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

Optimal. Leaf size=206 \[ \frac {b^2 \text {sech}(c+d x)}{a^3 d}-\frac {b^2 \tanh ^{-1}(\cosh (c+d x))}{a^3 d}+\frac {b \tanh (c+d x)}{a^2 d}+\frac {b \coth (c+d x)}{a^2 d}+\frac {2 b^5 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^3 d \left (a^2+b^2\right )^{3/2}}-\frac {b^3 \text {sech}(c+d x) (a \sinh (c+d x)+b)}{a^3 d \left (a^2+b^2\right )}-\frac {3 \text {sech}(c+d x)}{2 a d}+\frac {3 \tanh ^{-1}(\cosh (c+d x))}{2 a d}-\frac {\text {csch}^2(c+d x) \text {sech}(c+d x)}{2 a d} \]

[Out]

3/2*arctanh(cosh(d*x+c))/a/d-b^2*arctanh(cosh(d*x+c))/a^3/d+2*b^5*arctanh((b-a*tanh(1/2*d*x+1/2*c))/(a^2+b^2)^
(1/2))/a^3/(a^2+b^2)^(3/2)/d+b*coth(d*x+c)/a^2/d-3/2*sech(d*x+c)/a/d+b^2*sech(d*x+c)/a^3/d-1/2*csch(d*x+c)^2*s
ech(d*x+c)/a/d-b^3*sech(d*x+c)*(b+a*sinh(d*x+c))/a^3/(a^2+b^2)/d+b*tanh(d*x+c)/a^2/d

________________________________________________________________________________________

Rubi [A]  time = 0.42, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 12, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {2898, 2622, 321, 207, 2620, 14, 288, 2696, 12, 2660, 618, 204} \[ \frac {2 b^5 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^3 d \left (a^2+b^2\right )^{3/2}}+\frac {b^2 \text {sech}(c+d x)}{a^3 d}-\frac {b^2 \tanh ^{-1}(\cosh (c+d x))}{a^3 d}-\frac {b^3 \text {sech}(c+d x) (a \sinh (c+d x)+b)}{a^3 d \left (a^2+b^2\right )}+\frac {b \tanh (c+d x)}{a^2 d}+\frac {b \coth (c+d x)}{a^2 d}-\frac {3 \text {sech}(c+d x)}{2 a d}+\frac {3 \tanh ^{-1}(\cosh (c+d x))}{2 a d}-\frac {\text {csch}^2(c+d x) \text {sech}(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 207

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

Rule 288

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

Rule 321

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

Rule 618

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

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2898

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])
, x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/2, 0])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 2.43, size = 185, normalized size = 0.90 \[ \frac {\frac {8 \text {sech}(c+d x) (b \sinh (c+d x)-a)}{a^2+b^2}+\frac {4 b \tanh \left (\frac {1}{2} (c+d x)\right )}{a^2}+\frac {4 b \coth \left (\frac {1}{2} (c+d x)\right )}{a^2}-\frac {4 \left (3 a^2-2 b^2\right ) \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {16 b^5 \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{a^3 \left (-a^2-b^2\right )^{3/2}}-\frac {\text {csch}^2\left (\frac {1}{2} (c+d x)\right )}{a}-\frac {\text {sech}^2\left (\frac {1}{2} (c+d x)\right )}{a}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((16*b^5*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]])/(a^3*(-a^2 - b^2)^(3/2)) + (4*b*Coth[(c + d*x)/2]
)/a^2 - Csch[(c + d*x)/2]^2/a - (4*(3*a^2 - 2*b^2)*Log[Tanh[(c + d*x)/2]])/a^3 - Sech[(c + d*x)/2]^2/a + (8*Se
ch[c + d*x]*(-a + b*Sinh[c + d*x]))/(a^2 + b^2) + (4*b*Tanh[(c + d*x)/2])/a^2)/(8*d)

