### 3.80 $$\int \frac{\text{csch}^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx$$

Optimal. Leaf size=215 $\frac{\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 a^{4/3} d}+\frac{b \log \left (a+b \tanh ^3(c+d x)\right )}{3 a^2 d}-\frac{b \log (\tanh (c+d x))}{a^2 d}-\frac{\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 a^{4/3} d}-\frac{\sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{4/3} d}-\frac{\coth ^3(c+d x)}{3 a d}+\frac{\coth (c+d x)}{a d}$

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

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Rubi [A]  time = 0.238659, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 11, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.478, Rules used = {3663, 1834, 1871, 12, 292, 31, 634, 617, 204, 628, 260} $\frac{\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 a^{4/3} d}+\frac{b \log \left (a+b \tanh ^3(c+d x)\right )}{3 a^2 d}-\frac{b \log (\tanh (c+d x))}{a^2 d}-\frac{\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 a^{4/3} d}-\frac{\sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{4/3} d}-\frac{\coth ^3(c+d x)}{3 a d}+\frac{\coth (c+d x)}{a d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Csch[c + d*x]^4/(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) - Coth[c + d*x]^3/(3*a*d) - (b*Log[Tanh[c + d*x]])/(a^2*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)*Tanh[c + d*x] + b^(2/3)*Tanh[c + d*x]^2])/(6*a^(4/
3)*d) + (b*Log[a + b*Tanh[c + d*x]^3])/(3*a^2*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 1834

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[((c*x)^m*Pq)/(a + b*
x^n), x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IntegerQ[n] &&  !IGtQ[m, 0]

Rule 1871

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rule 260

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

Rubi steps

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

Mathematica [C]  time = 3.15715, size = 322, normalized size = 1.5 $\frac{b \text{RootSum}\left [\text{\#1}^3 a+3 \text{\#1}^2 a+\text{\#1}^3 b-3 \text{\#1}^2 b+3 \text{\#1} a+3 \text{\#1} b+a-b\& ,\frac{-\text{\#1}^2 a \log \left (e^{2 (c+d x)}-\text{\#1}\right )+2 \text{\#1}^2 a c+2 \text{\#1}^2 a d x-\text{\#1}^2 b \log \left (e^{2 (c+d x)}-\text{\#1}\right )+2 \text{\#1}^2 b c+2 \text{\#1}^2 b d x+4 \text{\#1} a \log \left (e^{2 (c+d x)}-\text{\#1}\right )+a \log \left (e^{2 (c+d x)}-\text{\#1}\right )-8 \text{\#1} a c-8 \text{\#1} a d x+2 \text{\#1} b \log \left (e^{2 (c+d x)}-\text{\#1}\right )-b \log \left (e^{2 (c+d x)}-\text{\#1}\right )-4 \text{\#1} b c-4 \text{\#1} b d x-2 a c-2 a d x+2 b c+2 b d x}{\text{\#1}^2 a-\text{\#1}^2 b+2 \text{\#1} a+2 \text{\#1} b+a-b}\& \right ]-a \coth (c+d x) \left (\text{csch}^2(c+d x)-2\right )+3 b (-\log (\sinh (c+d x))+c+d x)}{3 a^2 d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Csch[c + d*x]^4/(a + b*Tanh[c + d*x]^3),x]

[Out]

(-(a*Coth[c + d*x]*(-2 + Csch[c + d*x]^2)) + 3*b*(c + d*x - Log[Sinh[c + d*x]]) + b*RootSum[a - b + 3*a*#1 + 3
*b*#1 + 3*a*#1^2 - 3*b*#1^2 + a*#1^3 + b*#1^3 & , (-2*a*c + 2*b*c - 2*a*d*x + 2*b*d*x + a*Log[E^(2*(c + d*x))
- #1] - b*Log[E^(2*(c + d*x)) - #1] - 8*a*c*#1 - 4*b*c*#1 - 8*a*d*x*#1 - 4*b*d*x*#1 + 4*a*Log[E^(2*(c + d*x))
- #1]*#1 + 2*b*Log[E^(2*(c + d*x)) - #1]*#1 + 2*a*c*#1^2 + 2*b*c*#1^2 + 2*a*d*x*#1^2 + 2*b*d*x*#1^2 - a*Log[E^
(2*(c + d*x)) - #1]*#1^2 - b*Log[E^(2*(c + d*x)) - #1]*#1^2)/(a - b + 2*a*#1 + 2*b*#1 + a*#1^2 - b*#1^2) & ])/
(3*a^2*d)

