### 3.78 $$\int \frac{\text{csch}^2(c+d x)}{a+b \tanh ^3(c+d x)} \, dx$$

Optimal. Leaf size=157 $-\frac{\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 a^{4/3} d}+\frac{\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 a^{4/3} d}+\frac{\sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{4/3} d}-\frac{\coth (c+d x)}{a d}$

[Out]

(b^(1/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Tanh[c + d*x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(4/3)*d) - Coth[c + d*x]/(a
*d) + (b^(1/3)*Log[a^(1/3) + b^(1/3)*Tanh[c + d*x]])/(3*a^(4/3)*d) - (b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Ta
nh[c + d*x] + b^(2/3)*Tanh[c + d*x]^2])/(6*a^(4/3)*d)

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Rubi [A]  time = 0.140349, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.348, Rules used = {3663, 325, 292, 31, 634, 617, 204, 628} $-\frac{\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 a^{4/3} d}+\frac{\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 a^{4/3} d}+\frac{\sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{4/3} d}-\frac{\coth (c+d x)}{a d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Csch[c + d*x]^2/(a + b*Tanh[c + d*x]^3),x]

[Out]

(b^(1/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Tanh[c + d*x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(4/3)*d) - Coth[c + d*x]/(a
*d) + (b^(1/3)*Log[a^(1/3) + b^(1/3)*Tanh[c + d*x]])/(3*a^(4/3)*d) - (b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Ta
nh[c + d*x] + b^(2/3)*Tanh[c + d*x]^2])/(6*a^(4/3)*d)

Rule 3663

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]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\text{csch}^2(c+d x)}{a+b \tanh ^3(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^3\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{\coth (c+d x)}{a d}-\frac{b \operatorname{Subst}\left (\int \frac{x}{a+b x^3} \, dx,x,\tanh (c+d x)\right )}{a d}\\ &=-\frac{\coth (c+d x)}{a d}+\frac{b^{2/3} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,\tanh (c+d x)\right )}{3 a^{4/3} d}-\frac{b^{2/3} \operatorname{Subst}\left (\int \frac{\sqrt [3]{a}+\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\tanh (c+d x)\right )}{3 a^{4/3} d}\\ &=-\frac{\coth (c+d x)}{a d}+\frac{\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 a^{4/3} d}-\frac{\sqrt [3]{b} \operatorname{Subst}\left (\int \frac{-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\tanh (c+d x)\right )}{6 a^{4/3} d}-\frac{b^{2/3} \operatorname{Subst}\left (\int \frac{1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\tanh (c+d x)\right )}{2 a d}\\ &=-\frac{\coth (c+d x)}{a d}+\frac{\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 a^{4/3} d}-\frac{\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 a^{4/3} d}-\frac{\sqrt [3]{b} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt [3]{a}}\right )}{a^{4/3} d}\\ &=\frac{\sqrt [3]{b} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt{3} a^{4/3} d}-\frac{\coth (c+d x)}{a d}+\frac{\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 a^{4/3} d}-\frac{\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 a^{4/3} d}\\ \end{align*}

Mathematica [C]  time = 0.134613, size = 190, normalized size = 1.21 $-\frac{2 b \text{RootSum}\left [\text{\#1}^3 a+3 \text{\#1}^2 a+\text{\#1}^3 b-3 \text{\#1}^2 b+3 \text{\#1} a+3 \text{\#1} b+a-b\& ,\frac{-\log (-\text{\#1} \sinh (c+d x)+\text{\#1} \cosh (c+d x)-\sinh (c+d x)-\cosh (c+d x))+\text{\#1} \log (-\text{\#1} \sinh (c+d x)+\text{\#1} \cosh (c+d x)-\sinh (c+d x)-\cosh (c+d x))+\text{\#1} c+\text{\#1} d x-c-d x}{\text{\#1}^2 a+\text{\#1}^2 b+2 \text{\#1} a-2 \text{\#1} b+a+b}\& \right ]+3 \coth (c+d x)}{3 a d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Csch[c + d*x]^2/(a + b*Tanh[c + d*x]^3),x]

[Out]

