3.20 \(\int \frac {\csc ^4(x)}{(a+a \sin (x))^2} \, dx\)

Optimal. Leaf size=65 \[ -\frac {\cot ^3(x)}{3 a^2}-\frac {4 \cot (x)}{a^2}-\frac {13 \cos (x)}{3 a^2 (\sin (x)+1)}-\frac {\cos (x)}{3 a^2 (\sin (x)+1)^2}+\frac {5 \tanh ^{-1}(\cos (x))}{a^2}+\frac {\cot (x) \csc (x)}{a^2} \]

[Out]

5*arctanh(cos(x))/a^2-4*cot(x)/a^2-1/3*cot(x)^3/a^2+cot(x)*csc(x)/a^2-1/3*cos(x)/a^2/(1+sin(x))^2-13/3*cos(x)/
a^2/(1+sin(x))

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Rubi [A]  time = 0.15, antiderivative size = 71, normalized size of antiderivative = 1.09, 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 = {2766, 2978, 2748, 3767, 3768, 3770} \[ -\frac {4 \cot ^3(x)}{a^2}-\frac {12 \cot (x)}{a^2}+\frac {5 \tanh ^{-1}(\cos (x))}{a^2}+\frac {5 \cot (x) \csc (x)}{a^2}+\frac {10 \cot (x) \csc ^2(x)}{3 a^2 (\sin (x)+1)}+\frac {\cot (x) \csc ^2(x)}{3 (a \sin (x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^4/(a + a*Sin[x])^2,x]

[Out]

(5*ArcTanh[Cos[x]])/a^2 - (12*Cot[x])/a^2 - (4*Cot[x]^3)/a^2 + (5*Cot[x]*Csc[x])/a^2 + (10*Cot[x]*Csc[x]^2)/(3
*a^2*(1 + Sin[x])) + (Cot[x]*Csc[x]^2)/(3*(a + a*Sin[x])^2)

Rule 2748

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 2766

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 2978

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 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 {\csc ^4(x)}{(a+a \sin (x))^2} \, dx &=\frac {\cot (x) \csc ^2(x)}{3 (a+a \sin (x))^2}+\frac {\int \frac {\csc ^4(x) (6 a-4 a \sin (x))}{a+a \sin (x)} \, dx}{3 a^2}\\ &=\frac {10 \cot (x) \csc ^2(x)}{3 a^2 (1+\sin (x))}+\frac {\cot (x) \csc ^2(x)}{3 (a+a \sin (x))^2}+\frac {\int \csc ^4(x) \left (36 a^2-30 a^2 \sin (x)\right ) \, dx}{3 a^4}\\ &=\frac {10 \cot (x) \csc ^2(x)}{3 a^2 (1+\sin (x))}+\frac {\cot (x) \csc ^2(x)}{3 (a+a \sin (x))^2}-\frac {10 \int \csc ^3(x) \, dx}{a^2}+\frac {12 \int \csc ^4(x) \, dx}{a^2}\\ &=\frac {5 \cot (x) \csc (x)}{a^2}+\frac {10 \cot (x) \csc ^2(x)}{3 a^2 (1+\sin (x))}+\frac {\cot (x) \csc ^2(x)}{3 (a+a \sin (x))^2}-\frac {5 \int \csc (x) \, dx}{a^2}-\frac {12 \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (x)\right )}{a^2}\\ &=\frac {5 \tanh ^{-1}(\cos (x))}{a^2}-\frac {12 \cot (x)}{a^2}-\frac {4 \cot ^3(x)}{a^2}+\frac {5 \cot (x) \csc (x)}{a^2}+\frac {10 \cot (x) \csc ^2(x)}{3 a^2 (1+\sin (x))}+\frac {\cot (x) \csc ^2(x)}{3 (a+a \sin (x))^2}\\ \end {align*}

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Mathematica [B]  time = 3.59, size = 238, normalized size = 3.66 \[ \frac {\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right ) \left (16 \sin \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right ) \left (\tan \left (\frac {x}{2}\right )+1\right )^3+208 \sin \left (\frac {x}{2}\right ) \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^2-8 \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )-6 \cos \left (\frac {x}{2}\right ) \left (\tan \left (\frac {x}{2}\right )+1\right )^3-\cos \left (\frac {x}{2}\right ) \left (\cot \left (\frac {x}{2}\right )+1\right )^3+6 \sin \left (\frac {x}{2}\right ) \left (\cot \left (\frac {x}{2}\right )+1\right )^3+120 \log \left (\cos \left (\frac {x}{2}\right )\right ) \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^3-120 \log \left (\sin \left (\frac {x}{2}\right )\right ) \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^3+44 \tan \left (\frac {x}{2}\right ) \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^3-44 \cot \left (\frac {x}{2}\right ) \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^3\right )}{24 a^2 (\sin (x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^4/(a + a*Sin[x])^2,x]

