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

Optimal. Leaf size=113 \[ \frac {b \coth (c+d x)}{a^2 d}+\frac {\left (a^2-2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}+\frac {2 b^3 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^3 d \sqrt {a^2+b^2}}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2802, 3055, 3001, 3770, 2660, 618, 204} \[ \frac {2 b^3 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^3 d \sqrt {a^2+b^2}}+\frac {\left (a^2-2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}+\frac {b \coth (c+d x)}{a^2 d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 1.81, size = 145, normalized size = 1.28 \[ -\frac {4 \left (a^2-2 b^2\right ) \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+\frac {16 b^3 \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+a^2 \text {csch}^2\left (\frac {1}{2} (c+d x)\right )+a^2 \text {sech}^2\left (\frac {1}{2} (c+d x)\right )-4 a b \tanh \left (\frac {1}{2} (c+d x)\right )-4 a b \coth \left (\frac {1}{2} (c+d x)\right )}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((16*b^3*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - 4*a*b*Coth[(c + d*x)/2] +
a^2*Csch[(c + d*x)/2]^2 + 4*(a^2 - 2*b^2)*Log[Tanh[(c + d*x)/2]] + a^2*Sech[(c + d*x)/2]^2 - 4*a*b*Tanh[(c +
d*x)/2])/(a^3*d)

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fricas [B]  time = 0.66, size = 1203, normalized size = 10.65 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*a^3*b + 4*a*b^3 + 2*(a^4 + a^2*b^2)*cosh(d*x + c)^3 + 2*(a^4 + a^2*b^2)*sinh(d*x + c)^3 - 4*(a^3*b + a
*b^3)*cosh(d*x + c)^2 - 2*(2*a^3*b + 2*a*b^3 - 3*(a^4 + a^2*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 - 2*(b^3*cosh(
d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 - 2*b^3*cosh(d*x + c)^2 + b^3 + 2*(3*b^
3*cosh(d*x + c)^2 - b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 - b^3*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*s
inh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)) + 2*(a^4 + a^2*b^2)*cosh(d*x
+ c) - ((a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^4 + 4*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)*sinh(d*x + c)^3 + (a
^4 - a^2*b^2 - 2*b^4)*sinh(d*x + c)^4 + a^4 - a^2*b^2 - 2*b^4 - 2*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^2 - 2*
(a^4 - a^2*b^2 - 2*b^4 - 3*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^4 - a^2*b^2 - 2*b^
4)*cosh(d*x + c)^3 - (a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c))*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) +
1) + ((a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^4 + 4*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^
4 - a^2*b^2 - 2*b^4)*sinh(d*x + c)^4 + a^4 - a^2*b^2 - 2*b^4 - 2*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^2 - 2*(
a^4 - a^2*b^2 - 2*b^4 - 3*(a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^4 - a^2*b^2 - 2*b^4
)*cosh(d*x + c)^3 - (a^4 - a^2*b^2 - 2*b^4)*cosh(d*x + c))*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) -
1) + 2*(a^4 + a^2*b^2 + 3*(a^4 + a^2*b^2)*cosh(d*x + c)^2 - 4*(a^3*b + a*b^3)*cosh(d*x + c))*sinh(d*x + c))/((
a^5 + a^3*b^2)*d*cosh(d*x + c)^4 + 4*(a^5 + a^3*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^5 + a^3*b^2)*d*sinh(
d*x + c)^4 - 2*(a^5 + a^3*b^2)*d*cosh(d*x + c)^2 + 2*(3*(a^5 + a^3*b^2)*d*cosh(d*x + c)^2 - (a^5 + a^3*b^2)*d)
*sinh(d*x + c)^2 + (a^5 + a^3*b^2)*d + 4*((a^5 + a^3*b^2)*d*cosh(d*x + c)^3 - (a^5 + a^3*b^2)*d*cosh(d*x + c))
*sinh(d*x + c))

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giac [A]  time = 2.00, size = 176, normalized size = 1.56 \[ -\frac {\frac {2 \, b^{3} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{3}} + \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{3}} + \frac {2 \, {\left (a e^{\left (3 \, d x + 3 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a e^{\left (d x + c\right )} + 2 \, b\right )}}{a^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 164, normalized size = 1.45 \[ \frac {\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{2 d \,a^{2}}-\frac {2 b^{3} \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,a^{3} \sqrt {a^{2}+b^{2}}}-\frac {1}{8 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{3}}+\frac {b}{2 d \,a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^3/(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^3/(a^2+b^2)^(1/2)*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)^2-1/2/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)

