3.1.3 \(\int (a+b \coth ^2(c+d x))^3 \, dx\) [3]

Optimal. Leaf size=74 \[ (a+b)^3 x-\frac {b \left (3 a^2+3 a b+b^2\right ) \coth (c+d x)}{d}-\frac {b^2 (3 a+b) \coth ^3(c+d x)}{3 d}-\frac {b^3 \coth ^5(c+d x)}{5 d} \]

[Out]

(a+b)^3*x-b*(3*a^2+3*a*b+b^2)*coth(d*x+c)/d-1/3*b^2*(3*a+b)*coth(d*x+c)^3/d-1/5*b^3*coth(d*x+c)^5/d

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Rubi [A]
time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3742, 398, 212} \begin {gather*} -\frac {b \left (3 a^2+3 a b+b^2\right ) \coth (c+d x)}{d}-\frac {b^2 (3 a+b) \coth ^3(c+d x)}{3 d}+x (a+b)^3-\frac {b^3 \coth ^5(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Coth[c + d*x]^2)^3,x]

[Out]

(a + b)^3*x - (b*(3*a^2 + 3*a*b + b^2)*Coth[c + d*x])/d - (b^2*(3*a + b)*Coth[c + d*x]^3)/(3*d) - (b^3*Coth[c
+ d*x]^5)/(5*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 3742

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[c*(ff/f), Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \left (a+b \coth ^2(c+d x)\right )^3 \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^3}{1-x^2} \, dx,x,\coth (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (-b \left (3 a^2+3 a b+b^2\right )-b^2 (3 a+b) x^2-b^3 x^4+\frac {(a+b)^3}{1-x^2}\right ) \, dx,x,\coth (c+d x)\right )}{d}\\ &=-\frac {b \left (3 a^2+3 a b+b^2\right ) \coth (c+d x)}{d}-\frac {b^2 (3 a+b) \coth ^3(c+d x)}{3 d}-\frac {b^3 \coth ^5(c+d x)}{5 d}+\frac {(a+b)^3 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\coth (c+d x)\right )}{d}\\ &=(a+b)^3 x-\frac {b \left (3 a^2+3 a b+b^2\right ) \coth (c+d x)}{d}-\frac {b^2 (3 a+b) \coth ^3(c+d x)}{3 d}-\frac {b^3 \coth ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]
time = 0.98, size = 100, normalized size = 1.35 \begin {gather*} -\frac {b \coth (c+d x) \left (15 \left (3 a^2+3 a b+b^2\right )+5 b (3 a+b) \coth ^2(c+d x)+3 b^2 \coth ^4(c+d x)\right )}{15 d}+\frac {(a+b)^3 \tanh ^{-1}\left (\sqrt {\tanh ^2(c+d x)}\right ) \tanh (c+d x)}{d \sqrt {\tanh ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Coth[c + d*x]^2)^3,x]

[Out]

-1/15*(b*Coth[c + d*x]*(15*(3*a^2 + 3*a*b + b^2) + 5*b*(3*a + b)*Coth[c + d*x]^2 + 3*b^2*Coth[c + d*x]^4))/d +
 ((a + b)^3*ArcTanh[Sqrt[Tanh[c + d*x]^2]]*Tanh[c + d*x])/(d*Sqrt[Tanh[c + d*x]^2])

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Maple [A]
time = 0.33, size = 141, normalized size = 1.91

