3.3.59 \(\int \frac {\sinh ^8(c+d x)}{(a-b \sinh ^4(c+d x))^3} \, dx\) [259]

Optimal. Leaf size=319 \[ -\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} b^{3/2} d}+\frac {\left (2 \sqrt {a}+5 \sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} b^{3/2} d}-\frac {(a+5 b) \tanh (c+d x)}{32 a (a-b)^2 b d}-\frac {\tanh ^3(c+d x)}{32 a (a-b) b d}+\frac {\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac {\text {sech}^2(c+d x) \tanh ^5(c+d x)}{32 a b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )} \]

[Out]

-1/64*arctanh((a^(1/2)-b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))*(2*a^(1/2)-5*b^(1/2))/a^(3/4)/b^(3/2)/d/(a^(1/2)-b^
(1/2))^(5/2)+1/64*arctanh((a^(1/2)+b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))*(2*a^(1/2)+5*b^(1/2))/a^(3/4)/b^(3/2)/d
/(a^(1/2)+b^(1/2))^(5/2)-1/32*(a+5*b)*tanh(d*x+c)/a/(a-b)^2/b/d-1/32*tanh(d*x+c)^3/a/(a-b)/b/d+1/8*tanh(d*x+c)
^9/a/d/(a-2*a*tanh(d*x+c)^2+(a-b)*tanh(d*x+c)^4)^2-1/32*sech(d*x+c)^2*tanh(d*x+c)^5/a/b/d/(a-2*a*tanh(d*x+c)^2
+(a-b)*tanh(d*x+c)^4)

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Rubi [A]
time = 0.38, antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3296, 1289, 12, 1134, 1293, 1180, 214} \begin {gather*} -\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} b^{3/2} d \left (\sqrt {a}-\sqrt {b}\right )^{5/2}}+\frac {\left (2 \sqrt {a}+5 \sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} b^{3/2} d \left (\sqrt {a}+\sqrt {b}\right )^{5/2}}-\frac {\tanh ^3(c+d x)}{32 a b d (a-b)}+\frac {\tanh ^9(c+d x)}{8 a d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac {(a+5 b) \tanh (c+d x)}{32 a b d (a-b)^2}-\frac {\tanh ^5(c+d x) \text {sech}^2(c+d x)}{32 a b d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^8/(a - b*Sinh[c + d*x]^4)^3,x]

[Out]

-1/64*((2*Sqrt[a] - 5*Sqrt[b])*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(a^(3/4)*(Sqrt[a] - S
qrt[b])^(5/2)*b^(3/2)*d) + ((2*Sqrt[a] + 5*Sqrt[b])*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/
(64*a^(3/4)*(Sqrt[a] + Sqrt[b])^(5/2)*b^(3/2)*d) - ((a + 5*b)*Tanh[c + d*x])/(32*a*(a - b)^2*b*d) - Tanh[c + d
*x]^3/(32*a*(a - b)*b*d) + Tanh[c + d*x]^9/(8*a*d*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4)^2) - (Se
ch[c + d*x]^2*Tanh[c + d*x]^5)/(32*a*b*d*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4))

Rule 12

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

Rule 214

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

Rule 1134

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-d^3)*(d*x)^(m - 3)*(2*a
+ b*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*(p + 1)*(b^2 - 4*a*c))), x] + Dist[d^4/(2*(p + 1)*(b^2 - 4*a*c)), Int
[(d*x)^(m - 4)*(2*a*(m - 3) + b*(m + 4*p + 3)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 3] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1180

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 1289

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[f*(
f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1)*((b*d - 2*a*e - (b*e - 2*c*d)*x^2)/(2*(p + 1)*(b^2 - 4*a*c))), x] - D
ist[f^2/(2*(p + 1)*(b^2 - 4*a*c)), Int[(f*x)^(m - 2)*(a + b*x^2 + c*x^4)^(p + 1)*Simp[(m - 1)*(b*d - 2*a*e) -
(4*p + 4 + m + 1)*(b*e - 2*c*d)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[
p, -1] && GtQ[m, 1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1293

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[e*f*
(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 3))), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m -
2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[
{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (Inte
gerQ[p] || IntegerQ[m])

