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

Optimal. Leaf size=120 \[ -\frac{\cot ^3(c+d x)}{3 a^4 d}-\frac{9 \cot (c+d x)}{a^4 d}+\frac{14 \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{2 \cot (c+d x) \csc (c+d x)}{a^4 d}-\frac{44 \cot (c+d x)}{3 a^4 d (\csc (c+d x)+1)}+\frac{4 \cot (c+d x)}{3 a^4 d (\csc (c+d x)+1)^2} \]

[Out]

(14*ArcTanh[Cos[c + d*x]])/(a^4*d) - (9*Cot[c + d*x])/(a^4*d) - Cot[c + d*x]^3/(3*a^4*d) + (2*Cot[c + d*x]*Csc
[c + d*x])/(a^4*d) + (4*Cot[c + d*x])/(3*a^4*d*(1 + Csc[c + d*x])^2) - (44*Cot[c + d*x])/(3*a^4*d*(1 + Csc[c +
d*x]))

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Rubi [A]  time = 0.248752, antiderivative size = 120, normalized size of antiderivative = 1., 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, 3768, 3777, 3919, 3794} \[ -\frac{\cot ^3(c+d x)}{3 a^4 d}-\frac{9 \cot (c+d x)}{a^4 d}+\frac{14 \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{2 \cot (c+d x) \csc (c+d x)}{a^4 d}-\frac{44 \cot (c+d x)}{3 a^4 d (\csc (c+d x)+1)}+\frac{4 \cot (c+d x)}{3 a^4 d (\csc (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(14*ArcTanh[Cos[c + d*x]])/(a^4*d) - (9*Cot[c + d*x])/(a^4*d) - Cot[c + d*x]^3/(3*a^4*d) + (2*Cot[c + d*x]*Csc
[c + d*x])/(a^4*d) + (4*Cot[c + d*x])/(3*a^4*d*(1 + Csc[c + d*x])^2) - (44*Cot[c + d*x])/(3*a^4*d*(1 + Csc[c +
d*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 3770

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

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 8

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

Rule 3768

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

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 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 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]

Rubi steps

\begin{align*} \int \frac{\cot ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx &=\frac{\int \left (16-12 \csc (c+d x)+8 \csc ^2(c+d x)-4 \csc ^3(c+d x)+\csc ^4(c+d x)+\frac{4}{(1+\csc (c+d x))^2}-\frac{20}{1+\csc (c+d x)}\right ) \, dx}{a^4}\\ &=\frac{16 x}{a^4}+\frac{\int \csc ^4(c+d x) \, dx}{a^4}-\frac{4 \int \csc ^3(c+d x) \, dx}{a^4}+\frac{4 \int \frac{1}{(1+\csc (c+d x))^2} \, dx}{a^4}+\frac{8 \int \csc ^2(c+d x) \, dx}{a^4}-\frac{12 \int \csc (c+d x) \, dx}{a^4}-\frac{20 \int \frac{1}{1+\csc (c+d x)} \, dx}{a^4}\\ &=\frac{16 x}{a^4}+\frac{12 \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{2 \cot (c+d x) \csc (c+d x)}{a^4 d}+\frac{4 \cot (c+d x)}{3 a^4 d (1+\csc (c+d x))^2}-\frac{20 \cot (c+d x)}{a^4 d (1+\csc (c+d x))}-\frac{4 \int \frac{-3+\csc (c+d x)}{1+\csc (c+d x)} \, dx}{3 a^4}-\frac{2 \int \csc (c+d x) \, dx}{a^4}+\frac{20 \int -1 \, dx}{a^4}-\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^4 d}-\frac{8 \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^4 d}\\ &=\frac{14 \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac{9 \cot (c+d x)}{a^4 d}-\frac{\cot ^3(c+d x)}{3 a^4 d}+\frac{2 \cot (c+d x) \csc (c+d x)}{a^4 d}+\frac{4 \cot (c+d x)}{3 a^4 d (1+\csc (c+d x))^2}-\frac{20 \cot (c+d x)}{a^4 d (1+\csc (c+d x))}-\frac{16 \int \frac{\csc (c+d x)}{1+\csc (c+d x)} \, dx}{3 a^4}\\ &=\frac{14 \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac{9 \cot (c+d x)}{a^4 d}-\frac{\cot ^3(c+d x)}{3 a^4 d}+\frac{2 \cot (c+d x) \csc (c+d x)}{a^4 d}+\frac{4 \cot (c+d x)}{3 a^4 d (1+\csc (c+d x))^2}-\frac{44 \cot (c+d x)}{3 a^4 d (1+\csc (c+d x))}\\ \end{align*}

