3.24 \(\int \frac{c+d x}{(a+a \coth (e+f x))^2} \, dx\)

Optimal. Leaf size=133 \[ -\frac{c+d x}{4 f \left (a^2 \coth (e+f x)+a^2\right )}+\frac{x (c+d x)}{4 a^2}-\frac{3 d}{16 f^2 \left (a^2 \coth (e+f x)+a^2\right )}+\frac{3 d x}{16 a^2 f}-\frac{d x^2}{8 a^2}-\frac{c+d x}{4 f (a \coth (e+f x)+a)^2}-\frac{d}{16 f^2 (a \coth (e+f x)+a)^2} \]

[Out]

(3*d*x)/(16*a^2*f) - (d*x^2)/(8*a^2) + (x*(c + d*x))/(4*a^2) - d/(16*f^2*(a + a*Coth[e + f*x])^2) - (c + d*x)/
(4*f*(a + a*Coth[e + f*x])^2) - (3*d)/(16*f^2*(a^2 + a^2*Coth[e + f*x])) - (c + d*x)/(4*f*(a^2 + a^2*Coth[e +
f*x]))

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Rubi [A]  time = 0.129456, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3479, 8, 3730} \[ -\frac{c+d x}{4 f \left (a^2 \coth (e+f x)+a^2\right )}+\frac{x (c+d x)}{4 a^2}-\frac{3 d}{16 f^2 \left (a^2 \coth (e+f x)+a^2\right )}+\frac{3 d x}{16 a^2 f}-\frac{d x^2}{8 a^2}-\frac{c+d x}{4 f (a \coth (e+f x)+a)^2}-\frac{d}{16 f^2 (a \coth (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + a*Coth[e + f*x])^2,x]

[Out]

(3*d*x)/(16*a^2*f) - (d*x^2)/(8*a^2) + (x*(c + d*x))/(4*a^2) - d/(16*f^2*(a + a*Coth[e + f*x])^2) - (c + d*x)/
(4*f*(a + a*Coth[e + f*x])^2) - (3*d)/(16*f^2*(a^2 + a^2*Coth[e + f*x])) - (c + d*x)/(4*f*(a^2 + a^2*Coth[e +
f*x]))

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3730

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{u = IntHide[(a
+ b*Tan[e + f*x])^n, x]}, Dist[(c + d*x)^m, u, x] - Dist[d*m, Int[Dist[(c + d*x)^(m - 1), u, x], x], x]] /; Fr
eeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 + b^2, 0] && ILtQ[n, -1] && GtQ[m, 0]

Rubi steps

\begin{align*} \int \frac{c+d x}{(a+a \coth (e+f x))^2} \, dx &=\frac{x (c+d x)}{4 a^2}-\frac{c+d x}{4 f (a+a \coth (e+f x))^2}-\frac{c+d x}{4 f \left (a^2+a^2 \coth (e+f x)\right )}-d \int \left (\frac{x}{4 a^2}-\frac{1}{4 f (a+a \coth (e+f x))^2}-\frac{1}{4 f \left (a^2+a^2 \coth (e+f x)\right )}\right ) \, dx\\ &=-\frac{d x^2}{8 a^2}+\frac{x (c+d x)}{4 a^2}-\frac{c+d x}{4 f (a+a \coth (e+f x))^2}-\frac{c+d x}{4 f \left (a^2+a^2 \coth (e+f x)\right )}+\frac{d \int \frac{1}{(a+a \coth (e+f x))^2} \, dx}{4 f}+\frac{d \int \frac{1}{a^2+a^2 \coth (e+f x)} \, dx}{4 f}\\ &=-\frac{d x^2}{8 a^2}+\frac{x (c+d x)}{4 a^2}-\frac{d}{16 f^2 (a+a \coth (e+f x))^2}-\frac{c+d x}{4 f (a+a \coth (e+f x))^2}-\frac{d}{8 f^2 \left (a^2+a^2 \coth (e+f x)\right )}-\frac{c+d x}{4 f \left (a^2+a^2 \coth (e+f x)\right )}+\frac{d \int 1 \, dx}{8 a^2 f}+\frac{d \int \frac{1}{a+a \coth (e+f x)} \, dx}{8 a f}\\ &=\frac{d x}{8 a^2 f}-\frac{d x^2}{8 a^2}+\frac{x (c+d x)}{4 a^2}-\frac{d}{16 f^2 (a+a \coth (e+f x))^2}-\frac{c+d x}{4 f (a+a \coth (e+f x))^2}-\frac{3 d}{16 f^2 \left (a^2+a^2 \coth (e+f x)\right )}-\frac{c+d x}{4 f \left (a^2+a^2 \coth (e+f x)\right )}+\frac{d \int 1 \, dx}{16 a^2 f}\\ &=\frac{3 d x}{16 a^2 f}-\frac{d x^2}{8 a^2}+\frac{x (c+d x)}{4 a^2}-\frac{d}{16 f^2 (a+a \coth (e+f x))^2}-\frac{c+d x}{4 f (a+a \coth (e+f x))^2}-\frac{3 d}{16 f^2 \left (a^2+a^2 \coth (e+f x)\right )}-\frac{c+d x}{4 f \left (a^2+a^2 \coth (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.521503, size = 114, normalized size = 0.86 \[ \frac{\text{csch}^2(e+f x) \left (\left (4 c f (4 f x+1)+d \left (8 f^2 x^2+4 f x+1\right )\right ) \sinh (2 (e+f x))+\left (4 c f (4 f x-1)+d \left (8 f^2 x^2-4 f x-1\right )\right ) \cosh (2 (e+f x))+8 (2 c f+2 d f x+d)\right )}{64 a^2 f^2 (\coth (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)/(a + a*Coth[e + f*x])^2,x]

