3.6 \(\int \frac {1}{(a+b \coth ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=89 \[ -\frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {b}}\right )}{2 a^{3/2} d (a+b)^2}+\frac {b \coth (c+d x)}{2 a d (a+b) \left (a+b \coth ^2(c+d x)\right )}+\frac {x}{(a+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3661, 414, 522, 206, 205} \[ -\frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {b}}\right )}{2 a^{3/2} d (a+b)^2}+\frac {b \coth (c+d x)}{2 a d (a+b) \left (a+b \coth ^2(c+d x)\right )}+\frac {x}{(a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Coth[c + d*x]^2)^(-2),x]

[Out]

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

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 206

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

Rule 414

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

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \coth ^2(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\coth (c+d x)\right )}{d}\\ &=\frac {b \coth (c+d x)}{2 a (a+b) d \left (a+b \coth ^2(c+d x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {b-2 (a+b)+b x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\coth (c+d x)\right )}{2 a (a+b) d}\\ &=\frac {b \coth (c+d x)}{2 a (a+b) d \left (a+b \coth ^2(c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\coth (c+d x)\right )}{(a+b)^2 d}+\frac {(b (3 a+b)) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\coth (c+d x)\right )}{2 a (a+b)^2 d}\\ &=\frac {x}{(a+b)^2}-\frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {a} \tanh (c+d x)}{\sqrt {b}}\right )}{2 a^{3/2} (a+b)^2 d}+\frac {b \coth (c+d x)}{2 a (a+b) d \left (a+b \coth ^2(c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 97, normalized size = 1.09 \[ \frac {\frac {\sqrt {b} (3 a+b) \tan ^{-1}\left (\frac {\sqrt {b} \coth (c+d x)}{\sqrt {a}}\right )}{a^{3/2}}+\frac {b (a+b) \coth (c+d x)}{a \left (a+b \coth ^2(c+d x)\right )}-\log (1-\coth (c+d x))+\log (\coth (c+d x)+1)}{2 d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[b]*(3*a + b)*ArcTan[(Sqrt[b]*Coth[c + d*x])/Sqrt[a]])/a^(3/2) + (b*(a + b)*Coth[c + d*x])/(a*(a + b*Cot
h[c + d*x]^2)) - Log[1 - Coth[c + d*x]] + Log[1 + Coth[c + d*x]])/(2*(a + b)^2*d)

