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[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)

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Rubi [A]  time = 0.147012, antiderivative size = 80, normalized size of antiderivative = 1., 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 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 12

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

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 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 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])

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*}

Mathematica [A]  time = 0.734301, 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]

-(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])/(2*a^2*d)

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Maple [A]  time = 0.003, size = 105, normalized size = 1.3 \begin{align*} -{\frac{1}{2\,da}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{{b}^{2}}{d{a}^{2}\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{b}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 2.86922, size = 1200, normalized size = 15. \begin{align*} -\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} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}^{2}{\left (c + d x \right )}}{a + b \sinh{\left (c + d x \right )}}\, dx \end{align*}

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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Giac [A]  time = 1.62952, size = 177, normalized size = 2.21 \begin{align*} \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} d} + \frac{b \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{2} d} - \frac{b \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{2} d} - \frac{2}{a d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}} \end{align*}

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