[Out]

(Csch[e + f*x]^2*(8*(d + 2*c*f + 2*d*f*x) + (4*c*f*(-1 + 4*f*x) + d*(-1 - 4*f*x + 8*f^2*x^2))*Cosh[2*(e + f*x)
] + (4*c*f*(1 + 4*f*x) + d*(1 + 4*f*x + 8*f^2*x^2))*Sinh[2*(e + f*x)]))/(64*a^2*f^2*(1 + Coth[e + f*x])^2)

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Maple [A]  time = 0.122, size = 74, normalized size = 0.6 \begin{align*}{\frac{d{x}^{2}}{8\,{a}^{2}}}+{\frac{cx}{4\,{a}^{2}}}+{\frac{ \left ( 2\,dfx+2\,cf+d \right ){{\rm e}^{-2\,fx-2\,e}}}{8\,{a}^{2}{f}^{2}}}-{\frac{ \left ( 4\,dfx+4\,cf+d \right ){{\rm e}^{-4\,fx-4\,e}}}{64\,{a}^{2}{f}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/(a+a*coth(f*x+e))^2,x)

[Out]

1/8*d*x^2/a^2+1/4/a^2*c*x+1/8*(2*d*f*x+2*c*f+d)/a^2/f^2*exp(-2*f*x-2*e)-1/64*(4*d*f*x+4*c*f+d)/a^2/f^2*exp(-4*
f*x-4*e)

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Maxima [A]  time = 1.50981, size = 144, normalized size = 1.08 \begin{align*} \frac{1}{16} \, c{\left (\frac{4 \,{\left (f x + e\right )}}{a^{2} f} + \frac{4 \, e^{\left (-2 \, f x - 2 \, e\right )} - e^{\left (-4 \, f x - 4 \, e\right )}}{a^{2} f}\right )} + \frac{{\left (8 \, f^{2} x^{2} e^{\left (4 \, e\right )} + 8 \,{\left (2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} -{\left (4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} d e^{\left (-4 \, e\right )}}{64 \, a^{2} f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*c*(4*(f*x + e)/(a^2*f) + (4*e^(-2*f*x - 2*e) - e^(-4*f*x - 4*e))/(a^2*f)) + 1/64*(8*f^2*x^2*e^(4*e) + 8*(
2*f*x*e^(2*e) + e^(2*e))*e^(-2*f*x) - (4*f*x + 1)*e^(-4*f*x))*d*e^(-4*e)/(a^2*f^2)

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Fricas [A]  time = 2.03748, size = 454, normalized size = 3.41 \begin{align*} \frac{16 \, d f x +{\left (8 \, d f^{2} x^{2} - 4 \, c f + 4 \,{\left (4 \, c f^{2} - d f\right )} x - d\right )} \cosh \left (f x + e\right )^{2} + 2 \,{\left (8 \, d f^{2} x^{2} + 4 \, c f + 4 \,{\left (4 \, c f^{2} + d f\right )} x + d\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) +{\left (8 \, d f^{2} x^{2} - 4 \, c f + 4 \,{\left (4 \, c f^{2} - d f\right )} x - d\right )} \sinh \left (f x + e\right )^{2} + 16 \, c f + 8 \, d}{64 \,{\left (a^{2} f^{2} \cosh \left (f x + e\right )^{2} + 2 \, a^{2} f^{2} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + a^{2} f^{2} \sinh \left (f x + e\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(16*d*f*x + (8*d*f^2*x^2 - 4*c*f + 4*(4*c*f^2 - d*f)*x - d)*cosh(f*x + e)^2 + 2*(8*d*f^2*x^2 + 4*c*f + 4*
(4*c*f^2 + d*f)*x + d)*cosh(f*x + e)*sinh(f*x + e) + (8*d*f^2*x^2 - 4*c*f + 4*(4*c*f^2 - d*f)*x - d)*sinh(f*x
+ e)^2 + 16*c*f + 8*d)/(a^2*f^2*cosh(f*x + e)^2 + 2*a^2*f^2*cosh(f*x + e)*sinh(f*x + e) + a^2*f^2*sinh(f*x + e
)^2)

