3.45 \(\int \frac{c+d x}{(a+a \tanh (e+f x))^3} \, dx\)

Optimal. Leaf size=183 \[ -\frac{c+d x}{8 f \left (a^3 \tanh (e+f x)+a^3\right )}+\frac{x (c+d x)}{8 a^3}-\frac{11 d}{96 f^2 \left (a^3 \tanh (e+f x)+a^3\right )}+\frac{11 d x}{96 a^3 f}-\frac{d x^2}{16 a^3}-\frac{c+d x}{8 a f (a \tanh (e+f x)+a)^2}-\frac{c+d x}{6 f (a \tanh (e+f x)+a)^3}-\frac{5 d}{96 a f^2 (a \tanh (e+f x)+a)^2}-\frac{d}{36 f^2 (a \tanh (e+f x)+a)^3} \]

[Out]

(11*d*x)/(96*a^3*f) - (d*x^2)/(16*a^3) + (x*(c + d*x))/(8*a^3) - d/(36*f^2*(a + a*Tanh[e + f*x])^3) - (c + d*x
)/(6*f*(a + a*Tanh[e + f*x])^3) - (5*d)/(96*a*f^2*(a + a*Tanh[e + f*x])^2) - (c + d*x)/(8*a*f*(a + a*Tanh[e +
f*x])^2) - (11*d)/(96*f^2*(a^3 + a^3*Tanh[e + f*x])) - (c + d*x)/(8*f*(a^3 + a^3*Tanh[e + f*x]))

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Rubi [A]  time = 0.212382, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3479, 8, 3730} \[ -\frac{c+d x}{8 f \left (a^3 \tanh (e+f x)+a^3\right )}+\frac{x (c+d x)}{8 a^3}-\frac{11 d}{96 f^2 \left (a^3 \tanh (e+f x)+a^3\right )}+\frac{11 d x}{96 a^3 f}-\frac{d x^2}{16 a^3}-\frac{c+d x}{8 a f (a \tanh (e+f x)+a)^2}-\frac{c+d x}{6 f (a \tanh (e+f x)+a)^3}-\frac{5 d}{96 a f^2 (a \tanh (e+f x)+a)^2}-\frac{d}{36 f^2 (a \tanh (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + a*Tanh[e + f*x])^3,x]

[Out]

