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

Optimal. Leaf size=53 \[ \frac {\cot (c+d x)}{a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]

[Out]

-1/2*arctanh(cos(d*x+c))/a/d+cot(d*x+c)/a/d-1/2*cot(d*x+c)*csc(d*x+c)/a/d

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Rubi [A]  time = 0.10, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2839, 3768, 3770, 3767, 8} \[ \frac {\cot (c+d x)}{a d}-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Cos[c + d*x]]/(2*a*d) + Cot[c + d*x]/(a*d) - (Cot[c + d*x]*Csc[c + d*x])/(2*a*d)

Rule 8

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

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 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 {\cot ^2(c+d x) \csc (c+d x)}{a+a \sin (c+d x)} \, dx &=-\frac {\int \csc ^2(c+d x) \, dx}{a}+\frac {\int \csc ^3(c+d x) \, dx}{a}\\ &=-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\int \csc (c+d x) \, dx}{2 a}+\frac {\operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac {\cot (c+d x)}{a d}-\frac {\cot (c+d x) \csc (c+d x)}{2 a d}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 94, normalized size = 1.77 \[ \frac {\left (\csc \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (\sin (2 (c+d x))-\cos (c+d x)+\sin ^2(c+d x) \left (\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{8 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Csc[(c + d*x)/2] + Sec[(c + d*x)/2])^2*(-Cos[c + d*x] + (-Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]])*Sin
[c + d*x]^2 + Sin[2*(c + d*x)]))/(8*a*d*(1 + Sin[c + d*x]))

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fricas [A]  time = 0.50, size = 88, normalized size = 1.66 \[ -\frac {{\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (\cos \left (d x + c\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 4 \, \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right )^{2} - a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((cos(d*x + c)^2 - 1)*log(1/2*cos(d*x + c) + 1/2) - (cos(d*x + c)^2 - 1)*log(-1/2*cos(d*x + c) + 1/2) + 4
*cos(d*x + c)*sin(d*x + c) - 2*cos(d*x + c))/(a*d*cos(d*x + c)^2 - a*d)

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giac [A]  time = 0.20, size = 94, normalized size = 1.77 \[ \frac {\frac {4 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} - \frac {6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*log(abs(tan(1/2*d*x + 1/2*c)))/a + (a*tan(1/2*d*x + 1/2*c)^2 - 4*a*tan(1/2*d*x + 1/2*c))/a^2 - (6*tan(1
/2*d*x + 1/2*c)^2 - 4*tan(1/2*d*x + 1/2*c) + 1)/(a*tan(1/2*d*x + 1/2*c)^2))/d

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maple [A]  time = 0.42, size = 94, normalized size = 1.77 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {1}{2 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}-\frac {1}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^3/(a+a*sin(d*x+c)),x)

[Out]

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

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maxima [B]  time = 0.34, size = 115, normalized size = 2.17 \[ -\frac {\frac {\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a} - \frac {4 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {{\left (\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{a \sin \left (d x + c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.65, size = 87, normalized size = 1.64 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {1}{2}\right )}{4\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(sin(c + d*x)^3*(a + a*sin(c + d*x))),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cos ^{2}{\left (c + d x \right )} \csc ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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