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Maple [C]  time = 0.131, size = 187, normalized size = 0.9 \begin{align*} -{\frac{1}{24\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{3}{8\,da}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{24\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}+{\frac{3}{8\,da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{b}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{b}{3\,d{a}^{2}}\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}}^{5}a+4\,{{\it \_R}}^{2}b+3\,{\it \_R}\,a}{{{\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)^4/(a+b*tanh(d*x+c)^3),x)

[Out]

-1/24/d/a*tanh(1/2*d*x+1/2*c)^3+3/8/d/a*tanh(1/2*d*x+1/2*c)-1/24/d/a/tanh(1/2*d*x+1/2*c)^3+3/8/d/a/tanh(1/2*d*
x+1/2*c)-1/d/a^2*b*ln(tanh(1/2*d*x+1/2*c))+1/3/d/a^2*b*sum((_R^5*a+4*_R^2*b+3*_R*a)/(_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*} 2 \, a b{\left (\frac{-{\left (a - b\right )} \int \frac{1}{{\left (a e^{\left (6 \, c\right )} + b e^{\left (6 \, c\right )}\right )} e^{\left (6 \, d x\right )} + 3 \,{\left (a e^{\left (4 \, c\right )} - b e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 3 \,{\left (a e^{\left (2 \, c\right )} + b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} + a - b}\,{d x} + x}{a^{3} - a^{2} b} - \frac{d x + c}{{\left (a^{3} - a^{2} b\right )} d}\right )} - 2 \, b^{2}{\left (\frac{-{\left (a - b\right )} \int \frac{1}{{\left (a e^{\left (6 \, c\right )} + b e^{\left (6 \, c\right )}\right )} e^{\left (6 \, d x\right )} + 3 \,{\left (a e^{\left (4 \, c\right )} - b e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 3 \,{\left (a e^{\left (2 \, c\right )} + b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} + a - b}\,{d x} + x}{a^{3} - a^{2} b} - \frac{d x + c}{{\left (a^{3} - a^{2} b\right )} d}\right )} + \frac{0 \, }{a} + \frac{0 \, }{a^{2}} - \frac{0 \, }{a} - \frac{0 \, }{a^{2}} + \frac{2 \,{\left (3 \, b d x e^{\left (6 \, d x + 6 \, c\right )} - 9 \, b d x e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b d x + 3 \,{\left (3 \, b d x e^{\left (2 \, c\right )} - 2 \, a e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} + 2 \, a\right )}}{3 \,{\left (a^{2} d e^{\left (6 \, d x + 6 \, c\right )} - 3 \, a^{2} d e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a^{2} d e^{\left (2 \, d x + 2 \, c\right )} - a^{2} d\right )}} - \frac{b \log \left ({\left (e^{\left (d x + c\right )} + 1\right )} e^{\left (-c\right )}\right )}{a^{2} d} - \frac{b \log \left ({\left (e^{\left (d x + c\right )} - 1\right )} e^{\left (-c\right )}\right )}{a^{2} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4/(a+b*tanh(d*x+c)^3),x, algorithm="maxima")

[Out]

