln(tan(1/2*d*x+1/2*c))-16/5/a^4/d/(tan(1/2*d
*x+1/2*c)+1)^5+8/a^4/d/(tan(1/2*d*x+1/2*c)+1)^4-44/3/a^4/d/(tan(1/2*d*x+1/2*c)+1)^3+14/a^4/d/(tan(1/2*d*x+1/2*
c)+1)^2-18/a^4/d/(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.33, size = 288, normalized size = 2.67 \[ -\frac {\frac {\frac {491 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {1690 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2570 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1815 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {555 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 15}{\frac {a^{4} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{4} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {10 \, a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {5 \, a^{4} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a^{4} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {120 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}} - \frac {15 \, \sin \left (d x + c\right )}{a^{4} {\left (\cos \left (d x + c\right ) + 1\right )}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*((491*sin(d*x + c)/(cos(d*x + c) + 1) + 1690*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2570*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 + 1815*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 555*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15)
/(a^4*sin(d*x + c)/(cos(d*x + c) + 1) + 5*a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^4*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 10*a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 5*a^4*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + a^
4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + 120*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^4 - 15*sin(d*x + c)/(a^4*(
cos(d*x + c) + 1)))/d

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mupad [B]  time = 12.05, size = 203, normalized size = 1.88 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^4\,d}-\frac {37\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+121\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {514\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {338\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {491\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+1}{d\,\left (2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+20\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+20\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+10\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)/(2*a^4*d) - ((491*tan(c/2 + (d*x)/2))/15 + (338*tan(c/2 + (d*x)/2)^2)/3 + (514*tan(c/2 + (d
*x)/2)^3)/3 + 121*tan(c/2 + (d*x)/2)^4 + 37*tan(c/2 + (d*x)/2)^5 + 1)/(d*(10*a^4*tan(c/2 + (d*x)/2)^2 + 20*a^4
*tan(c/2 + (d*x)/2)^3 + 20*a^4*tan(c/2 + (d*x)/2)^4 + 10*a^4*tan(c/2 + (d*x)/2)^5 + 2*a^4*tan(c/2 + (d*x)/2)^6
+ 2*a^4*tan(c/2 + (d*x)/2))) - (4*log(tan(c/2 + (d*x)/2)))/(a^4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cot ^{2}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2/(a+a*sin(d*x+c))**4,x)

[Out]

Integral(cot(c + d*x)**2/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x) + 1), x)/a*
*4

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