\(\int \text {csch}^2(c+d x) (a+b \tanh ^2(c+d x)) \, dx\) [6]

Optimal result

Integrand size = 21, antiderivative size = 24 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {a \coth (c+d x)}{d}+\frac {b \tanh (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3744, 14} \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {b \tanh (c+d x)}{d}-\frac {a \coth (c+d x)}{d} \]

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 3744

Int[sin[(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^(m + 1)/f), Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)
^(m/2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {a \coth (c+d x)}{d}+\frac {b \tanh (c+d x)}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {b \tanh \left (d x +c \right )-\frac {a}{\tanh \left (d x +c \right )}}{d}\)	\(25\)
default	\(\frac {b \tanh \left (d x +c \right )-\frac {a}{\tanh \left (d x +c \right )}}{d}\)	\(25\)
risch	\(-\frac {2 \left ({\mathrm e}^{2 d x +2 c} a +b \,{\mathrm e}^{2 d x +2 c}+a -b \right )}{d \left ({\mathrm e}^{2 d x +2 c}-1\right ) \left ({\mathrm e}^{2 d x +2 c}+1\right )}\)	\(59\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (24) = 48\).

Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.67 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (a \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )}}{d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + d \sinh \left (d x + c\right )^{3} - d \cosh \left (d x + c\right ) + {\left (3 \, d \cosh \left (d x + c\right )^{2} + d\right )} \sinh \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {2 \, b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} + \frac {2 \, a}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b\right )}}{d {\left (e^{\left (4 \, d x + 4 \, c\right )} - 1\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {\frac {2\,\left (a-b\right )}{d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-1} \]

[In]

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

[Out]

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