3.75 \(\int \frac{1}{(a+b \text{csch}(c+d x))^2} \, dx\)

Optimal. Leaf size=101 \[ \frac{2 b \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a^2 d \left (a^2+b^2\right )^{3/2}}-\frac{b^2 \coth (c+d x)}{a d \left (a^2+b^2\right ) (a+b \text{csch}(c+d x))}+\frac{x}{a^2} \]

[Out]

x/a^2 + (2*b*(2*a^2 + b^2)*ArcTanh[(a - b*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2*(a^2 + b^2)^(3/2)*d) - (b^
2*Coth[c + d*x])/(a*(a^2 + b^2)*d*(a + b*Csch[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.157509, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3785, 3919, 3831, 2660, 618, 204} \[ \frac{2 b \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{a^2 d \left (a^2+b^2\right )^{3/2}}-\frac{b^2 \coth (c+d x)}{a d \left (a^2+b^2\right ) (a+b \text{csch}(c+d x))}+\frac{x}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csch[c + d*x])^(-2),x]

[Out]

x/a^2 + (2*b*(2*a^2 + b^2)*ArcTanh[(a - b*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2*(a^2 + b^2)^(3/2)*d) - (b^
2*Coth[c + d*x])/(a*(a^2 + b^2)*d*(a + b*Csch[c + d*x]))

Rule 3785

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n +
1))/(a*d*(n + 1)*(a^2 - b^2)), x] + Dist[1/(a*(n + 1)*(a^2 - b^2)), Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^
2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 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 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{1}{(a+b \text{csch}(c+d x))^2} \, dx &=-\frac{b^2 \coth (c+d x)}{a \left (a^2+b^2\right ) d (a+b \text{csch}(c+d x))}-\frac{\int \frac{-a^2-b^2+a b \text{csch}(c+d x)}{a+b \text{csch}(c+d x)} \, dx}{a \left (a^2+b^2\right )}\\ &=\frac{x}{a^2}-\frac{b^2 \coth (c+d x)}{a \left (a^2+b^2\right ) d (a+b \text{csch}(c+d x))}-\frac{\left (b \left (2 a^2+b^2\right )\right ) \int \frac{\text{csch}(c+d x)}{a+b \text{csch}(c+d x)} \, dx}{a^2 \left (a^2+b^2\right )}\\ &=\frac{x}{a^2}-\frac{b^2 \coth (c+d x)}{a \left (a^2+b^2\right ) d (a+b \text{csch}(c+d x))}-\frac{\left (2 a^2+b^2\right ) \int \frac{1}{1+\frac{a \sinh (c+d x)}{b}} \, dx}{a^2 \left (a^2+b^2\right )}\\ &=\frac{x}{a^2}-\frac{b^2 \coth (c+d x)}{a \left (a^2+b^2\right ) d (a+b \text{csch}(c+d x))}+\frac{\left (2 i \left (2 a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{2 i a x}{b}+x^2} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{a^2 \left (a^2+b^2\right ) d}\\ &=\frac{x}{a^2}-\frac{b^2 \coth (c+d x)}{a \left (a^2+b^2\right ) d (a+b \text{csch}(c+d x))}-\frac{\left (4 i \left (2 a^2+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (1+\frac{a^2}{b^2}\right )-x^2} \, dx,x,-\frac{2 i a}{b}+2 \tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{a^2 \left (a^2+b^2\right ) d}\\ &=\frac{x}{a^2}+\frac{2 b \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{b \left (\frac{a}{b}-\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2+b^2}}\right )}{a^2 \left (a^2+b^2\right )^{3/2} d}-\frac{b^2 \coth (c+d x)}{a \left (a^2+b^2\right ) d (a+b \text{csch}(c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.3813, size = 142, normalized size = 1.41 \[ \frac{\text{csch}(c+d x) (a \sinh (c+d x)+b) \left (-\frac{a b^2 \coth (c+d x)}{a^2+b^2}+\frac{2 b \left (2 a^2+b^2\right ) (a+b \text{csch}(c+d x)) \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+(c+d x) (a+b \text{csch}(c+d x))\right )}{a^2 d (a+b \text{csch}(c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csch[c + d*x]*(-((a*b^2*Coth[c + d*x])/(a^2 + b^2)) + (c + d*x)*(a + b*Csch[c + d*x]) + (2*b*(2*a^2 + b^2)*Ar
cTan[(a - b*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]]*(a + b*Csch[c + d*x]))/(-a^2 - b^2)^(3/2))*(b + a*Sinh[c + d*
x]))/(a^2*d*(a + b*Csch[c + d*x])^2)

