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 - (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))
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Rubi [A] time = 0.101391, antiderivative size = 89, normalized size of antiderivative = 1.,
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 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])
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 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 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]
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*}
Mathematica [A] time = 0.570929, 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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Maple [B] time = 0.028, size = 172, normalized size = 1.9 \begin{align*}{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ) }{2\,d \left ( a+b \right ) ^{2}}}+{\frac{b{\rm coth} \left (dx+c\right )}{2\,d \left ( a+b \right ) ^{2} \left ( a+b \left ({\rm coth} \left (dx+c\right ) \right ) ^{2} \right ) }}+{\frac{{b}^{2}{\rm coth} \left (dx+c\right )}{2\,d \left ( a+b \right ) ^{2}a \left ( a+b \left ({\rm coth} \left (dx+c\right ) \right ) ^{2} \right ) }}+{\frac{3\,b}{2\,d \left ( a+b \right ) ^{2}}\arctan \left ({b{\rm coth} \left (dx+c\right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{{b}^{2}}{2\,d \left ( a+b \right ) ^{2}a}\arctan \left ({b{\rm coth} \left (dx+c\right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ) }{2\,d \left ( a+b \right ) ^{2}}} \end{align*}
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*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))-1/2/d/(a+b)^2*ln(coth(d*x+c)-1)
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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(1/(a+b*coth(d*x+c)^2)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 3.71707, size = 4703, normalized size = 52.84 \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*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))]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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
________________________________________________________________________________________
Giac [B] time = 1.17083, size = 277, normalized size = 3.11 \begin{align*} -\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 )}{2 \,{\left (a^{3} d + 2 \, a^{2} b d + a b^{2} d\right )} \sqrt{a b}} + \frac{d x + c}{a^{2} d + 2 \, a b d + b^{2} d} + \frac{a b e^{\left (2 \, d x + 2 \, c\right )} - b^{2} e^{\left (2 \, d x + 2 \, c\right )} - a b - b^{2}}{{\left (a^{3} d + 2 \, a^{2} b d + a b^{2} d\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 )}} \end{align*}
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*d + 2*a^2*b*d +
a*b^2*d)*sqrt(a*b)) + (d*x + c)/(a^2*d + 2*a*b*d + b^2*d) + (a*b*e^(2*d*x + 2*c) - b^2*e^(2*d*x + 2*c) - a*b
- b^2)/((a^3*d + 2*a^2*b*d + a*b^2*d)*(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))