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

Optimal. Leaf size=80 \[ -\frac {2 b^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 d \sqrt {a^2+b^2}}+\frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac {\coth (c+d x)}{a d} \]

[Out]

b*arctanh(cosh(d*x+c))/a^2/d-coth(d*x+c)/a/d-2*b^2*arctanh((b-a*tanh(1/2*d*x+1/2*c))/(a^2+b^2)^(1/2))/a^2/d/(a
^2+b^2)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2802, 12, 2747, 3770, 2660, 618, 204} \[ -\frac {2 b^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 d \sqrt {a^2+b^2}}+\frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac {\coth (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTanh[Cosh[c + d*x]])/(a^2*d) - (2*b^2*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2*Sqrt[a^2
+ b^2]*d) - Coth[c + d*x]/(a*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2747

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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+b \sinh (c+d x)} \, dx &=-\frac {\coth (c+d x)}{a d}-\frac {\int \frac {b \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx}{a}\\ &=-\frac {\coth (c+d x)}{a d}-\frac {b \int \frac {\text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx}{a}\\ &=-\frac {\coth (c+d x)}{a d}-\frac {b \int \text {csch}(c+d x) \, dx}{a^2}+\frac {b^2 \int \frac {1}{a+b \sinh (c+d x)} \, dx}{a^2}\\ &=\frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac {\coth (c+d x)}{a d}-\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a-2 i b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{a^2 d}\\ &=\frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac {\coth (c+d x)}{a d}+\frac {\left (4 i b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,-2 i b+2 a \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{a^2 d}\\ &=\frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac {2 b^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^2 \sqrt {a^2+b^2} d}-\frac {\coth (c+d x)}{a d}\\ \end {align*}

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Mathematica [A]  time = 0.74, size = 100, normalized size = 1.25 \[ -\frac {2 b \left (\log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )-\frac {2 b \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}\right )+a \tanh \left (\frac {1}{2} (c+d x)\right )+a \coth \left (\frac {1}{2} (c+d x)\right )}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a*Coth[(c + d*x)/2] + 2*b*((-2*b*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] +
Log[Tanh[(c + d*x)/2]]) + a*Tanh[(c + d*x)/2])/(a^2*d)

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fricas [B]  time = 0.62, size = 479, normalized size = 5.99 \[ -\frac {2 \, a^{3} + 2 \, a b^{2} - {\left (b^{2} \cosh \left (d x + c\right )^{2} + 2 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b^{2} \sinh \left (d x + c\right )^{2} - b^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) + {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) - {\left (a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 1\right ) - {\left (a^{2} b + b^{3} - {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )^{2} - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) - {\left (a^{2} b + b^{3}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - 1\right )}{{\left (a^{4} + a^{2} b^{2}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{4} + a^{2} b^{2}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{4} + a^{2} b^{2}\right )} d \sinh \left (d x + c\right )^{2} - {\left (a^{4} + a^{2} b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^3 + 2*a*b^2 - (b^2*cosh(d*x + c)^2 + 2*b^2*cosh(d*x + c)*sinh(d*x + c) + b^2*sinh(d*x + c)^2 - b^2)*sqrt
(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + 2*a^2 + b^2 + 2*(b^2*cosh(d
*x + c) + a*b)*sinh(d*x + c) - 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a))/(b*cosh(d*x + c)^2 +
b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)) + (a^2*b + b^3 - (a^2*b +
b^3)*cosh(d*x + c)^2 - 2*(a^2*b + b^3)*cosh(d*x + c)*sinh(d*x + c) - (a^2*b + b^3)*sinh(d*x + c)^2)*log(cosh(
d*x + c) + sinh(d*x + c) + 1) - (a^2*b + b^3 - (a^2*b + b^3)*cosh(d*x + c)^2 - 2*(a^2*b + b^3)*cosh(d*x + c)*s
inh(d*x + c) - (a^2*b + b^3)*sinh(d*x + c)^2)*log(cosh(d*x + c) + sinh(d*x + c) - 1))/((a^4 + a^2*b^2)*d*cosh(
d*x + c)^2 + 2*(a^4 + a^2*b^2)*d*cosh(d*x + c)*sinh(d*x + c) + (a^4 + a^2*b^2)*d*sinh(d*x + c)^2 - (a^4 + a^2*
b^2)*d)

