3.1.47 \(\int \frac {1}{(a+a \tanh (c+d x))^3} \, dx\) [47]

Optimal. Leaf size=73 \[ \frac {x}{8 a^3}-\frac {1}{6 d (a+a \tanh (c+d x))^3}-\frac {1}{8 a d (a+a \tanh (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \tanh (c+d x)\right )} \]

[Out]

1/8*x/a^3-1/6/d/(a+a*tanh(d*x+c))^3-1/8/a/d/(a+a*tanh(d*x+c))^2-1/8/d/(a^3+a^3*tanh(d*x+c))

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Rubi [A]
time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3560, 8} \begin {gather*} -\frac {1}{8 d \left (a^3 \tanh (c+d x)+a^3\right )}+\frac {x}{8 a^3}-\frac {1}{8 a d (a \tanh (c+d x)+a)^2}-\frac {1}{6 d (a \tanh (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Tanh[c + d*x])^(-3),x]

[Out]

x/(8*a^3) - 1/(6*d*(a + a*Tanh[c + d*x])^3) - 1/(8*a*d*(a + a*Tanh[c + d*x])^2) - 1/(8*d*(a^3 + a^3*Tanh[c + d
*x]))

Rule 8

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

Rule 3560

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]

Rubi steps

\begin {align*} \int \frac {1}{(a+a \tanh (c+d x))^3} \, dx &=-\frac {1}{6 d (a+a \tanh (c+d x))^3}+\frac {\int \frac {1}{(a+a \tanh (c+d x))^2} \, dx}{2 a}\\ &=-\frac {1}{6 d (a+a \tanh (c+d x))^3}-\frac {1}{8 a d (a+a \tanh (c+d x))^2}+\frac {\int \frac {1}{a+a \tanh (c+d x)} \, dx}{4 a^2}\\ &=-\frac {1}{6 d (a+a \tanh (c+d x))^3}-\frac {1}{8 a d (a+a \tanh (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \tanh (c+d x)\right )}+\frac {\int 1 \, dx}{8 a^3}\\ &=\frac {x}{8 a^3}-\frac {1}{6 d (a+a \tanh (c+d x))^3}-\frac {1}{8 a d (a+a \tanh (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \tanh (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 83, normalized size = 1.14 \begin {gather*} \frac {\text {sech}^3(c+d x) (-27 \cosh (c+d x)+2 (-1+6 d x) \cosh (3 (c+d x))-9 \sinh (c+d x)+2 \sinh (3 (c+d x))+12 d x \sinh (3 (c+d x)))}{96 a^3 d (1+\tanh (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tanh[c + d*x])^(-3),x]

[Out]

(Sech[c + d*x]^3*(-27*Cosh[c + d*x] + 2*(-1 + 6*d*x)*Cosh[3*(c + d*x)] - 9*Sinh[c + d*x] + 2*Sinh[3*(c + d*x)]
+ 12*d*x*Sinh[3*(c + d*x)]))/(96*a^3*d*(1 + Tanh[c + d*x])^3)

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Maple [A]
time = 0.66, size = 67, normalized size = 0.92

method	result	size
risch	\(\frac {x}{8 a^{3}}-\frac {3 \,{\mathrm e}^{-2 d x -2 c}}{16 a^{3} d}-\frac {3 \,{\mathrm e}^{-4 d x -4 c}}{32 a^{3} d}-\frac {{\mathrm e}^{-6 d x -6 c}}{48 a^{3} d}\)	\(59\)
derivativedivides	\(\frac {-\frac {1}{6 \left (\tanh \left (d x +c \right )+1\right )^{3}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )^{2}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{16}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{16}}{d \,a^{3}}\)	\(67\)
default	\(\frac {-\frac {1}{6 \left (\tanh \left (d x +c \right )+1\right )^{3}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )^{2}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{16}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{16}}{d \,a^{3}}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*tanh(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d/a^3*(-1/6/(tanh(d*x+c)+1)^3-1/8/(tanh(d*x+c)+1)^2-1/8/(tanh(d*x+c)+1)+1/16*ln(tanh(d*x+c)+1)-1/16*ln(tanh(
d*x+c)-1))

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Maxima [A]
time = 0.27, size = 56, normalized size = 0.77 \begin {gather*} \frac {d x + c}{8 \, a^{3} d} - \frac {18 \, e^{\left (-2 \, d x - 2 \, c\right )} + 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2 \, e^{\left (-6 \, d x - 6 \, c\right )}}{96 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))^3,x, algorithm="maxima")

[Out]