________________________________________________________________________________________

fricas [B]  time = 1.23, size = 2653, normalized size = 12.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(8*a^5*b + 12*a^3*b^3 + 4*a*b^5 + 2*(3*a^6 + 4*a^4*b^2 + a^2*b^4)*cosh(d*x + c)^5 + 2*(3*a^6 + 4*a^4*b^2
+ a^2*b^4)*sinh(d*x + c)^5 - 4*(a^3*b^3 + a*b^5)*cosh(d*x + c)^4 - 2*(2*a^3*b^3 + 2*a*b^5 - 5*(3*a^6 + 4*a^4*b
^2 + a^2*b^4)*cosh(d*x + c))*sinh(d*x + c)^4 - 4*(a^6 - a^2*b^4)*cosh(d*x + c)^3 - 4*(a^6 - a^2*b^4 - 5*(3*a^6
+ 4*a^4*b^2 + a^2*b^4)*cosh(d*x + c)^2 + 4*(a^3*b^3 + a*b^5)*cosh(d*x + c))*sinh(d*x + c)^3 - 8*(a^5*b + a^3*
b^3)*cosh(d*x + c)^2 - 4*(2*a^5*b + 2*a^3*b^3 - 5*(3*a^6 + 4*a^4*b^2 + a^2*b^4)*cosh(d*x + c)^3 + 6*(a^3*b^3 +
a*b^5)*cosh(d*x + c)^2 + 3*(a^6 - a^2*b^4)*cosh(d*x + c))*sinh(d*x + c)^2 - 2*(b^5*cosh(d*x + c)^6 + 6*b^5*co
sh(d*x + c)*sinh(d*x + c)^5 + b^5*sinh(d*x + c)^6 - b^5*cosh(d*x + c)^4 - b^5*cosh(d*x + c)^2 + b^5 + (15*b^5*
cosh(d*x + c)^2 - b^5)*sinh(d*x + c)^4 + 4*(5*b^5*cosh(d*x + c)^3 - b^5*cosh(d*x + c))*sinh(d*x + c)^3 + (15*b
^5*cosh(d*x + c)^4 - 6*b^5*cosh(d*x + c)^2 - b^5)*sinh(d*x + c)^2 + 2*(3*b^5*cosh(d*x + c)^5 - 2*b^5*cosh(d*x
+ c)^3 - b^5*cosh(d*x + c))*sinh(d*x + c))*sqrt(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*
a*b*cosh(d*x + c) + 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(b*cosh(d*x +
c) + b*sinh(d*x + c) + a))/(b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a
)*sinh(d*x + c) - b)) + 2*(3*a^6 + 4*a^4*b^2 + a^2*b^4)*cosh(d*x + c) - ((3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)
*cosh(d*x + c)^6 + 6*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)*sinh(d*x + c)^5 + (3*a^6 + 4*a^4*b^2
- a^2*b^4 - 2*b^6)*sinh(d*x + c)^6 + 3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6 - (3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^
6)*cosh(d*x + c)^4 - (3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6 - 15*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x
+ c)^2)*sinh(d*x + c)^4 + 4*(5*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)^3 - (3*a^6 + 4*a^4*b^2 - a^
2*b^4 - 2*b^6)*cosh(d*x + c))*sinh(d*x + c)^3 - (3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)^2 - (3*a^6
+ 4*a^4*b^2 - a^2*b^4 - 2*b^6 - 15*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)^4 + 6*(3*a^6 + 4*a^4*b
^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 2*(3*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x +
c)^5 - 2*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)^3 - (3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d
*x + c))*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) + 1) + ((3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d
*x + c)^6 + 6*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)*sinh(d*x + c)^5 + (3*a^6 + 4*a^4*b^2 - a^2*b
^4 - 2*b^6)*sinh(d*x + c)^6 + 3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6 - (3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh
(d*x + c)^4 - (3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6 - 15*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)^2)
*sinh(d*x + c)^4 + 4*(5*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)^3 - (3*a^6 + 4*a^4*b^2 - a^2*b^4 -
2*b^6)*cosh(d*x + c))*sinh(d*x + c)^3 - (3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)^2 - (3*a^6 + 4*a^
4*b^2 - a^2*b^4 - 2*b^6 - 15*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)^4 + 6*(3*a^6 + 4*a^4*b^2 - a^
2*b^4 - 2*b^6)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 2*(3*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)^5 -
2*(3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)^3 - (3*a^6 + 4*a^4*b^2 - a^2*b^4 - 2*b^6)*cosh(d*x + c)
)*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 2*(3*a^6 + 4*a^4*b^2 + a^2*b^4 + 5*(3*a^6 + 4*a^4*b^
2 + a^2*b^4)*cosh(d*x + c)^4 - 8*(a^3*b^3 + a*b^5)*cosh(d*x + c)^3 - 6*(a^6 - a^2*b^4)*cosh(d*x + c)^2 - 8*(a^
5*b + a^3*b^3)*cosh(d*x + c))*sinh(d*x + c))/((a^7 + 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x + c)^6 + 6*(a^7 + 2*a^5*b
^2 + a^3*b^4)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^7 + 2*a^5*b^2 + a^3*b^4)*d*sinh(d*x + c)^6 - (a^7 + 2*a^5*b
^2 + a^3*b^4)*d*cosh(d*x + c)^4 + (15*(a^7 + 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x + c)^2 - (a^7 + 2*a^5*b^2 + a^3*b
^4)*d)*sinh(d*x + c)^4 - (a^7 + 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x + c)^2 + 4*(5*(a^7 + 2*a^5*b^2 + a^3*b^4)*d*co
sh(d*x + c)^3 - (a^7 + 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x + c))*sinh(d*x + c)^3 + (15*(a^7 + 2*a^5*b^2 + a^3*b^4)
*d*cosh(d*x + c)^4 - 6*(a^7 + 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x + c)^2 - (a^7 + 2*a^5*b^2 + a^3*b^4)*d)*sinh(d*x
+ c)^2 + (a^7 + 2*a^5*b^2 + a^3*b^4)*d + 2*(3*(a^7 + 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x + c)^5 - 2*(a^7 + 2*a^5*
b^2 + a^3*b^4)*d*cosh(d*x + c)^3 - (a^7 + 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x + c))*sinh(d*x + c))

