3.214 \(\int \frac {\text {csch}^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=57 \[ -\frac {2 \coth (c+d x)}{a d}+\frac {i \tanh ^{-1}(\cosh (c+d x))}{a d}+\frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2768, 2748, 3767, 8, 3770} \[ -\frac {2 \coth (c+d x)}{a d}+\frac {i \tanh ^{-1}(\cosh (c+d x))}{a d}+\frac {\coth (c+d x)}{d (a+i a \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^2/(a + I*a*Sinh[c + d*x]),x]

[Out]

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

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Mathematica [A]  time = 0.20, size = 61, normalized size = 1.07 \[ -\frac {\text {sech}(c+d x) \left (2 \sinh (c+d x)+\text {csch}(c+d x)-i \sqrt {\cosh ^2(c+d x)} \tanh ^{-1}\left (\sqrt {\cosh ^2(c+d x)}\right )+i\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^2/(a + I*a*Sinh[c + d*x]),x]

[Out]

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

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fricas [B]  time = 0.56, size = 146, normalized size = 2.56 \[ \frac {{\left (i \, e^{\left (3 \, d x + 3 \, c\right )} + e^{\left (2 \, d x + 2 \, c\right )} - i \, e^{\left (d x + c\right )} - 1\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) + {\left (-i \, e^{\left (3 \, d x + 3 \, c\right )} - e^{\left (2 \, d x + 2 \, c\right )} + i \, e^{\left (d x + c\right )} + 1\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) - 2 i \, e^{\left (2 \, d x + 2 \, c\right )} - 2 \, e^{\left (d x + c\right )} + 4 i}{a d e^{\left (3 \, d x + 3 \, c\right )} - i \, a d e^{\left (2 \, d x + 2 \, c\right )} - a d e^{\left (d x + c\right )} + i \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.45, size = 90, normalized size = 1.58 \[ -\frac {-\frac {i \, \log \left (e^{\left (d x + c\right )} + 1\right )}{a} + \frac {i \, \log \left (e^{\left (d x + c\right )} - 1\right )}{a} - \frac {2 \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - i \, e^{\left (d x + c\right )} - 2\right )}}{a {\left (i \, e^{\left (3 \, d x + 3 \, c\right )} + e^{\left (2 \, d x + 2 \, c\right )} - i \, e^{\left (d x + c\right )} - 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 79, normalized size = 1.39 \[ -\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}-\frac {2}{d a \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {1}{2 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x)

[Out]

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

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maxima [B]  time = 0.61, size = 110, normalized size = 1.93 \[ -\frac {4 \, {\left (e^{\left (-d x - c\right )} - i \, e^{\left (-2 \, d x - 2 \, c\right )} + 2 i\right )}}{{\left (2 \, a e^{\left (-d x - c\right )} - 2 i \, a e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a e^{\left (-3 \, d x - 3 \, c\right )} + 2 i \, a\right )} d} + \frac {i \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} - \frac {i \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.40, size = 122, normalized size = 2.14 \[ \frac {\frac {2\,{\mathrm {e}}^{c+d\,x}}{a\,d}-\frac {4{}\mathrm {i}}{a\,d}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,2{}\mathrm {i}}{a\,d}}{{\mathrm {e}}^{c+d\,x}+{\mathrm {e}}^{2\,c+2\,d\,x}\,1{}\mathrm {i}-{\mathrm {e}}^{3\,c+3\,d\,x}-\mathrm {i}}-\frac {\ln \left ({\mathrm {e}}^{c+d\,x}\,2{}\mathrm {i}-2{}\mathrm {i}\right )\,1{}\mathrm {i}}{a\,d}+\frac {\ln \left ({\mathrm {e}}^{c+d\,x}\,2{}\mathrm {i}+2{}\mathrm {i}\right )\,1{}\mathrm {i}}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*exp(c + d*x))/(a*d) - 4i/(a*d) + (exp(2*c + 2*d*x)*2i)/(a*d))/(exp(c + d*x) + exp(2*c + 2*d*x)*1i - exp(3*
c + 3*d*x) - 1i) - (log(exp(c + d*x)*2i - 2i)*1i)/(a*d) + (log(exp(c + d*x)*2i + 2i)*1i)/(a*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \int \frac {\operatorname {csch}^{2}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**2/(a+I*a*sinh(d*x+c)),x)

[Out]

-I*Integral(csch(c + d*x)**2/(sinh(c + d*x) - I), x)/a

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