method result size
derivativedivides \(\frac {-3 a \,b^{2} \coth \left (d x +c \right )-3 a^{2} b \coth \left (d x +c \right )-a \,b^{2} \left (\coth ^{3}\left (d x +c \right )\right )-\frac {b^{3} \left (\coth ^{3}\left (d x +c \right )\right )}{3}-b^{3} \coth \left (d x +c \right )+\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\coth \left (d x +c \right )+1\right )}{2}-\frac {b^{3} \left (\coth ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\coth \left (d x +c \right )-1\right )}{2}}{d}\) \(141\)
default \(\frac {-3 a \,b^{2} \coth \left (d x +c \right )-3 a^{2} b \coth \left (d x +c \right )-a \,b^{2} \left (\coth ^{3}\left (d x +c \right )\right )-\frac {b^{3} \left (\coth ^{3}\left (d x +c \right )\right )}{3}-b^{3} \coth \left (d x +c \right )+\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\coth \left (d x +c \right )+1\right )}{2}-\frac {b^{3} \left (\coth ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\coth \left (d x +c \right )-1\right )}{2}}{d}\) \(141\)
risch \(a^{3} x +3 a^{2} b x +3 a \,b^{2} x +b^{3} x -\frac {2 b \left (45 a^{2} {\mathrm e}^{8 d x +8 c}+90 a b \,{\mathrm e}^{8 d x +8 c}+45 b^{2} {\mathrm e}^{8 d x +8 c}-180 a^{2} {\mathrm e}^{6 d x +6 c}-270 a b \,{\mathrm e}^{6 d x +6 c}-90 b^{2} {\mathrm e}^{6 d x +6 c}+270 a^{2} {\mathrm e}^{4 d x +4 c}+330 a b \,{\mathrm e}^{4 d x +4 c}+140 b^{2} {\mathrm e}^{4 d x +4 c}-180 a^{2} {\mathrm e}^{2 d x +2 c}-210 a b \,{\mathrm e}^{2 d x +2 c}-70 b^{2} {\mathrm e}^{2 d x +2 c}+45 a^{2}+60 a b +23 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{5}}\) \(224\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*coth(d*x+c)^2)^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-3*a*b^2*coth(d*x+c)-3*a^2*b*coth(d*x+c)-a*b^2*coth(d*x+c)^3-1/3*b^3*coth(d*x+c)^3-b^3*coth(d*x+c)+1/2*(a
^3+3*a^2*b+3*a*b^2+b^3)*ln(coth(d*x+c)+1)-1/5*b^3*coth(d*x+c)^5-1/2*(a^3+3*a^2*b+3*a*b^2+b^3)*ln(coth(d*x+c)-1
))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (70) = 140\).
time = 0.27, size = 239, normalized size = 3.23 \begin {gather*} \frac {1}{15} \, b^{3} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} - 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} - 45 \, e^{\left (-8 \, d x - 8 \, c\right )} - 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + a b^{2} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + 3 \, a^{2} b {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + a^{3} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*coth(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/15*b^3*(15*x + 15*c/d - 2*(70*e^(-2*d*x - 2*c) - 140*e^(-4*d*x - 4*c) + 90*e^(-6*d*x - 6*c) - 45*e^(-8*d*x -
 8*c) - 23)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d
*x - 10*c) - 1))) + a*b^2*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) - 2)/(d*(3*e^(-2*d*x - 2*c
) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1))) + 3*a^2*b*(x + c/d + 2/(d*(e^(-2*d*x - 2*c) - 1))) + a^3*x

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 557 vs. \(2 (70) = 140\).
time = 0.37, size = 557, normalized size = 7.53 \begin {gather*} -\frac {{\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \sinh \left (d x + c\right )^{5} - 5 \, {\left (27 \, a^{2} b + 24 \, a b^{2} + 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x - 2 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - 3 \, {\left (27 \, a^{2} b + 24 \, a b^{2} + 5 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (9 \, a^{2} b + 6 \, a b^{2} + 5 \, b^{3}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{4} + 90 \, a^{2} b + 120 \, a b^{2} + 46 \, b^{3} + 30 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x - 3 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*coth(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-1/15*((45*a^2*b + 60*a*b^2 + 23*b^3)*cosh(d*x + c)^5 + 5*(45*a^2*b + 60*a*b^2 + 23*b^3)*cosh(d*x + c)*sinh(d*
x + c)^4 - (45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*sinh(d*x + c)^5 - 5*(27*a^2
*b + 24*a*b^2 + 5*b^3)*cosh(d*x + c)^3 + 5*(45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*
d*x - 2*(45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)
^3 + 5*(2*(45*a^2*b + 60*a*b^2 + 23*b^3)*cosh(d*x + c)^3 - 3*(27*a^2*b + 24*a*b^2 + 5*b^3)*cosh(d*x + c))*sinh
(d*x + c)^2 + 10*(9*a^2*b + 6*a*b^2 + 5*b^3)*cosh(d*x + c) - 5*((45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^
2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^4 + 90*a^2*b + 120*a*b^2 + 46*b^3 + 30*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)
*d*x - 3*(45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c
))/(d*sinh(d*x + c)^5 + 5*(2*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^3 + 5*(d*cosh(d*x + c)^4 - 3*d*cosh(d*x + c)
^2 + 2*d)*sinh(d*x + c))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 367 vs. \(2 (65) = 130\).
time = 3.34, size = 367, normalized size = 4.96 \begin {gather*} \begin {cases} - \frac {a^{3} \log {\left (- e^{- d x} \right )}}{d} - \frac {3 a^{2} b \log {\left (- e^{- d x} \right )} \coth ^{2}{\left (d x + \log {\left (- e^{- d x} \right )} \right )}}{d} - \frac {3 a b^{2} \log {\left (- e^{- d x} \right )} \coth ^{4}{\left (d x + \log {\left (- e^{- d x} \right )} \right )}}{d} - \frac {b^{3} \log {\left (- e^{- d x} \right )} \coth ^{6}{\left (d x + \log {\left (- e^{- d x} \right )} \right )}}{d} & \text {for}\: c = \log {\left (- e^{- d x} \right )} \\- \frac {a^{3} \log {\left (e^{- d x} \right )}}{d} - \frac {3 a^{2} b \log {\left (e^{- d x} \right )} \coth ^{2}{\left (d x + \log {\left (e^{- d x} \right )} \right )}}{d} - \frac {3 a b^{2} \log {\left (e^{- d x} \right )} \coth ^{4}{\left (d x + \log {\left (e^{- d x} \right )} \right )}}{d} - \frac {b^{3} \log {\left (e^{- d x} \right )} \coth ^{6}{\left (d x + \log {\left (e^{- d x} \right )} \right )}}{d} & \text {for}\: c = \log {\left (e^{- d x} \right )} \\x \left (a + b \coth ^{2}{\left (c \right )}\right )^{3} & \text {for}\: d = 0 \\a^{3} x + 3 a^{2} b x - \frac {3 a^{2} b}{d \tanh {\left (c + d x \right )}} + 3 a b^{2} x - \frac {3 a b^{2}}{d \tanh {\left (c + d x \right )}} - \frac {a b^{2}}{d \tanh ^{3}{\left (c + d x \right )}} + b^{3} x - \frac {b^{3}}{d \tanh {\left (c + d x \right )}} - \frac {b^{3}}{3 d \tanh ^{3}{\left (c + d x \right )}} - \frac {b^{3}}{5 d \tanh ^{5}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*coth(d*x+c)**2)**3,x)