[Out]

((Cos[x/2] + Sin[x/2])*(-(Cos[x/2]*(1 + Cot[x/2])^3) + 16*Sin[x/2] + 6*(1 + Cot[x/2])^3*Sin[x/2] - 8*(Cos[x/2]
+ Sin[x/2]) + 208*Sin[x/2]*(Cos[x/2] + Sin[x/2])^2 - 44*Cot[x/2]*(Cos[x/2] + Sin[x/2])^3 + 120*Log[Cos[x/2]]*
(Cos[x/2] + Sin[x/2])^3 - 120*Log[Sin[x/2]]*(Cos[x/2] + Sin[x/2])^3 + 44*(Cos[x/2] + Sin[x/2])^3*Tan[x/2] - 6*
Cos[x/2]*(1 + Tan[x/2])^3 + Sin[x/2]*(1 + Tan[x/2])^3))/(24*a^2*(1 + Sin[x])^2)

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fricas [B]  time = 0.49, size = 266, normalized size = 4.09 \[ -\frac {48 \, \cos \relax (x)^{5} - 18 \, \cos \relax (x)^{4} - 108 \, \cos \relax (x)^{3} + 22 \, \cos \relax (x)^{2} - 15 \, {\left (\cos \relax (x)^{5} + 2 \, \cos \relax (x)^{4} - 2 \, \cos \relax (x)^{3} - 4 \, \cos \relax (x)^{2} + {\left (\cos \relax (x)^{4} - \cos \relax (x)^{3} - 3 \, \cos \relax (x)^{2} + \cos \relax (x) + 2\right )} \sin \relax (x) + \cos \relax (x) + 2\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + 15 \, {\left (\cos \relax (x)^{5} + 2 \, \cos \relax (x)^{4} - 2 \, \cos \relax (x)^{3} - 4 \, \cos \relax (x)^{2} + {\left (\cos \relax (x)^{4} - \cos \relax (x)^{3} - 3 \, \cos \relax (x)^{2} + \cos \relax (x) + 2\right )} \sin \relax (x) + \cos \relax (x) + 2\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - 2 \, {\left (24 \, \cos \relax (x)^{4} + 33 \, \cos \relax (x)^{3} - 21 \, \cos \relax (x)^{2} - 32 \, \cos \relax (x) - 1\right )} \sin \relax (x) + 62 \, \cos \relax (x) - 2}{6 \, {\left (a^{2} \cos \relax (x)^{5} + 2 \, a^{2} \cos \relax (x)^{4} - 2 \, a^{2} \cos \relax (x)^{3} - 4 \, a^{2} \cos \relax (x)^{2} + a^{2} \cos \relax (x) + 2 \, a^{2} + {\left (a^{2} \cos \relax (x)^{4} - a^{2} \cos \relax (x)^{3} - 3 \, a^{2} \cos \relax (x)^{2} + a^{2} \cos \relax (x) + 2 \, a^{2}\right )} \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(48*cos(x)^5 - 18*cos(x)^4 - 108*cos(x)^3 + 22*cos(x)^2 - 15*(cos(x)^5 + 2*cos(x)^4 - 2*cos(x)^3 - 4*cos(
x)^2 + (cos(x)^4 - cos(x)^3 - 3*cos(x)^2 + cos(x) + 2)*sin(x) + cos(x) + 2)*log(1/2*cos(x) + 1/2) + 15*(cos(x)
^5 + 2*cos(x)^4 - 2*cos(x)^3 - 4*cos(x)^2 + (cos(x)^4 - cos(x)^3 - 3*cos(x)^2 + cos(x) + 2)*sin(x) + cos(x) +
2)*log(-1/2*cos(x) + 1/2) - 2*(24*cos(x)^4 + 33*cos(x)^3 - 21*cos(x)^2 - 32*cos(x) - 1)*sin(x) + 62*cos(x) - 2
)/(a^2*cos(x)^5 + 2*a^2*cos(x)^4 - 2*a^2*cos(x)^3 - 4*a^2*cos(x)^2 + a^2*cos(x) + 2*a^2 + (a^2*cos(x)^4 - a^2*
cos(x)^3 - 3*a^2*cos(x)^2 + a^2*cos(x) + 2*a^2)*sin(x))

