3.38 \(\int \frac {\text {csch}^3(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\)

Optimal. Leaf size=88 \[ \frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \cosh (c+d x)}{\sqrt {a-b}}\right )}{a^2 d \sqrt {a-b}}+\frac {(a+2 b) \tanh ^{-1}(\cosh (c+d x))}{2 a^2 d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d} \]

[Out]

1/2*(a+2*b)*arctanh(cosh(d*x+c))/a^2/d-1/2*coth(d*x+c)*csch(d*x+c)/a/d+b^(3/2)*arctan(cosh(d*x+c)*b^(1/2)/(a-b
)^(1/2))/a^2/d/(a-b)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3186, 414, 522, 206, 205} \[ \frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \cosh (c+d x)}{\sqrt {a-b}}\right )}{a^2 d \sqrt {a-b}}+\frac {(a+2 b) \tanh ^{-1}(\cosh (c+d x))}{2 a^2 d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^(3/2)*ArcTan[(Sqrt[b]*Cosh[c + d*x])/Sqrt[a - b]])/(a^2*Sqrt[a - b]*d) + ((a + 2*b)*ArcTanh[Cosh[c + d*x]])
/(2*a^2*d) - (Coth[c + d*x]*Csch[c + d*x])/(2*a*d)

Rule 205

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

Rule 206

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

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 3186

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.70, size = 201, normalized size = 2.28 \[ -\frac {\text {csch}^4(c+d x) (2 a+b \cosh (2 (c+d x))-b) \left (2 a \sqrt {a-b} \cosh (c+d x)-2 \sinh ^2(c+d x) \left (2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b}-i \sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a-b}}\right )+2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b}+i \sqrt {a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a-b}}\right )-\sqrt {a-b} (a+2 b) \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{8 a^2 d \sqrt {a-b} \left (a \text {csch}^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((2*a - b + b*Cosh[2*(c + d*x)])*Csch[c + d*x]^4*(2*a*Sqrt[a - b]*Cosh[c + d*x] - 2*(2*b^(3/2)*ArcTan[(Sq
rt[b] - I*Sqrt[a]*Tanh[(c + d*x)/2])/Sqrt[a - b]] + 2*b^(3/2)*ArcTan[(Sqrt[b] + I*Sqrt[a]*Tanh[(c + d*x)/2])/S
qrt[a - b]] - Sqrt[a - b]*(a + 2*b)*Log[Tanh[(c + d*x)/2]])*Sinh[c + d*x]^2))/(a^2*Sqrt[a - b]*d*(b + a*Csch[c
+ d*x]^2))

