3.32 \(\int \frac {\text {csch}^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=75 \[ -\frac {a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{d (a+b)^{5/2}}-\frac {\coth ^3(c+d x)}{3 d (a+b)}+\frac {a \coth (c+d x)}{d (a+b)^2} \]

[Out]

a*coth(d*x+c)/(a+b)^2/d-1/3*coth(d*x+c)^3/(a+b)/d-a*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))*b^(1/2)/(a+b)^(5/
2)/d

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Rubi [A]  time = 0.10, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4132, 453, 325, 208} \[ -\frac {a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{d (a+b)^{5/2}}-\frac {\coth ^3(c+d x)}{3 d (a+b)}+\frac {a \coth (c+d x)}{d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^4/(a + b*Sech[c + d*x]^2),x]

[Out]

-((a*Sqrt[b]*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/((a + b)^(5/2)*d)) + (a*Coth[c + d*x])/((a + b)^2*d
) - Coth[c + d*x]^3/(3*(a + b)*d)

Rule 208

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

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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps

\begin {align*} \int \frac {\text {csch}^4(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1-x^2}{x^4 \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac {\coth ^3(c+d x)}{3 (a+b) d}-\frac {a \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{(a+b) d}\\ &=\frac {a \coth (c+d x)}{(a+b)^2 d}-\frac {\coth ^3(c+d x)}{3 (a+b) d}-\frac {(a b) \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{(a+b)^2 d}\\ &=-\frac {a \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2} d}+\frac {a \coth (c+d x)}{(a+b)^2 d}-\frac {\coth ^3(c+d x)}{3 (a+b) d}\\ \end {align*}

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Mathematica [B]  time = 2.08, size = 216, normalized size = 2.88 \[ \frac {\text {sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (\frac {1}{4} \sqrt {a+b} \text {csch}(c) \sqrt {b (\cosh (c)-\sinh (c))^4} \text {csch}^3(c+d x) ((b-2 a) \sinh (2 c+3 d x)+6 a \sinh (d x)-3 b \sinh (2 c+d x))+3 a b (\sinh (2 c)-\cosh (2 c)) \tanh ^{-1}\left (\frac {(\cosh (2 c)-\sinh (2 c)) \text {sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )\right )}{6 d (a+b)^{5/2} \sqrt {b (\cosh (c)-\sinh (c))^4} \left (a+b \text {sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^4/(a + b*Sech[c + d*x]^2),x]

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(3*a*b*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*
Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(-Cosh[2*c] + Sinh[2*c]) + (Sqr
t[a + b]*Csch[c]*Csch[c + d*x]^3*Sqrt[b*(Cosh[c] - Sinh[c])^4]*(6*a*Sinh[d*x] - 3*b*Sinh[2*c + d*x] + (-2*a +
b)*Sinh[2*c + 3*d*x]))/4))/(6*(a + b)^(5/2)*d*(a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c] - Sinh[c])^4])

