3.181 \(\int \frac {\text {csch}^4(c+d x)}{a+b \sinh ^3(c+d x)} \, dx\)

Optimal. Leaf size=317 \[ \frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac {2 b^{4/3} \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 a^2 d \sqrt {a^{2/3}+b^{2/3}}}-\frac {2 b^{4/3} \tan ^{-1}\left (\frac {(-1)^{5/6} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 a^2 d \sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}-\frac {2 b^{4/3} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 a^2 d \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}-\frac {\coth ^3(c+d x)}{3 a d}+\frac {\coth (c+d x)}{a d} \]

[Out]

b*arctanh(cosh(d*x+c))/a^2/d+coth(d*x+c)/a/d-1/3*coth(d*x+c)^3/a/d-2/3*b^(4/3)*arctan((-1)^(1/6)*((-1)^(5/6)*b
^(1/3)+I*a^(1/3)*tanh(1/2*d*x+1/2*c))/((-1)^(1/3)*a^(2/3)-b^(2/3))^(1/2))/a^2/d/((-1)^(1/3)*a^(2/3)-b^(2/3))^(
1/2)-2/3*b^(4/3)*arctan((-1)^(5/6)*((-1)^(1/6)*b^(1/3)+I*a^(1/3)*tanh(1/2*d*x+1/2*c))/(-(-1)^(2/3)*a^(2/3)-b^(
2/3))^(1/2))/a^2/d/(-(-1)^(2/3)*a^(2/3)-b^(2/3))^(1/2)-2/3*b^(4/3)*arctanh((b^(1/3)-a^(1/3)*tanh(1/2*d*x+1/2*c
))/(a^(2/3)+b^(2/3))^(1/2))/a^2/d/(a^(2/3)+b^(2/3))^(1/2)

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Rubi [A]  time = 0.43, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3220, 3770, 3767, 2660, 618, 206, 204} \[ -\frac {2 b^{4/3} \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 a^2 d \sqrt {a^{2/3}+b^{2/3}}}-\frac {2 b^{4/3} \tan ^{-1}\left (\frac {(-1)^{5/6} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 a^2 d \sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}-\frac {2 b^{4/3} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 a^2 d \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}+\frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac {\coth ^3(c+d x)}{3 a d}+\frac {\coth (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^4/(a + b*Sinh[c + d*x]^3),x]

[Out]

