3.80 \(\int (a+b \coth (c+d x))^2 \, dx\)

Optimal. Leaf size=38 \[ x \left (a^2+b^2\right )+\frac{2 a b \log (\sinh (c+d x))}{d}-\frac{b^2 \coth (c+d x)}{d} \]

[Out]

(a^2 + b^2)*x - (b^2*Coth[c + d*x])/d + (2*a*b*Log[Sinh[c + d*x]])/d

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Rubi [A]  time = 0.024292, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3477, 3475} \[ x \left (a^2+b^2\right )+\frac{2 a b \log (\sinh (c+d x))}{d}-\frac{b^2 \coth (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Coth[c + d*x])^2,x]

[Out]

(a^2 + b^2)*x - (b^2*Coth[c + d*x])/d + (2*a*b*Log[Sinh[c + d*x]])/d

Rule 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int (a+b \coth (c+d x))^2 \, dx &=\left (a^2+b^2\right ) x-\frac{b^2 \coth (c+d x)}{d}+(2 a b) \int \coth (c+d x) \, dx\\ &=\left (a^2+b^2\right ) x-\frac{b^2 \coth (c+d x)}{d}+\frac{2 a b \log (\sinh (c+d x))}{d}\\ \end{align*}

Mathematica [A]  time = 0.127926, size = 65, normalized size = 1.71 \[ \frac{(a-b)^2 \log (\tanh (c+d x)+1)-(a+b)^2 \log (1-\tanh (c+d x))+4 a b \log (\tanh (c+d x))-2 b^2 \coth (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Coth[c + d*x])^2,x]

[Out]

(-2*b^2*Coth[c + d*x] - (a + b)^2*Log[1 - Tanh[c + d*x]] + 4*a*b*Log[Tanh[c + d*x]] + (a - b)^2*Log[1 + Tanh[c
+ d*x]])/(2*d)

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Maple [B]  time = 0.005, size = 116, normalized size = 3.1 \begin{align*} -{\frac{{b}^{2}{\rm coth} \left (dx+c\right )}{d}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ){a}^{2}}{2\,d}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ) ab}{d}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ){b}^{2}}{2\,d}}+{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ){a}^{2}}{2\,d}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ) ab}{d}}+{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ){b}^{2}}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01268, size = 66, normalized size = 1.74 \begin{align*} b^{2}{\left (x + \frac{c}{d} + \frac{2}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + a^{2} x + \frac{2 \, a b \log \left (\sinh \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*(x + c/d + 2/(d*(e^(-2*d*x - 2*c) - 1))) + a^2*x + 2*a*b*log(sinh(d*x + c))/d

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Fricas [B]  time = 2.85372, size = 535, normalized size = 14.08 \begin{align*} \frac{{\left (a^{2} - 2 \, a b + b^{2}\right )} d x \cosh \left (d x + c\right )^{2} + 2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a^{2} - 2 \, a b + b^{2}\right )} d x \sinh \left (d x + c\right )^{2} -{\left (a^{2} - 2 \, a b + b^{2}\right )} d x - 2 \, b^{2} + 2 \,{\left (a b \cosh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a b \sinh \left (d x + c\right )^{2} - a b\right )} \log \left (\frac{2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{d \cosh \left (d x + c\right )^{2} + 2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + d \sinh \left (d x + c\right )^{2} - d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2 - 2*a*b + b^2)*d*x*cosh(d*x + c)^2 + 2*(a^2 - 2*a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c) + (a^2 - 2*a*
b + b^2)*d*x*sinh(d*x + c)^2 - (a^2 - 2*a*b + b^2)*d*x - 2*b^2 + 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)*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))))/(d*cosh(d*x +
c)^2 + 2*d*cosh(d*x + c)*sinh(d*x + c) + d*sinh(d*x + c)^2 - d)

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Sympy [A]  time = 1.86648, size = 104, normalized size = 2.74 \begin{align*} \begin{cases} x \left (a + b \coth{\left (c \right )}\right )^{2} & \text{for}\: d = 0 \\a^{2} x + \tilde{\infty } a b x + \tilde{\infty } b^{2} x & \text{for}\: c = \log{\left (- e^{- d x} \right )} \vee c = \log{\left (e^{- d x} \right )} \\a^{2} x + 2 a b x - \frac{2 a b \log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{d} + \frac{2 a b \log{\left (\tanh{\left (c + d x \right )} \right )}}{d} + b^{2} x - \frac{b^{2}}{d \tanh{\left (c + d x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*(a + b*coth(c))**2, Eq(d, 0)), (a**2*x + zoo*a*b*x + zoo*b**2*x, Eq(c, log(exp(-d*x))) | Eq(c, lo
g(-exp(-d*x)))), (a**2*x + 2*a*b*x - 2*a*b*log(tanh(c + d*x) + 1)/d + 2*a*b*log(tanh(c + d*x))/d + b**2*x - b*
*2/(d*tanh(c + d*x)), True))

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Giac [A]  time = 1.12428, size = 84, normalized size = 2.21 \begin{align*} \frac{2 \, a b \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{d} + \frac{{\left (a^{2} - 2 \, a b + b^{2}\right )}{\left (d x + c\right )}}{d} - \frac{2 \, b^{2}}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*b*log(abs(e^(2*d*x + 2*c) - 1))/d + (a^2 - 2*a*b + b^2)*(d*x + c)/d - 2*b^2/(d*(e^(2*d*x + 2*c) - 1))