3.2 \(\int \frac {\csc ^4(x)}{a+a \csc (x)} \, dx\)

Optimal. Leaf size=44 \[ \frac {2 \cot (x)}{a}-\frac {3 \tanh ^{-1}(\cos (x))}{2 a}+\frac {\cot (x) \csc ^2(x)}{a \csc (x)+a}-\frac {3 \cot (x) \csc (x)}{2 a} \]

[Out]

-3/2*arctanh(cos(x))/a+2*cot(x)/a-3/2*cot(x)*csc(x)/a+cot(x)*csc(x)^2/(a+a*csc(x))

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Rubi [A]  time = 0.07, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3818, 3787, 3767, 8, 3768, 3770} \[ \frac {2 \cot (x)}{a}-\frac {3 \tanh ^{-1}(\cos (x))}{2 a}+\frac {\cot (x) \csc ^2(x)}{a \csc (x)+a}-\frac {3 \cot (x) \csc (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^4/(a + a*Csc[x]),x]

[Out]

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

Rule 8

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3818

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(d^2*Cot[e
+ f*x]*(d*Csc[e + f*x])^(n - 2))/(f*(a + b*Csc[e + f*x])), x] - Dist[d^2/(a*b), Int[(d*Csc[e + f*x])^(n - 2)*(
b*(n - 2) - a*(n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1]

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 83, normalized size = 1.89 \[ \frac {-4 \tan \left (\frac {x}{2}\right )+4 \cot \left (\frac {x}{2}\right )-\csc ^2\left (\frac {x}{2}\right )+\sec ^2\left (\frac {x}{2}\right )+12 \log \left (\sin \left (\frac {x}{2}\right )\right )-12 \log \left (\cos \left (\frac {x}{2}\right )\right )-\frac {16 \sin \left (\frac {x}{2}\right )}{\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )}}{8 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^4/(a + a*Csc[x]),x]

[Out]

(4*Cot[x/2] - Csc[x/2]^2 - 12*Log[Cos[x/2]] + 12*Log[Sin[x/2]] + Sec[x/2]^2 - (16*Sin[x/2])/(Cos[x/2] + Sin[x/
2]) - 4*Tan[x/2])/(8*a)

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fricas [B]  time = 0.47, size = 134, normalized size = 3.05 \[ \frac {8 \, \cos \relax (x)^{3} + 6 \, \cos \relax (x)^{2} - 3 \, {\left (\cos \relax (x)^{3} + \cos \relax (x)^{2} + {\left (\cos \relax (x)^{2} - 1\right )} \sin \relax (x) - \cos \relax (x) - 1\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + 3 \, {\left (\cos \relax (x)^{3} + \cos \relax (x)^{2} + {\left (\cos \relax (x)^{2} - 1\right )} \sin \relax (x) - \cos \relax (x) - 1\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - 2 \, {\left (4 \, \cos \relax (x)^{2} + \cos \relax (x) - 2\right )} \sin \relax (x) - 6 \, \cos \relax (x) - 4}{4 \, {\left (a \cos \relax (x)^{3} + a \cos \relax (x)^{2} - a \cos \relax (x) + {\left (a \cos \relax (x)^{2} - a\right )} \sin \relax (x) - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*cos(x)^3 + 6*cos(x)^2 - 3*(cos(x)^3 + cos(x)^2 + (cos(x)^2 - 1)*sin(x) - cos(x) - 1)*log(1/2*cos(x) + 1
/2) + 3*(cos(x)^3 + cos(x)^2 + (cos(x)^2 - 1)*sin(x) - cos(x) - 1)*log(-1/2*cos(x) + 1/2) - 2*(4*cos(x)^2 + co
s(x) - 2)*sin(x) - 6*cos(x) - 4)/(a*cos(x)^3 + a*cos(x)^2 - a*cos(x) + (a*cos(x)^2 - a)*sin(x) - a)

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giac [A]  time = 0.37, size = 73, normalized size = 1.66 \[ \frac {3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, a} + \frac {a \tan \left (\frac {1}{2} \, x\right )^{2} - 4 \, a \tan \left (\frac {1}{2} \, x\right )}{8 \, a^{2}} + \frac {2}{a {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}} - \frac {18 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 4 \, \tan \left (\frac {1}{2} \, x\right ) + 1}{8 \, a \tan \left (\frac {1}{2} \, x\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*log(abs(tan(1/2*x)))/a + 1/8*(a*tan(1/2*x)^2 - 4*a*tan(1/2*x))/a^2 + 2/(a*(tan(1/2*x) + 1)) - 1/8*(18*tan(
1/2*x)^2 - 4*tan(1/2*x) + 1)/(a*tan(1/2*x)^2)

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maple [A]  time = 0.27, size = 67, normalized size = 1.52 \[ \frac {\tan ^{2}\left (\frac {x}{2}\right )}{8 a}-\frac {\tan \left (\frac {x}{2}\right )}{2 a}-\frac {1}{8 a \tan \left (\frac {x}{2}\right )^{2}}+\frac {1}{2 a \tan \left (\frac {x}{2}\right )}+\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 a}+\frac {2}{a \left (\tan \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/a*tan(1/2*x)^2-1/2/a*tan(1/2*x)-1/8/a/tan(1/2*x)^2+1/2/a/tan(1/2*x)+3/2/a*ln(tan(1/2*x))+2/a/(tan(1/2*x)+1
)

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maxima [B]  time = 0.33, size = 97, normalized size = 2.20 \[ -\frac {\frac {4 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}}{8 \, a} + \frac {\frac {3 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {20 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - 1}{8 \, {\left (\frac {a \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {a \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}\right )}} + \frac {3 \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*sin(x)/(cos(x) + 1) - sin(x)^2/(cos(x) + 1)^2)/a + 1/8*(3*sin(x)/(cos(x) + 1) + 20*sin(x)^2/(cos(x) +
1)^2 - 1)/(a*sin(x)^2/(cos(x) + 1)^2 + a*sin(x)^3/(cos(x) + 1)^3) + 3/2*log(sin(x)/(cos(x) + 1))/a

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mupad [B]  time = 0.26, size = 69, normalized size = 1.57 \[ \frac {10\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\frac {3\,\mathrm {tan}\left (\frac {x}{2}\right )}{2}-\frac {1}{2}}{4\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+4\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{2\,a}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{8\,a}+\frac {3\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^4*(a + a/sin(x))),x)

[Out]

((3*tan(x/2))/2 + 10*tan(x/2)^2 - 1/2)/(4*a*tan(x/2)^2 + 4*a*tan(x/2)^3) - tan(x/2)/(2*a) + tan(x/2)^2/(8*a) +
(3*log(tan(x/2)))/(2*a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)**4/(csc(x) + 1), x)/a

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