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

Optimal. Leaf size=46 \[ \frac{x}{a+b}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{b}}\right )}{\sqrt{a} d (a+b)} \]

[Out]

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

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Rubi [A]  time = 0.0774405, antiderivative size = 46, normalized size of antiderivative = 1., 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 = {3660, 3675, 205} \[ \frac{x}{a+b}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{b}}\right )}{\sqrt{a} d (a+b)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Coth[c + d*x]^2)^(-1),x]

[Out]

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

Rule 3660

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

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

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

Mathematica [A]  time = 0.0575124, size = 47, normalized size = 1.02 \[ \frac{\tanh ^{-1}(\tanh (c+d x))-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{a} \tanh (c+d x)}{\sqrt{b}}\right )}{\sqrt{a}}}{d (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 76, normalized size = 1.7 \begin{align*}{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ) }{d \left ( 2\,b+2\,a \right ) }}+{\frac{b}{d \left ( a+b \right ) }\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 ) }{d \left ( 2\,b+2\,a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/(2*b+2*a)*ln(coth(d*x+c)+1)+1/d*b/(a+b)/(a*b)^(1/2)*arctan(coth(d*x+c)*b/(a*b)^(1/2))-1/d/(2*b+2*a)*ln(cot
h(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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.2201, size = 1249, normalized size = 27.15 \begin{align*} \left [\frac{2 \, d x + \sqrt{-\frac{b}{a}} \log \left (\frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} + 4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{4} - 2 \,{\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} - a^{2} + b^{2}\right )} \sinh \left (d x + c\right )^{2} + a^{2} - 6 \, a b + b^{2} + 4 \,{\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{3} -{\left (a^{2} - b^{2}\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} + a b\right )} \sqrt{-\frac{b}{a}}}{{\left (a + b\right )} \cosh \left (d x + c\right )^{4} + 4 \,{\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (a + b\right )} \sinh \left (d x + c\right )^{4} - 2 \,{\left (a - b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \,{\left (a + b\right )} \cosh \left (d x + c\right )^{2} - a + b\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{3} -{\left (a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a + b}\right )}{2 \,{\left (a + b\right )} d}, \frac{d x - \sqrt{\frac{b}{a}} \arctan \left (\frac{{\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a + b\right )} \sinh \left (d x + c\right )^{2} - a + b\right )} \sqrt{\frac{b}{a}}}{2 \, b}\right )}{{\left (a + b\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*d*x + 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)*cos
h(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)*s
inh(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)*s
inh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 - (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)))/((a + b)*d), (d*
x - 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))/((a + b)*d)]

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Sympy [A]  time = 27.7931, size = 294, normalized size = 6.39 \begin{align*} \begin{cases} \frac{\tilde{\infty } x}{\coth ^{2}{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{x - \frac{\tanh{\left (c + d x \right )}}{d}}{b} & \text{for}\: a = 0 \\- \frac{d x \tanh ^{2}{\left (c + d x \right )}}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} + \frac{d x}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} - \frac{\tanh{\left (c + d x \right )}}{2 b d \tanh ^{2}{\left (c + d x \right )} - 2 b d} & \text{for}\: a = - b \\\frac{x}{a} & \text{for}\: b = 0 \\\frac{x}{a + b \coth ^{2}{\left (c \right )}} & \text{for}\: d = 0 \\\frac{2 i a \sqrt{b} d x \sqrt{\frac{1}{a}}}{2 i a^{2} \sqrt{b} d \sqrt{\frac{1}{a}} + 2 i a b^{\frac{3}{2}} d \sqrt{\frac{1}{a}}} - \frac{b \log{\left (- i \sqrt{b} \sqrt{\frac{1}{a}} + \tanh{\left (c + d x \right )} \right )}}{2 i a^{2} \sqrt{b} d \sqrt{\frac{1}{a}} + 2 i a b^{\frac{3}{2}} d \sqrt{\frac{1}{a}}} + \frac{b \log{\left (i \sqrt{b} \sqrt{\frac{1}{a}} + \tanh{\left (c + d x \right )} \right )}}{2 i a^{2} \sqrt{b} d \sqrt{\frac{1}{a}} + 2 i a b^{\frac{3}{2}} d \sqrt{\frac{1}{a}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x/coth(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((x - tanh(c + d*x)/d)/b, Eq(a, 0)), (-d*x*tanh(
c + d*x)**2/(2*b*d*tanh(c + d*x)**2 - 2*b*d) + d*x/(2*b*d*tanh(c + d*x)**2 - 2*b*d) - tanh(c + d*x)/(2*b*d*tan
h(c + d*x)**2 - 2*b*d), Eq(a, -b)), (x/a, Eq(b, 0)), (x/(a + b*coth(c)**2), Eq(d, 0)), (2*I*a*sqrt(b)*d*x*sqrt
(1/a)/(2*I*a**2*sqrt(b)*d*sqrt(1/a) + 2*I*a*b**(3/2)*d*sqrt(1/a)) - b*log(-I*sqrt(b)*sqrt(1/a) + tanh(c + d*x)
)/(2*I*a**2*sqrt(b)*d*sqrt(1/a) + 2*I*a*b**(3/2)*d*sqrt(1/a)) + b*log(I*sqrt(b)*sqrt(1/a) + tanh(c + d*x))/(2*
I*a**2*sqrt(b)*d*sqrt(1/a) + 2*I*a*b**(3/2)*d*sqrt(1/a)), True))

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Giac [A]  time = 1.14612, size = 92, normalized size = 2. \begin{align*} -\frac{b \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 )}{\sqrt{a b}{\left (a d + b d\right )}} + \frac{d x + c}{a d + b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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