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

Optimal. Leaf size=50 \[ \frac {x}{a}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {b}}\right )}{a d \sqrt {a-b}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4127, 3181, 205} \[ \frac {x}{a}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {b}}\right )}{a d \sqrt {a-b}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csch[c + d*x]^2)^(-1),x]

[Out]

x/a - (Sqrt[b]*ArcTan[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[b]])/(a*Sqrt[a - b]*d)

Rule 205

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

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 4127

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> Simp[x/a, x] - Dist[b/a, Int[1/(b + a*Cos[e +
f*x]^2), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0]

Rubi steps

\begin {align*} \int \frac {1}{a+b \text {csch}^2(c+d x)} \, dx &=\frac {x}{a}+\frac {b \int \frac {1}{-b-a \sinh ^2(c+d x)} \, dx}{a}\\ &=\frac {x}{a}+\frac {b \operatorname {Subst}\left (\int \frac {1}{-b-(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{a d}\\ &=\frac {x}{a}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {b}}\right )}{a \sqrt {a-b} d}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 52, normalized size = 1.04 \[ \frac {-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {b}}\right )}{d \sqrt {a-b}}+\frac {c}{d}+x}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csch[c + d*x]^2)^(-1),x]

[Out]

(c/d + x - (Sqrt[b]*ArcTan[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[b]])/(Sqrt[a - b]*d))/a

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fricas [A]  time = 0.45, size = 457, normalized size = 9.14 \[ \left [\frac {2 \, d x + \sqrt {-\frac {b}{a - b}} \log \left (\frac {a^{2} \cosh \left (d x + c\right )^{4} + 4 \, a^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{2} \sinh \left (d x + c\right )^{4} - 2 \, {\left (a^{2} - 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} - a^{2} + 2 \, a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} - 8 \, a b + 8 \, b^{2} + 4 \, {\left (a^{2} \cosh \left (d x + c\right )^{3} - {\left (a^{2} - 2 \, a b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 4 \, {\left ({\left (a^{2} - a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{2} - a b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{2} - a b\right )} \sinh \left (d x + c\right )^{2} - a^{2} + 3 \, a b - 2 \, b^{2}\right )} \sqrt {-\frac {b}{a - b}}}{a \cosh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a \sinh \left (d x + c\right )^{4} - 2 \, {\left (a - 2 \, b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a \cosh \left (d x + c\right )^{2} - a + 2 \, b\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (a \cosh \left (d x + c\right )^{3} - {\left (a - 2 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a}\right )}{2 \, a d}, \frac {d x - \sqrt {\frac {b}{a - b}} \arctan \left (\frac {{\left (a \cosh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a \sinh \left (d x + c\right )^{2} - a + 2 \, b\right )} \sqrt {\frac {b}{a - b}}}{2 \, b}\right )}{a d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*d*x + sqrt(-b/(a - b))*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x +
c)^4 - 2*(a^2 - 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 - a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 - 8*a*
b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 - (a^2 - 2*a*b)*cosh(d*x + c))*sinh(d*x + c) - 4*((a^2 - a*b)*cosh(d*x + c)
^2 + 2*(a^2 - a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 - a*b)*sinh(d*x + c)^2 - a^2 + 3*a*b - 2*b^2)*sqrt(-b/(a
- b)))/(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 - 2*(a - 2*b)*cosh(d*x + c)
^2 + 2*(3*a*cosh(d*x + c)^2 - a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 - (a - 2*b)*cosh(d*x + c))*sinh(
d*x + c) + a)))/(a*d), (d*x - sqrt(b/(a - b))*arctan(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c)
+ a*sinh(d*x + c)^2 - a + 2*b)*sqrt(b/(a - b))/b))/(a*d)]

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giac [A]  time = 0.38, size = 64, normalized size = 1.28 \[ -\frac {\frac {b \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} - a + 2 \, b}{2 \, \sqrt {a b - b^{2}}}\right )}{\sqrt {a b - b^{2}} a} - \frac {d x + c}{a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.30, size = 302, normalized size = 6.04 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}+\frac {b \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {\left (2 \sqrt {a \left (a -b \right )}+2 a -b \right ) b}}\right )}{d a \sqrt {\left (2 \sqrt {a \left (a -b \right )}+2 a -b \right ) b}}+\frac {b \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {\left (2 \sqrt {a \left (a -b \right )}+2 a -b \right ) b}}\right )}{d \sqrt {a \left (a -b \right )}\, \sqrt {\left (2 \sqrt {a \left (a -b \right )}+2 a -b \right ) b}}-\frac {b \arctanh \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {\left (2 \sqrt {a \left (a -b \right )}-2 a +b \right ) b}}\right )}{d a \sqrt {\left (2 \sqrt {a \left (a -b \right )}-2 a +b \right ) b}}+\frac {b \arctanh \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {\left (2 \sqrt {a \left (a -b \right )}-2 a +b \right ) b}}\right )}{d \sqrt {a \left (a -b \right )}\, \sqrt {\left (2 \sqrt {a \left (a -b \right )}-2 a +b \right ) b}} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