-(3*Coth[c + d*x] + 2*b*RootSum[a - b + 3*a*#1 + 3*b*#1 + 3*a*#1^2 - 3*b*#1^2 + a*#1^3 + b*#1^3 & , (-c - d*x
- Log[-Cosh[c + d*x] - Sinh[c + d*x] + Cosh[c + d*x]*#1 - Sinh[c + d*x]*#1] + c*#1 + d*x*#1 + Log[-Cosh[c + d*
x] - Sinh[c + d*x] + Cosh[c + d*x]*#1 - Sinh[c + d*x]*#1]*#1)/(a + b + 2*a*#1 - 2*b*#1 + a*#1^2 + b*#1^2) & ])
/(3*a*d)

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Maple [C]  time = 0.112, size = 121, normalized size = 0.8 \begin{align*} -{\frac{1}{2\,da}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{2\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}+{\frac{2\,b}{3\,da}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{6}+3\,a{{\it \_Z}}^{4}+8\,b{{\it \_Z}}^{3}+3\,a{{\it \_Z}}^{2}+a \right ) }{\frac{{{\it \_R}}^{3}-{\it \_R}}{{{\it \_R}}^{5}a+2\,{{\it \_R}}^{3}a+4\,{{\it \_R}}^{2}b+{\it \_R}\,a}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -{\it \_R} \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^2/(a+b*tanh(d*x+c)^3),x)

[Out]

-1/2/d/a*tanh(1/2*d*x+1/2*c)-1/2/d/a/tanh(1/2*d*x+1/2*c)+2/3/d/a*b*sum((_R^3-_R)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*
a)*ln(tanh(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2}{a d e^{\left (2 \, d x + 2 \, c\right )} - a d} - 4 \, \int \frac{b e^{\left (4 \, d x + 4 \, c\right )} - b e^{\left (2 \, d x + 2 \, c\right )}}{a^{2} - a b +{\left (a^{2} e^{\left (6 \, c\right )} + a b e^{\left (6 \, c\right )}\right )} e^{\left (6 \, d x\right )} + 3 \,{\left (a^{2} e^{\left (4 \, c\right )} - a b e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 3 \,{\left (a^{2} e^{\left (2 \, c\right )} + a b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2/(a+b*tanh(d*x+c)^3),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.66535, size = 1801, normalized size = 11.47 \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)^2/(a+b*tanh(d*x+c)^3),x, algorithm="fricas")

[Out]

-1/6*(2*(sqrt(3)*cosh(d*x + c)^2 + 2*sqrt(3)*cosh(d*x + c)*sinh(d*x + c) + sqrt(3)*sinh(d*x + c)^2 - sqrt(3))*
(b/a)^(1/3)*arctan(-1/3*(sqrt(3)*b*cosh(d*x + c)^2 + 2*sqrt(3)*b*cosh(d*x + c)*sinh(d*x + c) + sqrt(3)*b*sinh(
d*x + c)^2 - (sqrt(3)*a*cosh(d*x + c)^2 + 2*sqrt(3)*a*cosh(d*x + c)*sinh(d*x + c) + sqrt(3)*a*sinh(d*x + c)^2
+ sqrt(3)*a)*(b/a)^(2/3) - (sqrt(3)*b*cosh(d*x + c)^2 + 2*sqrt(3)*b*cosh(d*x + c)*sinh(d*x + c) + sqrt(3)*b*si
nh(d*x + c)^2 - sqrt(3)*b)*(b/a)^(1/3))/b) + (cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^
2 - 1)*(b/a)^(1/3)*log((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x +
c)^4 + 2*(a - b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x
+ c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) - 2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*si
nh(d*x + c)^2 - a)*(b/a)^(2/3) + 2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 +
a)*(b/a)^(1/3) + a + b) - 2*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*(b/a)^(1/3
)*log((a + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2 + 2*a*(b/a)^(2
/3) - 2*a*(b/a)^(1/3) + a - b) + 12)/(a*d*cosh(d*x + c)^2 + 2*a*d*cosh(d*x + c)*sinh(d*x + c) + a*d*sinh(d*x +
c)^2 - a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}^{2}{\left (c + d x \right )}}{a + b \tanh ^{3}{\left (c + d x \right )}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**2/(a+b*tanh(d*x+c)**3),x)

[Out]

Integral(csch(c + d*x)**2/(a + b*tanh(c + d*x)**3), x)

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Giac [A]  time = 1.33049, size = 28, normalized size = 0.18 \begin{align*} -\frac{2}{a d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2/(a+b*tanh(d*x+c)^3),x, algorithm="giac")

[Out]

-2/(a*d*(e^(2*d*x + 2*c) - 1))