3.48 \(\int \frac{1}{(a+a \tanh (c+d x))^4} \, dx\)

Optimal. Leaf size=96 \[ -\frac{1}{16 d \left (a^4 \tanh (c+d x)+a^4\right )}-\frac{1}{16 d \left (a^2 \tanh (c+d x)+a^2\right )^2}+\frac{x}{16 a^4}-\frac{1}{12 a d (a \tanh (c+d x)+a)^3}-\frac{1}{8 d (a \tanh (c+d x)+a)^4} \]

[Out]

x/(16*a^4) - 1/(8*d*(a + a*Tanh[c + d*x])^4) - 1/(12*a*d*(a + a*Tanh[c + d*x])^3) - 1/(16*d*(a^2 + a^2*Tanh[c
+ d*x])^2) - 1/(16*d*(a^4 + a^4*Tanh[c + d*x]))

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Rubi [A]  time = 0.0621694, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3479, 8} \[ -\frac{1}{16 d \left (a^4 \tanh (c+d x)+a^4\right )}-\frac{1}{16 d \left (a^2 \tanh (c+d x)+a^2\right )^2}+\frac{x}{16 a^4}-\frac{1}{12 a d (a \tanh (c+d x)+a)^3}-\frac{1}{8 d (a \tanh (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tanh[c + d*x])^(-4),x]

[Out]

x/(16*a^4) - 1/(8*d*(a + a*Tanh[c + d*x])^4) - 1/(12*a*d*(a + a*Tanh[c + d*x])^3) - 1/(16*d*(a^2 + a^2*Tanh[c
+ d*x])^2) - 1/(16*d*(a^4 + a^4*Tanh[c + d*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]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \tanh (c+d x))^4} \, dx &=-\frac{1}{8 d (a+a \tanh (c+d x))^4}+\frac{\int \frac{1}{(a+a \tanh (c+d x))^3} \, dx}{2 a}\\ &=-\frac{1}{8 d (a+a \tanh (c+d x))^4}-\frac{1}{12 a d (a+a \tanh (c+d x))^3}+\frac{\int \frac{1}{(a+a \tanh (c+d x))^2} \, dx}{4 a^2}\\ &=-\frac{1}{8 d (a+a \tanh (c+d x))^4}-\frac{1}{12 a d (a+a \tanh (c+d x))^3}-\frac{1}{16 d \left (a^2+a^2 \tanh (c+d x)\right )^2}+\frac{\int \frac{1}{a+a \tanh (c+d x)} \, dx}{8 a^3}\\ &=-\frac{1}{8 d (a+a \tanh (c+d x))^4}-\frac{1}{12 a d (a+a \tanh (c+d x))^3}-\frac{1}{16 d \left (a^2+a^2 \tanh (c+d x)\right )^2}-\frac{1}{16 d \left (a^4+a^4 \tanh (c+d x)\right )}+\frac{\int 1 \, dx}{16 a^4}\\ &=\frac{x}{16 a^4}-\frac{1}{8 d (a+a \tanh (c+d x))^4}-\frac{1}{12 a d (a+a \tanh (c+d x))^3}-\frac{1}{16 d \left (a^2+a^2 \tanh (c+d x)\right )^2}-\frac{1}{16 d \left (a^4+a^4 \tanh (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.249169, size = 88, normalized size = 0.92 \[ \frac{\text{sech}^4(c+d x) (-32 \sinh (2 (c+d x))+24 d x \sinh (4 (c+d x))+3 \sinh (4 (c+d x))-64 \cosh (2 (c+d x))+3 (8 d x-1) \cosh (4 (c+d x))-36)}{384 a^4 d (\tanh (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tanh[c + d*x])^(-4),x]

[Out]

(Sech[c + d*x]^4*(-36 - 64*Cosh[2*(c + d*x)] + 3*(-1 + 8*d*x)*Cosh[4*(c + d*x)] - 32*Sinh[2*(c + d*x)] + 3*Sin
h[4*(c + d*x)] + 24*d*x*Sinh[4*(c + d*x)]))/(384*a^4*d*(1 + Tanh[c + d*x])^4)

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Maple [A]  time = 0.02, size = 108, normalized size = 1.1 \begin{align*} -{\frac{1}{8\,d{a}^{4} \left ( \tanh \left ( dx+c \right ) +1 \right ) ^{4}}}-{\frac{1}{12\,d{a}^{4} \left ( \tanh \left ( dx+c \right ) +1 \right ) ^{3}}}-{\frac{1}{16\,d{a}^{4} \left ( \tanh \left ( dx+c \right ) +1 \right ) ^{2}}}-{\frac{1}{16\,d{a}^{4} \left ( \tanh \left ( dx+c \right ) +1 \right ) }}+{\frac{\ln \left ( \tanh \left ( dx+c \right ) +1 \right ) }{32\,d{a}^{4}}}-{\frac{\ln \left ( \tanh \left ( dx+c \right ) -1 \right ) }{32\,d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*tanh(d*x+c))^4,x)

[Out]