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fricas [B]  time = 0.48, size = 1952, normalized size = 21.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*(a^2 + a*b)*d*x*cosh(d*x + c)^4 + 16*(a^2 + a*b)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 4*(a^2 + a*b)*d*x
*sinh(d*x + c)^4 + 4*(a^2 + a*b)*d*x - 4*(2*(a^2 - a*b)*d*x - a*b + b^2)*cosh(d*x + c)^2 + 4*(6*(a^2 + a*b)*d*
x*cosh(d*x + c)^2 - 2*(a^2 - a*b)*d*x + a*b - b^2)*sinh(d*x + c)^2 + ((3*a^2 + 4*a*b + b^2)*cosh(d*x + c)^4 +
4*(3*a^2 + 4*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (3*a^2 + 4*a*b + b^2)*sinh(d*x + c)^4 - 2*(3*a^2 - 2*a
*b - b^2)*cosh(d*x + c)^2 + 2*(3*(3*a^2 + 4*a*b + b^2)*cosh(d*x + c)^2 - 3*a^2 + 2*a*b + b^2)*sinh(d*x + c)^2
+ 3*a^2 + 4*a*b + b^2 + 4*((3*a^2 + 4*a*b + b^2)*cosh(d*x + c)^3 - (3*a^2 - 2*a*b - b^2)*cosh(d*x + c))*sinh(d
*x + c))*sqrt(-b/a)*log(((a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x +
c)^3 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 - 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x
+ c)^2 - a^2 + b^2)*sinh(d*x + c)^2 + a^2 - 6*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 - (a^2 - b^2
)*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 + a*b)*sqrt(-b/a))/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d
*x + c)^3 + (a + b)*sinh(d*x + c)^4 - 2*(a - b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 - a + b)*sinh(d
*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 - (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)) - 4*a*b - 4*b^2 + 8*(2
*(a^2 + a*b)*d*x*cosh(d*x + c)^3 - (2*(a^2 - a*b)*d*x - a*b + b^2)*cosh(d*x + c))*sinh(d*x + c))/((a^4 + 3*a^3
*b + 3*a^2*b^2 + a*b^3)*d*cosh(d*x + c)^4 + 4*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*cosh(d*x + c)*sinh(d*x + c
)^3 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*sinh(d*x + c)^4 - 2*(a^4 + a^3*b - a^2*b^2 - a*b^3)*d*cosh(d*x + c
)^2 + 2*(3*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*cosh(d*x + c)^2 - (a^4 + a^3*b - a^2*b^2 - a*b^3)*d)*sinh(d*x
+ c)^2 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d + 4*((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*cosh(d*x + c)^3 - (
a^4 + a^3*b - a^2*b^2 - a*b^3)*d*cosh(d*x + c))*sinh(d*x + c)), 1/2*(2*(a^2 + a*b)*d*x*cosh(d*x + c)^4 + 8*(a^
2 + a*b)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a^2 + a*b)*d*x*sinh(d*x + c)^4 + 2*(a^2 + a*b)*d*x - 2*(2*(a^2
- a*b)*d*x - a*b + b^2)*cosh(d*x + c)^2 + 2*(6*(a^2 + a*b)*d*x*cosh(d*x + c)^2 - 2*(a^2 - a*b)*d*x + a*b - b^
2)*sinh(d*x + c)^2 - ((3*a^2 + 4*a*b + b^2)*cosh(d*x + c)^4 + 4*(3*a^2 + 4*a*b + b^2)*cosh(d*x + c)*sinh(d*x +
c)^3 + (3*a^2 + 4*a*b + b^2)*sinh(d*x + c)^4 - 2*(3*a^2 - 2*a*b - b^2)*cosh(d*x + c)^2 + 2*(3*(3*a^2 + 4*a*b
+ b^2)*cosh(d*x + c)^2 - 3*a^2 + 2*a*b + b^2)*sinh(d*x + c)^2 + 3*a^2 + 4*a*b + b^2 + 4*((3*a^2 + 4*a*b + b^2)
*cosh(d*x + c)^3 - (3*a^2 - 2*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(b/a)*arctan(1/2*((a + b)*cosh(d*x
+ c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2 - a + b)*sqrt(b/a)/b) - 2*a*b - 2*b^2
+ 4*(2*(a^2 + a*b)*d*x*cosh(d*x + c)^3 - (2*(a^2 - a*b)*d*x - a*b + b^2)*cosh(d*x + c))*sinh(d*x + c))/((a^4
+ 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*cosh(d*x + c)^4 + 4*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*cosh(d*x + c)*sinh(
d*x + c)^3 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*sinh(d*x + c)^4 - 2*(a^4 + a^3*b - a^2*b^2 - a*b^3)*d*cosh(
d*x + c)^2 + 2*(3*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*cosh(d*x + c)^2 - (a^4 + a^3*b - a^2*b^2 - a*b^3)*d)*s
inh(d*x + c)^2 + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d + 4*((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*cosh(d*x + c
)^3 - (a^4 + a^3*b - a^2*b^2 - a*b^3)*d*cosh(d*x + c))*sinh(d*x + c))]