Rule 3296

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*((a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*
x^2)^(m/2 + 2*p + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\sinh ^8(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx &=\frac {\text {Subst}\left (\int \frac {x^8 \left (1-x^2\right )}{\left (a-2 a x^2+(a-b) x^4\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}+\frac {\text {Subst}\left (\int -\frac {2 b x^8}{\left (a-2 a x^2+(a-b) x^4\right )^2} \, dx,x,\tanh (c+d x)\right )}{16 a b d}\\ &=\frac {\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac {\text {Subst}\left (\int \frac {x^8}{\left (a-2 a x^2+(a-b) x^4\right )^2} \, dx,x,\tanh (c+d x)\right )}{8 a d}\\ &=\frac {\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac {\text {sech}^2(c+d x) \tanh ^5(c+d x)}{32 a b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {x^4 \left (10 a-6 a x^2\right )}{a-2 a x^2+(a-b) x^4} \, dx,x,\tanh (c+d x)\right )}{64 a^2 b d}\\ &=-\frac {\tanh ^3(c+d x)}{32 a (a-b) b d}+\frac {\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac {\text {sech}^2(c+d x) \tanh ^5(c+d x)}{32 a b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {x^2 \left (-18 a^2+6 a (a+5 b) x^2\right )}{a-2 a x^2+(a-b) x^4} \, dx,x,\tanh (c+d x)\right )}{192 a^2 (a-b) b d}\\ &=-\frac {(a+5 b) \tanh (c+d x)}{32 a (a-b)^2 b d}-\frac {\tanh ^3(c+d x)}{32 a (a-b) b d}+\frac {\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac {\text {sech}^2(c+d x) \tanh ^5(c+d x)}{32 a b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {6 a^2 (a+5 b)+6 a^2 (a-13 b) x^2}{a-2 a x^2+(a-b) x^4} \, dx,x,\tanh (c+d x)\right )}{192 a^2 (a-b)^2 b d}\\ &=-\frac {(a+5 b) \tanh (c+d x)}{32 a (a-b)^2 b d}-\frac {\tanh ^3(c+d x)}{32 a (a-b) b d}+\frac {\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac {\text {sech}^2(c+d x) \tanh ^5(c+d x)}{32 a b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}-\frac {\left (2 a+3 \sqrt {a} \sqrt {b}-5 b\right ) \text {Subst}\left (\int \frac {1}{-a+\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{64 \sqrt {a} \left (\sqrt {a}+\sqrt {b}\right )^2 b^{3/2} d}+\frac {\left (\left (2 \sqrt {a}-5 \sqrt {b}\right ) \left (\sqrt {a}+\sqrt {b}\right )^3\right ) \text {Subst}\left (\int \frac {1}{-a-\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{64 \sqrt {a} (a-b)^2 b^{3/2} d}\\ &=-\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} b^{3/2} d}+\frac {\left (2 \sqrt {a}+5 \sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} b^{3/2} d}-\frac {(a+5 b) \tanh (c+d x)}{32 a (a-b)^2 b d}-\frac {\tanh ^3(c+d x)}{32 a (a-b) b d}+\frac {\tanh ^9(c+d x)}{8 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}-\frac {\text {sech}^2(c+d x) \tanh ^5(c+d x)}{32 a b d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 2.89, size = 331, normalized size = 1.04 \begin {gather*} \frac {\frac {\left (2 \sqrt {a}-5 \sqrt {b}\right ) \left (\sqrt {a}+\sqrt {b}\right )^2 \sqrt {b} \text {ArcTan}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}}}+\frac {\left (2 a^{3/2} \sqrt {b}+a b-8 \sqrt {a} b^{3/2}+5 b^2\right ) \tanh ^{-1}\left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}}}+\frac {8 b (5 a-14 b+(-2 a+5 b) \cosh (2 (c+d x))) \sinh (2 (c+d x))}{8 a-3 b+4 b \cosh (2 (c+d x))-b \cosh (4 (c+d x))}+\frac {64 a (a-b) b (-6 \sinh (2 (c+d x))+\sinh (4 (c+d x)))}{(-8 a+3 b-4 b \cosh (2 (c+d x))+b \cosh (4 (c+d x)))^2}}{64 (a-b)^2 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^8/(a - b*Sinh[c + d*x]^4)^3,x]

[Out]

(((2*Sqrt[a] - 5*Sqrt[b])*(Sqrt[a] + Sqrt[b])^2*Sqrt[b]*ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt[-a + S
qrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[-a + Sqrt[a]*Sqrt[b]]) + ((2*a^(3/2)*Sqrt[b] + a*b - 8*Sqrt[a]*b^(3/2) + 5*b^2
)*ArcTanh[((Sqrt[a] + Sqrt[b])*Tanh[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[a + Sqrt[a]*Sqrt[b]])
+ (8*b*(5*a - 14*b + (-2*a + 5*b)*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)])/(8*a - 3*b + 4*b*Cosh[2*(c + d*x)] - b
*Cosh[4*(c + d*x)]) + (64*a*(a - b)*b*(-6*Sinh[2*(c + d*x)] + Sinh[4*(c + d*x)]))/(-8*a + 3*b - 4*b*Cosh[2*(c
+ d*x)] + b*Cosh[4*(c + d*x)])^2)/(64*(a - b)^2*b^2*d)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 2.50, size = 535, normalized size = 1.68