-1/8/d/a^4/(tanh(d*x+c)+1)^4-1/12/d/a^4/(tanh(d*x+c)+1)^3-1/16/d/a^4/(tanh(d*x+c)+1)^2-1/16/d/a^4/(tanh(d*x+c)
+1)+1/32/d/a^4*ln(tanh(d*x+c)+1)-1/32/d/a^4*ln(tanh(d*x+c)-1)

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Maxima [A]  time = 1.07848, size = 90, normalized size = 0.94 \begin{align*} \frac{d x + c}{16 \, a^{4} d} - \frac{48 \, e^{\left (-2 \, d x - 2 \, c\right )} + 36 \, e^{\left (-4 \, d x - 4 \, c\right )} + 16 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3 \, e^{\left (-8 \, d x - 8 \, c\right )}}{384 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))^4,x, algorithm="maxima")

[Out]

1/16*(d*x + c)/(a^4*d) - 1/384*(48*e^(-2*d*x - 2*c) + 36*e^(-4*d*x - 4*c) + 16*e^(-6*d*x - 6*c) + 3*e^(-8*d*x
- 8*c))/(a^4*d)

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Fricas [B]  time = 2.15626, size = 594, normalized size = 6.19 \begin{align*} \frac{3 \,{\left (8 \, d x - 1\right )} \cosh \left (d x + c\right )^{4} + 12 \,{\left (8 \, d x + 1\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 3 \,{\left (8 \, d x - 1\right )} \sinh \left (d x + c\right )^{4} + 2 \,{\left (9 \,{\left (8 \, d x - 1\right )} \cosh \left (d x + c\right )^{2} - 32\right )} \sinh \left (d x + c\right )^{2} - 64 \, \cosh \left (d x + c\right )^{2} + 4 \,{\left (3 \,{\left (8 \, d x + 1\right )} \cosh \left (d x + c\right )^{3} - 16 \, \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 36}{384 \,{\left (a^{4} d \cosh \left (d x + c\right )^{4} + 4 \, a^{4} d \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, a^{4} d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, a^{4} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{4} d \sinh \left (d x + c\right )^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))^4,x, algorithm="fricas")

[Out]

1/384*(3*(8*d*x - 1)*cosh(d*x + c)^4 + 12*(8*d*x + 1)*cosh(d*x + c)*sinh(d*x + c)^3 + 3*(8*d*x - 1)*sinh(d*x +
c)^4 + 2*(9*(8*d*x - 1)*cosh(d*x + c)^2 - 32)*sinh(d*x + c)^2 - 64*cosh(d*x + c)^2 + 4*(3*(8*d*x + 1)*cosh(d*
x + c)^3 - 16*cosh(d*x + c))*sinh(d*x + c) - 36)/(a^4*d*cosh(d*x + c)^4 + 4*a^4*d*cosh(d*x + c)^3*sinh(d*x + c
) + 6*a^4*d*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*a^4*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^4*d*sinh(d*x + c)^4)

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Sympy [A]  time = 2.60847, size = 694, normalized size = 7.23 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))**4,x)

[Out]

Piecewise((3*d*x*tanh(c + d*x)**4/(48*a**4*d*tanh(c + d*x)**4 + 192*a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(
c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4*d) + 12*d*x*tanh(c + d*x)**3/(48*a**4*d*tanh(c + d*x)**4 + 19
2*a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4*d) + 18*d*x*tanh(
c + d*x)**2/(48*a**4*d*tanh(c + d*x)**4 + 192*a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4
*d*tanh(c + d*x) + 48*a**4*d) + 12*d*x*tanh(c + d*x)/(48*a**4*d*tanh(c + d*x)**4 + 192*a**4*d*tanh(c + d*x)**3
+ 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4*d) + 3*d*x/(48*a**4*d*tanh(c + d*x)**4 + 1
92*a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4*d) - 3*tanh(c +
d*x)**3/(48*a**4*d*tanh(c + d*x)**4 + 192*a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*t
anh(c + d*x) + 48*a**4*d) - 12*tanh(c + d*x)**2/(48*a**4*d*tanh(c + d*x)**4 + 192*a**4*d*tanh(c + d*x)**3 + 28
8*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4*d) - 19*tanh(c + d*x)/(48*a**4*d*tanh(c + d*x)*
*4 + 192*a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4*d) - 16/(4
8*a**4*d*tanh(c + d*x)**4 + 192*a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*
x) + 48*a**4*d), Ne(d, 0)), (x/(a*tanh(c) + a)**4, True))

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Giac [A]  time = 1.18606, size = 84, normalized size = 0.88 \begin{align*} \frac{24 \, d x -{\left (48 \, e^{\left (6 \, d x + 6 \, c\right )} + 36 \, e^{\left (4 \, d x + 4 \, c\right )} + 16 \, e^{\left (2 \, d x + 2 \, c\right )} + 3\right )} e^{\left (-8 \, d x - 8 \, c\right )} + 24 \, c}{384 \, a^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))^4,x, algorithm="giac")

[Out]

1/384*(24*d*x - (48*e^(6*d*x + 6*c) + 36*e^(4*d*x + 4*c) + 16*e^(2*d*x + 2*c) + 3)*e^(-8*d*x - 8*c) + 24*c)/(a
^4*d)