________________________________________________________________________________________

giac [A]  time = 0.57, size = 224, normalized size = 1.09 \[ -\frac {\frac {2 \, b^{5} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{5} + a^{3} b^{2}\right )} \sqrt {a^{2} + b^{2}}} + \frac {4 \, {\left (a e^{\left (d x + c\right )} + b\right )}}{{\left (a^{2} + b^{2}\right )} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}} - \frac {{\left (3 \, a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{3}} + \frac {{\left (3 \, a^{2} - 2 \, b^{2}\right )} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{3}} + \frac {2 \, {\left (a e^{\left (3 \, d x + 3 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a e^{\left (d x + c\right )} + 2 \, b\right )}}{a^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b^5*log(abs(2*b*e^(d*x + c) + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^(d*x + c) + 2*a + 2*sqrt(a^2 + b^2)))
/((a^5 + a^3*b^2)*sqrt(a^2 + b^2)) + 4*(a*e^(d*x + c) + b)/((a^2 + b^2)*(e^(2*d*x + 2*c) + 1)) - (3*a^2 - 2*b^
2)*log(e^(d*x + c) + 1)/a^3 + (3*a^2 - 2*b^2)*log(abs(e^(d*x + c) - 1))/a^3 + 2*(a*e^(3*d*x + 3*c) - 2*b*e^(2*
d*x + 2*c) + a*e^(d*x + c) + 2*b)/(a^2*(e^(2*d*x + 2*c) - 1)^2))/d

________________________________________________________________________________________

maple [A]  time = 0.00, size = 233, normalized size = 1.13 \[ \frac {\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{2 d \,a^{2}}-\frac {2 b^{5} \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,a^{3} \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}-\frac {1}{8 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {3 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{3}}+\frac {b}{2 d \,a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{d \left (a^{2}+b^{2}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a}{d \left (a^{2}+b^{2}\right ) \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d/a*tanh(1/2*d*x+1/2*c)^2+1/2/d/a^2*tanh(1/2*d*x+1/2*c)*b-2/d/a^3*b^5/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tan
h(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-1/8/d/a/tanh(1/2*d*x+1/2*c)^2-3/2/d/a*ln(tanh(1/2*d*x+1/2*c))+1/d/a^3*l
n(tanh(1/2*d*x+1/2*c))*b^2+1/2/d*b/a^2/tanh(1/2*d*x+1/2*c)+2/d/(a^2+b^2)/(tanh(1/2*d*x+1/2*c)^2+1)*tanh(1/2*d*
x+1/2*c)*b-2/d/(a^2+b^2)/(tanh(1/2*d*x+1/2*c)^2+1)*a

