3.1.29 \(\int \frac {\csc ^2(x)}{(a+a \sin (x))^3} \, dx\) [29]

Optimal. Leaf size=65 \[ \frac {3 \tanh ^{-1}(\cos (x))}{a^3}-\frac {24 \cot (x)}{5 a^3}+\frac {\cot (x)}{5 (a+a \sin (x))^3}+\frac {3 \cot (x)}{5 a (a+a \sin (x))^2}+\frac {3 \cot (x)}{a^3+a^3 \sin (x)} \]

[Out]

3*arctanh(cos(x))/a^3-24/5*cot(x)/a^3+1/5*cot(x)/(a+a*sin(x))^3+3/5*cot(x)/a/(a+a*sin(x))^2+3*cot(x)/(a^3+a^3*
sin(x))

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Rubi [A]
time = 0.15, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2845, 3057, 2827, 3852, 8, 3855} \begin {gather*} -\frac {24 \cot (x)}{5 a^3}+\frac {3 \tanh ^{-1}(\cos (x))}{a^3}+\frac {3 \cot (x)}{a^3 \sin (x)+a^3}+\frac {3 \cot (x)}{5 a (a \sin (x)+a)^2}+\frac {\cot (x)}{5 (a \sin (x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[x]^2/(a + a*Sin[x])^3,x]

[Out]

(3*ArcTanh[Cos[x]])/a^3 - (24*Cot[x])/(5*a^3) + Cot[x]/(5*(a + a*Sin[x])^3) + (3*Cot[x])/(5*a*(a + a*Sin[x])^2
) + (3*Cot[x])/(a^3 + a^3*Sin[x])

Rule 8

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

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2845

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 3852

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 3855

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 {\csc ^2(x)}{(a+a \sin (x))^3} \, dx &=\frac {\cot (x)}{5 (a+a \sin (x))^3}+\frac {\int \frac {\csc ^2(x) (6 a-3 a \sin (x))}{(a+a \sin (x))^2} \, dx}{5 a^2}\\ &=\frac {\cot (x)}{5 (a+a \sin (x))^3}+\frac {3 \cot (x)}{5 a (a+a \sin (x))^2}+\frac {\int \frac {\csc ^2(x) \left (27 a^2-18 a^2 \sin (x)\right )}{a+a \sin (x)} \, dx}{15 a^4}\\ &=\frac {\cot (x)}{5 (a+a \sin (x))^3}+\frac {3 \cot (x)}{5 a (a+a \sin (x))^2}+\frac {3 \cot (x)}{a^3+a^3 \sin (x)}+\frac {\int \csc ^2(x) \left (72 a^3-45 a^3 \sin (x)\right ) \, dx}{15 a^6}\\ &=\frac {\cot (x)}{5 (a+a \sin (x))^3}+\frac {3 \cot (x)}{5 a (a+a \sin (x))^2}+\frac {3 \cot (x)}{a^3+a^3 \sin (x)}-\frac {3 \int \csc (x) \, dx}{a^3}+\frac {24 \int \csc ^2(x) \, dx}{5 a^3}\\ &=\frac {3 \tanh ^{-1}(\cos (x))}{a^3}+\frac {\cot (x)}{5 (a+a \sin (x))^3}+\frac {3 \cot (x)}{5 a (a+a \sin (x))^2}+\frac {3 \cot (x)}{a^3+a^3 \sin (x)}-\frac {24 \text {Subst}(\int 1 \, dx,x,\cot (x))}{5 a^3}\\ &=\frac {3 \tanh ^{-1}(\cos (x))}{a^3}-\frac {24 \cot (x)}{5 a^3}+\frac {\cot (x)}{5 (a+a \sin (x))^3}+\frac {3 \cot (x)}{5 a (a+a \sin (x))^2}+\frac {3 \cot (x)}{a^3+a^3 \sin (x)}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(206\) vs. \(2(65)=130\).
time = 0.11, size = 206, normalized size = 3.17 \begin {gather*} \frac {\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right ) \left (4 \sin \left (\frac {x}{2}\right )-2 \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+16 \sin \left (\frac {x}{2}\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2-8 \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^3+76 \sin \left (\frac {x}{2}\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^4-5 \cot \left (\frac {x}{2}\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^5+30 \log \left (\cos \left (\frac {x}{2}\right )\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^5-30 \log \left (\sin \left (\frac {x}{2}\right )\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^5+5 \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^5 \tan \left (\frac {x}{2}\right )\right )}{10 (a+a \sin (x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^2/(a + a*Sin[x])^3,x]

[Out]