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Sympy [A]  time = 2.84553, size = 700, normalized size = 5.26 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+a*coth(f*x+e))**2,x)

[Out]

Piecewise((4*c*f**2*x*tanh(e + f*x)**2/(16*a**2*f**2*tanh(e + f*x)**2 + 32*a**2*f**2*tanh(e + f*x) + 16*a**2*f
**2) + 8*c*f**2*x*tanh(e + f*x)/(16*a**2*f**2*tanh(e + f*x)**2 + 32*a**2*f**2*tanh(e + f*x) + 16*a**2*f**2) +
4*c*f**2*x/(16*a**2*f**2*tanh(e + f*x)**2 + 32*a**2*f**2*tanh(e + f*x) + 16*a**2*f**2) + 12*c*f*tanh(e + f*x)/
(16*a**2*f**2*tanh(e + f*x)**2 + 32*a**2*f**2*tanh(e + f*x) + 16*a**2*f**2) + 8*c*f/(16*a**2*f**2*tanh(e + f*x
)**2 + 32*a**2*f**2*tanh(e + f*x) + 16*a**2*f**2) + 2*d*f**2*x**2*tanh(e + f*x)**2/(16*a**2*f**2*tanh(e + f*x)
**2 + 32*a**2*f**2*tanh(e + f*x) + 16*a**2*f**2) + 4*d*f**2*x**2*tanh(e + f*x)/(16*a**2*f**2*tanh(e + f*x)**2
+ 32*a**2*f**2*tanh(e + f*x) + 16*a**2*f**2) + 2*d*f**2*x**2/(16*a**2*f**2*tanh(e + f*x)**2 + 32*a**2*f**2*tan
h(e + f*x) + 16*a**2*f**2) - 5*d*f*x*tanh(e + f*x)**2/(16*a**2*f**2*tanh(e + f*x)**2 + 32*a**2*f**2*tanh(e + f
*x) + 16*a**2*f**2) + 2*d*f*x*tanh(e + f*x)/(16*a**2*f**2*tanh(e + f*x)**2 + 32*a**2*f**2*tanh(e + f*x) + 16*a
**2*f**2) + 3*d*f*x/(16*a**2*f**2*tanh(e + f*x)**2 + 32*a**2*f**2*tanh(e + f*x) + 16*a**2*f**2) + 5*d*tanh(e +
f*x)/(16*a**2*f**2*tanh(e + f*x)**2 + 32*a**2*f**2*tanh(e + f*x) + 16*a**2*f**2) + 4*d/(16*a**2*f**2*tanh(e +
f*x)**2 + 32*a**2*f**2*tanh(e + f*x) + 16*a**2*f**2), Ne(f, 0)), ((c*x + d*x**2/2)/(a*coth(e) + a)**2, True))

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Giac [A]  time = 1.145, size = 147, normalized size = 1.11 \begin{align*} \frac{{\left (8 \, d f^{2} x^{2} e^{\left (4 \, f x + 4 \, e\right )} + 16 \, c f^{2} x e^{\left (4 \, f x + 4 \, e\right )} + 16 \, d f x e^{\left (2 \, f x + 2 \, e\right )} - 4 \, d f x + 16 \, c f e^{\left (2 \, f x + 2 \, e\right )} - 4 \, c f + 8 \, d e^{\left (2 \, f x + 2 \, e\right )} - d\right )} e^{\left (-4 \, f x - 4 \, e\right )}}{64 \, a^{2} f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(8*d*f^2*x^2*e^(4*f*x + 4*e) + 16*c*f^2*x*e^(4*f*x + 4*e) + 16*d*f*x*e^(2*f*x + 2*e) - 4*d*f*x + 16*c*f*e
^(2*f*x + 2*e) - 4*c*f + 8*d*e^(2*f*x + 2*e) - d)*e^(-4*f*x - 4*e)/(a^2*f^2)