[Out]

Piecewise((-a**3*log(-exp(-d*x))/d - 3*a**2*b*log(-exp(-d*x))*coth(d*x + log(-exp(-d*x)))**2/d - 3*a*b**2*log(
-exp(-d*x))*coth(d*x + log(-exp(-d*x)))**4/d - b**3*log(-exp(-d*x))*coth(d*x + log(-exp(-d*x)))**6/d, Eq(c, lo
g(-exp(-d*x)))), (-a**3*log(exp(-d*x))/d - 3*a**2*b*log(exp(-d*x))*coth(d*x + log(exp(-d*x)))**2/d - 3*a*b**2*
log(exp(-d*x))*coth(d*x + log(exp(-d*x)))**4/d - b**3*log(exp(-d*x))*coth(d*x + log(exp(-d*x)))**6/d, Eq(c, lo
g(exp(-d*x)))), (x*(a + b*coth(c)**2)**3, Eq(d, 0)), (a**3*x + 3*a**2*b*x - 3*a**2*b/(d*tanh(c + d*x)) + 3*a*b
**2*x - 3*a*b**2/(d*tanh(c + d*x)) - a*b**2/(d*tanh(c + d*x)**3) + b**3*x - b**3/(d*tanh(c + d*x)) - b**3/(3*d
*tanh(c + d*x)**3) - b**3/(5*d*tanh(c + d*x)**5), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (70) = 140\).
time = 0.41, size = 241, normalized size = 3.26 \begin {gather*} \frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 45 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} - 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 270 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 90 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 330 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 140 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 210 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 70 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*coth(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/15*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(d*x + c) - 2*(45*a^2*b*e^(8*d*x + 8*c) + 90*a*b^2*e^(8*d*x + 8*c) +
45*b^3*e^(8*d*x + 8*c) - 180*a^2*b*e^(6*d*x + 6*c) - 270*a*b^2*e^(6*d*x + 6*c) - 90*b^3*e^(6*d*x + 6*c) + 270*
a^2*b*e^(4*d*x + 4*c) + 330*a*b^2*e^(4*d*x + 4*c) + 140*b^3*e^(4*d*x + 4*c) - 180*a^2*b*e^(2*d*x + 2*c) - 210*
a*b^2*e^(2*d*x + 2*c) - 70*b^3*e^(2*d*x + 2*c) + 45*a^2*b + 60*a*b^2 + 23*b^3)/(e^(2*d*x + 2*c) - 1)^5)/d

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Mupad [B]
time = 1.28, size = 72, normalized size = 0.97 \begin {gather*} x\,{\left (a+b\right )}^3-\frac {{\mathrm {coth}\left (c+d\,x\right )}^3\,\left (b^3+3\,a\,b^2\right )}{3\,d}-\frac {b^3\,{\mathrm {coth}\left (c+d\,x\right )}^5}{5\,d}-\frac {b\,\mathrm {coth}\left (c+d\,x\right )\,\left (3\,a^2+3\,a\,b+b^2\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*coth(c + d*x)^2)^3,x)

[Out]

x*(a + b)^3 - (coth(c + d*x)^3*(3*a*b^2 + b^3))/(3*d) - (b^3*coth(c + d*x)^5)/(5*d) - (b*coth(c + d*x)*(3*a*b
+ 3*a^2 + b^2))/d

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