Mathematica [B]  time = 6.0893, size = 589, normalized size = 4.91 \[ \frac{80 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^7}{3 d (a \sin (c+d x)+a)^4}-\frac{4 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6}{3 d (a \sin (c+d x)+a)^4}+\frac{8 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5}{3 d (a \sin (c+d x)+a)^4}+\frac{14 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8}{d (a \sin (c+d x)+a)^4}-\frac{14 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8}{d (a \sin (c+d x)+a)^4}+\frac{13 \tan \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8}{3 d (a \sin (c+d x)+a)^4}-\frac{13 \cot \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8}{3 d (a \sin (c+d x)+a)^4}+\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8}{2 d (a \sin (c+d x)+a)^4}-\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8}{2 d (a \sin (c+d x)+a)^4}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8}{24 d (a \sin (c+d x)+a)^4}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8}{24 d (a \sin (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)/(3*d*(a + a*Sin[c + d*x])^4) - (4*(Cos[(c + d*x)/
2] + Sin[(c + d*x)/2])^6)/(3*d*(a + a*Sin[c + d*x])^4) + (80*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2])^7)/(3*d*(a + a*Sin[c + d*x])^4) - (13*Cot[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8)/(3*d*(a
+ a*Sin[c + d*x])^4) + (Csc[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8)/(2*d*(a + a*Sin[c + d*x])^
4) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8)/(24*d*(a + a*Sin[c + d*x])^
4) + (14*Log[Cos[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8)/(d*(a + a*Sin[c + d*x])^4) - (14*Log[S
in[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8)/(d*(a + a*Sin[c + d*x])^4) - (Sec[(c + d*x)/2]^2*(Co
s[(c + d*x)/2] + Sin[(c + d*x)/2])^8)/(2*d*(a + a*Sin[c + d*x])^4) + (13*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])
^8*Tan[(c + d*x)/2])/(3*d*(a + a*Sin[c + d*x])^4) + (Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^
8*Tan[(c + d*x)/2])/(24*d*(a + a*Sin[c + d*x])^4)

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Maple [A]  time = 0.138, size = 195, normalized size = 1.6 \begin{align*}{\frac{1}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{1}{2\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{35}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{16}{3\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+8\,{\frac{1}{d{a}^{4} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}-32\,{\frac{1}{d{a}^{4} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-{\frac{1}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{1}{2\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{35}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-14\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24/d/a^4*tan(1/2*d*x+1/2*c)^3-1/2/d/a^4*tan(1/2*d*x+1/2*c)^2+35/8/d/a^4*tan(1/2*d*x+1/2*c)-16/3/d/a^4/(tan(1
/2*d*x+1/2*c)+1)^3+8/d/a^4/(tan(1/2*d*x+1/2*c)+1)^2-32/d/a^4/(tan(1/2*d*x+1/2*c)+1)-1/24/d/a^4/tan(1/2*d*x+1/2
*c)^3+1/2/d/a^4/tan(1/2*d*x+1/2*c)^2-35/8/d/a^4/tan(1/2*d*x+1/2*c)-14/d/a^4*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 3.06899, size = 385, normalized size = 3.21 \begin{align*} \frac{\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{72 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{984 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{1647 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{873 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 1}{\frac{a^{4} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{3 \, 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{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{12 \, \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^{4}} - \frac{336 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((9*sin(d*x + c)/(cos(d*x + c) + 1) - 72*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 984*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 - 1647*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 873*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1)/(a^4*s
in(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 3*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) + (105*sin(d*x + c)/(cos(d*x + c) + 1) - 12*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^4 - 336*log(sin(d*x + c)/(cos(d*x + c) + 1)
)/a^4)/d