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mupad [B]  time = 1.93, size = 471, normalized size = 9.42 \[ \frac {x}{a}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\left (a^5\,\sqrt {a^3\,d^2-a^2\,b\,d^2}-a^4\,b\,\sqrt {a^3\,d^2-a^2\,b\,d^2}\right )\,\left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {2\,\left (a^2-8\,a\,b+8\,b^2\right )\,\left (8\,b^{5/2}\,\sqrt {a^3\,d^2-a^2\,b\,d^2}-8\,a\,b^{3/2}\,\sqrt {a^3\,d^2-a^2\,b\,d^2}+a^2\,\sqrt {b}\,\sqrt {a^3\,d^2-a^2\,b\,d^2}\right )}{a^8\,d\,{\left (a-b\right )}^2\,\sqrt {a^3\,d^2-a^2\,b\,d^2}}+\frac {4\,\sqrt {b}\,\left (2\,a-4\,b\right )\,\left (4\,d\,a^3\,b-12\,d\,a^2\,b^2+8\,d\,a\,b^3\right )}{a^7\,\left (a-b\right )\,\sqrt {a^3\,d^2-a^2\,b\,d^2}\,\sqrt {a^2\,d^2\,\left (a-b\right )}}\right )+\frac {2\,\left (2\,a\,b^{3/2}\,\sqrt {a^3\,d^2-a^2\,b\,d^2}-a^2\,\sqrt {b}\,\sqrt {a^3\,d^2-a^2\,b\,d^2}\right )\,\left (a^2-8\,a\,b+8\,b^2\right )}{a^8\,d\,{\left (a-b\right )}^2\,\sqrt {a^3\,d^2-a^2\,b\,d^2}}+\frac {4\,\sqrt {b}\,\left (2\,a^2\,b^2\,d-2\,a^3\,b\,d\right )\,\left (2\,a-4\,b\right )}{a^7\,\left (a-b\right )\,\sqrt {a^3\,d^2-a^2\,b\,d^2}\,\sqrt {a^2\,d^2\,\left (a-b\right )}}\right )}{4\,b}\right )}{\sqrt {a^3\,d^2-a^2\,b\,d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/sinh(c + d*x)^2),x)

[Out]

x/a - (b^(1/2)*atan(((a^5*(a^3*d^2 - a^2*b*d^2)^(1/2) - a^4*b*(a^3*d^2 - a^2*b*d^2)^(1/2))*(exp(2*c)*exp(2*d*x
)*((2*(a^2 - 8*a*b + 8*b^2)*(8*b^(5/2)*(a^3*d^2 - a^2*b*d^2)^(1/2) - 8*a*b^(3/2)*(a^3*d^2 - a^2*b*d^2)^(1/2) +
a^2*b^(1/2)*(a^3*d^2 - a^2*b*d^2)^(1/2)))/(a^8*d*(a - b)^2*(a^3*d^2 - a^2*b*d^2)^(1/2)) + (4*b^(1/2)*(2*a - 4
*b)*(8*a*b^3*d - 12*a^2*b^2*d + 4*a^3*b*d))/(a^7*(a - b)*(a^3*d^2 - a^2*b*d^2)^(1/2)*(a^2*d^2*(a - b))^(1/2)))
+ (2*(2*a*b^(3/2)*(a^3*d^2 - a^2*b*d^2)^(1/2) - a^2*b^(1/2)*(a^3*d^2 - a^2*b*d^2)^(1/2))*(a^2 - 8*a*b + 8*b^2
))/(a^8*d*(a - b)^2*(a^3*d^2 - a^2*b*d^2)^(1/2)) + (4*b^(1/2)*(2*a^2*b^2*d - 2*a^3*b*d)*(2*a - 4*b))/(a^7*(a -
b)*(a^3*d^2 - a^2*b*d^2)^(1/2)*(a^2*d^2*(a - b))^(1/2))))/(4*b)))/(a^3*d^2 - a^2*b*d^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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