3.1.28 \(\int \frac {\csc (x)}{(a+a \sin (x))^3} \, dx\) [28]

Optimal. Leaf size=58 \[ -\frac {\tanh ^{-1}(\cos (x))}{a^3}+\frac {\cos (x)}{5 (a+a \sin (x))^3}+\frac {7 \cos (x)}{15 a (a+a \sin (x))^2}+\frac {22 \cos (x)}{15 \left (a^3+a^3 \sin (x)\right )} \]

[Out]

-arctanh(cos(x))/a^3+1/5*cos(x)/(a+a*sin(x))^3+7/15*cos(x)/a/(a+a*sin(x))^2+22/15*cos(x)/(a^3+a^3*sin(x))

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Rubi [A]
time = 0.11, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2845, 3057, 12, 3855} \begin {gather*} \frac {22 \cos (x)}{15 \left (a^3 \sin (x)+a^3\right )}-\frac {\tanh ^{-1}(\cos (x))}{a^3}+\frac {7 \cos (x)}{15 a (a \sin (x)+a)^2}+\frac {\cos (x)}{5 (a \sin (x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[x]/(a + a*Sin[x])^3,x]

[Out]

-(ArcTanh[Cos[x]]/a^3) + Cos[x]/(5*(a + a*Sin[x])^3) + (7*Cos[x])/(15*a*(a + a*Sin[x])^2) + (22*Cos[x])/(15*(a
^3 + a^3*Sin[x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 (x)}{(a+a \sin (x))^3} \, dx &=\frac {\cos (x)}{5 (a+a \sin (x))^3}+\frac {\int \frac {\csc (x) (5 a-2 a \sin (x))}{(a+a \sin (x))^2} \, dx}{5 a^2}\\ &=\frac {\cos (x)}{5 (a+a \sin (x))^3}+\frac {7 \cos (x)}{15 a (a+a \sin (x))^2}+\frac {\int \frac {\csc (x) \left (15 a^2-7 a^2 \sin (x)\right )}{a+a \sin (x)} \, dx}{15 a^4}\\ &=\frac {\cos (x)}{5 (a+a \sin (x))^3}+\frac {7 \cos (x)}{15 a (a+a \sin (x))^2}+\frac {22 \cos (x)}{15 \left (a^3+a^3 \sin (x)\right )}+\frac {\int 15 a^3 \csc (x) \, dx}{15 a^6}\\ &=\frac {\cos (x)}{5 (a+a \sin (x))^3}+\frac {7 \cos (x)}{15 a (a+a \sin (x))^2}+\frac {22 \cos (x)}{15 \left (a^3+a^3 \sin (x)\right )}+\frac {\int \csc (x) \, dx}{a^3}\\ &=-\frac {\tanh ^{-1}(\cos (x))}{a^3}+\frac {\cos (x)}{5 (a+a \sin (x))^3}+\frac {7 \cos (x)}{15 a (a+a \sin (x))^2}+\frac {22 \cos (x)}{15 \left (a^3+a^3 \sin (x)\right )}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(160\) vs. \(2(58)=116\).
time = 0.05, size = 160, normalized size = 2.76 \begin {gather*} \frac {\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right ) \left (-6 \sin \left (\frac {x}{2}\right )+3 \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )-14 \sin \left (\frac {x}{2}\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2+7 \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^3-44 \sin \left (\frac {x}{2}\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^4-15 \log \left (\cos \left (\frac {x}{2}\right )\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^5+15 \log \left (\sin \left (\frac {x}{2}\right )\right ) \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^5\right )}{15 (a+a \sin (x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]/(a + a*Sin[x])^3,x]

[Out]

((Cos[x/2] + Sin[x/2])*(-6*Sin[x/2] + 3*(Cos[x/2] + Sin[x/2]) - 14*Sin[x/2]*(Cos[x/2] + Sin[x/2])^2 + 7*(Cos[x
/2] + Sin[x/2])^3 - 44*Sin[x/2]*(Cos[x/2] + Sin[x/2])^4 - 15*Log[Cos[x/2]]*(Cos[x/2] + Sin[x/2])^5 + 15*Log[Si
n[x/2]]*(Cos[x/2] + Sin[x/2])^5))/(15*(a + a*Sin[x])^3)

