### 3.13 $$\int \text{csch}(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx$$

Optimal. Leaf size=51 $-\frac{a^2 \tanh ^{-1}(\cosh (c+d x))}{d}-\frac{b (2 a+b) \text{sech}(c+d x)}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d}$

[Out]

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

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Rubi [A]  time = 0.0645143, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {3664, 390, 207} $-\frac{a^2 \tanh ^{-1}(\cosh (c+d x))}{d}-\frac{b (2 a+b) \text{sech}(c+d x)}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 390

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 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 \text{csch}(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b-b x^2\right )^2}{-1+x^2} \, dx,x,\text{sech}(c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-b (2 a+b)+b^2 x^2+\frac{a^2}{-1+x^2}\right ) \, dx,x,\text{sech}(c+d x)\right )}{d}\\ &=-\frac{b (2 a+b) \text{sech}(c+d x)}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\text{sech}(c+d x)\right )}{d}\\ &=-\frac{a^2 \tanh ^{-1}(\cosh (c+d x))}{d}-\frac{b (2 a+b) \text{sech}(c+d x)}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.155003, size = 50, normalized size = 0.98 $\frac{3 a^2 \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )-3 b (2 a+b) \text{sech}(c+d x)+b^2 \text{sech}^3(c+d x)}{3 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 97, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ( -2\,{a}^{2}{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) +2\,ab \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{\cosh \left ( dx+c \right ) }}-\cosh \left ( dx+c \right ) \right ) +{b}^{2} \left ( -{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+{\frac{2\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3\,\cosh \left ( dx+c \right ) }}-{\frac{2\,\cosh \left ( dx+c \right ) }{3}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.04153, size = 265, normalized size = 5.2 \begin{align*} -\frac{2}{3} \, b^{2}{\left (\frac{3 \, e^{\left (-d x - c\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 )}} + \frac{2 \, e^{\left (-3 \, d x - 3 \, c\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 )}} + \frac{3 \, e^{\left (-5 \, d x - 5 \, c\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 )} + \frac{a^{2} \log \left (\tanh \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{d} - \frac{4 \, a b}{d{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*b^2*(3*e^(-d*x - c)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 2*e^(-3*d*x -
3*c)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 3*e^(-5*d*x - 5*c)/(d*(3*e^(-2*d*x
- 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1))) + a^2*log(tanh(1/2*d*x + 1/2*c))/d - 4*a*b/(d*(e^(d*x +
c) + e^(-d*x - c)))

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Fricas [B]  time = 2.00274, size = 2331, normalized size = 45.71 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*(2*a*b + b^2)*cosh(d*x + c)^5 + 30*(2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^4 + 6*(2*a*b + b^2)*sinh(
d*x + c)^5 + 4*(6*a*b + b^2)*cosh(d*x + c)^3 + 4*(15*(2*a*b + b^2)*cosh(d*x + c)^2 + 6*a*b + b^2)*sinh(d*x + c
)^3 + 12*(5*(2*a*b + b^2)*cosh(d*x + c)^3 + (6*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 6*(2*a*b + b^2)*cos
h(d*x + c) + 3*(a^2*cosh(d*x + c)^6 + 6*a^2*cosh(d*x + c)*sinh(d*x + c)^5 + a^2*sinh(d*x + c)^6 + 3*a^2*cosh(d
*x + c)^4 + 3*(5*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^4 + 3*a^2*cosh(d*x + c)^2 + 4*(5*a^2*cosh(d*x + c)^3
+ 3*a^2*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*a^2*cosh(d*x + c)^4 + 6*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c
)^2 + a^2 + 6*(a^2*cosh(d*x + c)^5 + 2*a^2*cosh(d*x + c)^3 + a^2*cosh(d*x + c))*sinh(d*x + c))*log(cosh(d*x +
c) + sinh(d*x + c) + 1) - 3*(a^2*cosh(d*x + c)^6 + 6*a^2*cosh(d*x + c)*sinh(d*x + c)^5 + a^2*sinh(d*x + c)^6 +
3*a^2*cosh(d*x + c)^4 + 3*(5*a^2*cosh(d*x + c)^2 + a^2)*sinh(d*x + c)^4 + 3*a^2*cosh(d*x + c)^2 + 4*(5*a^2*co
sh(d*x + c)^3 + 3*a^2*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*a^2*cosh(d*x + c)^4 + 6*a^2*cosh(d*x + c)^2 + a^2)
*sinh(d*x + c)^2 + a^2 + 6*(a^2*cosh(d*x + c)^5 + 2*a^2*cosh(d*x + c)^3 + a^2*cosh(d*x + c))*sinh(d*x + c))*lo
g(cosh(d*x + c) + sinh(d*x + c) - 1) + 6*(5*(2*a*b + b^2)*cosh(d*x + c)^4 + 2*(6*a*b + b^2)*cosh(d*x + c)^2 +
2*a*b + b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^6 + 6*d*cosh(d*x + c)*sinh(d*x + c)^5 + d*sinh(d*x + c)^6 + 3*d*c
osh(d*x + c)^4 + 3*(5*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sin
h(d*x + c)^3 + 3*d*cosh(d*x + c)^2 + 3*(5*d*cosh(d*x + c)^4 + 6*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^2 + 6*(d*
cosh(d*x + c)^5 + 2*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \operatorname{csch}{\left (c + d x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.33456, size = 170, normalized size = 3.33 \begin{align*} -\frac{3 \, a^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) - 3 \, a^{2} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right ) + \frac{2 \,{\left (6 \, a b e^{\left (5 \, d x + 5 \, c\right )} + 3 \, b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 12 \, a b e^{\left (3 \, d x + 3 \, c\right )} + 2 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 6 \, a b e^{\left (d x + c\right )} + 3 \, b^{2} e^{\left (d x + c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{3 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*a^2*log(e^(d*x + c) + 1) - 3*a^2*log(abs(e^(d*x + c) - 1)) + 2*(6*a*b*e^(5*d*x + 5*c) + 3*b^2*e^(5*d*x
+ 5*c) + 12*a*b*e^(3*d*x + 3*c) + 2*b^2*e^(3*d*x + 3*c) + 6*a*b*e^(d*x + c) + 3*b^2*e^(d*x + c))/(e^(2*d*x +
2*c) + 1)^3)/d