________________________________________________________________________________________

fricas [B]  time = 0.53, size = 1837, normalized size = 20.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*a*cosh(d*x + c)^3 + 6*a*cosh(d*x + c)*sinh(d*x + c)^2 + 2*a*sinh(d*x + c)^3 - (b*cosh(d*x + c)^4 + 4*
b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*b*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 - b)*sinh(d
*x + c)^2 + 4*(b*cosh(d*x + c)^3 - b*cosh(d*x + c))*sinh(d*x + c) + b)*sqrt(-b/(a - b))*log((b*cosh(d*x + c)^4
+ 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*(2*a - 3*b)*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c
)^2 - 2*a + 3*b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 - (2*a - 3*b)*cosh(d*x + c))*sinh(d*x + c) + 4*((a - b
)*cosh(d*x + c)^3 + 3*(a - b)*cosh(d*x + c)*sinh(d*x + c)^2 + (a - b)*sinh(d*x + c)^3 + (a - b)*cosh(d*x + c)
+ (3*(a - b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c))*sqrt(-b/(a - b)) + b)/(b*cosh(d*x + c)^4 + 4*b*cosh(d*x +
c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 2*(2*a - b)*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 + 2*a - b)*sinh
(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 + (2*a - b)*cosh(d*x + c))*sinh(d*x + c) + b)) + 2*a*cosh(d*x + c) - ((a +
2*b)*cosh(d*x + c)^4 + 4*(a + 2*b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + 2*b)*sinh(d*x + c)^4 - 2*(a + 2*b)*cos
h(d*x + c)^2 + 2*(3*(a + 2*b)*cosh(d*x + c)^2 - a - 2*b)*sinh(d*x + c)^2 + 4*((a + 2*b)*cosh(d*x + c)^3 - (a +
2*b)*cosh(d*x + c))*sinh(d*x + c) + a + 2*b)*log(cosh(d*x + c) + sinh(d*x + c) + 1) + ((a + 2*b)*cosh(d*x + c
)^4 + 4*(a + 2*b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + 2*b)*sinh(d*x + c)^4 - 2*(a + 2*b)*cosh(d*x + c)^2 + 2*
(3*(a + 2*b)*cosh(d*x + c)^2 - a - 2*b)*sinh(d*x + c)^2 + 4*((a + 2*b)*cosh(d*x + c)^3 - (a + 2*b)*cosh(d*x +
c))*sinh(d*x + c) + a + 2*b)*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 2*(3*a*cosh(d*x + c)^2 + a)*sinh(d*x + c
))/(a^2*d*cosh(d*x + c)^4 + 4*a^2*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*d*sinh(d*x + c)^4 - 2*a^2*d*cosh(d*x +
c)^2 + a^2*d + 2*(3*a^2*d*cosh(d*x + c)^2 - a^2*d)*sinh(d*x + c)^2 + 4*(a^2*d*cosh(d*x + c)^3 - a^2*d*cosh(d*
x + c))*sinh(d*x + c)), -1/2*(2*a*cosh(d*x + c)^3 + 6*a*cosh(d*x + c)*sinh(d*x + c)^2 + 2*a*sinh(d*x + c)^3 -
2*(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*b*cosh(d*x + c)^2 + 2*(3*b*co
sh(d*x + c)^2 - b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 - b*cosh(d*x + c))*sinh(d*x + c) + b)*sqrt(b/(a - b)
)*arctan(1/2*sqrt(b/(a - b))*(cosh(d*x + c) + sinh(d*x + c))) + 2*(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(
d*x + c)^3 + b*sinh(d*x + c)^4 - 2*b*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 - b)*sinh(d*x + c)^2 + 4*(b*cosh
(d*x + c)^3 - b*cosh(d*x + c))*sinh(d*x + c) + b)*sqrt(b/(a - b))*arctan(1/2*(b*cosh(d*x + c)^3 + 3*b*cosh(d*x
+ c)*sinh(d*x + c)^2 + b*sinh(d*x + c)^3 + (4*a - 3*b)*cosh(d*x + c) + (3*b*cosh(d*x + c)^2 + 4*a - 3*b)*sinh
(d*x + c))*sqrt(b/(a - b))/b) + 2*a*cosh(d*x + c) - ((a + 2*b)*cosh(d*x + c)^4 + 4*(a + 2*b)*cosh(d*x + c)*sin
h(d*x + c)^3 + (a + 2*b)*sinh(d*x + c)^4 - 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*(a + 2*b)*cosh(d*x + c)^2 - a -
2*b)*sinh(d*x + c)^2 + 4*((a + 2*b)*cosh(d*x + c)^3 - (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a + 2*b)*log(co
sh(d*x + c) + sinh(d*x + c) + 1) + ((a + 2*b)*cosh(d*x + c)^4 + 4*(a + 2*b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a
+ 2*b)*sinh(d*x + c)^4 - 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*(a + 2*b)*cosh(d*x + c)^2 - a - 2*b)*sinh(d*x + c
)^2 + 4*((a + 2*b)*cosh(d*x + c)^3 - (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a + 2*b)*log(cosh(d*x + c) + sin
h(d*x + c) - 1) + 2*(3*a*cosh(d*x + c)^2 + a)*sinh(d*x + c))/(a^2*d*cosh(d*x + c)^4 + 4*a^2*d*cosh(d*x + c)*si
nh(d*x + c)^3 + a^2*d*sinh(d*x + c)^4 - 2*a^2*d*cosh(d*x + c)^2 + a^2*d + 2*(3*a^2*d*cosh(d*x + c)^2 - a^2*d)*
sinh(d*x + c)^2 + 4*(a^2*d*cosh(d*x + c)^3 - a^2*d*cosh(d*x + c))*sinh(d*x + c))]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [a,b]=[31,78]Warning, need to choose a branch for the root of a polynomial with parameters. This
might be wrong.The choice was done assuming [a,b]=[85,31]Warning, need to choose a branch for the root of a p
olynomial with parameters. This might be wrong.The choice was done assuming [a,b]=[46,18]Warning, need to choo
se a branch for the root of a polynomial with parameters. This might be wrong.The choice was done assuming [a,
b]=[-27,57]Undef/Unsigned Inf encountered in limitEvaluation time: 0.63Limit: Max order reached or unable to m
ake series expansion Error: Bad Argument Value

