\(\int \coth ^4(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\) [152]

Optimal result

Integrand size = 23, antiderivative size = 43 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=(a+b)^2 x-\frac {a (a+2 b) \coth (c+d x)}{d}-\frac {a^2 \coth ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3751, 472, 213} \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {a^2 \coth ^3(c+d x)}{3 d}-\frac {a (a+2 b) \coth (c+d x)}{d}+x (a+b)^2 \]

[In]

Int[Coth[c + d*x]^4*(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

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

Rule 213

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

Rule 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 3751

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((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[(d*ff*(x/c))^m*((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, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

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

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.51 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {\coth (c+d x) \left (a \left (3 a+6 b+a \coth ^2(c+d x)\right )-3 (a+b)^2 \text {arctanh}\left (\sqrt {\tanh ^2(c+d x)}\right ) \sqrt {\tanh ^2(c+d x)}\right )}{3 d} \]

[In]

Integrate[Coth[c + d*x]^4*(a + b*Tanh[c + d*x]^2)^2,x]

[Out]

-1/3*(Coth[c + d*x]*(a*(3*a + 6*b + a*Coth[c + d*x]^2) - 3*(a + b)^2*ArcTanh[Sqrt[Tanh[c + d*x]^2]]*Sqrt[Tanh[
c + d*x]^2]))/d

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {-\coth \left (d x +c \right )^{3} a^{2}-3 a \left (a +2 b \right ) \coth \left (d x +c \right )+3 d x \left (a +b \right )^{2}}{3 d}\)	\(43\)
derivativedivides	\(-\frac {\frac {a^{2}}{3 \tanh \left (d x +c \right )^{3}}+\frac {a \left (a +2 b \right )}{\tanh \left (d x +c \right )}+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )+\left (-\frac {1}{2} a^{2}-a b -\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{d}\)	\(84\)
default	\(-\frac {\frac {a^{2}}{3 \tanh \left (d x +c \right )^{3}}+\frac {a \left (a +2 b \right )}{\tanh \left (d x +c \right )}+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )+\left (-\frac {1}{2} a^{2}-a b -\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{d}\)	\(84\)
risch	\(a^{2} x +2 a b x +b^{2} x -\frac {4 a \left (3 a \,{\mathrm e}^{4 d x +4 c}+3 b \,{\mathrm e}^{4 d x +4 c}-3 \,{\mathrm e}^{2 d x +2 c} a -6 b \,{\mathrm e}^{2 d x +2 c}+2 a +3 b \right )}{3 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\)	\(91\)

[In]

int(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (41) = 82\).

Time = 0.24 (sec) , antiderivative size = 197, normalized size of antiderivative = 4.58 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {2 \, {\left (2 \, a^{2} + 3 \, a b\right )} \cosh \left (d x + c\right )^{3} + 6 \, {\left (2 \, a^{2} + 3 \, a b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} - {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 4 \, a^{2} + 6 \, a b\right )} \sinh \left (d x + c\right )^{3} - 6 \, a b \cosh \left (d x + c\right ) + 3 \, {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x - {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 4 \, a^{2} + 6 \, a b\right )} \cosh \left (d x + c\right )^{2} + 4 \, a^{2} + 6 \, a b\right )} \sinh \left (d x + c\right )}{3 \, {\left (d \sinh \left (d x + c\right )^{3} + 3 \, {\left (d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )\right )}} \]

[In]

integrate(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

-1/3*(2*(2*a^2 + 3*a*b)*cosh(d*x + c)^3 + 6*(2*a^2 + 3*a*b)*cosh(d*x + c)*sinh(d*x + c)^2 - (3*(a^2 + 2*a*b +
b^2)*d*x + 4*a^2 + 6*a*b)*sinh(d*x + c)^3 - 6*a*b*cosh(d*x + c) + 3*(3*(a^2 + 2*a*b + b^2)*d*x - (3*(a^2 + 2*a
*b + b^2)*d*x + 4*a^2 + 6*a*b)*cosh(d*x + c)^2 + 4*a^2 + 6*a*b)*sinh(d*x + c))/(d*sinh(d*x + c)^3 + 3*(d*cosh(
d*x + c)^2 - d)*sinh(d*x + c))

Sympy [F]

\[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \coth ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(coth(d*x+c)**4*(a+b*tanh(d*x+c)**2)**2,x)

[Out]

Integral((a + b*tanh(c + d*x)**2)**2*coth(c + d*x)**4, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (41) = 82\).

Time = 0.21 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.65 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {1}{3} \, a^{2} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + 2 \, a b {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + b^{2} x \]

[In]

integrate(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (41) = 82\).

Time = 0.37 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.40 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (d x + c\right )} - \frac {4 \, {\left (3 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 6 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} + 3 \, a b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \]

[In]

integrate(coth(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/3*(3*(a^2 + 2*a*b + b^2)*(d*x + c) - 4*(3*a^2*e^(4*d*x + 4*c) + 3*a*b*e^(4*d*x + 4*c) - 3*a^2*e^(2*d*x + 2*c
) - 6*a*b*e^(2*d*x + 2*c) + 2*a^2 + 3*a*b)/(e^(2*d*x + 2*c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 175, normalized size of antiderivative = 4.07 \[ \int \coth ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=x\,{\left (a+b\right )}^2-\frac {\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2+b\,a\right )}{3\,d}-\frac {4\,a\,b}{3\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {4\,\left (a^2+b\,a\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2+b\,a\right )}{3\,d}-\frac {8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {4\,\left (a^2+b\,a\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

[In]

int(coth(c + d*x)^4*(a + b*tanh(c + d*x)^2)^2,x)

[Out]

x*(a + b)^2 - ((4*exp(2*c + 2*d*x)*(a*b + a^2))/(3*d) - (4*a*b)/(3*d))/(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x)
+ 1) - ((4*(a*b + a^2))/(3*d) + (4*exp(4*c + 4*d*x)*(a*b + a^2))/(3*d) - (8*a*b*exp(2*c + 2*d*x))/(3*d))/(3*ex
p(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1) - (4*(a*b + a^2))/(3*d*(exp(2*c + 2*d*x) - 1))