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

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

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

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Rubi [A]  time = 0.131447, 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} \[ -\frac{c+d x}{4 f \left (a^2 \tanh (e+f x)+a^2\right )}+\frac{x (c+d x)}{4 a^2}-\frac{3 d}{16 f^2 \left (a^2 \tanh (e+f x)+a^2\right )}+\frac{3 d x}{16 a^2 f}-\frac{d x^2}{8 a^2}-\frac{c+d x}{4 f (a \tanh (e+f x)+a)^2}-\frac{d}{16 f^2 (a \tanh (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + a*Tanh[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*Tanh[e + f*x])^2) - (c + d*x)/
(4*f*(a + a*Tanh[e + f*x])^2) - (3*d)/(16*f^2*(a^2 + a^2*Tanh[e + f*x])) - (c + d*x)/(4*f*(a^2 + a^2*Tanh[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 \tanh (e+f x))^2} \, dx &=\frac{x (c+d x)}{4 a^2}-\frac{c+d x}{4 f (a+a \tanh (e+f x))^2}-\frac{c+d x}{4 f \left (a^2+a^2 \tanh (e+f x)\right )}-d \int \left (\frac{x}{4 a^2}-\frac{1}{4 f (a+a \tanh (e+f x))^2}-\frac{1}{4 f \left (a^2+a^2 \tanh (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 \tanh (e+f x))^2}-\frac{c+d x}{4 f \left (a^2+a^2 \tanh (e+f x)\right )}+\frac{d \int \frac{1}{(a+a \tanh (e+f x))^2} \, dx}{4 f}+\frac{d \int \frac{1}{a^2+a^2 \tanh (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 \tanh (e+f x))^2}-\frac{c+d x}{4 f (a+a \tanh (e+f x))^2}-\frac{d}{8 f^2 \left (a^2+a^2 \tanh (e+f x)\right )}-\frac{c+d x}{4 f \left (a^2+a^2 \tanh (e+f x)\right )}+\frac{d \int 1 \, dx}{8 a^2 f}+\frac{d \int \frac{1}{a+a \tanh (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 \tanh (e+f x))^2}-\frac{c+d x}{4 f (a+a \tanh (e+f x))^2}-\frac{3 d}{16 f^2 \left (a^2+a^2 \tanh (e+f x)\right )}-\frac{c+d x}{4 f \left (a^2+a^2 \tanh (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 \tanh (e+f x))^2}-\frac{c+d x}{4 f (a+a \tanh (e+f x))^2}-\frac{3 d}{16 f^2 \left (a^2+a^2 \tanh (e+f x)\right )}-\frac{c+d x}{4 f \left (a^2+a^2 \tanh (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.496152, size = 114, normalized size = 0.86 \[ \frac{\text{sech}^2(e+f x) \left (\left (4 c f (4 f x+1)+d \left (8 f^2 x^2+4 f x+1\right )\right ) \sinh (2 (e+f x))+\left (4 c f (4 f x-1)+d \left (8 f^2 x^2-4 f x-1\right )\right ) \cosh (2 (e+f x))-8 (2 c f+2 d f x+d)\right )}{64 a^2 f^2 (\tanh (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sech[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 + Tanh[e + f*x])^2)