method	result	size
derivativedivides	\(\frac {-\frac {512 \left (\frac {a \left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a +49 b \right ) a \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 \left (3 a^{2}+55 a b -48 b^{2}\right ) \left (\tanh ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a^{2}+377 a b -784 b^{2}\right ) \left (\tanh ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a^{2}+377 a b -784 b^{2}\right ) \left (\tanh ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 \left (3 a^{2}+55 a b -48 b^{2}\right ) \left (\tanh ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a +49 b \right ) a \left (\tanh ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {a \left (a +5 b \right ) \left (\tanh ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (a \left (\tanh ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 a \left (\tanh ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 a \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 a \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a \right )^{2}}-\frac {\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{8}-4 a \,\textit {\_Z}^{6}+\left (6 a -16 b \right ) \textit {\_Z}^{4}-4 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\left (a +5 b \right ) \textit {\_R}^{6}+\left (5 a -47 b \right ) \textit {\_R}^{4}+\left (-5 a +47 b \right ) \textit {\_R}^{2}-a -5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7} a -3 \textit {\_R}^{5} a +3 \textit {\_R}^{3} a -8 \textit {\_R}^{3} b -\textit {\_R} a}}{128 b \left (a^{2}-2 a b +b^{2}\right )}}{d}\)	\(535\)
default	\(\frac {-\frac {512 \left (\frac {a \left (a +5 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a +49 b \right ) a \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 \left (3 a^{2}+55 a b -48 b^{2}\right ) \left (\tanh ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a^{2}+377 a b -784 b^{2}\right ) \left (\tanh ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a^{2}+377 a b -784 b^{2}\right ) \left (\tanh ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 \left (3 a^{2}+55 a b -48 b^{2}\right ) \left (\tanh ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a +49 b \right ) a \left (\tanh ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}+\frac {a \left (a +5 b \right ) \left (\tanh ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8192 b \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (a \left (\tanh ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 a \left (\tanh ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 a \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 a \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a \right )^{2}}-\frac {\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{8}-4 a \,\textit {\_Z}^{6}+\left (6 a -16 b \right ) \textit {\_Z}^{4}-4 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\left (a +5 b \right ) \textit {\_R}^{6}+\left (5 a -47 b \right ) \textit {\_R}^{4}+\left (-5 a +47 b \right ) \textit {\_R}^{2}-a -5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{7} a -3 \textit {\_R}^{5} a +3 \textit {\_R}^{3} a -8 \textit {\_R}^{3} b -\textit {\_R} a}}{128 b \left (a^{2}-2 a b +b^{2}\right )}}{d}\)	\(535\)
risch	\(\text {Expression too large to display}\)	\(1574\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-512*(1/8192*a*(a+5*b)/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)-1/8192*(5*a+49*b)*a/b/(a^2-2*a*b+b^2)*tanh(1
/2*d*x+1/2*c)^3+3/8192/b*(3*a^2+55*a*b-48*b^2)/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^5-1/8192*(5*a^2+377*a*b-784
*b^2)/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^7-1/8192*(5*a^2+377*a*b-784*b^2)/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/
2*c)^9+3/8192/b*(3*a^2+55*a*b-48*b^2)/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^11-1/8192*(5*a+49*b)*a/b/(a^2-2*a*b+
b^2)*tanh(1/2*d*x+1/2*c)^13+1/8192*a*(a+5*b)/b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^15)/(a*tanh(1/2*d*x+1/2*c)^
8-4*a*tanh(1/2*d*x+1/2*c)^6+6*a*tanh(1/2*d*x+1/2*c)^4-16*b*tanh(1/2*d*x+1/2*c)^4-4*a*tanh(1/2*d*x+1/2*c)^2+a)^
2-1/128/b/(a^2-2*a*b+b^2)*sum(((a+5*b)*_R^6+(5*a-47*b)*_R^4+(-5*a+47*b)*_R^2-a-5*b)/(_R^7*a-3*_R^5*a+3*_R^3*a-
8*_R^3*b-_R*a)*ln(tanh(1/2*d*x+1/2*c)-_R),_R=RootOf(a*_Z^8-4*a*_Z^6+(6*a-16*b)*_Z^4-4*a*_Z^2+a)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x, algorithm="maxima")

[Out]