((Cos[x/2] + Sin[x/2])*(4*Sin[x/2] - 2*(Cos[x/2] + Sin[x/2]) + 16*Sin[x/2]*(Cos[x/2] + Sin[x/2])^2 - 8*(Cos[x/
2] + Sin[x/2])^3 + 76*Sin[x/2]*(Cos[x/2] + Sin[x/2])^4 - 5*Cot[x/2]*(Cos[x/2] + Sin[x/2])^5 + 30*Log[Cos[x/2]]
*(Cos[x/2] + Sin[x/2])^5 - 30*Log[Sin[x/2]]*(Cos[x/2] + Sin[x/2])^5 + 5*(Cos[x/2] + Sin[x/2])^5*Tan[x/2]))/(10
*(a + a*Sin[x])^3)

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Maple [A]
time = 0.19, size = 76, normalized size = 1.17

method	result	size
default	\(\frac {\tan \left (\frac {x}{2}\right )-\frac {16}{5 \left (\tan \left (\frac {x}{2}\right )+1\right )^{5}}+\frac {8}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {16}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {16}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {24}{\tan \left (\frac {x}{2}\right )+1}-\frac {1}{\tan \left (\frac {x}{2}\right )}-6 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 a^{3}}\)	\(76\)
risch	\(-\frac {2 \left (-160 \,{\mathrm e}^{4 i x}+75 i {\mathrm e}^{5 i x}+189 \,{\mathrm e}^{2 i x}-200 i {\mathrm e}^{3 i x}-24+105 i {\mathrm e}^{i x}+15 \,{\mathrm e}^{6 i x}\right )}{5 \left ({\mathrm e}^{2 i x}-1\right ) \left ({\mathrm e}^{i x}+i\right )^{5} a^{3}}-\frac {3 \ln \left ({\mathrm e}^{i x}-1\right )}{a^{3}}+\frac {3 \ln \left ({\mathrm e}^{i x}+1\right )}{a^{3}}\)	\(99\)
norman	\(\frac {-\frac {53 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{a}-\frac {1}{2 a}+\frac {\tan ^{7}\left (\frac {x}{2}\right )}{2 a}-\frac {20 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{a}-\frac {125 \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{2 a}-\frac {73 \tan \left (\frac {x}{2}\right )}{5 a}-\frac {167 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2 a}}{\tan \left (\frac {x}{2}\right ) a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{5}}-\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{3}}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^2/(a+a*sin(x))^3,x,method=_RETURNVERBOSE)

[Out]

1/2/a^3*(tan(1/2*x)-16/5/(tan(1/2*x)+1)^5+8/(tan(1/2*x)+1)^4-16/(tan(1/2*x)+1)^3+16/(tan(1/2*x)+1)^2-24/(tan(1
/2*x)+1)-1/tan(1/2*x)-6*ln(tan(1/2*x)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (59) = 118\).
time = 0.58, size = 180, normalized size = 2.77 \begin {gather*} -\frac {\frac {121 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {410 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {610 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {425 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {125 \, \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + 5}{10 \, {\left (\frac {a^{3} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {5 \, a^{3} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {10 \, a^{3} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {5 \, a^{3} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac {a^{3} \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}}\right )}} - \frac {3 \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{3}} + \frac {\sin \left (x\right )}{2 \, a^{3} {\left (\cos \left (x\right ) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^2/(a+a*sin(x))^3,x, algorithm="maxima")

[Out]

-1/10*(121*sin(x)/(cos(x) + 1) + 410*sin(x)^2/(cos(x) + 1)^2 + 610*sin(x)^3/(cos(x) + 1)^3 + 425*sin(x)^4/(cos
(x) + 1)^4 + 125*sin(x)^5/(cos(x) + 1)^5 + 5)/(a^3*sin(x)/(cos(x) + 1) + 5*a^3*sin(x)^2/(cos(x) + 1)^2 + 10*a^
3*sin(x)^3/(cos(x) + 1)^3 + 10*a^3*sin(x)^4/(cos(x) + 1)^4 + 5*a^3*sin(x)^5/(cos(x) + 1)^5 + a^3*sin(x)^6/(cos
(x) + 1)^6) - 3*log(sin(x)/(cos(x) + 1))/a^3 + 1/2*sin(x)/(a^3*(cos(x) + 1))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (59) = 118\).
time = 0.39, size = 225, normalized size = 3.46 \begin {gather*} \frac {48 \, \cos \left (x\right )^{4} + 114 \, \cos \left (x\right )^{3} - 60 \, \cos \left (x\right )^{2} + 15 \, {\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{3} - 5 \, \cos \left (x\right )^{2} - {\left (\cos \left (x\right )^{3} + 3 \, \cos \left (x\right )^{2} - 2 \, \cos \left (x\right ) - 4\right )} \sin \left (x\right ) + 2 \, \cos \left (x\right ) + 4\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - 15 \, {\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{3} - 5 \, \cos \left (x\right )^{2} - {\left (\cos \left (x\right )^{3} + 3 \, \cos \left (x\right )^{2} - 2 \, \cos \left (x\right ) - 4\right )} \sin \left (x\right ) + 2 \, \cos \left (x\right ) + 4\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + 2 \, {\left (24 \, \cos \left (x\right )^{3} - 33 \, \cos \left (x\right )^{2} - 63 \, \cos \left (x\right ) - 1\right )} \sin \left (x\right ) - 124 \, \cos \left (x\right ) + 2}{10 \, {\left (a^{3} \cos \left (x\right )^{4} - 2 \, a^{3} \cos \left (x\right )^{3} - 5 \, a^{3} \cos \left (x\right )^{2} + 2 \, a^{3} \cos \left (x\right ) + 4 \, a^{3} - {\left (a^{3} \cos \left (x\right )^{3} + 3 \, a^{3} \cos \left (x\right )^{2} - 2 \, a^{3} \cos \left (x\right ) - 4 \, a^{3}\right )} \sin \left (x\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^2/(a+a*sin(x))^3,x, algorithm="fricas")