2*a*b*(integrate(((a + b)*e^(4*d*x + 4*c) + 3*(a - b)*e^(2*d*x + 2*c) + 3*a + 3*b)*e^(2*d*x + 2*c)/((a + b)*e^
(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^3 - a^2*b) - (d*x + c)/(
(a^3 - a^2*b)*d)) - 2*b^2*(integrate(((a + b)*e^(4*d*x + 4*c) + 3*(a - b)*e^(2*d*x + 2*c) + 3*a + 3*b)*e^(2*d*
x + 2*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^3 -
a^2*b) - (d*x + c)/((a^3 - a^2*b)*d)) + 2*b*integrate(e^(4*d*x + 4*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(
4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/a + 2*b^2*integrate(e^(4*d*x + 4*c)/((a + b)*e^(6*d*x +
6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/a^2 - 8*b*integrate(e^(2*d*x + 2*c)/
((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/a - 4*b^2*integr
ate(e^(2*d*x + 2*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b),
x)/a^2 + 2/3*(3*b*d*x*e^(6*d*x + 6*c) - 9*b*d*x*e^(4*d*x + 4*c) - 3*b*d*x + 3*(3*b*d*x*e^(2*c) - 2*a*e^(2*c))
*e^(2*d*x) + 2*a)/(a^2*d*e^(6*d*x + 6*c) - 3*a^2*d*e^(4*d*x + 4*c) + 3*a^2*d*e^(2*d*x + 2*c) - a^2*d) - b*log(
(e^(d*x + c) + 1)*e^(-c))/(a^2*d) - b*log((e^(d*x + c) - 1)*e^(-c))/(a^2*d)

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Fricas [C]  time = 17.1023, size = 4415, normalized size = 20.53 \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)^4/(a+b*tanh(d*x+c)^3),x, algorithm="fricas")

[Out]

1/12*(12*sqrt(1/3)*(a^2*d*e^(6*d*x + 6*c) - 3*a^2*d*e^(4*d*x + 4*c) + 3*a^2*d*e^(2*d*x + 2*c) - a^2*d)*sqrt(((
(1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2*b/(a^2*d))^2*a^4
*d^2 + 4*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2*b/(a^2
*d))*a^2*b*d + 4*b^2)/(a^4*d^2))*arctan(-1/8*sqrt(1/3)*((a^6 + a^5*b)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3
) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2*b/(a^2*d))^2*d^3*e^(2*d*x + 2*c) - 4*(2*a^3*b + a^2*b^2
- a*b^3)*d*e^(2*d*x + 2*c) - 2*((a^5 - a^4*b - 2*a^3*b^2)*d^2*e^(2*d*x + 2*c) + (a^5 + a^4*b)*d^2)*((1/2)^(1/
3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2*b/(a^2*d)) - (((1/2)^(1/3
)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2*b/(a^2*d))^2*a^4*d^3 - 2*(
a^3 - 2*a^2*b)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2*
b/(a^2*d))*d^2 - 4*(2*a*b - b^2)*d)*sqrt(a^4 + 4*a^3*b + 5*a^2*b^2 + 2*a*b^3 - 1/2*((a^6 + a^5*b)*d^2*e^(2*d*x
+ 2*c) - (a^6 + a^5*b)*d^2)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^
3))^(1/3) - 2*b/(a^2*d))^2 + ((a^5 - a^4*b - 2*a^3*b^2)*d*e^(2*d*x + 2*c) + (a^5 + 3*a^4*b + 2*a^3*b^2)*d)*((1
/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2*b/(a^2*d)) + (a^4
+ 2*a^3*b + a^2*b^2)*e^(4*d*x + 4*c) + 2*(a^4 + a^3*b - a^2*b^2 - a*b^3)*e^(2*d*x + 2*c)))*sqrt((((1/2)^(1/3)*
(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2*b/(a^2*d))^2*a^4*d^2 + 4*((1
/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2*b/(a^2*d))*a^2*b*d
+ 4*b^2)/(a^4*d^2))/(a^2*b + a*b^2)) - 2*(a^2*d*e^(6*d*x + 6*c) - 3*a^2*d*e^(4*d*x + 4*c) + 3*a^2*d*e^(2*d*x
+ 2*c) - a^2*d)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2
*b/(a^2*d))*log(1/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3)
- 2*b/(a^2*d))^2*a^4*d^2 - (a^3 - 2*a^2*b)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b
- b^3)/(a^6*d^3))^(1/3) - 2*b/(a^2*d))*d + a^2 - 3*a*b + 2*b^2 + (a^2 + a*b)*e^(2*d*x + 2*c)) - 48*a*e^(2*d*x
+ 2*c) + ((a^2*d*e^(6*d*x + 6*c) - 3*a^2*d*e^(4*d*x + 4*c) + 3*a^2*d*e^(2*d*x + 2*c) - a^2*d)*((1/2)^(1/3)*(I
*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2*b/(a^2*d)) + 6*b*e^(6*d*x + 6*
c) - 18*b*e^(4*d*x + 4*c) + 18*b*e^(2*d*x + 2*c) - 6*b)*log(a^4 + 4*a^3*b + 5*a^2*b^2 + 2*a*b^3 - 1/2*((a^6 +
a^5*b)*d^2*e^(2*d*x + 2*c) - (a^6 + a^5*b)*d^2)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a
^2*b - b^3)/(a^6*d^3))^(1/3) - 2*b/(a^2*d))^2 + ((a^5 - a^4*b - 2*a^3*b^2)*d*e^(2*d*x + 2*c) + (a^5 + 3*a^4*b
+ 2*a^3*b^2)*d)*((1/2)^(1/3)*(I*sqrt(3) + 1)*(b/(a^4*d^3) - b^3/(a^6*d^3) - (a^2*b - b^3)/(a^6*d^3))^(1/3) - 2
*b/(a^2*d)) + (a^4 + 2*a^3*b + a^2*b^2)*e^(4*d*x + 4*c) + 2*(a^4 + a^3*b - a^2*b^2 - a*b^3)*e^(2*d*x + 2*c)) -
12*(b*e^(6*d*x + 6*c) - 3*b*e^(4*d*x + 4*c) + 3*b*e^(2*d*x + 2*c) - b)*log(e^(2*d*x + 2*c) - 1) + 16*a)/(a^2*
d*e^(6*d*x + 6*c) - 3*a^2*d*e^(4*d*x + 4*c) + 3*a^2*d*e^(2*d*x + 2*c) - a^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}^{4}{\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)**4/(a+b*tanh(d*x+c)**3),x)