(-2*b^(4/3)*ArcTan[((-1)^(5/6)*((-1)^(1/6)*b^(1/3) + I*a^(1/3)*Tanh[(c + d*x)/2]))/Sqrt[-((-1)^(2/3)*a^(2/3))
- b^(2/3)]])/(3*a^2*Sqrt[-((-1)^(2/3)*a^(2/3)) - b^(2/3)]*d) - (2*b^(4/3)*ArcTan[((-1)^(1/6)*((-1)^(5/6)*b^(1/
3) + I*a^(1/3)*Tanh[(c + d*x)/2]))/Sqrt[(-1)^(1/3)*a^(2/3) - b^(2/3)]])/(3*a^2*Sqrt[(-1)^(1/3)*a^(2/3) - b^(2/
3)]*d) + (b*ArcTanh[Cosh[c + d*x]])/(a^2*d) - (2*b^(4/3)*ArcTanh[(b^(1/3) - a^(1/3)*Tanh[(c + d*x)/2])/Sqrt[a^
(2/3) + b^(2/3)]])/(3*a^2*Sqrt[a^(2/3) + b^(2/3)]*d) + Coth[c + d*x]/(a*d) - Coth[c + d*x]^3/(3*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 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 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 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 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}^4(c+d x)}{a+b \sinh ^3(c+d x)} \, dx &=\int \left (-\frac {b \text {csch}(c+d x)}{a^2}+\frac {\text {csch}^4(c+d x)}{a}-\frac {b^2 \sinh ^2(c+d x)}{a^2 \left (-a-b \sinh ^3(c+d x)\right )}\right ) \, dx\\ &=\frac {\int \text {csch}^4(c+d x) \, dx}{a}-\frac {b \int \text {csch}(c+d x) \, dx}{a^2}-\frac {b^2 \int \frac {\sinh ^2(c+d x)}{-a-b \sinh ^3(c+d x)} \, dx}{a^2}\\ &=\frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}+\frac {b^2 \int \left (-\frac {i}{3 b^{2/3} \left (-i \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)\right )}-\frac {i}{3 b^{2/3} \left (\sqrt [6]{-1} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)\right )}-\frac {i}{3 b^{2/3} \left ((-1)^{5/6} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)\right )}\right ) \, dx}{a^2}+\frac {i \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-i \coth (c+d x)\right )}{a d}\\ &=\frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}+\frac {\coth (c+d x)}{a d}-\frac {\coth ^3(c+d x)}{3 a d}-\frac {\left (i b^{4/3}\right ) \int \frac {1}{-i \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 a^2}-\frac {\left (i b^{4/3}\right ) \int \frac {1}{\sqrt [6]{-1} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 a^2}-\frac {\left (i b^{4/3}\right ) \int \frac {1}{(-1)^{5/6} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 a^2}\\ &=\frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}+\frac {\coth (c+d x)}{a d}-\frac {\coth ^3(c+d x)}{3 a d}-\frac {\left (2 b^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-i \sqrt [3]{a}-2 \sqrt [3]{b} x-i \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^2 d}-\frac {\left (2 b^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [6]{-1} \sqrt [3]{a}-2 \sqrt [3]{b} x+\sqrt [6]{-1} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^2 d}-\frac {\left (2 b^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{(-1)^{5/6} \sqrt [3]{a}-2 \sqrt [3]{b} x+(-1)^{5/6} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^2 d}\\ &=\frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}+\frac {\coth (c+d x)}{a d}-\frac {\coth ^3(c+d x)}{3 a d}+\frac {\left (4 b^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (\sqrt [3]{-1} a^{2/3}-b^{2/3}\right )-x^2} \, dx,x,-2 \sqrt [3]{b}+2 \sqrt [6]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^2 d}+\frac {\left (4 b^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^{2/3}+b^{2/3}\right )-x^2} \, dx,x,-2 \sqrt [3]{b}-2 i \sqrt [3]{a} \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^2 d}+\frac {\left (4 b^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left ((-1)^{2/3} a^{2/3}+b^{2/3}\right )-x^2} \, dx,x,-2 \sqrt [3]{b}+2 (-1)^{5/6} \sqrt [3]{a} \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^2 d}\\ &=\frac {2 b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{b}+\sqrt [3]{-1} \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 a^2 \sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}} d}+\frac {2 b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{b}-(-1)^{2/3} \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 a^2 \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}} d}+\frac {b \tanh ^{-1}(\cosh (c+d x))}{a^2 d}-\frac {2 b^{4/3} \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 a^2 \sqrt {a^{2/3}+b^{2/3}} d}+\frac {\coth (c+d x)}{a d}-\frac {\coth ^3(c+d x)}{3 a d}\\ \end {align*}

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Mathematica [C]  time = 5.79, size = 370, normalized size = 1.17 \[ \frac {4 b^2 \text {RootSum}\left [\text {$\#$1}^6 b-3 \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 \text {$\#$1}^2 b-b\& ,\frac {2 \text {$\#$1}^4 \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )+\text {$\#$1}^4 c+\text {$\#$1}^4 d x-4 \text {$\#$1}^2 \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )-2 \text {$\#$1}^2 c-2 \text {$\#$1}^2 d x+2 \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )+c+d x}{\text {$\#$1}^5 b-2 \text {$\#$1}^3 b+4 \text {$\#$1}^2 a+\text {$\#$1} b}\& \right ]+8 a \tanh \left (\frac {1}{2} (c+d x)\right )+8 a \coth \left (\frac {1}{2} (c+d x)\right )-\frac {1}{2} a \sinh (c+d x) \text {csch}^4\left (\frac {1}{2} (c+d x)\right )+8 a \sinh ^4\left (\frac {1}{2} (c+d x)\right ) \text {csch}^3(c+d x)-24 b \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )}{24 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^4/(a + b*Sinh[c + d*x]^3),x]

