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

Optimal. Leaf size=51 \[ -\frac {2 \cot (c+d x)}{a d}+\frac {\tanh ^{-1}(\cos (c+d x))}{a d}+\frac {\cot (c+d x)}{d (a \sin (c+d x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2768, 2748, 3767, 8, 3770} \[ -\frac {2 \cot (c+d x)}{a d}+\frac {\tanh ^{-1}(\cos (c+d x))}{a d}+\frac {\cot (c+d x)}{d (a \sin (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

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 2768

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b
^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x])), x] + Dist[d/(a*(b*c - a*
d)), Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 0] && (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 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 ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\cot (c+d x)}{d (a+a \sin (c+d x))}-\frac {\int \csc ^2(c+d x) (-2 a+a \sin (c+d x)) \, dx}{a^2}\\ &=\frac {\cot (c+d x)}{d (a+a \sin (c+d x))}-\frac {\int \csc (c+d x) \, dx}{a}+\frac {2 \int \csc ^2(c+d x) \, dx}{a}\\ &=\frac {\tanh ^{-1}(\cos (c+d x))}{a d}+\frac {\cot (c+d x)}{d (a+a \sin (c+d x))}-\frac {2 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a d}\\ &=\frac {\tanh ^{-1}(\cos (c+d x))}{a d}-\frac {2 \cot (c+d x)}{a d}+\frac {\cot (c+d x)}{d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 57, normalized size = 1.12 \[ \frac {\sec (c+d x) \left (2 \sin (c+d x)-\csc (c+d x)+\sqrt {\cos ^2(c+d x)} \tanh ^{-1}\left (\sqrt {\cos ^2(c+d x)}\right )-1\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 156, normalized size = 3.06 \[ \frac {4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) - 2}{2 \, {\left (a d \cos \left (d x + c\right )^{2} - a d - {\left (a d \cos \left (d x + c\right ) + a d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.38, size = 88, normalized size = 1.73 \[ -\frac {\frac {2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 77, normalized size = 1.51 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {1}{2 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.36, size = 112, normalized size = 2.20 \[ -\frac {\frac {\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1}{\frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac {2 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((5*sin(d*x + c)/(cos(d*x + c) + 1) + 1)/(a*sin(d*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*x +
c) + 1)^2) + 2*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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mupad [B]  time = 1.27, size = 83, normalized size = 1.63 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a\,d}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1}{d\,\left (2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^2*(a + a*sin(c + d*x))),x)

[Out]

tan(c/2 + (d*x)/2)/(2*a*d) - log(tan(c/2 + (d*x)/2))/(a*d) - (5*tan(c/2 + (d*x)/2) + 1)/(d*(2*a*tan(c/2 + (d*x
)/2) + 2*a*tan(c/2 + (d*x)/2)^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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