[Out]

Integral(csch(c + d*x)**4/(a + b*tanh(c + d*x)**3), x)

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Giac [A]  time = 1.42274, size = 243, normalized size = 1.13 \begin{align*} \frac{\frac{2 \, b \log \left ({\left | a e^{\left (6 \, d x + 6 \, c\right )} + b e^{\left (6 \, d x + 6 \, c\right )} + 3 \, a e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b e^{\left (2 \, d x + 2 \, c\right )} + a - b \right |}\right )}{a^{2}} - \frac{6 \, b \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{a^{2}} + \frac{11 \, b e^{\left (6 \, d x + 6 \, c\right )} - 33 \, b e^{\left (4 \, d x + 4 \, c\right )} - 24 \, a e^{\left (2 \, d x + 2 \, c\right )} + 33 \, b e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a - 11 \, b}{a^{2}{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{6 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4/(a+b*tanh(d*x+c)^3),x, algorithm="giac")

[Out]

1/6*(2*b*log(abs(a*e^(6*d*x + 6*c) + b*e^(6*d*x + 6*c) + 3*a*e^(4*d*x + 4*c) - 3*b*e^(4*d*x + 4*c) + 3*a*e^(2*
d*x + 2*c) + 3*b*e^(2*d*x + 2*c) + a - b))/a^2 - 6*b*log(abs(e^(2*d*x + 2*c) - 1))/a^2 + (11*b*e^(6*d*x + 6*c)
- 33*b*e^(4*d*x + 4*c) - 24*a*e^(2*d*x + 2*c) + 33*b*e^(2*d*x + 2*c) + 8*a - 11*b)/(a^2*(e^(2*d*x + 2*c) - 1)
^3))/d