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fricas [B]  time = 0.52, size = 1753, normalized size = 23.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4/(a+b*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/6*(12*b*cosh(d*x + c)^4 + 48*b*cosh(d*x + c)*sinh(d*x + c)^3 + 12*b*sinh(d*x + c)^4 + 24*a*cosh(d*x + c)^2
+ 24*(3*b*cosh(d*x + c)^2 + a)*sinh(d*x + c)^2 - 3*(a*cosh(d*x + c)^6 + 6*a*cosh(d*x + c)*sinh(d*x + c)^5 + a
*sinh(d*x + c)^6 - 3*a*cosh(d*x + c)^4 + 3*(5*a*cosh(d*x + c)^2 - a)*sinh(d*x + c)^4 + 4*(5*a*cosh(d*x + c)^3
- 3*a*cosh(d*x + c))*sinh(d*x + c)^3 + 3*a*cosh(d*x + c)^2 + 3*(5*a*cosh(d*x + c)^4 - 6*a*cosh(d*x + c)^2 + a)
*sinh(d*x + c)^2 + 6*(a*cosh(d*x + c)^5 - 2*a*cosh(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) - a)*sqrt(b/(a
+ b))*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*c
osh(d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d
*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) + 4*((a^2 + a*b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d
*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 + 3*a*b + 2*b^2)*sqrt(b/(a + b)))/(a*cosh(d*x + c)^4
+ 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^
2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) + 48*(b*cos
h(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) - 8*a + 4*b)/((a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^6 + 6*(a^2 + 2
*a*b + b^2)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^2 + 2*a*b + b^2)*d*sinh(d*x + c)^6 - 3*(a^2 + 2*a*b + b^2)*d*
cosh(d*x + c)^4 + 3*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d)*sinh(d*x + c)^4 + 3*(a^2
+ 2*a*b + b^2)*d*cosh(d*x + c)^2 + 4*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^3 - 3*(a^2 + 2*a*b + b^2)*d*cosh(
d*x + c))*sinh(d*x + c)^3 + 3*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^4 - 6*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)
^2 + (a^2 + 2*a*b + b^2)*d)*sinh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d + 6*((a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^5
- 2*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*d*cosh(d*x + c))*sinh(d*x + c)), -1/3*(6*b*co
sh(d*x + c)^4 + 24*b*cosh(d*x + c)*sinh(d*x + c)^3 + 6*b*sinh(d*x + c)^4 + 12*a*cosh(d*x + c)^2 + 12*(3*b*cosh
(d*x + c)^2 + a)*sinh(d*x + c)^2 + 3*(a*cosh(d*x + c)^6 + 6*a*cosh(d*x + c)*sinh(d*x + c)^5 + a*sinh(d*x + c)^
6 - 3*a*cosh(d*x + c)^4 + 3*(5*a*cosh(d*x + c)^2 - a)*sinh(d*x + c)^4 + 4*(5*a*cosh(d*x + c)^3 - 3*a*cosh(d*x
+ c))*sinh(d*x + c)^3 + 3*a*cosh(d*x + c)^2 + 3*(5*a*cosh(d*x + c)^4 - 6*a*cosh(d*x + c)^2 + a)*sinh(d*x + c)^
2 + 6*(a*cosh(d*x + c)^5 - 2*a*cosh(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) - a)*sqrt(-b/(a + b))*arctan(1
/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-b/(a + b))/b) + 2
4*(b*cosh(d*x + c)^3 + a*cosh(d*x + c))*sinh(d*x + c) - 4*a + 2*b)/((a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^6 + 6*
(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^2 + 2*a*b + b^2)*d*sinh(d*x + c)^6 - 3*(a^2 + 2*a*b +
b^2)*d*cosh(d*x + c)^4 + 3*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d)*sinh(d*x + c)^4
+ 3*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^2 + 4*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^3 - 3*(a^2 + 2*a*b + b^2)
*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^4 - 6*(a^2 + 2*a*b + b^2)*d*cosh(
d*x + c)^2 + (a^2 + 2*a*b + b^2)*d)*sinh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d + 6*((a^2 + 2*a*b + b^2)*d*cosh(d*
x + c)^5 - 2*(a^2 + 2*a*b + b^2)*d*cosh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*d*cosh(d*x + c))*sinh(d*x + c))]

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giac [A]  time = 0.67, size = 123, normalized size = 1.64 \[ -\frac {\frac {3 \, a b \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a b - b^{2}}} + \frac {2 \, {\left (3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a + b\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4/(a+b*sech(d*x+c)^2),x, algorithm="giac")

[Out]

-1/3*(3*a*b*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/((a^2 + 2*a*b + b^2)*sqrt(-a*b - b^2))
+ 2*(3*b*e^(4*d*x + 4*c) + 6*a*e^(2*d*x + 2*c) - 2*a + b)/((a^2 + 2*a*b + b^2)*(e^(2*d*x + 2*c) - 1)^3))/d

