3.220 \(\int \frac {\text {csch}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=87 \[ \frac {2 i \coth (c+d x)}{a d}+\frac {3 \tanh ^{-1}(\cosh (c+d x))}{2 a d}-\frac {3 \coth (c+d x) \text {csch}(c+d x)}{2 a d}+\frac {\coth (c+d x) \text {csch}(c+d x)}{d (a+i a \sinh (c+d x))} \]

[Out]

3/2*arctanh(cosh(d*x+c))/a/d+2*I*coth(d*x+c)/a/d-3/2*coth(d*x+c)*csch(d*x+c)/a/d+coth(d*x+c)*csch(d*x+c)/d/(a+
I*a*sinh(d*x+c))

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Rubi [A]  time = 0.12, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2768, 2748, 3768, 3770, 3767, 8} \[ \frac {2 i \coth (c+d x)}{a d}+\frac {3 \tanh ^{-1}(\cosh (c+d x))}{2 a d}-\frac {3 \coth (c+d x) \text {csch}(c+d x)}{2 a d}+\frac {\coth (c+d x) \text {csch}(c+d x)}{d (a+i a \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^3/(a + I*a*Sinh[c + d*x]),x]

[Out]

(3*ArcTanh[Cosh[c + d*x]])/(2*a*d) + ((2*I)*Coth[c + d*x])/(a*d) - (3*Coth[c + d*x]*Csch[c + d*x])/(2*a*d) + (
Coth[c + d*x]*Csch[c + d*x])/(d*(a + I*a*Sinh[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 {\text {csch}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx &=\frac {\coth (c+d x) \text {csch}(c+d x)}{d (a+i a \sinh (c+d x))}-\frac {\int \text {csch}^3(c+d x) (-3 a+2 i a \sinh (c+d x)) \, dx}{a^2}\\ &=\frac {\coth (c+d x) \text {csch}(c+d x)}{d (a+i a \sinh (c+d x))}-\frac {(2 i) \int \text {csch}^2(c+d x) \, dx}{a}+\frac {3 \int \text {csch}^3(c+d x) \, dx}{a}\\ &=-\frac {3 \coth (c+d x) \text {csch}(c+d x)}{2 a d}+\frac {\coth (c+d x) \text {csch}(c+d x)}{d (a+i a \sinh (c+d x))}-\frac {3 \int \text {csch}(c+d x) \, dx}{2 a}-\frac {2 \operatorname {Subst}(\int 1 \, dx,x,-i \coth (c+d x))}{a d}\\ &=\frac {3 \tanh ^{-1}(\cosh (c+d x))}{2 a d}+\frac {2 i \coth (c+d x)}{a d}-\frac {3 \coth (c+d x) \text {csch}(c+d x)}{2 a d}+\frac {\coth (c+d x) \text {csch}(c+d x)}{d (a+i a \sinh (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 90, normalized size = 1.03 \[ \frac {4 i \tanh (c+d x)+4 i \text {csch}(2 (c+d x))-3 \text {sech}(c+d x)+\text {csch}^2(c+d x) (-\text {sech}(c+d x))+3 \sqrt {\cosh ^2(c+d x)} \text {sech}(c+d x) \tanh ^{-1}\left (\sqrt {\cosh ^2(c+d x)}\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^3/(a + I*a*Sinh[c + d*x]),x]

[Out]

((4*I)*Csch[2*(c + d*x)] - 3*Sech[c + d*x] + 3*ArcTanh[Sqrt[Cosh[c + d*x]^2]]*Sqrt[Cosh[c + d*x]^2]*Sech[c + d
*x] - Csch[c + d*x]^2*Sech[c + d*x] + (4*I)*Tanh[c + d*x])/(2*a*d)

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fricas [B]  time = 0.58, size = 242, normalized size = 2.78 \[ \frac {{\left (3 \, e^{\left (5 \, d x + 5 \, c\right )} - 3 i \, e^{\left (4 \, d x + 4 \, c\right )} - 6 \, e^{\left (3 \, d x + 3 \, c\right )} + 6 i \, e^{\left (2 \, d x + 2 \, c\right )} + 3 \, e^{\left (d x + c\right )} - 3 i\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) - {\left (3 \, e^{\left (5 \, d x + 5 \, c\right )} - 3 i \, e^{\left (4 \, d x + 4 \, c\right )} - 6 \, e^{\left (3 \, d x + 3 \, c\right )} + 6 i \, e^{\left (2 \, d x + 2 \, c\right )} + 3 \, e^{\left (d x + c\right )} - 3 i\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) - 6 \, e^{\left (4 \, d x + 4 \, c\right )} + 6 i \, e^{\left (3 \, d x + 3 \, c\right )} + 10 \, e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, e^{\left (d x + c\right )} - 8}{2 \, a d e^{\left (5 \, d x + 5 \, c\right )} - 2 i \, a d e^{\left (4 \, d x + 4 \, c\right )} - 4 \, a d e^{\left (3 \, d x + 3 \, c\right )} + 4 i \, a d e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a d e^{\left (d x + c\right )} - 2 i \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