-1/8*(2*a*b^2 - 5*b^3 + (a*b^2*e^(14*c) - 4*b^3*e^(14*c))*e^(14*d*x) - (32*a^2*b*e^(12*c) - 58*a*b^2*e^(12*c)
- b^3*e^(12*c))*e^(12*d*x) + 3*(48*a^2*b*e^(10*c) - 73*a*b^2*e^(10*c) + 20*b^3*e^(10*c))*e^(10*d*x) + (256*a^3
*e^(8*c) - 832*a^2*b*e^(8*c) + 550*a*b^2*e^(8*c) - 175*b^3*e^(8*c))*e^(8*d*x) + (112*a^2*b*e^(6*c) - 533*a*b^2
*e^(6*c) + 220*b^3*e^(6*c))*e^(6*d*x) - (32*a^2*b*e^(4*c) - 158*a*b^2*e^(4*c) + 141*b^3*e^(4*c))*e^(4*d*x) - (
17*a*b^2*e^(2*c) - 44*b^3*e^(2*c))*e^(2*d*x))/(a^2*b^4*d - 2*a*b^5*d + b^6*d + (a^2*b^4*d*e^(16*c) - 2*a*b^5*d
*e^(16*c) + b^6*d*e^(16*c))*e^(16*d*x) - 8*(a^2*b^4*d*e^(14*c) - 2*a*b^5*d*e^(14*c) + b^6*d*e^(14*c))*e^(14*d*
x) - 4*(8*a^3*b^3*d*e^(12*c) - 23*a^2*b^4*d*e^(12*c) + 22*a*b^5*d*e^(12*c) - 7*b^6*d*e^(12*c))*e^(12*d*x) + 8*
(16*a^3*b^3*d*e^(10*c) - 39*a^2*b^4*d*e^(10*c) + 30*a*b^5*d*e^(10*c) - 7*b^6*d*e^(10*c))*e^(10*d*x) + 2*(128*a
^4*b^2*d*e^(8*c) - 352*a^3*b^3*d*e^(8*c) + 355*a^2*b^4*d*e^(8*c) - 166*a*b^5*d*e^(8*c) + 35*b^6*d*e^(8*c))*e^(
8*d*x) + 8*(16*a^3*b^3*d*e^(6*c) - 39*a^2*b^4*d*e^(6*c) + 30*a*b^5*d*e^(6*c) - 7*b^6*d*e^(6*c))*e^(6*d*x) - 4*
(8*a^3*b^3*d*e^(4*c) - 23*a^2*b^4*d*e^(4*c) + 22*a*b^5*d*e^(4*c) - 7*b^6*d*e^(4*c))*e^(4*d*x) - 8*(a^2*b^4*d*e
^(2*c) - 2*a*b^5*d*e^(2*c) + b^6*d*e^(2*c))*e^(2*d*x)) - 1/256*integrate(64*((a*e^(6*c) - 4*b*e^(6*c))*e^(6*d*
x) + (a*e^(2*c) - 4*b*e^(2*c))*e^(2*d*x) + 18*b*e^(4*d*x + 4*c))/(a^2*b^2 - 2*a*b^3 + b^4 + (a^2*b^2*e^(8*c) -
2*a*b^3*e^(8*c) + b^4*e^(8*c))*e^(8*d*x) - 4*(a^2*b^2*e^(6*c) - 2*a*b^3*e^(6*c) + b^4*e^(6*c))*e^(6*d*x) - 2*
(8*a^3*b*e^(4*c) - 19*a^2*b^2*e^(4*c) + 14*a*b^3*e^(4*c) - 3*b^4*e^(4*c))*e^(4*d*x) - 4*(a^2*b^2*e^(2*c) - 2*a
*b^3*e^(2*c) + b^4*e^(2*c))*e^(2*d*x)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 20486 vs. \(2 (263) = 526\).
time = 0.91, size = 20486, normalized size = 64.22 \begin {gather*} \text {too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x, algorithm="fricas")

[Out]

