3.212 \(\int \frac {\csc ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=82 \[ \frac {2 \cot (c+d x)}{a d}-\frac {3 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\cot (c+d x) \csc (c+d x)}{d (a \sin (c+d x)+a)} \]

[Out]

-3/2*arctanh(cos(d*x+c))/a/d+2*cot(d*x+c)/a/d-3/2*cot(d*x+c)*csc(d*x+c)/a/d+cot(d*x+c)*csc(d*x+c)/d/(a+a*sin(d
*x+c))

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Rubi [A]  time = 0.09, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2768, 2748, 3768, 3770, 3767, 8} \[ \frac {2 \cot (c+d x)}{a d}-\frac {3 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\cot (c+d x) \csc (c+d x)}{d (a \sin (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^3/(a + a*Sin[c + d*x]),x]

[Out]

(-3*ArcTanh[Cos[c + d*x]])/(2*a*d) + (2*Cot[c + d*x])/(a*d) - (3*Cot[c + d*x]*Csc[c + d*x])/(2*a*d) + (Cot[c +
d*x]*Csc[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 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 ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\cot (c+d x) \csc (c+d x)}{d (a+a \sin (c+d x))}-\frac {\int \csc ^3(c+d x) (-3 a+2 a \sin (c+d x)) \, dx}{a^2}\\ &=\frac {\cot (c+d x) \csc (c+d x)}{d (a+a \sin (c+d x))}-\frac {2 \int \csc ^2(c+d x) \, dx}{a}+\frac {3 \int \csc ^3(c+d x) \, dx}{a}\\ &=-\frac {3 \cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\cot (c+d x) \csc (c+d x)}{d (a+a \sin (c+d x))}+\frac {3 \int \csc (c+d x) \, dx}{2 a}+\frac {2 \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a d}\\ &=-\frac {3 \tanh ^{-1}(\cos (c+d x))}{2 a d}+\frac {2 \cot (c+d x)}{a d}-\frac {3 \cot (c+d x) \csc (c+d x)}{2 a d}+\frac {\cot (c+d x) \csc (c+d x)}{d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 85, normalized size = 1.04 \[ -\frac {4 \tan (c+d x)-4 \csc (2 (c+d x))-3 \sec (c+d x)+\csc ^2(c+d x) \sec (c+d x)+3 \sqrt {\cos ^2(c+d x)} \sec (c+d x) \tanh ^{-1}\left (\sqrt {\cos ^2(c+d x)}\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^3/(a + a*Sin[c + d*x]),x]

[Out]

-1/2*(-4*Csc[2*(c + d*x)] - 3*Sec[c + d*x] + 3*ArcTanh[Sqrt[Cos[c + d*x]^2]]*Sqrt[Cos[c + d*x]^2]*Sec[c + d*x]
+ Csc[c + d*x]^2*Sec[c + d*x] + 4*Tan[c + d*x])/(a*d)

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fricas [B]  time = 0.49, size = 232, normalized size = 2.83 \[ \frac {8 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (4 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - 6 \, \cos \left (d x + c\right ) - 4}{4 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d + {\left (a d \cos \left (d x + c\right )^{2} - a d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.66, size = 112, normalized size = 1.37 \[ \frac {\frac {12 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2}} + \frac {16}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}} - \frac {18 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(12*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 + 16/(a*(
tan(1/2*d*x + 1/2*c) + 1)) - (18*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 = 0.11, size = 115, normalized size = 1.40 \[ \frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {1}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {1}{2 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}+\frac {2}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^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)-1/8/a/d/tan(1/2*d*x+1/2*c)^2+1/2/a/d/tan(1/2*d*x+1/2*c
)+3/2/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 = 0.57, size = 157, normalized size = 1.91 \[ -\frac {\frac {\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a} - \frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}{\frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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 - (3*sin(d*x + c)/(cos(d*x +
c) + 1) + 20*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)/(a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)
^3/(cos(d*x + c) + 1)^3) - 12*log(sin(d*x + c)/(cos(d*x + c) + 1))/a)/d

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mupad [B]  time = 1.39, size = 116, normalized size = 1.41 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}+\frac {3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a\,d}+\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {1}{2}}{d\,\left (4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^3*(a + a*sin(c + d*x))),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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