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giac [A]  time = 0.41, size = 123, normalized size = 1.54 \[ \frac {\frac {b^{2} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{2}} + \frac {b \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{2}} - \frac {b \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{2}} - \frac {2}{a {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*log(abs(2*b*e^(d*x + c) + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^(d*x + c) + 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(
a^2 + b^2)*a^2) + b*log(e^(d*x + c) + 1)/a^2 - b*log(abs(e^(d*x + c) - 1))/a^2 - 2/(a*(e^(2*d*x + 2*c) - 1)))/
d

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maple [A]  time = 0.00, size = 105, normalized size = 1.31 \[ -\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {2 b^{2} \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,a^{2} \sqrt {a^{2}+b^{2}}}-\frac {1}{2 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d/a*tanh(1/2*d*x+1/2*c)+2/d/a^2*b^2/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(
1/2))-1/2/d/a/tanh(1/2*d*x+1/2*c)-1/d/a^2*b*ln(tanh(1/2*d*x+1/2*c))

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maxima [A]  time = 0.52, size = 137, normalized size = 1.71 \[ \frac {b^{2} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{2} d} + \frac {b \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{2} d} - \frac {b \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{2} d} + \frac {2}{{\left (a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*log((b*e^(-d*x - c) - a - sqrt(a^2 + b^2))/(b*e^(-d*x - c) - a + sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^2*d)
+ b*log(e^(-d*x - c) + 1)/(a^2*d) - b*log(e^(-d*x - c) - 1)/(a^2*d) + 2/((a*e^(-2*d*x - 2*c) - a)*d)

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mupad [B]  time = 0.39, size = 360, normalized size = 4.50 \[ \frac {2}{a\,d-a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}}+\frac {b^2\,\ln \left (128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-64\,a\,b^3-64\,a^3\,b-32\,b^3\,\sqrt {a^2+b^2}+32\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-64\,a^2\,b\,\sqrt {a^2+b^2}+160\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+128\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}+96\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{d\,a^4+d\,a^2\,b^2}-\frac {b^2\,\ln \left (32\,b^3\,\sqrt {a^2+b^2}-64\,a\,b^3-64\,a^3\,b+128\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+64\,a^2\,b\,\sqrt {a^2+b^2}+160\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-128\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}-96\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{d\,a^4+d\,a^2\,b^2}-\frac {b\,\ln \left (32\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-32\right )}{a^2\,d}+\frac {b\,\ln \left (32\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\right )}{a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(a*d - a*d*exp(2*c + 2*d*x)) + (b^2*log(128*a^4*exp(d*x)*exp(c) - 64*a*b^3 - 64*a^3*b - 32*b^3*(a^2 + b^2)^(
1/2) + 32*b^4*exp(d*x)*exp(c) - 64*a^2*b*(a^2 + b^2)^(1/2) + 160*a^2*b^2*exp(d*x)*exp(c) + 128*a^3*exp(d*x)*ex
p(c)*(a^2 + b^2)^(1/2) + 96*a*b^2*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a^4*d + a^2*b^2*d) -
(b^2*log(32*b^3*(a^2 + b^2)^(1/2) - 64*a*b^3 - 64*a^3*b + 128*a^4*exp(d*x)*exp(c) + 32*b^4*exp(d*x)*exp(c) + 6
4*a^2*b*(a^2 + b^2)^(1/2) + 160*a^2*b^2*exp(d*x)*exp(c) - 128*a^3*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2) - 96*a*b^2
*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a^4*d + a^2*b^2*d) - (b*log(32*exp(d*x)*exp(c) - 32))/
(a^2*d) + (b*log(32*exp(d*x)*exp(c) + 32))/(a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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