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Maple [A]
time = 0.18, size = 61, normalized size = 1.05

method	result	size
default	\(\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )+\frac {8}{5 \left (\tan \left (\frac {x}{2}\right )+1\right )^{5}}-\frac {4}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {20}{3 \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {6}{\left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {6}{\tan \left (\frac {x}{2}\right )+1}}{a^{3}}\)	\(61\)
norman	\(\frac {\frac {6 \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{a}+\frac {64}{15 a}+\frac {18 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{a}+\frac {74 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{3 a}+\frac {46 \tan \left (\frac {x}{2}\right )}{3 a}}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{5}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{3}}\)	\(71\)
risch	\(\frac {10 i {\mathrm e}^{3 i x}+2 \,{\mathrm e}^{4 i x}-\frac {38 i {\mathrm e}^{i x}}{3}-\frac {58 \,{\mathrm e}^{2 i x}}{3}+\frac {44}{15}}{\left ({\mathrm e}^{i x}+i\right )^{5} a^{3}}-\frac {\ln \left ({\mathrm e}^{i x}+1\right )}{a^{3}}+\frac {\ln \left ({\mathrm e}^{i x}-1\right )}{a^{3}}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^3*(ln(tan(1/2*x))+8/5/(tan(1/2*x)+1)^5-4/(tan(1/2*x)+1)^4+20/3/(tan(1/2*x)+1)^3-6/(tan(1/2*x)+1)^2+6/(tan(
1/2*x)+1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (52) = 104\).
time = 0.42, size = 143, normalized size = 2.47 \begin {gather*} \frac {2 \, {\left (\frac {115 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {185 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {135 \, \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {45 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + 32\right )}}{15 \, {\left (a^{3} + \frac {5 \, a^{3} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {10 \, 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 {5 \, a^{3} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}}\right )}} + \frac {\log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(115*sin(x)/(cos(x) + 1) + 185*sin(x)^2/(cos(x) + 1)^2 + 135*sin(x)^3/(cos(x) + 1)^3 + 45*sin(x)^4/(cos(x
) + 1)^4 + 32)/(a^3 + 5*a^3*sin(x)/(cos(x) + 1) + 10*a^3*sin(x)^2/(cos(x) + 1)^2 + 10*a^3*sin(x)^3/(cos(x) + 1
)^3 + 5*a^3*sin(x)^4/(cos(x) + 1)^4 + a^3*sin(x)^5/(cos(x) + 1)^5) + log(sin(x)/(cos(x) + 1))/a^3

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (52) = 104\).
time = 0.34, size = 168, normalized size = 2.90 \begin {gather*} \frac {44 \, \cos \left (x\right )^{3} - 58 \, \cos \left (x\right )^{2} - 15 \, {\left (\cos \left (x\right )^{3} + 3 \, \cos \left (x\right )^{2} + {\left (\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 )^{3} + 3 \, \cos \left (x\right )^{2} + {\left (\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 (22 \, \cos \left (x\right )^{2} + 51 \, \cos \left (x\right ) - 3\right )} \sin \left (x\right ) - 108 \, \cos \left (x\right ) - 6}{30 \, {\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} + {\left (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)/(a+a*sin(x))^3,x, algorithm="fricas")

[Out]

1/30*(44*cos(x)^3 - 58*cos(x)^2 - 15*(cos(x)^3 + 3*cos(x)^2 + (cos(x)^2 - 2*cos(x) - 4)*sin(x) - 2*cos(x) - 4)
*log(1/2*cos(x) + 1/2) + 15*(cos(x)^3 + 3*cos(x)^2 + (cos(x)^2 - 2*cos(x) - 4)*sin(x) - 2*cos(x) - 4)*log(-1/2
*cos(x) + 1/2) - 2*(22*cos(x)^2 + 51*cos(x) - 3)*sin(x) - 108*cos(x) - 6)/(a^3*cos(x)^3 + 3*a^3*cos(x)^2 - 2*a
^3*cos(x) - 4*a^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 {\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)/(a+a*sin(x))**3,x)

[Out]

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

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Giac [A]
time = 0.62, size = 56, normalized size = 0.97 \begin {gather*} \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{3}} + \frac {2 \, {\left (45 \, \tan \left (\frac {1}{2} \, x\right )^{4} + 135 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 185 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 115 \, \tan \left (\frac {1}{2} \, x\right ) + 32\right )}}{15 \, 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)/(a+a*sin(x))^3,x, algorithm="giac")

[Out]

log(abs(tan(1/2*x)))/a^3 + 2/15*(45*tan(1/2*x)^4 + 135*tan(1/2*x)^3 + 185*tan(1/2*x)^2 + 115*tan(1/2*x) + 32)/
(a^3*(tan(1/2*x) + 1)^5)

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Mupad [B]
time = 6.65, size = 54, normalized size = 0.93 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^3}+\frac {6\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+18\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\frac {74\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{3}+\frac {46\,\mathrm {tan}\left (\frac {x}{2}\right )}{3}+\frac {64}{15}}{a^3\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)*(a + a*sin(x))^3),x)

[Out]

log(tan(x/2))/a^3 + ((46*tan(x/2))/3 + (74*tan(x/2)^2)/3 + 18*tan(x/2)^3 + 6*tan(x/2)^4 + 64/15)/(a^3*(tan(x/2
) + 1)^5)

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