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 334, normalized size = 1.62 \[ -\frac {b^{5} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{5} + a^{3} b^{2}\right )} \sqrt {a^{2} + b^{2}} d} - \frac {4 \, a^{2} b e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, b^{3} e^{\left (-4 \, d x - 4 \, c\right )} - 4 \, a^{2} b - 2 \, b^{3} + {\left (3 \, a^{3} + a b^{2}\right )} e^{\left (-d x - c\right )} - 2 \, {\left (a^{3} - a b^{2}\right )} e^{\left (-3 \, d x - 3 \, c\right )} + {\left (3 \, a^{3} + a b^{2}\right )} e^{\left (-5 \, d x - 5 \, c\right )}}{{\left (a^{4} + a^{2} b^{2} - {\left (a^{4} + a^{2} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - {\left (a^{4} + a^{2} b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (a^{4} + a^{2} b^{2}\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )} d} + \frac {{\left (3 \, a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{2 \, a^{3} d} - \frac {{\left (3 \, a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{2 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.47, size = 531, normalized size = 2.58 \[ \frac {b^5\,\ln \left (2\,a^4\,b-4\,a^5\,{\mathrm {e}}^{c+d\,x}+b^5+3\,a^2\,b^3+4\,a^2\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}+b^2\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}-7\,a^3\,b^2\,{\mathrm {e}}^{c+d\,x}-2\,a\,b\,\sqrt {{\left (a^2+b^2\right )}^3}-3\,a\,b^4\,{\mathrm {e}}^{c+d\,x}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{d\,a^9+3\,d\,a^7\,b^2+3\,d\,a^5\,b^4+d\,a^3\,b^6}-\frac {\frac {{\mathrm {e}}^{c+d\,x}}{a\,d}-\frac {2\,\left (a^2\,b+b^3\right )}{a^2\,d\,\left (a^2+b^2\right )}}{{\mathrm {e}}^{2\,c+2\,d\,x}-1}-\frac {\frac {2\,b}{d\,\left (a^2+b^2\right )}+\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2+b^2\right )}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\ln \left ({\mathrm {e}}^{c+d\,x}-1\right )\,\left (3\,a^2-2\,b^2\right )}{2\,a^3\,d}+\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1\right )\,\left (3\,a^2-2\,b^2\right )}{2\,a^3\,d}-\frac {b^5\,\ln \left (4\,a^5\,{\mathrm {e}}^{c+d\,x}-2\,a^4\,b-b^5-3\,a^2\,b^3+4\,a^2\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}+b^2\,{\mathrm {e}}^{c+d\,x}\,\sqrt {{\left (a^2+b^2\right )}^3}+7\,a^3\,b^2\,{\mathrm {e}}^{c+d\,x}-2\,a\,b\,\sqrt {{\left (a^2+b^2\right )}^3}+3\,a\,b^4\,{\mathrm {e}}^{c+d\,x}\right )\,\sqrt {{\left (a^2+b^2\right )}^3}}{d\,a^9+3\,d\,a^7\,b^2+3\,d\,a^5\,b^4+d\,a^3\,b^6}-\frac {2\,{\mathrm {e}}^{c+d\,x}}{a\,d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^5*log(2*a^4*b - 4*a^5*exp(c + d*x) + b^5 + 3*a^2*b^3 + 4*a^2*exp(c + d*x)*((a^2 + b^2)^3)^(1/2) + b^2*exp(c
+ d*x)*((a^2 + b^2)^3)^(1/2) - 7*a^3*b^2*exp(c + d*x) - 2*a*b*((a^2 + b^2)^3)^(1/2) - 3*a*b^4*exp(c + d*x))*(
(a^2 + b^2)^3)^(1/2))/(a^9*d + a^3*b^6*d + 3*a^5*b^4*d + 3*a^7*b^2*d) - (exp(c + d*x)/(a*d) - (2*(a^2*b + b^3)
)/(a^2*d*(a^2 + b^2)))/(exp(2*c + 2*d*x) - 1) - ((2*b)/(d*(a^2 + b^2)) + (2*a*exp(c + d*x))/(d*(a^2 + b^2)))/(
exp(2*c + 2*d*x) + 1) - (log(exp(c + d*x) - 1)*(3*a^2 - 2*b^2))/(2*a^3*d) + (log(exp(c + d*x) + 1)*(3*a^2 - 2*
b^2))/(2*a^3*d) - (b^5*log(4*a^5*exp(c + d*x) - 2*a^4*b - b^5 - 3*a^2*b^3 + 4*a^2*exp(c + d*x)*((a^2 + b^2)^3)
^(1/2) + b^2*exp(c + d*x)*((a^2 + b^2)^3)^(1/2) + 7*a^3*b^2*exp(c + d*x) - 2*a*b*((a^2 + b^2)^3)^(1/2) + 3*a*b
^4*exp(c + d*x))*((a^2 + b^2)^3)^(1/2))/(a^9*d + a^3*b^6*d + 3*a^5*b^4*d + 3*a^7*b^2*d) - (2*exp(c + d*x))/(a*
d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________