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]

-coth(d*x+c)/a/d+1/3*b^(1/3)*ln(a^(1/3)+b^(1/3)*tanh(d*x+c))/a^(4/3)/d-1/6*b^(1/3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*
tanh(d*x+c)+b^(2/3)*tanh(d*x+c)^2)/a^(4/3)/d+1/3*b^(1/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*tanh(d*x+c))/a^(1/3)*3^
(1/2))/a^(4/3)/d*3^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 157, normalized size of antiderivative = 1.00, 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 31

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

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 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.15, size = 190, normalized size = 1.21 \[ -\frac {2 b \text {RootSum}\left [\text {$\#$1}^3 a+\text {$\#$1}^3 b+3 \text {$\#$1}^2 a-3 \text {$\#$1}^2 b+3 \text {$\#$1} a+3 \text {$\#$1} b+a-b\& ,\frac {-\log (-\text {$\#$1} \sinh (c+d x)+\text {$\#$1} \cosh (c+d x)-\sinh (c+d x)-\cosh (c+d x))+\text {$\#$1} \log (-\text {$\#$1} \sinh (c+d x)+\text {$\#$1} \cosh (c+d x)-\sinh (c+d x)-\cosh (c+d x))+\text {$\#$1} c+\text {$\#$1} d x-c-d x}{\text {$\#$1}^2 a+\text {$\#$1}^2 b+2 \text {$\#$1} a-2 \text {$\#$1} b+a+b}\& \right ]+3 \coth (c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(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)
& ])/(a*d)

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fricas [B]  time = 0.72, size = 640, normalized size = 4.08 \[ -\frac {2 \, {\left (\sqrt {3} \cosh \left (d x + c\right )^{2} + 2 \, \sqrt {3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sqrt {3} \sinh \left (d x + c\right )^{2} - \sqrt {3}\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} b \cosh \left (d x + c\right )^{2} + 2 \, \sqrt {3} b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sqrt {3} b \sinh \left (d x + c\right )^{2} - {\left (\sqrt {3} a \cosh \left (d x + c\right )^{2} + 2 \, \sqrt {3} a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sqrt {3} a \sinh \left (d x + c\right )^{2} + \sqrt {3} a\right )} \left (\frac {b}{a}\right )^{\frac {2}{3}} - {\left (\sqrt {3} b \cosh \left (d x + c\right )^{2} + 2 \, \sqrt {3} b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sqrt {3} b \sinh \left (d x + c\right )^{2} - \sqrt {3} b\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}}}{3 \, b}\right ) + {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left ({\left (a + b\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a + b\right )} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a - b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a + b\right )} \cosh \left (d x + c\right )^{2} + a - b\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{3} + {\left (a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 2 \, {\left (a \cosh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a \sinh \left (d x + c\right )^{2} - a\right )} \left (\frac {b}{a}\right )^{\frac {2}{3}} + 2 \, {\left (a \cosh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a \sinh \left (d x + c\right )^{2} + a\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} + a + b\right ) - 2 \, {\left (\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2} - 1\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a + b\right )} \sinh \left (d x + c\right )^{2} + 2 \, a \left (\frac {b}{a}\right )^{\frac {2}{3}} - 2 \, a \left (\frac {b}{a}\right )^{\frac {1}{3}} + a - b\right ) + 12}{6 \, {\left (a d \cosh \left (d x + c\right )^{2} + 2 \, a d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a d \sinh \left (d x + c\right )^{2} - a d\right )}} \]

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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giac [A]  time = 0.25, size = 21, normalized size = 0.13 \[ -\frac {2}{a d {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}} \]

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))

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maple [C]  time = 0.57, size = 121, normalized size = 0.77 \[ -\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {2 b \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{3}-\textit {\_R} \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d a}-\frac {1}{2 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2}{a d e^{\left (2 \, d x + 2 \, c\right )} - a d} - 4 \, \int \frac {b e^{\left (4 \, d x + 4 \, c\right )} - b e^{\left (2 \, d x + 2 \, c\right )}}{a^{2} - a b + {\left (a^{2} e^{\left (6 \, c\right )} + a b e^{\left (6 \, c\right )}\right )} e^{\left (6 \, d x\right )} + 3 \, {\left (a^{2} e^{\left (4 \, c\right )} - a b e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 3 \, {\left (a^{2} e^{\left (2 \, c\right )} + a b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}}\,{d x} \]