(11*d*x)/(96*a^3*f) - (d*x^2)/(16*a^3) + (x*(c + d*x))/(8*a^3) - d/(36*f^2*(a + a*Tanh[e + f*x])^3) - (c + d*x
)/(6*f*(a + a*Tanh[e + f*x])^3) - (5*d)/(96*a*f^2*(a + a*Tanh[e + f*x])^2) - (c + d*x)/(8*a*f*(a + a*Tanh[e +
f*x])^2) - (11*d)/(96*f^2*(a^3 + a^3*Tanh[e + f*x])) - (c + d*x)/(8*f*(a^3 + a^3*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))^3} \, dx &=\frac{x (c+d x)}{8 a^3}-\frac{c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac{c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac{c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}-d \int \left (\frac{x}{8 a^3}-\frac{1}{6 f (a+a \tanh (e+f x))^3}-\frac{1}{8 a f (a+a \tanh (e+f x))^2}-\frac{1}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}\right ) \, dx\\ &=-\frac{d x^2}{16 a^3}+\frac{x (c+d x)}{8 a^3}-\frac{c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac{c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac{c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}+\frac{d \int \frac{1}{a^3+a^3 \tanh (e+f x)} \, dx}{8 f}+\frac{d \int \frac{1}{(a+a \tanh (e+f x))^3} \, dx}{6 f}+\frac{d \int \frac{1}{(a+a \tanh (e+f x))^2} \, dx}{8 a f}\\ &=-\frac{d x^2}{16 a^3}+\frac{x (c+d x)}{8 a^3}-\frac{d}{36 f^2 (a+a \tanh (e+f x))^3}-\frac{c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac{d}{32 a f^2 (a+a \tanh (e+f x))^2}-\frac{c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac{d}{16 f^2 \left (a^3+a^3 \tanh (e+f x)\right )}-\frac{c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}+\frac{d \int 1 \, dx}{16 a^3 f}+\frac{d \int \frac{1}{a+a \tanh (e+f x)} \, dx}{16 a^2 f}+\frac{d \int \frac{1}{(a+a \tanh (e+f x))^2} \, dx}{12 a f}\\ &=\frac{d x}{16 a^3 f}-\frac{d x^2}{16 a^3}+\frac{x (c+d x)}{8 a^3}-\frac{d}{36 f^2 (a+a \tanh (e+f x))^3}-\frac{c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac{5 d}{96 a f^2 (a+a \tanh (e+f x))^2}-\frac{c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac{3 d}{32 f^2 \left (a^3+a^3 \tanh (e+f x)\right )}-\frac{c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}+\frac{d \int 1 \, dx}{32 a^3 f}+\frac{d \int \frac{1}{a+a \tanh (e+f x)} \, dx}{24 a^2 f}\\ &=\frac{3 d x}{32 a^3 f}-\frac{d x^2}{16 a^3}+\frac{x (c+d x)}{8 a^3}-\frac{d}{36 f^2 (a+a \tanh (e+f x))^3}-\frac{c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac{5 d}{96 a f^2 (a+a \tanh (e+f x))^2}-\frac{c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac{11 d}{96 f^2 \left (a^3+a^3 \tanh (e+f x)\right )}-\frac{c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}+\frac{d \int 1 \, dx}{48 a^3 f}\\ &=\frac{11 d x}{96 a^3 f}-\frac{d x^2}{16 a^3}+\frac{x (c+d x)}{8 a^3}-\frac{d}{36 f^2 (a+a \tanh (e+f x))^3}-\frac{c+d x}{6 f (a+a \tanh (e+f x))^3}-\frac{5 d}{96 a f^2 (a+a \tanh (e+f x))^2}-\frac{c+d x}{8 a f (a+a \tanh (e+f x))^2}-\frac{11 d}{96 f^2 \left (a^3+a^3 \tanh (e+f x)\right )}-\frac{c+d x}{8 f \left (a^3+a^3 \tanh (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.775518, size = 185, normalized size = 1.01 \[ \frac{\text{sech}^3(e+f x) \left (4 \left (6 c f (6 f x-1)+d \left (18 f^2 x^2-6 f x-1\right )\right ) \cosh (3 (e+f x))-27 (12 c f+d (12 f x+5)) \cosh (e+f x)+144 c f^2 x \sinh (3 (e+f x))-108 c f \sinh (e+f x)+24 c f \sinh (3 (e+f x))+72 d f^2 x^2 \sinh (3 (e+f x))-108 d f x \sinh (e+f x)+24 d f x \sinh (3 (e+f x))-81 d \sinh (e+f x)+4 d \sinh (3 (e+f x))\right )}{1152 a^3 f^2 (\tanh (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)/(a + a*Tanh[e + f*x])^3,x]

[Out]

(Sech[e + f*x]^3*(-27*(12*c*f + d*(5 + 12*f*x))*Cosh[e + f*x] + 4*(6*c*f*(-1 + 6*f*x) + d*(-1 - 6*f*x + 18*f^2
*x^2))*Cosh[3*(e + f*x)] - 81*d*Sinh[e + f*x] - 108*c*f*Sinh[e + f*x] - 108*d*f*x*Sinh[e + f*x] + 4*d*Sinh[3*(
e + f*x)] + 24*c*f*Sinh[3*(e + f*x)] + 24*d*f*x*Sinh[3*(e + f*x)] + 144*c*f^2*x*Sinh[3*(e + f*x)] + 72*d*f^2*x
^2*Sinh[3*(e + f*x)]))/(1152*a^3*f^2*(1 + Tanh[e + f*x])^3)

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Maple [B]  time = 0.049, size = 745, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/(a+a*tanh(f*x+e))^3,x)

[Out]