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giac [A]  time = 0.28, size = 114, normalized size = 1.75 \[ -\frac {5 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{2}} + \frac {110 \, \tan \left (\frac {1}{2} \, x\right )^{6} + 45 \, \tan \left (\frac {1}{2} \, x\right )^{5} - 231 \, \tan \left (\frac {1}{2} \, x\right )^{4} - 232 \, \tan \left (\frac {1}{2} \, x\right )^{3} - 30 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, \tan \left (\frac {1}{2} \, x\right ) - 1}{24 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + \tan \left (\frac {1}{2} \, x\right )\right )}^{3} a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, x\right )^{3} - 6 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{2} + 45 \, a^{4} \tan \left (\frac {1}{2} \, x\right )}{24 \, a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*log(abs(tan(1/2*x)))/a^2 + 1/24*(110*tan(1/2*x)^6 + 45*tan(1/2*x)^5 - 231*tan(1/2*x)^4 - 232*tan(1/2*x)^3 -
30*tan(1/2*x)^2 + 3*tan(1/2*x) - 1)/((tan(1/2*x)^2 + tan(1/2*x))^3*a^2) + 1/24*(a^4*tan(1/2*x)^3 - 6*a^4*tan(
1/2*x)^2 + 45*a^4*tan(1/2*x))/a^6

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maple [A]  time = 0.14, size = 115, normalized size = 1.77 \[ \frac {\tan ^{3}\left (\frac {x}{2}\right )}{24 a^{2}}-\frac {\tan ^{2}\left (\frac {x}{2}\right )}{4 a^{2}}+\frac {15 \tan \left (\frac {x}{2}\right )}{8 a^{2}}-\frac {1}{24 a^{2} \tan \left (\frac {x}{2}\right )^{3}}+\frac {1}{4 a^{2} \tan \left (\frac {x}{2}\right )^{2}}-\frac {15}{8 a^{2} \tan \left (\frac {x}{2}\right )}-\frac {5 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}-\frac {4}{3 a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {2}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {10}{a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^4/(a+a*sin(x))^2,x)

[Out]

1/24/a^2*tan(1/2*x)^3-1/4/a^2*tan(1/2*x)^2+15/8/a^2*tan(1/2*x)-1/24/a^2/tan(1/2*x)^3+1/4/a^2/tan(1/2*x)^2-15/8
/a^2/tan(1/2*x)-5/a^2*ln(tan(1/2*x))-4/3/a^2/(tan(1/2*x)+1)^3+2/a^2/(tan(1/2*x)+1)^2-10/a^2/(tan(1/2*x)+1)

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maxima [B]  time = 1.04, size = 178, normalized size = 2.74 \[ \frac {\frac {3 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {30 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {342 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {561 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} - \frac {285 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} - 1}{24 \, {\left (\frac {a^{2} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {3 \, a^{2} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {3 \, a^{2} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {a^{2} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}}\right )}} + \frac {\frac {45 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {6 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {\sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}}{24 \, a^{2}} - \frac {5 \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*sin(x)/(cos(x) + 1) - 30*sin(x)^2/(cos(x) + 1)^2 - 342*sin(x)^3/(cos(x) + 1)^3 - 561*sin(x)^4/(cos(x)
+ 1)^4 - 285*sin(x)^5/(cos(x) + 1)^5 - 1)/(a^2*sin(x)^3/(cos(x) + 1)^3 + 3*a^2*sin(x)^4/(cos(x) + 1)^4 + 3*a^2
*sin(x)^5/(cos(x) + 1)^5 + a^2*sin(x)^6/(cos(x) + 1)^6) + 1/24*(45*sin(x)/(cos(x) + 1) - 6*sin(x)^2/(cos(x) +
1)^2 + sin(x)^3/(cos(x) + 1)^3)/a^2 - 5*log(sin(x)/(cos(x) + 1))/a^2

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mupad [B]  time = 6.40, size = 101, normalized size = 1.55 \[ \frac {15\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a^2}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{4\,a^2}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{24\,a^2}-\frac {5\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^2}-\frac {\frac {95\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{8}+\frac {187\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{8}+\frac {57\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{4}+\frac {5\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{4}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{8}+\frac {1}{24}}{a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^4*(a + a*sin(x))^2),x)

[Out]

(15*tan(x/2))/(8*a^2) - tan(x/2)^2/(4*a^2) + tan(x/2)^3/(24*a^2) - (5*log(tan(x/2)))/a^2 - ((5*tan(x/2)^2)/4 -
tan(x/2)/8 + (57*tan(x/2)^3)/4 + (187*tan(x/2)^4)/8 + (95*tan(x/2)^5)/8 + 1/24)/(a^2*tan(x/2)^3*(tan(x/2) + 1
)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\csc ^{4}{\relax (x )}}{\sin ^{2}{\relax (x )} + 2 \sin {\relax (x )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)**4/(sin(x)**2 + 2*sin(x) + 1), x)/a**2

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