((3*e^(5*d*x + 5*c) - 3*I*e^(4*d*x + 4*c) - 6*e^(3*d*x + 3*c) + 6*I*e^(2*d*x + 2*c) + 3*e^(d*x + c) - 3*I)*log
(e^(d*x + c) + 1) - (3*e^(5*d*x + 5*c) - 3*I*e^(4*d*x + 4*c) - 6*e^(3*d*x + 3*c) + 6*I*e^(2*d*x + 2*c) + 3*e^(
d*x + c) - 3*I)*log(e^(d*x + c) - 1) - 6*e^(4*d*x + 4*c) + 6*I*e^(3*d*x + 3*c) + 10*e^(2*d*x + 2*c) - 2*I*e^(d
*x + c) - 8)/(2*a*d*e^(5*d*x + 5*c) - 2*I*a*d*e^(4*d*x + 4*c) - 4*a*d*e^(3*d*x + 3*c) + 4*I*a*d*e^(2*d*x + 2*c
) + 2*a*d*e^(d*x + c) - 2*I*a*d)

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giac [A]  time = 0.75, size = 97, normalized size = 1.11 \[ \frac {\frac {3 \, \log \left (e^{\left (d x + c\right )} + 1\right )}{a} - \frac {3 \, \log \left (e^{\left (d x + c\right )} - 1\right )}{a} - \frac {2 \, {\left (e^{\left (3 \, d x + 3 \, c\right )} - 2 i \, e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (d x + c\right )} + 2 i\right )}}{a {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}} - \frac {4 i}{a {\left (i \, e^{\left (d x + c\right )} + 1\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

1/2*(3*log(e^(d*x + c) + 1)/a - 3*log(e^(d*x + c) - 1)/a - 2*(e^(3*d*x + 3*c) - 2*I*e^(2*d*x + 2*c) + e^(d*x +
c) + 2*I)/(a*(e^(2*d*x + 2*c) - 1)^2) - 4*I/(a*(I*e^(d*x + c) + 1)))/d

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maple [A]  time = 0.11, size = 119, normalized size = 1.37 \[ \frac {i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {2 i}{d a \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {1}{8 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {i}{2 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^3/(a+I*a*sinh(d*x+c)),x)

[Out]

1/2*I/d/a*tanh(1/2*d*x+1/2*c)+1/8/d/a*tanh(1/2*d*x+1/2*c)^2+2*I/d/a/(-I+tanh(1/2*d*x+1/2*c))-1/8/d/a/tanh(1/2*
d*x+1/2*c)^2+1/2*I/d/a/tanh(1/2*d*x+1/2*c)-3/2/d/a*ln(tanh(1/2*d*x+1/2*c))

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maxima [A]  time = 0.32, size = 158, normalized size = 1.82 \[ -\frac {8 \, {\left (-i \, e^{\left (-d x - c\right )} - 5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 i \, e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4\right )}}{{\left (8 \, a e^{\left (-d x - c\right )} - 16 i \, a e^{\left (-2 \, d x - 2 \, c\right )} - 16 \, a e^{\left (-3 \, d x - 3 \, c\right )} + 8 i \, a e^{\left (-4 \, d x - 4 \, c\right )} + 8 \, a e^{\left (-5 \, d x - 5 \, c\right )} + 8 i \, a\right )} d} + \frac {3 \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{2 \, a d} - \frac {3 \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{2 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-8*(-I*e^(-d*x - c) - 5*e^(-2*d*x - 2*c) + 3*I*e^(-3*d*x - 3*c) + 3*e^(-4*d*x - 4*c) + 4)/((8*a*e^(-d*x - c) -
16*I*a*e^(-2*d*x - 2*c) - 16*a*e^(-3*d*x - 3*c) + 8*I*a*e^(-4*d*x - 4*c) + 8*a*e^(-5*d*x - 5*c) + 8*I*a)*d) +
3/2*log(e^(-d*x - c) + 1)/(a*d) - 3/2*log(e^(-d*x - c) - 1)/(a*d)

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mupad [B]  time = 0.61, size = 132, normalized size = 1.52 \[ \frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-a^2\,d^2}}{a\,d}\right )}{\sqrt {-a^2\,d^2}}-\frac {2}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )}-\frac {{\mathrm {e}}^{c+d\,x}}{a\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {2\,{\mathrm {e}}^{c+d\,x}}{a\,d\,{\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}^2}+\frac {2{}\mathrm {i}}{a\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*atan((exp(d*x)*exp(c)*(-a^2*d^2)^(1/2))/(a*d)))/(-a^2*d^2)^(1/2) - 2/(a*d*(exp(c + d*x) - 1i)) + 2i/(a*d*(e
xp(2*c + 2*d*x) - 1)) - exp(c + d*x)/(a*d*(exp(2*c + 2*d*x) - 1)) - (2*exp(c + d*x))/(a*d*(exp(2*c + 2*d*x) -
1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \int \frac {\operatorname {csch}^{3}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**3/(a+I*a*sinh(d*x+c)),x)

[Out]

-I*Integral(csch(c + d*x)**3/(sinh(c + d*x) - I), x)/a

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