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maple [B]  time = 0.36, size = 258, normalized size = 3.44 \[ -\frac {a \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \left (a +b \right )^{2}}-\frac {\left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{24 d \left (a +b \right )^{2}}+\frac {3 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \left (a +b \right )^{2}}-\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{8 d \left (a +b \right )^{2}}+\frac {a \sqrt {b}\, \ln \left (-\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a +b}\right )}{2 d \left (a +b \right )^{\frac {5}{2}}}-\frac {a \sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \left (a +b \right )^{\frac {5}{2}}}-\frac {1}{24 d \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {3 a}{8 d \left (a +b \right )^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b}{8 d \left (a +b \right )^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^4/(a+b*sech(d*x+c)^2),x)

[Out]

-1/24/d/(a+b)^2*a*tanh(1/2*d*x+1/2*c)^3-1/24/d/(a+b)^2*tanh(1/2*d*x+1/2*c)^3*b+3/8/d/(a+b)^2*a*tanh(1/2*d*x+1/
2*c)-1/8/d/(a+b)^2*tanh(1/2*d*x+1/2*c)*b+1/2/d*a*b^(1/2)/(a+b)^(5/2)*ln(-(a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*b
^(1/2)*tanh(1/2*d*x+1/2*c)-(a+b)^(1/2))-1/2/d*a*b^(1/2)/(a+b)^(5/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*b^(
1/2)*tanh(1/2*d*x+1/2*c)+(a+b)^(1/2))-1/24/d/(a+b)/tanh(1/2*d*x+1/2*c)^3+3/8/d/(a+b)^2/tanh(1/2*d*x+1/2*c)*a-1
/8/d/(a+b)^2/tanh(1/2*d*x+1/2*c)*b

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maxima [B]  time = 0.46, size = 195, normalized size = 2.60 \[ \frac {a b \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {2 \, {\left (6 \, a e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, b e^{\left (-4 \, d x - 4 \, c\right )} - 2 \, a + b\right )}}{3 \, {\left (a^{2} + 2 \, a b + b^{2} - 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - {\left (a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4/(a+b*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.17, size = 248, normalized size = 3.31 \[ \frac {a\,\sqrt {b}\,\ln \left (\frac {4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{{\left (a+b\right )}^2}-\frac {2\,\sqrt {b}\,\left (a+a\,{\mathrm {e}}^{2\,c+2\,d\,x}+2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{{\left (a+b\right )}^{5/2}}\right )}{2\,d\,{\left (a+b\right )}^{5/2}}-\frac {8}{3\,\left (a\,d+b\,d\right )\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )}-\frac {2\,b}{\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )\,\left (a+b\right )\,\left (a\,d+b\,d\right )}-\frac {4}{\left (a\,d+b\,d\right )\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {a\,\sqrt {b}\,\ln \left (\frac {4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{{\left (a+b\right )}^2}+\frac {2\,\sqrt {b}\,\left (a+a\,{\mathrm {e}}^{2\,c+2\,d\,x}+2\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{{\left (a+b\right )}^{5/2}}\right )}{2\,d\,{\left (a+b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)^4*(a + b/cosh(c + d*x)^2)),x)

[Out]

(a*b^(1/2)*log((4*b*exp(2*c + 2*d*x))/(a + b)^2 - (2*b^(1/2)*(a + a*exp(2*c + 2*d*x) + 2*b*exp(2*c + 2*d*x)))/
(a + b)^(5/2)))/(2*d*(a + b)^(5/2)) - 8/(3*(a*d + b*d)*(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*
d*x) - 1)) - (2*b)/((exp(2*c + 2*d*x) - 1)*(a + b)*(a*d + b*d)) - 4/((a*d + b*d)*(exp(4*c + 4*d*x) - 2*exp(2*c
+ 2*d*x) + 1)) - (a*b^(1/2)*log((4*b*exp(2*c + 2*d*x))/(a + b)^2 + (2*b^(1/2)*(a + a*exp(2*c + 2*d*x) + 2*b*e
xp(2*c + 2*d*x)))/(a + b)^(5/2)))/(2*d*(a + b)^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {csch}^{4}{\left (c + d x \right )}}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**4/(a+b*sech(d*x+c)**2),x)

[Out]

Integral(csch(c + d*x)**4/(a + b*sech(c + d*x)**2), x)

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