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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mupad [B]  time = 8.71, size = 669, normalized size = 4.26 \[ \frac {b^{1/3}\,\ln \left (a^{1/3}-b^{1/3}+a^{1/3}\,{\mathrm {e}}^{2\,c+2\,d\,x}+b^{1/3}\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{3\,a^{4/3}\,d}-\frac {2}{a\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}+\frac {b^{1/3}\,\ln \left (\frac {256\,b^3\,\left (19\,a^2\,b-24\,a\,b^2+6\,a^3-b^3+8\,a^3\,{\mathrm {e}}^{2\,c+2\,d\,x}+b^3\,{\mathrm {e}}^{2\,c+2\,d\,x}+70\,a\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+113\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^4\,{\left (a+b\right )}^6}+\frac {b^{1/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1536\,b^3\,d\,\left (8\,a^2-8\,b^2+15\,a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+15\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+66\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^2\,{\left (a+b\right )}^6}+\frac {768\,b^{7/3}\,d\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (24\,a^2\,b-19\,a\,b^2+a^3-6\,b^3+a^3\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^3\,{\mathrm {e}}^{2\,c+2\,d\,x}+113\,a\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+70\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^{7/3}\,{\left (a+b\right )}^6}\right )}{3\,a^{4/3}\,d}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,a^{4/3}\,d}-\frac {b^{1/3}\,\ln \left (\frac {256\,b^3\,\left (19\,a^2\,b-24\,a\,b^2+6\,a^3-b^3+8\,a^3\,{\mathrm {e}}^{2\,c+2\,d\,x}+b^3\,{\mathrm {e}}^{2\,c+2\,d\,x}+70\,a\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+113\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^4\,{\left (a+b\right )}^6}-\frac {b^{1/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1536\,b^3\,d\,\left (8\,a^2-8\,b^2+15\,a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+15\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+66\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^2\,{\left (a+b\right )}^6}-\frac {768\,b^{7/3}\,d\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (24\,a^2\,b-19\,a\,b^2+a^3-6\,b^3+a^3\,{\mathrm {e}}^{2\,c+2\,d\,x}+8\,b^3\,{\mathrm {e}}^{2\,c+2\,d\,x}+113\,a\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}+70\,a^2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{a^{7/3}\,{\left (a+b\right )}^6}\right )}{3\,a^{4/3}\,d}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,a^{4/3}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)^2*(a + b*tanh(c + d*x)^3)),x)

[Out]

(b^(1/3)*log(a^(1/3) - b^(1/3) + a^(1/3)*exp(2*c + 2*d*x) + b^(1/3)*exp(2*c + 2*d*x)))/(3*a^(4/3)*d) - 2/(a*d*
(exp(2*c + 2*d*x) - 1)) + (b^(1/3)*log((256*b^3*(19*a^2*b - 24*a*b^2 + 6*a^3 - b^3 + 8*a^3*exp(2*c + 2*d*x) +
b^3*exp(2*c + 2*d*x) + 70*a*b^2*exp(2*c + 2*d*x) + 113*a^2*b*exp(2*c + 2*d*x)))/(a^4*(a + b)^6) + (b^(1/3)*((3
^(1/2)*1i)/2 - 1/2)*((1536*b^3*d*(8*a^2 - 8*b^2 + 15*a^2*exp(2*c + 2*d*x) + 15*b^2*exp(2*c + 2*d*x) + 66*a*b*e
xp(2*c + 2*d*x)))/(a^2*(a + b)^6) + (768*b^(7/3)*d*((3^(1/2)*1i)/2 - 1/2)*(24*a^2*b - 19*a*b^2 + a^3 - 6*b^3 +
a^3*exp(2*c + 2*d*x) + 8*b^3*exp(2*c + 2*d*x) + 113*a*b^2*exp(2*c + 2*d*x) + 70*a^2*b*exp(2*c + 2*d*x)))/(a^(
7/3)*(a + b)^6)))/(3*a^(4/3)*d))*((3^(1/2)*1i)/2 - 1/2))/(3*a^(4/3)*d) - (b^(1/3)*log((256*b^3*(19*a^2*b - 24*
a*b^2 + 6*a^3 - b^3 + 8*a^3*exp(2*c + 2*d*x) + b^3*exp(2*c + 2*d*x) + 70*a*b^2*exp(2*c + 2*d*x) + 113*a^2*b*ex
p(2*c + 2*d*x)))/(a^4*(a + b)^6) - (b^(1/3)*((3^(1/2)*1i)/2 + 1/2)*((1536*b^3*d*(8*a^2 - 8*b^2 + 15*a^2*exp(2*
c + 2*d*x) + 15*b^2*exp(2*c + 2*d*x) + 66*a*b*exp(2*c + 2*d*x)))/(a^2*(a + b)^6) - (768*b^(7/3)*d*((3^(1/2)*1i
)/2 + 1/2)*(24*a^2*b - 19*a*b^2 + a^3 - 6*b^3 + a^3*exp(2*c + 2*d*x) + 8*b^3*exp(2*c + 2*d*x) + 113*a*b^2*exp(
2*c + 2*d*x) + 70*a^2*b*exp(2*c + 2*d*x)))/(a^(7/3)*(a + b)^6)))/(3*a^(4/3)*d))*((3^(1/2)*1i)/2 + 1/2))/(3*a^(
4/3)*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {csch}^{2}{\left (c + d x \right )}}{a + b \tanh ^{3}{\left (c + d x \right )}}\, dx \]

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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