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maxima [A]  time = 0.53, size = 211, normalized size = 1.87 \[ -\frac {b^{3} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3} d} + \frac {a e^{\left (-d x - c\right )} + 2 \, b e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-3 \, d x - 3 \, c\right )} - 2 \, b}{{\left (2 \, a^{2} e^{\left (-2 \, d x - 2 \, c\right )} - a^{2} e^{\left (-4 \, d x - 4 \, c\right )} - a^{2}\right )} d} + \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{2 \, a^{3} d} - \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{2 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.75, size = 776, normalized size = 6.87 \[ \frac {{\mathrm {e}}^{c+d\,x}}{a\,d-a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}}-\frac {2\,{\mathrm {e}}^{c+d\,x}}{a\,d-2\,a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+a\,d\,{\mathrm {e}}^{4\,c+4\,d\,x}}-\frac {2\,b}{a^2\,d-a^2\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}}-\frac {\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-24\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+20\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{2\,a\,d}+\frac {\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2+4\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+24\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-20\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{2\,a\,d}-\frac {b^3\,\ln \left (16\,a^5\,b-48\,a\,b^5-24\,b^5\,\sqrt {a^2+b^2}-32\,a^3\,b^3-40\,a^2\,b^3\,\sqrt {a^2+b^2}-32\,a^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+24\,b^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+16\,a^4\,b\,\sqrt {a^2+b^2}+112\,a^2\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+56\,a^4\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-32\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}+72\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}+72\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{d\,a^5+d\,a^3\,b^2}+\frac {b^3\,\ln \left (24\,b^5\,\sqrt {a^2+b^2}-48\,a\,b^5+16\,a^5\,b-32\,a^3\,b^3+40\,a^2\,b^3\,\sqrt {a^2+b^2}-32\,a^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+24\,b^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-16\,a^4\,b\,\sqrt {a^2+b^2}+112\,a^2\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+56\,a^4\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\,a^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}-72\,a\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}-72\,a^3\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{d\,a^5+d\,a^3\,b^2}+\frac {b^2\,\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-24\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+20\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a^3\,d}-\frac {b^2\,\ln \left (4\,a^4+24\,b^4-20\,a^2\,b^2+4\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+24\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-20\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(c + d*x)/(a*d - a*d*exp(2*c + 2*d*x)) - (2*exp(c + d*x))/(a*d - 2*a*d*exp(2*c + 2*d*x) + a*d*exp(4*c + 4*d
*x)) - (2*b)/(a^2*d - a^2*d*exp(2*c + 2*d*x)) - log(4*a^4 + 24*b^4 - 20*a^2*b^2 - 4*a^4*exp(d*x)*exp(c) - 24*b
^4*exp(d*x)*exp(c) + 20*a^2*b^2*exp(d*x)*exp(c))/(2*a*d) + log(4*a^4 + 24*b^4 - 20*a^2*b^2 + 4*a^4*exp(d*x)*ex
p(c) + 24*b^4*exp(d*x)*exp(c) - 20*a^2*b^2*exp(d*x)*exp(c))/(2*a*d) - (b^3*log(16*a^5*b - 48*a*b^5 - 24*b^5*(a
^2 + b^2)^(1/2) - 32*a^3*b^3 - 40*a^2*b^3*(a^2 + b^2)^(1/2) - 32*a^6*exp(d*x)*exp(c) + 24*b^6*exp(d*x)*exp(c)
+ 16*a^4*b*(a^2 + b^2)^(1/2) + 112*a^2*b^4*exp(d*x)*exp(c) + 56*a^4*b^2*exp(d*x)*exp(c) - 32*a^5*exp(d*x)*exp(
c)*(a^2 + b^2)^(1/2) + 72*a*b^4*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2) + 72*a^3*b^2*exp(d*x)*exp(c)*(a^2 + b^2)^(1/
2))*(a^2 + b^2)^(1/2))/(a^5*d + a^3*b^2*d) + (b^3*log(24*b^5*(a^2 + b^2)^(1/2) - 48*a*b^5 + 16*a^5*b - 32*a^3*
b^3 + 40*a^2*b^3*(a^2 + b^2)^(1/2) - 32*a^6*exp(d*x)*exp(c) + 24*b^6*exp(d*x)*exp(c) - 16*a^4*b*(a^2 + b^2)^(1
/2) + 112*a^2*b^4*exp(d*x)*exp(c) + 56*a^4*b^2*exp(d*x)*exp(c) + 32*a^5*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2) - 72
*a*b^4*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2) - 72*a^3*b^2*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a
^5*d + a^3*b^2*d) + (b^2*log(4*a^4 + 24*b^4 - 20*a^2*b^2 - 4*a^4*exp(d*x)*exp(c) - 24*b^4*exp(d*x)*exp(c) + 20
*a^2*b^2*exp(d*x)*exp(c)))/(a^3*d) - (b^2*log(4*a^4 + 24*b^4 - 20*a^2*b^2 + 4*a^4*exp(d*x)*exp(c) + 24*b^4*exp
(d*x)*exp(c) - 20*a^2*b^2*exp(d*x)*exp(c)))/(a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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