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

Optimal. Leaf size=66 \[ -\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d} \]

[Out]

arctanh(cos(d*x+c))/a^2/d-2*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 = 0.13, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2708, 2757, 3767, 8, 3768, 3770} \[ -\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {2 \cot (c+d x)}{a^2 d}+\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\cot (c+d x) \csc (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Cos[c + d*x]]/(a^2*d) - (2*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 2708

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

Rule 2757

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

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 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 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 {\cot ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \csc ^4(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \csc ^2(c+d x)-2 a^2 \csc ^3(c+d x)+a^2 \csc ^4(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \csc ^2(c+d x) \, dx}{a^2}+\frac {\int \csc ^4(c+d x) \, dx}{a^2}-\frac {2 \int \csc ^3(c+d x) \, dx}{a^2}\\ &=\frac {\cot (c+d x) \csc (c+d x)}{a^2 d}-\frac {\int \csc (c+d x) \, dx}{a^2}-\frac {\operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^2 d}-\frac {\operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^2 d}\\ &=\frac {\tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {2 \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 = 0.91, size = 121, normalized size = 1.83 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \left (\cot \left (\frac {1}{2} (c+d x)\right )+1\right )^4 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-9 \cos (c+d x)+5 \cos (3 (c+d x))+6 \left (\sin (2 (c+d x))+2 \sin ^3(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )\right )}{96 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + Cot[(c + d*x)/2])^4*Sec[(c + d*x)/2]^2*(-9*Cos[c + d*x] + 5*Cos[3*(c + d*x)] + 6*(2*(Log[Cos[(c + d*x)/2
]] - Log[Sin[(c + d*x)/2]])*Sin[c + d*x]^3 + Sin[2*(c + d*x)]))*Tan[(c + d*x)/2])/(96*a^2*d*(1 + Sin[c + d*x])
^2)

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fricas [A]  time = 0.45, size = 123, normalized size = 1.86 \[ -\frac {10 \, \cos \left (d x + c\right )^{3} - 3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 6 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 12 \, \cos \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(10*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*cos(d*x + c)*sin(d*x + c) - 12*cos(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 = 0.22, size = 128, normalized size = 1.94 \[ -\frac {\frac {24 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {44 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 21 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(24*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - (44*tan(1/2*d*x + 1/2*c)^3 - 21*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 + 21*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [B]  time = 0.61, size = 132, normalized size = 2.00 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{2}}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2} d}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{2}}-\frac {7}{8 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {1}{4 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {1}{24 a^{2} d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}} \]

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))^2,x)

[Out]

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+7/8/d/a^2*tan(1/2*d*x+1/2*c)-7/8/d/a^2/tan(1/2*
d*x+1/2*c)-1/d/a^2*ln(tan(1/2*d*x+1/2*c))+1/4/a^2/d/tan(1/2*d*x+1/2*c)^2-1/24/a^2/d/tan(1/2*d*x+1/2*c)^3

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maxima [B]  time = 0.34, size = 153, normalized size = 2.32 \[ \frac {\frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {6 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {{\left (\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {21 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a^{2} \sin \left (d x + c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((21*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 - 24*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + (6*sin(d*x + c)/(cos(d*x + c) + 1) - 21*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 = 8.67, size = 119, normalized size = 1.80 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^2\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {1}{3}\right )}{8\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(c/2 + (d*x)/2)^3/(24*a^2*d) - tan(c/2 + (d*x)/2)^2/(4*a^2*d) - log(tan(c/2 + (d*x)/2))/(a^2*d) + (7*tan(c/
2 + (d*x)/2))/(8*a^2*d) - (cot(c/2 + (d*x)/2)^3*(7*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2) + 1/3))/(8*a^2*
d)

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

[Out]

Timed out

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