[Out]

(8*a*Coth[(c + d*x)/2] - 24*b*Log[Tanh[(c + d*x)/2]] + 4*b^2*RootSum[-b + 3*b*#1^2 + 8*a*#1^3 - 3*b*#1^4 + b*#
1^6 & , (c + d*x + 2*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1]
- 2*c*#1^2 - 2*d*x*#1^2 - 4*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x
)/2]*#1]*#1^2 + c*#1^4 + d*x*#1^4 + 2*Log[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh
[(c + d*x)/2]*#1]*#1^4)/(b*#1 + 4*a*#1^2 - 2*b*#1^3 + b*#1^5) & ] + 8*a*Csch[c + d*x]^3*Sinh[(c + d*x)/2]^4 -
(a*Csch[(c + d*x)/2]^4*Sinh[c + d*x])/2 + 8*a*Tanh[(c + d*x)/2])/(24*a^2*d)

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fricas [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)^4/(a+b*sinh(d*x+c)^3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {csch}\left (d x + c\right )^{4}}{b \sinh \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4/(a+b*sinh(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(csch(d*x + c)^4/(b*sinh(d*x + c)^3 + a), x)

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maple [C]  time = 0.18, size = 178, normalized size = 0.56 \[ -\frac {\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}+\frac {3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {4 b^{2} \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}-3 a \,\textit {\_Z}^{4}-8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}-a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a -2 \textit {\_R}^{3} a -4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d \,a^{2}}-\frac {1}{24 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {3}{8 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^4/(a+b*sinh(d*x+c)^3),x)

[Out]

-1/24/d/a*tanh(1/2*d*x+1/2*c)^3+3/8/d/a*tanh(1/2*d*x+1/2*c)-4/3/d*b^2/a^2*sum(_R^2/(_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))-1/24/d/a/tanh(1/2*d*x+1/2*c)^3
+3/8/d/a/tanh(1/2*d*x+1/2*c)-1/d/a^2*b*ln(tanh(1/2*d*x+1/2*c))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {4 \, {\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}{3 \, {\left (a d e^{\left (6 \, d x + 6 \, c\right )} - 3 \, a d e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a d e^{\left (2 \, d x + 2 \, c\right )} - a d\right )}} + \frac {b \log \left ({\left (e^{\left (d x + c\right )} + 1\right )} e^{\left (-c\right )}\right )}{a^{2} d} - \frac {b \log \left ({\left (e^{\left (d x + c\right )} - 1\right )} e^{\left (-c\right )}\right )}{a^{2} d} + 16 \, \int \frac {b^{2} e^{\left (5 \, d x + 5 \, c\right )} - 2 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )} + b^{2} e^{\left (d x + c\right )}}{8 \, {\left (a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 3 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} e^{\left (3 \, d x + 3 \, c\right )} + 3 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - a^{2} b\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4/(a+b*sinh(d*x+c)^3),x, algorithm="maxima")

[Out]

-4/3*(3*e^(2*d*x + 2*c) - 1)/(a*d*e^(6*d*x + 6*c) - 3*a*d*e^(4*d*x + 4*c) + 3*a*d*e^(2*d*x + 2*c) - a*d) + b*l
og((e^(d*x + c) + 1)*e^(-c))/(a^2*d) - b*log((e^(d*x + c) - 1)*e^(-c))/(a^2*d) + 16*integrate(1/8*(b^2*e^(5*d*
x + 5*c) - 2*b^2*e^(3*d*x + 3*c) + b^2*e^(d*x + c))/(a^2*b*e^(6*d*x + 6*c) - 3*a^2*b*e^(4*d*x + 4*c) + 8*a^3*e
^(3*d*x + 3*c) + 3*a^2*b*e^(2*d*x + 2*c) - a^2*b), x)

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mupad [B]  time = 59.69, size = 3086, normalized size = 9.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)^4*(a + b*sinh(c + d*x)^3)),x)