1/8*(d*x + c)/(a^3*d) - 1/96*(18*e^(-2*d*x - 2*c) + 9*e^(-4*d*x - 4*c) + 2*e^(-6*d*x - 6*c))/(a^3*d)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (65) = 130\).
time = 0.44, size = 160, normalized size = 2.19 \begin {gather*} \frac {2 \, {\left (6 \, d x - 1\right )} \cosh \left (d x + c\right )^{3} + 6 \, {\left (6 \, d x - 1\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 2 \, {\left (6 \, d x + 1\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (2 \, {\left (6 \, d x + 1\right )} \cosh \left (d x + c\right )^{2} - 3\right )} \sinh \left (d x + c\right ) - 27 \, \cosh \left (d x + c\right )}{96 \, {\left (a^{3} d \cosh \left (d x + c\right )^{3} + 3 \, a^{3} d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, a^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + a^{3} d \sinh \left (d x + c\right )^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*(2*(6*d*x - 1)*cosh(d*x + c)^3 + 6*(6*d*x - 1)*cosh(d*x + c)*sinh(d*x + c)^2 + 2*(6*d*x + 1)*sinh(d*x + c
)^3 + 3*(2*(6*d*x + 1)*cosh(d*x + c)^2 - 3)*sinh(d*x + c) - 27*cosh(d*x + c))/(a^3*d*cosh(d*x + c)^3 + 3*a^3*d
*cosh(d*x + c)^2*sinh(d*x + c) + 3*a^3*d*cosh(d*x + c)*sinh(d*x + c)^2 + a^3*d*sinh(d*x + c)^3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 430 vs. \(2 (60) = 120\).
time = 0.77, size = 430, normalized size = 5.89 \begin {gather*} \begin {cases} \frac {3 d x \tanh ^{3}{\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} + \frac {9 d x \tanh ^{2}{\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} + \frac {9 d x \tanh {\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} + \frac {3 d x}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} - \frac {3 \tanh ^{2}{\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} - \frac {9 \tanh {\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} - \frac {10}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x}{\left (a \tanh {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))**3,x)

[Out]

Piecewise((3*d*x*tanh(c + d*x)**3/(24*a**3*d*tanh(c + d*x)**3 + 72*a**3*d*tanh(c + d*x)**2 + 72*a**3*d*tanh(c
+ d*x) + 24*a**3*d) + 9*d*x*tanh(c + d*x)**2/(24*a**3*d*tanh(c + d*x)**3 + 72*a**3*d*tanh(c + d*x)**2 + 72*a**
3*d*tanh(c + d*x) + 24*a**3*d) + 9*d*x*tanh(c + d*x)/(24*a**3*d*tanh(c + d*x)**3 + 72*a**3*d*tanh(c + d*x)**2
+ 72*a**3*d*tanh(c + d*x) + 24*a**3*d) + 3*d*x/(24*a**3*d*tanh(c + d*x)**3 + 72*a**3*d*tanh(c + d*x)**2 + 72*a
**3*d*tanh(c + d*x) + 24*a**3*d) - 3*tanh(c + d*x)**2/(24*a**3*d*tanh(c + d*x)**3 + 72*a**3*d*tanh(c + d*x)**2
+ 72*a**3*d*tanh(c + d*x) + 24*a**3*d) - 9*tanh(c + d*x)/(24*a**3*d*tanh(c + d*x)**3 + 72*a**3*d*tanh(c + d*x
)**2 + 72*a**3*d*tanh(c + d*x) + 24*a**3*d) - 10/(24*a**3*d*tanh(c + d*x)**3 + 72*a**3*d*tanh(c + d*x)**2 + 72
*a**3*d*tanh(c + d*x) + 24*a**3*d), Ne(d, 0)), (x/(a*tanh(c) + a)**3, True))

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Giac [A]
time = 0.42, size = 53, normalized size = 0.73 \begin {gather*} -\frac {\frac {{\left (18 \, e^{\left (4 \, d x + 4 \, c\right )} + 9 \, e^{\left (2 \, d x + 2 \, c\right )} + 2\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{a^{3}} - \frac {12 \, {\left (d x + c\right )}}{a^{3}}}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*tanh(d*x+c))^3,x, algorithm="giac")

[Out]

-1/96*((18*e^(4*d*x + 4*c) + 9*e^(2*d*x + 2*c) + 2)*e^(-6*d*x - 6*c)/a^3 - 12*(d*x + c)/a^3)/d

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Mupad [B]
time = 1.10, size = 58, normalized size = 0.79 \begin {gather*} \frac {x}{8\,a^3}-\frac {3\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{16\,a^3\,d}-\frac {3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{32\,a^3\,d}-\frac {{\mathrm {e}}^{-6\,c-6\,d\,x}}{48\,a^3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*tanh(c + d*x))^3,x)

[Out]

x/(8*a^3) - (3*exp(- 2*c - 2*d*x))/(16*a^3*d) - (3*exp(- 4*c - 4*d*x))/(32*a^3*d) - exp(- 6*c - 6*d*x)/(48*a^3
*d)

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