[Out]

1/10*(48*cos(x)^4 + 114*cos(x)^3 - 60*cos(x)^2 + 15*(cos(x)^4 - 2*cos(x)^3 - 5*cos(x)^2 - (cos(x)^3 + 3*cos(x)
^2 - 2*cos(x) - 4)*sin(x) + 2*cos(x) + 4)*log(1/2*cos(x) + 1/2) - 15*(cos(x)^4 - 2*cos(x)^3 - 5*cos(x)^2 - (co
s(x)^3 + 3*cos(x)^2 - 2*cos(x) - 4)*sin(x) + 2*cos(x) + 4)*log(-1/2*cos(x) + 1/2) + 2*(24*cos(x)^3 - 33*cos(x)
^2 - 63*cos(x) - 1)*sin(x) - 124*cos(x) + 2)/(a^3*cos(x)^4 - 2*a^3*cos(x)^3 - 5*a^3*cos(x)^2 + 2*a^3*cos(x) +
4*a^3 - (a^3*cos(x)^3 + 3*a^3*cos(x)^2 - 2*a^3*cos(x) - 4*a^3)*sin(x))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\csc ^{2}{\left (x \right )}}{\sin ^{3}{\left (x \right )} + 3 \sin ^{2}{\left (x \right )} + 3 \sin {\left (x \right )} + 1}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**2/(a+a*sin(x))**3,x)

[Out]

Integral(csc(x)**2/(sin(x)**3 + 3*sin(x)**2 + 3*sin(x) + 1), x)/a**3

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Giac [A]
time = 0.48, size = 85, normalized size = 1.31 \begin {gather*} -\frac {3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{3}} + \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, a^{3}} + \frac {6 \, \tan \left (\frac {1}{2} \, x\right ) - 1}{2 \, a^{3} \tan \left (\frac {1}{2} \, x\right )} - \frac {4 \, {\left (15 \, \tan \left (\frac {1}{2} \, x\right )^{4} + 50 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 70 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 45 \, \tan \left (\frac {1}{2} \, x\right ) + 12\right )}}{5 \, a^{3} {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^2/(a+a*sin(x))^3,x, algorithm="giac")

[Out]

-3*log(abs(tan(1/2*x)))/a^3 + 1/2*tan(1/2*x)/a^3 + 1/2*(6*tan(1/2*x) - 1)/(a^3*tan(1/2*x)) - 4/5*(15*tan(1/2*x
)^4 + 50*tan(1/2*x)^3 + 70*tan(1/2*x)^2 + 45*tan(1/2*x) + 12)/(a^3*(tan(1/2*x) + 1)^5)

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Mupad [B]
time = 6.72, size = 129, normalized size = 1.98 \begin {gather*} \frac {\mathrm {tan}\left (\frac {x}{2}\right )}{2\,a^3}-\frac {25\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5+85\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+122\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+82\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\frac {121\,\mathrm {tan}\left (\frac {x}{2}\right )}{5}+1}{2\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+10\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5+20\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+20\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,a^3\,\mathrm {tan}\left (\frac {x}{2}\right )}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^2*(a + a*sin(x))^3),x)

[Out]

tan(x/2)/(2*a^3) - ((121*tan(x/2))/5 + 82*tan(x/2)^2 + 122*tan(x/2)^3 + 85*tan(x/2)^4 + 25*tan(x/2)^5 + 1)/(2*
a^3*tan(x/2) + 10*a^3*tan(x/2)^2 + 20*a^3*tan(x/2)^3 + 20*a^3*tan(x/2)^4 + 10*a^3*tan(x/2)^5 + 2*a^3*tan(x/2)^
6) - (3*log(tan(x/2)))/a^3

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