3.438 \(\int \frac{\cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=96 \[ -\frac{\cot ^3(c+d x)}{3 a^3 d}-\frac{5 \cot (c+d x)}{a^3 d}+\frac{11 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac{4 \cot (c+d x)}{a^3 d (\csc (c+d x)+1)} \]

[Out]

(11*ArcTanh[Cos[c + d*x]])/(2*a^3*d) - (5*Cot[c + d*x])/(a^3*d) - Cot[c + d*x]^3/(3*a^3*d) + (3*Cot[c + d*x]*C
sc[c + d*x])/(2*a^3*d) - (4*Cot[c + d*x])/(a^3*d*(1 + Csc[c + d*x]))

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Rubi [A]  time = 0.154857, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2709, 3770, 3767, 8, 3768, 3777} \[ -\frac{\cot ^3(c+d x)}{3 a^3 d}-\frac{5 \cot (c+d x)}{a^3 d}+\frac{11 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac{4 \cot (c+d x)}{a^3 d (\csc (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4/(a + a*Sin[c + d*x])^3,x]

[Out]

(11*ArcTanh[Cos[c + d*x]])/(2*a^3*d) - (5*Cot[c + d*x])/(a^3*d) - Cot[c + d*x]^3/(3*a^3*d) + (3*Cot[c + d*x]*C
sc[c + d*x])/(2*a^3*d) - (4*Cot[c + d*x])/(a^3*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]

Rubi steps

\begin{align*} \int \frac{\cot ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \left (4 a-4 a \csc (c+d x)+4 a \csc ^2(c+d x)-3 a \csc ^3(c+d x)+a \csc ^4(c+d x)-\frac{4 a}{1+\csc (c+d x)}\right ) \, dx}{a^4}\\ &=\frac{4 x}{a^3}+\frac{\int \csc ^4(c+d x) \, dx}{a^3}-\frac{3 \int \csc ^3(c+d x) \, dx}{a^3}-\frac{4 \int \csc (c+d x) \, dx}{a^3}+\frac{4 \int \csc ^2(c+d x) \, dx}{a^3}-\frac{4 \int \frac{1}{1+\csc (c+d x)} \, dx}{a^3}\\ &=\frac{4 x}{a^3}+\frac{4 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac{4 \cot (c+d x)}{a^3 d (1+\csc (c+d x))}-\frac{3 \int \csc (c+d x) \, dx}{2 a^3}+\frac{4 \int -1 \, dx}{a^3}-\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^3 d}-\frac{4 \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=\frac{11 \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{5 \cot (c+d x)}{a^3 d}-\frac{\cot ^3(c+d x)}{3 a^3 d}+\frac{3 \cot (c+d x) \csc (c+d x)}{2 a^3 d}-\frac{4 \cot (c+d x)}{a^3 d (1+\csc (c+d x))}\\ \end{align*}

Mathematica [B]  time = 4.82126, size = 251, normalized size = 2.61 \[ -\frac{\sin ^2\left (\frac{1}{2} (c+d x)\right ) \left (\cot \left (\frac{1}{2} (c+d x)\right )+1\right )^5 \csc ^3(c+d x) \left (-4 \sin ^8\left (\frac{1}{2} (c+d x)\right )-8 \sin (c+d x) (7 \sin (c+d x)-2) \sin ^6\left (\frac{1}{2} (c+d x)\right )+\frac{1}{4} \sin ^4(c+d x) \left (28 \sin (c+d x)+\cot \left (\frac{1}{2} (c+d x)\right )-8\right )+\sin ^2(c+d x) \sin ^4\left (\frac{1}{2} (c+d x)\right ) \left (9-2 \sin (c+d x) \left (-33 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+33 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+62\right )\right )-\frac{1}{2} \sin ^3(c+d x) \sin ^2\left (\frac{1}{2} (c+d x)\right ) \left (\sin (c+d x) \left (-66 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+66 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-28\right )+9\right )\right )}{12 a^3 d (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4/(a + a*Sin[c + d*x])^3,x]

[Out]