1/f^2/a^3*(4*d*(1/6*(f*x+e)*sinh(f*x+e)*cosh(f*x+e)^5+5/24*(f*x+e)*sinh(f*x+e)*cosh(f*x+e)^3+5/16*(f*x+e)*cosh
(f*x+e)*sinh(f*x+e)+5/32*(f*x+e)^2-1/36*sinh(f*x+e)^2*cosh(f*x+e)^4-23/288*sinh(f*x+e)^2*cosh(f*x+e)^2-17/72*c
osh(f*x+e)^2)+4*c*f*((1/6*cosh(f*x+e)^5+5/24*cosh(f*x+e)^3+5/16*cosh(f*x+e))*sinh(f*x+e)+5/16*f*x+5/16*e)-4*d*
e*((1/6*cosh(f*x+e)^5+5/24*cosh(f*x+e)^3+5/16*cosh(f*x+e))*sinh(f*x+e)+5/16*f*x+5/16*e)-4*d*(1/6*(f*x+e)*sinh(
f*x+e)^2*cosh(f*x+e)^4+1/6*(f*x+e)*sinh(f*x+e)^2*cosh(f*x+e)^2+1/6*(f*x+e)*cosh(f*x+e)^2-1/36*sinh(f*x+e)*cosh
(f*x+e)^5-5/144*cosh(f*x+e)^3*sinh(f*x+e)-5/96*cosh(f*x+e)*sinh(f*x+e)-5/96*f*x-5/96*e)-4*c*f*(1/6*sinh(f*x+e)
^2*cosh(f*x+e)^4+1/6*sinh(f*x+e)^2*cosh(f*x+e)^2+1/6*cosh(f*x+e)^2)+4*d*e*(1/6*sinh(f*x+e)^2*cosh(f*x+e)^4+1/6
*sinh(f*x+e)^2*cosh(f*x+e)^2+1/6*cosh(f*x+e)^2)-3*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)-3*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)+3*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)+d*(1/4*(f*x+e)*sinh(f*x+e)^2*cosh(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*c
osh(f*x+e)*sinh(f*x+e)-3/32*f*x-3/32*e)+c*f*(1/4*sinh(f*x+e)^2*cosh(f*x+e)^2+1/4*cosh(f*x+e)^2)-d*e*(1/4*sinh(
f*x+e)^2*cosh(f*x+e)^2+1/4*cosh(f*x+e)^2))

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Maxima [A]  time = 3.40281, size = 188, normalized size = 1.03 \begin{align*} \frac{1}{96} \, c{\left (\frac{12 \,{\left (f x + e\right )}}{a^{3} f} - \frac{18 \, e^{\left (-2 \, f x - 2 \, e\right )} + 9 \, e^{\left (-4 \, f x - 4 \, e\right )} + 2 \, e^{\left (-6 \, f x - 6 \, e\right )}}{a^{3} f}\right )} + \frac{{\left (72 \, f^{2} x^{2} e^{\left (6 \, e\right )} - 108 \,{\left (2 \, f x e^{\left (4 \, e\right )} + e^{\left (4 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - 27 \,{\left (4 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-4 \, f x\right )} - 4 \,{\left (6 \, f x + 1\right )} e^{\left (-6 \, f x\right )}\right )} d e^{\left (-6 \, e\right )}}{1152 \, a^{3} 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))^3,x, algorithm="maxima")

[Out]

1/96*c*(12*(f*x + e)/(a^3*f) - (18*e^(-2*f*x - 2*e) + 9*e^(-4*f*x - 4*e) + 2*e^(-6*f*x - 6*e))/(a^3*f)) + 1/11
52*(72*f^2*x^2*e^(6*e) - 108*(2*f*x*e^(4*e) + e^(4*e))*e^(-2*f*x) - 27*(4*f*x*e^(2*e) + e^(2*e))*e^(-4*f*x) -
4*(6*f*x + 1)*e^(-6*f*x))*d*e^(-6*e)/(a^3*f^2)

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Fricas [A]  time = 2.17172, size = 709, normalized size = 3.87 \begin{align*} \frac{4 \,{\left (18 \, d f^{2} x^{2} - 6 \, c f + 6 \,{\left (6 \, c f^{2} - d f\right )} x - d\right )} \cosh \left (f x + e\right )^{3} + 12 \,{\left (18 \, d f^{2} x^{2} - 6 \, c f + 6 \,{\left (6 \, c f^{2} - d f\right )} x - d\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{2} + 4 \,{\left (18 \, d f^{2} x^{2} + 6 \, c f + 6 \,{\left (6 \, c f^{2} + d f\right )} x + d\right )} \sinh \left (f x + e\right )^{3} - 27 \,{\left (12 \, d f x + 12 \, c f + 5 \, d\right )} \cosh \left (f x + e\right ) - 3 \,{\left (36 \, d f x - 4 \,{\left (18 \, d f^{2} x^{2} + 6 \, c f + 6 \,{\left (6 \, c f^{2} + d f\right )} x + d\right )} \cosh \left (f x + e\right )^{2} + 36 \, c f + 27 \, d\right )} \sinh \left (f x + e\right )}{1152 \,{\left (a^{3} f^{2} \cosh \left (f x + e\right )^{3} + 3 \, a^{3} f^{2} \cosh \left (f x + e\right )^{2} \sinh \left (f x + e\right ) + 3 \, a^{3} f^{2} \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{2} + a^{3} f^{2} \sinh \left (f x + e\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+a*tanh(f*x+e))^3,x, algorithm="fricas")