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Maple [B]  time = 0.043, size = 404, normalized size = 3. \begin{align*}{\frac{1}{{a}^{2}{f}^{2}} \left ( 2\,d \left ( 1/4\, \left ( fx+e \right ) \sinh \left ( fx+e \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{3}+3/8\, \left ( fx+e \right ) \cosh \left ( fx+e \right ) \sinh \left ( fx+e \right ) +3/16\, \left ( fx+e \right ) ^{2}-1/16\, \left ( \sinh \left ( fx+e \right ) \right ) ^{2} \left ( \cosh \left ( fx+e \right ) \right ) ^{2}-1/4\, \left ( \cosh \left ( fx+e \right ) \right ) ^{2} \right ) +2\,cf \left ( \left ( 1/4\, \left ( \cosh \left ( fx+e \right ) \right ) ^{3}+3/8\,\cosh \left ( fx+e \right ) \right ) \sinh \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -2\,de \left ( \left ( 1/4\, \left ( \cosh \left ( fx+e \right ) \right ) ^{3}+3/8\,\cosh \left ( fx+e \right ) \right ) \sinh \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -2\,d \left ( 1/4\, \left ( fx+e \right ) \left ( \sinh \left ( fx+e \right ) \right ) ^{2} \left ( \cosh \left ( fx+e \right ) \right ) ^{2}+1/4\, \left ( fx+e \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}-1/16\, \left ( \cosh \left ( fx+e \right ) \right ) ^{3}\sinh \left ( fx+e \right ) -{\frac{3\,\cosh \left ( fx+e \right ) \sinh \left ( fx+e \right ) }{32}}-{\frac{3\,fx}{32}}-{\frac{3\,e}{32}} \right ) -2\,cf \left ( 1/4\, \left ( \sinh \left ( fx+e \right ) \right ) ^{2} \left ( \cosh \left ( fx+e \right ) \right ) ^{2}+1/4\, \left ( \cosh \left ( fx+e \right ) \right ) ^{2} \right ) +2\,de \left ( 1/4\, \left ( \sinh \left ( fx+e \right ) \right ) ^{2} \left ( \cosh \left ( fx+e \right ) \right ) ^{2}+1/4\, \left ( \cosh \left ( fx+e \right ) \right ) ^{2} \right ) -d \left ({\frac{ \left ( fx+e \right ) \cosh \left ( fx+e \right ) \sinh \left ( fx+e \right ) }{2}}+{\frac{ \left ( fx+e \right ) ^{2}}{4}}-{\frac{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{4}} \right ) -cf \left ({\frac{\cosh \left ( fx+e \right ) \sinh \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) +de \left ({\frac{\cosh \left ( fx+e \right ) \sinh \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f^2/a^2*(2*d*(1/4*(f*x+e)*sinh(f*x+e)*cosh(f*x+e)^3+3/8*(f*x+e)*cosh(f*x+e)*sinh(f*x+e)+3/16*(f*x+e)^2-1/16*
sinh(f*x+e)^2*cosh(f*x+e)^2-1/4*cosh(f*x+e)^2)+2*c*f*((1/4*cosh(f*x+e)^3+3/8*cosh(f*x+e))*sinh(f*x+e)+3/8*f*x+
3/8*e)-2*d*e*((1/4*cosh(f*x+e)^3+3/8*cosh(f*x+e))*sinh(f*x+e)+3/8*f*x+3/8*e)-2*d*(1/4*(f*x+e)*sinh(f*x+e)^2*co
sh(f*x+e)^2+1/4*(f*x+e)*cosh(f*x+e)^2-1/16*cosh(f*x+e)^3*sinh(f*x+e)-3/32*cosh(f*x+e)*sinh(f*x+e)-3/32*f*x-3/3
2*e)-2*c*f*(1/4*sinh(f*x+e)^2*cosh(f*x+e)^2+1/4*cosh(f*x+e)^2)+2*d*e*(1/4*sinh(f*x+e)^2*cosh(f*x+e)^2+1/4*cosh
(f*x+e)^2)-d*(1/2*(f*x+e)*cosh(f*x+e)*sinh(f*x+e)+1/4*(f*x+e)^2-1/4*cosh(f*x+e)^2)-c*f*(1/2*cosh(f*x+e)*sinh(f
*x+e)+1/2*f*x+1/2*e)+d*e*(1/2*cosh(f*x+e)*sinh(f*x+e)+1/2*f*x+1/2*e))

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Maxima [A]  time = 1.55577, size = 143, 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*tanh(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 = 2.30716, size = 455, normalized size = 3.42 \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*tanh(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh{\left (e + f x \right )} + 1}\, dx + \int \frac{d x}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh{\left (e + f x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(c/(tanh(e + f*x)**2 + 2*tanh(e + f*x) + 1), x) + Integral(d*x/(tanh(e + f*x)**2 + 2*tanh(e + f*x) +
1), x))/a**2

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Giac [A]  time = 1.22751, 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*tanh(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)