-((1 + Cot[(c + d*x)/2])^5*Csc[c + d*x]^3*Sin[(c + d*x)/2]^2*(-4*Sin[(c + d*x)/2]^8 - 8*Sin[(c + d*x)/2]^6*Sin
[c + d*x]*(-2 + 7*Sin[c + d*x]) + (Sin[c + d*x]^4*(-8 + Cot[(c + d*x)/2] + 28*Sin[c + d*x]))/4 - (Sin[(c + d*x
)/2]^2*Sin[c + d*x]^3*(9 + (-28 + 66*Log[Cos[(c + d*x)/2]] - 66*Log[Sin[(c + d*x)/2]])*Sin[c + d*x]))/2 + Sin[
(c + d*x)/2]^4*Sin[c + d*x]^2*(9 - 2*(62 + 33*Log[Cos[(c + d*x)/2]] - 33*Log[Sin[(c + d*x)/2]])*Sin[c + d*x]))
)/(12*a^3*d*(1 + Sin[c + d*x])^3)

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Maple [A]  time = 0.193, size = 153, normalized size = 1.6 \begin{align*}{\frac{1}{24\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{3}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+{\frac{19}{8\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-8\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-{\frac{1}{24\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{3}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{19}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{11}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x)

[Out]

1/24/d/a^3*tan(1/2*d*x+1/2*c)^3-3/8/d/a^3*tan(1/2*d*x+1/2*c)^2+19/8/d/a^3*tan(1/2*d*x+1/2*c)-8/d/a^3/(tan(1/2*
d*x+1/2*c)+1)-1/24/d/a^3/tan(1/2*d*x+1/2*c)^3+3/8/d/a^3/tan(1/2*d*x+1/2*c)^2-19/8/d/a^3/tan(1/2*d*x+1/2*c)-11/
2/d/a^3*ln(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.10518, size = 269, normalized size = 2.8 \begin{align*} \frac{\frac{\frac{8 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{48 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{249 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 1}{\frac{a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{57 \, \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}} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{3}} - \frac{132 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/24*((8*sin(d*x + c)/(cos(d*x + c) + 1) - 48*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 249*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 - 1)/(a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (57*s
in(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/
a^3 - 132*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [B]  time = 1.24088, size = 810, normalized size = 8.44 \begin{align*} \frac{104 \, \cos \left (d x + c\right )^{4} + 38 \, \cos \left (d x + c\right )^{3} - 156 \, \cos \left (d x + c\right )^{2} + 33 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 33 \,{\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 2 \,{\left (52 \, \cos \left (d x + c\right )^{3} + 33 \, \cos \left (d x + c\right )^{2} - 45 \, \cos \left (d x + c\right ) - 24\right )} \sin \left (d x + c\right ) - 42 \, \cos \left (d x + c\right ) + 48}{12 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d -{\left (a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d \cos \left (d x + c\right ) - a^{3} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/12*(104*cos(d*x + c)^4 + 38*cos(d*x + c)^3 - 156*cos(d*x + c)^2 + 33*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (c
os(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*sin(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) - 33*(cos(d*x
+ c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*sin(d*x + c) + 1)*log(-1/2*c
os(d*x + c) + 1/2) + 2*(52*cos(d*x + c)^3 + 33*cos(d*x + c)^2 - 45*cos(d*x + c) - 24)*sin(d*x + c) - 42*cos(d*
x + c) + 48)/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d - (a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x +
c)^2 - a^3*d*cos(d*x + c) - a^3*d)*sin(d*x + c))

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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(cos(d*x+c)**4*csc(d*x+c)**4/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.3999, size = 197, normalized size = 2.05 \begin{align*} -\frac{\frac{132 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{192}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}} - \frac{242 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 57 \, \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}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}} - \frac{a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 57 \, a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{9}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/24*(132*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 192/(a^3*(tan(1/2*d*x + 1/2*c) + 1)) - (242*tan(1/2*d*x + 1/2*
c)^3 - 57*tan(1/2*d*x + 1/2*c)^2 + 9*tan(1/2*d*x + 1/2*c) - 1)/(a^3*tan(1/2*d*x + 1/2*c)^3) - (a^6*tan(1/2*d*x
+ 1/2*c)^3 - 9*a^6*tan(1/2*d*x + 1/2*c)^2 + 57*a^6*tan(1/2*d*x + 1/2*c))/a^9)/d