[Out]

1/1152*(4*(18*d*f^2*x^2 - 6*c*f + 6*(6*c*f^2 - d*f)*x - d)*cosh(f*x + e)^3 + 12*(18*d*f^2*x^2 - 6*c*f + 6*(6*c
*f^2 - d*f)*x - d)*cosh(f*x + e)*sinh(f*x + e)^2 + 4*(18*d*f^2*x^2 + 6*c*f + 6*(6*c*f^2 + d*f)*x + d)*sinh(f*x
+ e)^3 - 27*(12*d*f*x + 12*c*f + 5*d)*cosh(f*x + e) - 3*(36*d*f*x - 4*(18*d*f^2*x^2 + 6*c*f + 6*(6*c*f^2 + d*
f)*x + d)*cosh(f*x + e)^2 + 36*c*f + 27*d)*sinh(f*x + e))/(a^3*f^2*cosh(f*x + e)^3 + 3*a^3*f^2*cosh(f*x + e)^2
*sinh(f*x + e) + 3*a^3*f^2*cosh(f*x + e)*sinh(f*x + e)^2 + a^3*f^2*sinh(f*x + e)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c}{\tanh ^{3}{\left (e + f x \right )} + 3 \tanh ^{2}{\left (e + f x \right )} + 3 \tanh{\left (e + f x \right )} + 1}\, dx + \int \frac{d x}{\tanh ^{3}{\left (e + f x \right )} + 3 \tanh ^{2}{\left (e + f x \right )} + 3 \tanh{\left (e + f x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+a*tanh(f*x+e))**3,x)

[Out]

(Integral(c/(tanh(e + f*x)**3 + 3*tanh(e + f*x)**2 + 3*tanh(e + f*x) + 1), x) + Integral(d*x/(tanh(e + f*x)**3
+ 3*tanh(e + f*x)**2 + 3*tanh(e + f*x) + 1), x))/a**3

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Giac [A]  time = 1.22028, size = 204, normalized size = 1.11 \begin{align*} \frac{{\left (72 \, d f^{2} x^{2} e^{\left (6 \, f x + 6 \, e\right )} + 144 \, c f^{2} x e^{\left (6 \, f x + 6 \, e\right )} - 216 \, d f x e^{\left (4 \, f x + 4 \, e\right )} - 108 \, d f x e^{\left (2 \, f x + 2 \, e\right )} - 24 \, d f x - 216 \, c f e^{\left (4 \, f x + 4 \, e\right )} - 108 \, c f e^{\left (2 \, f x + 2 \, e\right )} - 24 \, c f - 108 \, d e^{\left (4 \, f x + 4 \, e\right )} - 27 \, d e^{\left (2 \, f x + 2 \, e\right )} - 4 \, d\right )} e^{\left (-6 \, f x - 6 \, e\right )}}{1152 \, a^{3} 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))^3,x, algorithm="giac")

[Out]

1/1152*(72*d*f^2*x^2*e^(6*f*x + 6*e) + 144*c*f^2*x*e^(6*f*x + 6*e) - 216*d*f*x*e^(4*f*x + 4*e) - 108*d*f*x*e^(
2*f*x + 2*e) - 24*d*f*x - 216*c*f*e^(4*f*x + 4*e) - 108*c*f*e^(2*f*x + 2*e) - 24*c*f - 108*d*e^(4*f*x + 4*e) -
27*d*e^(2*f*x + 2*e) - 4*d)*e^(-6*f*x - 6*e)/(a^3*f^2)