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Fricas [B]  time = 1.58112, size = 1187, normalized size = 9.89 \begin{align*} -\frac{66 \, \cos \left (d x + c\right )^{5} - 24 \, \cos \left (d x + c\right )^{4} - 147 \, \cos \left (d x + c\right )^{3} + 29 \, \cos \left (d x + c\right )^{2} - 21 \,{\left (\cos \left (d x + c\right )^{5} + 2 \, \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) + 2\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 21 \,{\left (\cos \left (d x + c\right )^{5} + 2 \, \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) + 2\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (66 \, \cos \left (d x + c\right )^{4} + 90 \, \cos \left (d x + c\right )^{3} - 57 \, \cos \left (d x + c\right )^{2} - 86 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 82 \, \cos \left (d x + c\right ) - 4}{3 \,{\left (a^{4} d \cos \left (d x + c\right )^{5} + 2 \, a^{4} d \cos \left (d x + c\right )^{4} - 2 \, a^{4} d \cos \left (d x + c\right )^{3} - 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right ) + 2 \, a^{4} d +{\left (a^{4} d \cos \left (d x + c\right )^{4} - a^{4} d \cos \left (d x + c\right )^{3} - 3 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right ) + 2 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(66*cos(d*x + c)^5 - 24*cos(d*x + c)^4 - 147*cos(d*x + c)^3 + 29*cos(d*x + c)^2 - 21*(cos(d*x + c)^5 + 2*
cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 - cos(d*x + c)^3 - 3*cos(d*x + c)^2 + c
os(d*x + c) + 2)*sin(d*x + c) + cos(d*x + c) + 2)*log(1/2*cos(d*x + c) + 1/2) + 21*(cos(d*x + c)^5 + 2*cos(d*x
+ c)^4 - 2*cos(d*x + c)^3 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 - cos(d*x + c)^3 - 3*cos(d*x + c)^2 + cos(d*x
+ c) + 2)*sin(d*x + c) + cos(d*x + c) + 2)*log(-1/2*cos(d*x + c) + 1/2) - (66*cos(d*x + c)^4 + 90*cos(d*x + c)
^3 - 57*cos(d*x + c)^2 - 86*cos(d*x + c) - 4)*sin(d*x + c) + 82*cos(d*x + c) - 4)/(a^4*d*cos(d*x + c)^5 + 2*a^
4*d*cos(d*x + c)^4 - 2*a^4*d*cos(d*x + c)^3 - 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c) + 2*a^4*d + (a^4*d*c
os(d*x + c)^4 - a^4*d*cos(d*x + c)^3 - 3*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c) + 2*a^4*d)*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cot ^{4}{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(c + d*x)**4/(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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Giac [A]  time = 2.40079, size = 242, normalized size = 2.02 \begin{align*} -\frac{\frac{336 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac{308 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 51 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 723 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 676 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 72 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{3} a^{4}} - \frac{a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 105 \, a^{8} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{12}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(336*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 - (308*tan(1/2*d*x + 1/2*c)^6 + 51*tan(1/2*d*x + 1/2*c)^5 - 723*
tan(1/2*d*x + 1/2*c)^4 - 676*tan(1/2*d*x + 1/2*c)^3 - 72*tan(1/2*d*x + 1/2*c)^2 + 9*tan(1/2*d*x + 1/2*c) - 1)/
((tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c))^3*a^4) - (a^8*tan(1/2*d*x + 1/2*c)^3 - 12*a^8*tan(1/2*d*x + 1
/2*c)^2 + 105*a^8*tan(1/2*d*x + 1/2*c))/a^12)/d