________________________________________________________________________________________

Maple [B]  time = 0.057, size = 238, normalized size = 2.4 \begin{align*}{\frac{1}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+2\,{\frac{b\tanh \left ( 1/2\,dx+c/2 \right ) }{d \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -b \right ) \left ({a}^{2}+{b}^{2} \right ) }}+2\,{\frac{{b}^{2}}{da \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -b \right ) \left ({a}^{2}+{b}^{2} \right ) }}-4\,{\frac{b}{d \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,b\tanh \left ( 1/2\,dx+c/2 \right ) -2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-2\,{\frac{{b}^{3}}{d{a}^{2} \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,b\tanh \left ( 1/2\,dx+c/2 \right ) -2\,a}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a^2*ln(tanh(1/2*d*x+1/2*c)+1)+2/d*b/(tanh(1/2*d*x+1/2*c)^2*b-2*a*tanh(1/2*d*x+1/2*c)-b)/(a^2+b^2)*tanh(1/2
*d*x+1/2*c)+2/d/a*b^2/(tanh(1/2*d*x+1/2*c)^2*b-2*a*tanh(1/2*d*x+1/2*c)-b)/(a^2+b^2)-4/d*b/(a^2+b^2)^(3/2)*arct
anh(1/2*(2*b*tanh(1/2*d*x+1/2*c)-2*a)/(a^2+b^2)^(1/2))-2/d/a^2*b^3/(a^2+b^2)^(3/2)*arctanh(1/2*(2*b*tanh(1/2*d
*x+1/2*c)-2*a)/(a^2+b^2)^(1/2))-1/d/a^2*ln(tanh(1/2*d*x+1/2*c)-1)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.9289, size = 1503, normalized size = 14.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csch(d*x+c))^2,x, algorithm="fricas")

[Out]

-(2*a^3*b^2 + 2*a*b^4 - (a^5 + 2*a^3*b^2 + a*b^4)*d*x*cosh(d*x + c)^2 - (a^5 + 2*a^3*b^2 + a*b^4)*d*x*sinh(d*x
+ c)^2 + (a^5 + 2*a^3*b^2 + a*b^4)*d*x + (2*a^3*b + a*b^3 - (2*a^3*b + a*b^3)*cosh(d*x + c)^2 - (2*a^3*b + a*
b^3)*sinh(d*x + c)^2 - 2*(2*a^2*b^2 + b^4)*cosh(d*x + c) - 2*(2*a^2*b^2 + b^4 + (2*a^3*b + a*b^3)*cosh(d*x + c
))*sinh(d*x + c))*sqrt(a^2 + b^2)*log((a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + a^2 +
2*b^2 + 2*(a^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(a*cosh(d*x + c) + a*sinh(d*x + c) + b)
)/(a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + 2*b*cosh(d*x + c) + 2*(a*cosh(d*x + c) + b)*sinh(d*x + c) - a)) - 2
*(a^2*b^3 + b^5 + (a^4*b + 2*a^2*b^3 + b^5)*d*x)*cosh(d*x + c) - 2*(a^2*b^3 + b^5 + (a^5 + 2*a^3*b^2 + a*b^4)*
d*x*cosh(d*x + c) + (a^4*b + 2*a^2*b^3 + b^5)*d*x)*sinh(d*x + c))/((a^7 + 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x + c)
^2 + (a^7 + 2*a^5*b^2 + a^3*b^4)*d*sinh(d*x + c)^2 + 2*(a^6*b + 2*a^4*b^3 + a^2*b^5)*d*cosh(d*x + c) - (a^7 +
2*a^5*b^2 + a^3*b^4)*d + 2*((a^7 + 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x + c) + (a^6*b + 2*a^4*b^3 + a^2*b^5)*d)*sin
h(d*x + c))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \operatorname{csch}{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csch(d*x+c))**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.22093, size = 223, normalized size = 2.21 \begin{align*} -\frac{{\left (2 \, a^{2} b + b^{3}\right )} \log \left (\frac{{\left | 2 \, a e^{\left (d x + c\right )} + 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{\left (d x + c\right )} + 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} d + a^{2} b^{2} d\right )} \sqrt{a^{2} + b^{2}}} + \frac{2 \,{\left (b^{3} e^{\left (d x + c\right )} - a b^{2}\right )}}{{\left (a^{4} d + a^{2} b^{2} d\right )}{\left (a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (d x + c\right )} - a\right )}} + \frac{d x + c}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csch(d*x+c))^2,x, algorithm="giac")

[Out]

-(2*a^2*b + b^3)*log(abs(2*a*e^(d*x + c) + 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*e^(d*x + c) + 2*b + 2*sqrt(a^2 + b
^2)))/((a^4*d + a^2*b^2*d)*sqrt(a^2 + b^2)) + 2*(b^3*e^(d*x + c) - a*b^2)/((a^4*d + a^2*b^2*d)*(a*e^(2*d*x + 2
*c) + 2*b*e^(d*x + c) - a)) + (d*x + c)/(a^2*d)