________________________________________________________________________________________

maple [A]  time = 0.13, size = 133, normalized size = 1.51 \[ \frac {\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {b^{2} \arctan \left (\frac {2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 a +4 b}{4 \sqrt {a b -b^{2}}}\right )}{d \,a^{2} \sqrt {a b -b^{2}}}-\frac {1}{8 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d/a*tanh(1/2*d*x+1/2*c)^2+1/d*b^2/a^2/(a*b-b^2)^(1/2)*arctan(1/4*(2*tanh(1/2*d*x+1/2*c)^2*a-2*a+4*b)/(a*b-
b^2)^(1/2))-1/8/d/a/tanh(1/2*d*x+1/2*c)^2-1/2/d/a*ln(tanh(1/2*d*x+1/2*c))-1/d/a^2*b*ln(tanh(1/2*d*x+1/2*c))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {e^{\left (3 \, d x + 3 \, c\right )} + e^{\left (d x + c\right )}}{a d e^{\left (4 \, d x + 4 \, c\right )} - 2 \, a d e^{\left (2 \, d x + 2 \, c\right )} + a d} + \frac {{\left (a + 2 \, b\right )} \log \left ({\left (e^{\left (d x + c\right )} + 1\right )} e^{\left (-c\right )}\right )}{2 \, a^{2} d} - \frac {{\left (a + 2 \, b\right )} \log \left ({\left (e^{\left (d x + c\right )} - 1\right )} e^{\left (-c\right )}\right )}{2 \, a^{2} d} + 8 \, \int \frac {b^{2} e^{\left (3 \, d x + 3 \, c\right )} - b^{2} e^{\left (d x + c\right )}}{4 \, {\left (a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + a^{2} b + 2 \, {\left (2 \, a^{3} e^{\left (2 \, c\right )} - a^{2} b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.40, size = 571, normalized size = 6.49 \[ \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (a^7\,\sqrt {-a^4\,d^2}+18\,b^7\,\sqrt {-a^4\,d^2}-36\,a^2\,b^5\,\sqrt {-a^4\,d^2}-30\,a^3\,b^4\,\sqrt {-a^4\,d^2}+12\,a^4\,b^3\,\sqrt {-a^4\,d^2}+21\,a^5\,b^2\,\sqrt {-a^4\,d^2}+9\,a\,b^6\,\sqrt {-a^4\,d^2}+8\,a^6\,b\,\sqrt {-a^4\,d^2}\right )}{a^8\,d\,\sqrt {a^2+4\,a\,b+4\,b^2}+9\,a^2\,b^6\,d\,\sqrt {a^2+4\,a\,b+4\,b^2}-18\,a^4\,b^4\,d\,\sqrt {a^2+4\,a\,b+4\,b^2}-6\,a^5\,b^3\,d\,\sqrt {a^2+4\,a\,b+4\,b^2}+9\,a^6\,b^2\,d\,\sqrt {a^2+4\,a\,b+4\,b^2}+6\,a^7\,b\,d\,\sqrt {a^2+4\,a\,b+4\,b^2}}\right )\,\sqrt {a^2+4\,a\,b+4\,b^2}}{\sqrt {-a^4\,d^2}}-\frac {{\mathrm {e}}^{c+d\,x}}{a\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {2\,{\mathrm {e}}^{c+d\,x}}{a\,d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\left (-b\right )}^{3/2}\,\ln \left (\frac {64\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )\,\left (a^3+3\,a^2\,b-3\,b^3\right )}{a^5\,{\left (a-b\right )}^2}-\frac {128\,{\mathrm {e}}^{c+d\,x}\,\left (a^3+3\,a^2\,b-3\,b^3\right )}{a^5\,\sqrt {-b}\,{\left (a-b\right )}^{3/2}}\right )}{2\,a^2\,d\,\sqrt {a-b}}+\frac {{\left (-b\right )}^{3/2}\,\ln \left (\frac {64\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )\,\left (a^3+3\,a^2\,b-3\,b^3\right )}{a^5\,{\left (a-b\right )}^2}+\frac {128\,{\mathrm {e}}^{c+d\,x}\,\left (a^3+3\,a^2\,b-3\,b^3\right )}{a^5\,\sqrt {-b}\,{\left (a-b\right )}^{3/2}}\right )}{2\,a^2\,d\,\sqrt {a-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((exp(d*x)*exp(c)*(a^7*(-a^4*d^2)^(1/2) + 18*b^7*(-a^4*d^2)^(1/2) - 36*a^2*b^5*(-a^4*d^2)^(1/2) - 30*a^3*
b^4*(-a^4*d^2)^(1/2) + 12*a^4*b^3*(-a^4*d^2)^(1/2) + 21*a^5*b^2*(-a^4*d^2)^(1/2) + 9*a*b^6*(-a^4*d^2)^(1/2) +
8*a^6*b*(-a^4*d^2)^(1/2)))/(a^8*d*(4*a*b + a^2 + 4*b^2)^(1/2) + 9*a^2*b^6*d*(4*a*b + a^2 + 4*b^2)^(1/2) - 18*a
^4*b^4*d*(4*a*b + a^2 + 4*b^2)^(1/2) - 6*a^5*b^3*d*(4*a*b + a^2 + 4*b^2)^(1/2) + 9*a^6*b^2*d*(4*a*b + a^2 + 4*
b^2)^(1/2) + 6*a^7*b*d*(4*a*b + a^2 + 4*b^2)^(1/2)))*(4*a*b + a^2 + 4*b^2)^(1/2))/(-a^4*d^2)^(1/2) - exp(c + d
*x)/(a*d*(exp(2*c + 2*d*x) - 1)) - (2*exp(c + d*x))/(a*d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) - ((-b)^
(3/2)*log((64*(exp(2*c + 2*d*x) + 1)*(3*a^2*b + a^3 - 3*b^3))/(a^5*(a - b)^2) - (128*exp(c + d*x)*(3*a^2*b + a
^3 - 3*b^3))/(a^5*(-b)^(1/2)*(a - b)^(3/2))))/(2*a^2*d*(a - b)^(1/2)) + ((-b)^(3/2)*log((64*(exp(2*c + 2*d*x)
+ 1)*(3*a^2*b + a^3 - 3*b^3))/(a^5*(a - b)^2) + (128*exp(c + d*x)*(3*a^2*b + a^3 - 3*b^3))/(a^5*(-b)^(1/2)*(a
- b)^(3/2))))/(2*a^2*d*(a - b)^(1/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________