[Out]

symsum(log((18589155328*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^
2 - b^8, z, k)^2*a^4*b^7*d^2 - 134217728*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 +
27*a^4*b^6*d^2*z^2 - b^8, z, k)*a*b^9*d - 1073741824*b^9 + 2818572288*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6
*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^3*a^5*b^7*d^3 + 17716740096*root(729*a^12*b^2*d^6
*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^3*a^7*b^5*d^3 - 88181047296*ro
ot(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^4*a^8*b^5*d
^4 - 86973087744*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8
, z, k)^4*a^10*b^3*d^4 - 18119393280*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a
^4*b^6*d^2*z^2 - b^8, z, k)^5*a^9*b^5*d^5 - 30802968576*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8
*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^5*a^11*b^3*d^5 + 70665633792*root(729*a^12*b^2*d^6*z^6 + 729*a^
14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^6*a^12*b^3*d^6 + 32614907904*root(729*a^12*
b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^7*a^13*b^3*d^7 - 322122
5472*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)*a^3*
b^7*d + 2147483648*a*b^8*exp(d*x)*exp(root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*
a^4*b^6*d^2*z^2 - b^8, z, k)) + 86973087744*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4
+ 27*a^4*b^6*d^2*z^2 - b^8, z, k)^6*a^14*b*d^6 + 40768634880*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 2
43*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^7*a^15*b*d^7 - 391378894848*root(729*a^12*b^2*d^6*z^6 + 7
29*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^7*a^16*d^7*exp(d*x)*exp(root(729*a^12*
b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)) + 268435456*root(729*a
^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^2*a^3*b^8*d^2*exp(d
*x)*exp(root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k))
- 16642998272*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z
, k)^2*a^5*b^6*d^2*exp(d*x)*exp(root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^
6*d^2*z^2 - b^8, z, k)) - 100763959296*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27
*a^4*b^6*d^2*z^2 - b^8, z, k)^3*a^6*b^6*d^3*exp(d*x)*exp(root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^
8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)) + 57982058496*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 -
243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^3*a^8*b^4*d^3*exp(d*x)*exp(root(729*a^12*b^2*d^6*z^6 + 7
29*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)) - 4831838208*root(729*a^12*b^2*d^6*z^
6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^4*a^7*b^6*d^4*exp(d*x)*exp(root(7
29*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)) + 36238786560*
root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^4*a^9*b^4
*d^4*exp(d*x)*exp(root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^
8, z, k)) + 494659436544*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z
^2 - b^8, z, k)^5*a^10*b^4*d^5*exp(d*x)*exp(root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4
+ 27*a^4*b^6*d^2*z^2 - b^8, z, k)) + 333396836352*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*
d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^5*a^12*b^2*d^5*exp(d*x)*exp(root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6
*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)) + 21743271936*root(729*a^12*b^2*d^6*z^6 + 729*a^
14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^6*a^11*b^4*d^6*exp(d*x)*exp(root(729*a^12*b
^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)) + 2717908992*root(729*a
^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)^6*a^13*b^2*d^6*exp(
d*x)*exp(root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k))
- 399532621824*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8,
z, k)^7*a^14*b^2*d^7*exp(d*x)*exp(root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 27*a^4
*b^6*d^2*z^2 - b^8, z, k)) + 5637144576*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b^4*d^4*z^4 + 2
7*a^4*b^6*d^2*z^2 - b^8, z, k)*a^2*b^8*d*exp(d*x)*exp(root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a^8*b
^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k)))/(a^7*b^9))*root(729*a^12*b^2*d^6*z^6 + 729*a^14*d^6*z^6 - 243*a
^8*b^4*d^4*z^4 + 27*a^4*b^6*d^2*z^2 - b^8, z, k), k, 1, 6) + 8/(3*(a*d - 3*a*d*exp(2*c + 2*d*x) + 3*a*d*exp(4*
c + 4*d*x) - a*d*exp(6*c + 6*d*x))) - 4/(a*d - 2*a*d*exp(2*c + 2*d*x) + a*d*exp(4*c + 4*d*x)) + (b*log(exp(d*x
+ b/(a^2*d)) + 1))/(a^2*d) - (b*log(exp(d*x - b/(a^2*d)) - 1))/(a^2*d)

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

[Out]

Timed out

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