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

Optimal. Leaf size=36 \[ -\frac{a^2 \coth (c+d x)}{d}+x (a+b)^2-\frac{b^2 \tanh (c+d x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0671894, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3670, 461, 207} \[ -\frac{a^2 \coth (c+d x)}{d}+x (a+b)^2-\frac{b^2 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[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 3670

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]))

Rule 461

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 207

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

Rubi steps

\begin{align*} \int \coth ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{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{\operatorname{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 \operatorname{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]  time = 0.0959733, size = 64, normalized size = 1.78 \[ -\frac{a^2 \coth (c+d x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\tanh ^2(c+d x)\right )}{d}+2 a b x+\frac{b^2 \tanh ^{-1}(\tanh (c+d x))}{d}-\frac{b^2 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[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]  time = 0.043, size = 49, normalized size = 1.4 \begin{align*}{\frac{{a}^{2} \left ( dx+c-{\rm coth} \left (dx+c\right ) \right ) +2\, \left ( dx+c \right ) ab+{b}^{2} \left ( dx+c-\tanh \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.07441, size = 86, normalized size = 2.39 \begin{align*} 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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [B]  time = 1.90056, size = 243, normalized size = 6.75 \begin{align*} -\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.30827, size = 99, normalized size = 2.75 \begin{align*} \frac{{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (d x + c\right )}}{d} - \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 )}}{d{\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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