-1/128*(16*(a*b^2 - 4*b^3)*cosh(d*x + c)^14 + 224*(a*b^2 - 4*b^3)*cosh(d*x + c)*sinh(d*x + c)^13 + 16*(a*b^2 -
4*b^3)*sinh(d*x + c)^14 - 16*(32*a^2*b - 58*a*b^2 - b^3)*cosh(d*x + c)^12 - 16*(32*a^2*b - 58*a*b^2 - b^3 - 9
1*(a*b^2 - 4*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^12 + 64*(91*(a*b^2 - 4*b^3)*cosh(d*x + c)^3 - 3*(32*a^2*b - 5
8*a*b^2 - b^3)*cosh(d*x + c))*sinh(d*x + c)^11 + 48*(48*a^2*b - 73*a*b^2 + 20*b^3)*cosh(d*x + c)^10 + 16*(1001
*(a*b^2 - 4*b^3)*cosh(d*x + c)^4 + 144*a^2*b - 219*a*b^2 + 60*b^3 - 66*(32*a^2*b - 58*a*b^2 - b^3)*cosh(d*x +
c)^2)*sinh(d*x + c)^10 + 32*(1001*(a*b^2 - 4*b^3)*cosh(d*x + c)^5 - 110*(32*a^2*b - 58*a*b^2 - b^3)*cosh(d*x +
c)^3 + 15*(48*a^2*b - 73*a*b^2 + 20*b^3)*cosh(d*x + c))*sinh(d*x + c)^9 + 16*(256*a^3 - 832*a^2*b + 550*a*b^2
- 175*b^3)*cosh(d*x + c)^8 + 16*(3003*(a*b^2 - 4*b^3)*cosh(d*x + c)^6 - 495*(32*a^2*b - 58*a*b^2 - b^3)*cosh(
d*x + c)^4 + 256*a^3 - 832*a^2*b + 550*a*b^2 - 175*b^3 + 135*(48*a^2*b - 73*a*b^2 + 20*b^3)*cosh(d*x + c)^2)*s
inh(d*x +  ...

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**8/(a-b*sinh(d*x+c)**4)**3,x)

[Out]

Timed out

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Giac [A]
time = 1.49, size = 389, normalized size = 1.22 \begin {gather*} -\frac {a b^{2} e^{\left (14 \, d x + 14 \, c\right )} - 4 \, b^{3} e^{\left (14 \, d x + 14 \, c\right )} - 32 \, a^{2} b e^{\left (12 \, d x + 12 \, c\right )} + 58 \, a b^{2} e^{\left (12 \, d x + 12 \, c\right )} + b^{3} e^{\left (12 \, d x + 12 \, c\right )} + 144 \, a^{2} b e^{\left (10 \, d x + 10 \, c\right )} - 219 \, a b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 60 \, b^{3} e^{\left (10 \, d x + 10 \, c\right )} + 256 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} - 832 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 550 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 175 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 112 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 533 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 220 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 32 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 158 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 141 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 17 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 44 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b^{2} - 5 \, b^{3}}{8 \, {\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} {\left (b e^{\left (8 \, d x + 8 \, c\right )} - 4 \, b e^{\left (6 \, d x + 6 \, c\right )} - 16 \, a e^{\left (4 \, d x + 4 \, c\right )} + 6 \, b e^{\left (4 \, d x + 4 \, c\right )} - 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + b\right )}^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^8/(a-b*sinh(d*x+c)^4)^3,x, algorithm="giac")

[Out]

-1/8*(a*b^2*e^(14*d*x + 14*c) - 4*b^3*e^(14*d*x + 14*c) - 32*a^2*b*e^(12*d*x + 12*c) + 58*a*b^2*e^(12*d*x + 12
*c) + b^3*e^(12*d*x + 12*c) + 144*a^2*b*e^(10*d*x + 10*c) - 219*a*b^2*e^(10*d*x + 10*c) + 60*b^3*e^(10*d*x + 1
0*c) + 256*a^3*e^(8*d*x + 8*c) - 832*a^2*b*e^(8*d*x + 8*c) + 550*a*b^2*e^(8*d*x + 8*c) - 175*b^3*e^(8*d*x + 8*
c) + 112*a^2*b*e^(6*d*x + 6*c) - 533*a*b^2*e^(6*d*x + 6*c) + 220*b^3*e^(6*d*x + 6*c) - 32*a^2*b*e^(4*d*x + 4*c
) + 158*a*b^2*e^(4*d*x + 4*c) - 141*b^3*e^(4*d*x + 4*c) - 17*a*b^2*e^(2*d*x + 2*c) + 44*b^3*e^(2*d*x + 2*c) +
2*a*b^2 - 5*b^3)/((a^2*b^2 - 2*a*b^3 + b^4)*(b*e^(8*d*x + 8*c) - 4*b*e^(6*d*x + 6*c) - 16*a*e^(4*d*x + 4*c) +
6*b*e^(4*d*x + 4*c) - 4*b*e^(2*d*x + 2*c) + b)^2*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {sinh}\left (c+d\,x\right )}^8}{{\left (a-b\,{\mathrm {sinh}\left (c+d\,x\right )}^4\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(c + d*x)^8/(a - b*sinh(c + d*x)^4)^3,x)

[Out]

int(sinh(c + d*x)^8/(a - b*sinh(c + d*x)^4)^3, x)

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