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(1/2*d*x+1/2*c))/(a^2+b^2)^(1/2))/a^2/(a^2+b^2)^(3/2)/d-b^2*coth(d*x+c)
/a/(a^2+b^2)/d/(a+b*csch(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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 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 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 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]

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 = 0.39, 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)

________________________________________________________________________________________

fricas [B]  time = 0.42, size = 645, normalized size = 6.39 \[ -\frac {2 \, a^{3} b^{2} + 2 \, a b^{4} - {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cosh \left (d x + c\right )^{2} - {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d x \sinh \left (d x + c\right )^{2} + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d x + {\left (2 \, a^{3} b + a b^{3} - {\left (2 \, a^{3} b + a b^{3}\right )} \cosh \left (d x + c\right )^{2} - {\left (2 \, a^{3} b + a b^{3}\right )} \sinh \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (d x + c\right ) - 2 \, {\left (2 \, a^{2} b^{2} + b^{4} + {\left (2 \, a^{3} b + a b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {a^{2} \cosh \left (d x + c\right )^{2} + a^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + b\right )}}{a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) + 2 \, {\left (a \cosh \left (d x + c\right ) + b\right )} \sinh \left (d x + c\right ) - a}\right ) - 2 \, {\left (a^{2} b^{3} + b^{5} + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d x\right )} \cosh \left (d x + c\right ) - 2 \, {\left (a^{2} b^{3} + b^{5} + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cosh \left (d x + c\right ) + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d x\right )} \sinh \left (d x + c\right )}{{\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cosh \left (d x + c\right )^{2} + {\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sinh \left (d x + c\right )^{2} + 2 \, {\left (a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d \cosh \left (d x + c\right ) - {\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d + 2 \, {\left ({\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cosh \left (d x + c\right ) + {\left (a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )} \sinh \left (d x + c\right )} \]

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))

________________________________________________________________________________________

giac [A]  time = 0.15, size = 161, normalized size = 1.59 \[ -\frac {\frac {{\left (2 \, a^{2} b + b^{3}\right )} \log \left (\frac {{\left | 2 \, a e^{\left (d x + c\right )} + 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a e^{\left (d x + c\right )} + 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + a^{2} b^{2}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (b^{3} e^{\left (d x + c\right )} - a b^{2}\right )}}{{\left (a^{4} + a^{2} b^{2}\right )} {\left (a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (d x + c\right )} - a\right )}} - \frac {d x + c}{a^{2}}}{d} \]

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

________________________________________________________________________________________

maple [B]  time = 0.32, size = 238, normalized size = 2.36 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}+\frac {2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-b \right ) \left (a^{2}+b^{2}\right )}+\frac {2 b^{2}}{d a \left (\left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-b \right ) \left (a^{2}+b^{2}\right )}-\frac {4 b \arctanh \left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}-\frac {2 b^{3} \arctanh \left (\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,a^{2} \left (a^{2}+b^{2}\right )^{\frac {3}{2}}} \]

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

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 187, normalized size = 1.85 \[ -\frac {{\left (2 \, a^{2} b + b^{3}\right )} \log \left (\frac {a e^{\left (-d x - c\right )} - b - \sqrt {a^{2} + b^{2}}}{a e^{\left (-d x - c\right )} - b + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + a^{2} b^{2}\right )} \sqrt {a^{2} + b^{2}} d} - \frac {2 \, {\left (b^{3} e^{\left (-d x - c\right )} + a b^{2}\right )}}{{\left (a^{5} + a^{3} b^{2} + 2 \, {\left (a^{4} b + a^{2} b^{3}\right )} e^{\left (-d x - c\right )} - {\left (a^{5} + a^{3} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} + \frac {d x + c}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^2*b + b^3)*log((a*e^(-d*x - c) - b - sqrt(a^2 + b^2))/(a*e^(-d*x - c) - b + sqrt(a^2 + b^2)))/((a^4 + a^
2*b^2)*sqrt(a^2 + b^2)*d) - 2*(b^3*e^(-d*x - c) + a*b^2)/((a^5 + a^3*b^2 + 2*(a^4*b + a^2*b^3)*e^(-d*x - c) -
(a^5 + a^3*b^2)*e^(-2*d*x - 2*c))*d) + (d*x + c)/(a^2*d)

________________________________________________________________________________________

mupad [B]  time = 1.95, size = 269, normalized size = 2.66 \[ \frac {x}{a^2}-\frac {\frac {2\,b^2}{d\,\left (a^3+a\,b^2\right )}-\frac {2\,b^3\,{\mathrm {e}}^{c+d\,x}}{a\,d\,\left (a^3+a\,b^2\right )}}{2\,b\,{\mathrm {e}}^{c+d\,x}-a+a\,{\mathrm {e}}^{2\,c+2\,d\,x}}-\frac {b\,\ln \left (\frac {2\,{\mathrm {e}}^{c+d\,x}\,\left (2\,a^2\,b+b^3\right )}{a^3\,\left (a^2+b^2\right )}-\frac {2\,b\,\left (2\,a^2+b^2\right )\,\left (a-b\,{\mathrm {e}}^{c+d\,x}\right )}{a^3\,{\left (a^2+b^2\right )}^{3/2}}\right )\,\left (2\,a^2+b^2\right )}{a^2\,d\,{\left (a^2+b^2\right )}^{3/2}}+\frac {b\,\ln \left (\frac {2\,{\mathrm {e}}^{c+d\,x}\,\left (2\,a^2\,b+b^3\right )}{a^3\,\left (a^2+b^2\right )}+\frac {2\,b\,\left (2\,a^2+b^2\right )\,\left (a-b\,{\mathrm {e}}^{c+d\,x}\right )}{a^3\,{\left (a^2+b^2\right )}^{3/2}}\right )\,\left (2\,a^2+b^2\right )}{a^2\,d\,{\left (a^2+b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/sinh(c + d*x))^2,x)

[Out]

x/a^2 - ((2*b^2)/(d*(a*b^2 + a^3)) - (2*b^3*exp(c + d*x))/(a*d*(a*b^2 + a^3)))/(2*b*exp(c + d*x) - a + a*exp(2
*c + 2*d*x)) - (b*log((2*exp(c + d*x)*(2*a^2*b + b^3))/(a^3*(a^2 + b^2)) - (2*b*(2*a^2 + b^2)*(a - b*exp(c + d
*x)))/(a^3*(a^2 + b^2)^(3/2)))*(2*a^2 + b^2))/(a^2*d*(a^2 + b^2)^(3/2)) + (b*log((2*exp(c + d*x)*(2*a^2*b + b^
3))/(a^3*(a^2 + b^2)) + (2*b*(2*a^2 + b^2)*(a - b*exp(c + d*x)))/(a^3*(a^2 + b^2)^(3/2)))*(2*a^2 + b^2))/(a^2*
d*(a^2 + b^2)^(3/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {csch}{\left (c + d x \right )}\right )^{2}}\, dx \]

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)

________________________________________________________________________________________