\(\int \coth ^2(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\) [150]

Optimal result

Integrand size = 23, antiderivative size = 36 \[ \int \coth ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=(a+b)^2 x-\frac {a^2 \coth (c+d x)}{d}-\frac {b^2 \tanh (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 36, 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} \[ \int \coth ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {a^2 \coth (c+d x)}{d}+x (a+b)^2-\frac {b^2 \tanh (c+d x)}{d} \]

[In]

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

[Out]

(a + b)^2*x - (a^2*Coth[c + d*x])/d - (b^2*Tanh[c + d*x])/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^2 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (-b^2+\frac {a^2}{x^2}-\frac {(a+b)^2}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d} \\ & = -\frac {a^2 \coth (c+d x)}{d}-\frac {b^2 \tanh (c+d x)}{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^2 \coth (c+d x)}{d}-\frac {b^2 \tanh (c+d x)}{d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.78 \[ \int \coth ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=2 a b x+\frac {b^2 \text {arctanh}(\tanh (c+d x))}{d}-\frac {a^2 \coth (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\tanh ^2(c+d x)\right )}{d}-\frac {b^2 \tanh (c+d x)}{d} \]

[In]

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

[Out]

2*a*b*x + (b^2*ArcTanh[Tanh[c + d*x]])/d - (a^2*Coth[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, Tanh[c + d*x]^2]
)/d - (b^2*Tanh[c + d*x])/d

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {-\coth \left (d x +c \right ) a^{2}-b^{2} \tanh \left (d x +c \right )+d x \left (a +b \right )^{2}}{d}\)	\(36\)
derivativedivides	\(\frac {a^{2} \left (d x +c -\coth \left (d x +c \right )\right )+2 a b \left (d x +c \right )+b^{2} \left (d x +c -\tanh \left (d x +c \right )\right )}{d}\)	\(49\)
default	\(\frac {a^{2} \left (d x +c -\coth \left (d x +c \right )\right )+2 a b \left (d x +c \right )+b^{2} \left (d x +c -\tanh \left (d x +c \right )\right )}{d}\)	\(49\)
risch	\(a^{2} x +2 a b x +b^{2} x -\frac {2 \left (a^{2} {\mathrm e}^{2 d x +2 c}-{\mathrm e}^{2 d x +2 c} b^{2}+a^{2}+b^{2}\right )}{d \left ({\mathrm e}^{2 d x +2 c}+1\right ) \left ({\mathrm e}^{2 d x +2 c}-1\right )}\)	\(82\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (36) = 72\).

Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.69 \[ \int \coth ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {{\left (a^{2} + b^{2}\right )} \cosh \left (d x + c\right )^{2} - 2 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} d x + a^{2} + b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{2} + b^{2}\right )} \sinh \left (d x + c\right )^{2} + a^{2} - b^{2}}{2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.78 \[ \int \coth ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=b^{2} {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + a^{2} {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + 2 \, a b x \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.97 \[ \int \coth ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (a^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + b^{2}\right )}}{e^{\left (4 \, d x + 4 \, c\right )} - 1}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.83 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.64 \[ \int \coth ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=x\,{\left (a+b\right )}^2-\frac {\frac {2\,\left (a^2+b^2\right )}{d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2-b^2\right )}{d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-1} \]

[In]

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

[Out]

x*(a + b)^2 - ((2*(a^2 + b^2))/d + (2*exp(2*c + 2*d*x)*(a^2 - b^2))/d)/(exp(4*c + 4*d*x) - 1)