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giac [B]  time = 0.17, size = 198, normalized size = 2.22 \[ -\frac {\frac {{\left (3 \, a b + b^{2}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} - a + b}{2 \, \sqrt {a b}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b}} - \frac {2 \, {\left (d x + c\right )}}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (a b e^{\left (2 \, d x + 2 \, c\right )} - b^{2} e^{\left (2 \, d x + 2 \, c\right )} - a b - b^{2}\right )}}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} - 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((3*a*b + b^2)*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) - a + b)/sqrt(a*b))/((a^3 + 2*a^2*b + a*
b^2)*sqrt(a*b)) - 2*(d*x + c)/(a^2 + 2*a*b + b^2) - 2*(a*b*e^(2*d*x + 2*c) - b^2*e^(2*d*x + 2*c) - a*b - b^2)/
((a^3 + 2*a^2*b + a*b^2)*(a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) - 2*a*e^(2*d*x + 2*c) + 2*b*e^(2*d*x + 2*c) +
a + b)))/d

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maple [B]  time = 0.13, size = 172, normalized size = 1.93 \[ -\frac {\ln \left (\coth \left (d x +c \right )-1\right )}{2 d \left (a +b \right )^{2}}+\frac {\ln \left (\coth \left (d x +c \right )+1\right )}{2 d \left (a +b \right )^{2}}+\frac {b \coth \left (d x +c \right )}{2 d \left (a +b \right )^{2} \left (a +b \left (\coth ^{2}\left (d x +c \right )\right )\right )}+\frac {b^{2} \coth \left (d x +c \right )}{2 d \left (a +b \right )^{2} a \left (a +b \left (\coth ^{2}\left (d x +c \right )\right )\right )}+\frac {3 b \arctan \left (\frac {\coth \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d \left (a +b \right )^{2} \sqrt {a b}}+\frac {b^{2} \arctan \left (\frac {\coth \left (d x +c \right ) b}{\sqrt {a b}}\right )}{2 d \left (a +b \right )^{2} a \sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.47, size = 207, normalized size = 2.33 \[ \frac {{\left (3 \, a b + b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} - a + b}{2 \, \sqrt {a b}}\right )}{2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b} d} + \frac {a b + b^{2} - {\left (a b - b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3} - 2 \, {\left (a^{4} + a^{3} b - a^{2} b^{2} - a b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*a*b + b^2)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) - a + b)/sqrt(a*b))/((a^3 + 2*a^2*b + a*b^2)*sqrt(a*b)*
d) + (a*b + b^2 - (a*b - b^2)*e^(-2*d*x - 2*c))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 - 2*(a^4 + a^3*b - a^2*b^2
- a*b^3)*e^(-2*d*x - 2*c) + (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*e^(-4*d*x - 4*c))*d) + (d*x + c)/((a^2 + 2*a*
b + b^2)*d)

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mupad [B]  time = 1.58, size = 110, normalized size = 1.24 \[ \frac {\frac {a\,x}{{\left (a+b\right )}^2}+\frac {b\,x\,{\mathrm {coth}\left (c+d\,x\right )}^2}{{\left (a+b\right )}^2}+\frac {b\,\mathrm {coth}\left (c+d\,x\right )}{2\,a\,d\,\left (a+b\right )}}{b\,{\mathrm {coth}\left (c+d\,x\right )}^2+a}+\frac {\mathrm {atan}\left (\frac {b\,\mathrm {coth}\left (c+d\,x\right )}{\sqrt {a\,b}}\right )\,\left (b^2+3\,a\,b\right )}{\sqrt {a\,b}\,\left (2\,a^3\,d+a\,b\,\left (4\,a\,d+2\,b\,d\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*coth(c + d*x)^2)^2,x)

[Out]

((a*x)/(a + b)^2 + (b*x*coth(c + d*x)^2)/(a + b)^2 + (b*coth(c + d*x))/(2*a*d*(a + b)))/(a + b*coth(c + d*x)^2
) + (atan((b*coth(c + d*x))/(a*b)^(1/2))*(3*a*b + b^2))/((a*b)^(1/2)*(2*a^3*d + a*b*(4*a*d + 2*b*d)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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