3.3.62 \(\int \frac {\sinh ^2(c+d x)}{(a-b \sinh ^4(c+d x))^3} \, dx\) [262]

Optimal. Leaf size=348 \[ -\frac {\left (12 a-14 \sqrt {a} \sqrt {b}+5 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} \sqrt {b} d}+\frac {\left (12 a+14 \sqrt {a} \sqrt {b}+5 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} \sqrt {b} d}+\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}+\frac {\tanh (c+d x) \left (\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}-\frac {5 \left (2 a^2+3 a b-b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a^2 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )} \]

[Out]

-1/64*arctanh((a^(1/2)-b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))*(12*a+5*b-14*a^(1/2)*b^(1/2))/a^(9/4)/d/(a^(1/2)-b^
(1/2))^(5/2)/b^(1/2)+1/64*arctanh((a^(1/2)+b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))*(12*a+5*b+14*a^(1/2)*b^(1/2))/a
^(9/4)/d/b^(1/2)/(a^(1/2)+b^(1/2))^(5/2)+1/8*b*tanh(d*x+c)*(a*(a+3*b)-(a^2+6*a*b+b^2)*tanh(d*x+c)^2)/a/(a-b)^3
/d/(a-2*a*tanh(d*x+c)^2+(a-b)*tanh(d*x+c)^4)^2+1/32*tanh(d*x+c)*(2*a*(5*a^2-9*a*b-4*b^2)/(a-b)^3-5*(2*a^2+3*a*
b-b^2)*tanh(d*x+c)^2/(a-b)^2)/a^2/d/(a-2*a*tanh(d*x+c)^2+(a-b)*tanh(d*x+c)^4)

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Rubi [A]
time = 0.49, antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3296, 1347, 1692, 1180, 214} \begin {gather*} -\frac {\left (-14 \sqrt {a} \sqrt {b}+12 a+5 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \sqrt {b} d \left (\sqrt {a}-\sqrt {b}\right )^{5/2}}+\frac {\left (14 \sqrt {a} \sqrt {b}+12 a+5 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \sqrt {b} d \left (\sqrt {a}+\sqrt {b}\right )^{5/2}}+\frac {\tanh (c+d x) \left (\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}-\frac {5 \left (2 a^2+3 a b-b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a^2 d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}+\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a d (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^2/(a - b*Sinh[c + d*x]^4)^3,x]

[Out]

-1/64*((12*a - 14*Sqrt[a]*Sqrt[b] + 5*b)*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(a^(9/4)*(S
qrt[a] - Sqrt[b])^(5/2)*Sqrt[b]*d) + ((12*a + 14*Sqrt[a]*Sqrt[b] + 5*b)*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[
c + d*x])/a^(1/4)])/(64*a^(9/4)*(Sqrt[a] + Sqrt[b])^(5/2)*Sqrt[b]*d) + (b*Tanh[c + d*x]*(a*(a + 3*b) - (a^2 +
6*a*b + b^2)*Tanh[c + d*x]^2))/(8*a*(a - b)^3*d*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4)^2) + (Tanh
[c + d*x]*((2*a*(5*a^2 - 9*a*b - 4*b^2))/(a - b)^3 - (5*(2*a^2 + 3*a*b - b^2)*Tanh[c + d*x]^2)/(a - b)^2))/(32
*a^2*d*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4))

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 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 1347

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{f = Coe
ff[PolynomialRemainder[x^m*(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[x^m*(d +
e*x^2)^q, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^4)^(p + 1)*((a*b*g - f*(b^2 - 2*a*c) - c*(b*
f - 2*a*g)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)
^(p + 1)*Simp[ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[x^m*(d + e*x^2)^q, a + b*x^2 + c*x^4, x
] + b^2*f*(2*p + 3) - 2*a*c*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)*x^2, x], x], x], x]] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IGtQ[q, 1] && IGtQ[m/2, 0]

Rule 1692

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainder[Pq, a +
b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[Pq, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 +
c*x^4)^(p + 1)*((a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Dist[1/(2*
a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuot
ient[Pq, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4*p + 7)*(b*d - 2*a*e)*x^2,
x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

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 ^2(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx &=\frac {\text {Subst}\left (\int \frac {x^2 \left (1-x^2\right )^4}{\left (a-2 a x^2+(a-b) x^4\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 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 {\frac {2 a^2 b^2 (a+3 b)}{(a-b)^3}-\frac {2 a b \left (8 a^3-29 a^2 b+18 a b^2-5 b^3\right ) x^2}{(a-b)^3}+\frac {32 a^2 (a-2 b) b x^4}{(a-b)^2}-\frac {16 a^2 b x^6}{a-b}}{\left (a-2 a x^2+(a-b) x^4\right )^2} \, dx,x,\tanh (c+d x)\right )}{16 a^2 b d}\\ &=\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}+\frac {\tanh (c+d x) \left (\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}-\frac {5 \left (2 a^2+3 a b-b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a^2 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {-\frac {8 a^3 (5 a-2 b) b^2}{(a-b)^2}+\frac {4 a^2 b^2 \left (22 a^2-15 a b+5 b^2\right ) x^2}{(a-b)^2}}{a-2 a x^2+(a-b) x^4} \, dx,x,\tanh (c+d x)\right )}{128 a^4 b^2 d}\\ &=\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}+\frac {\tanh (c+d x) \left (\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}-\frac {5 \left (2 a^2+3 a b-b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a^2 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}+\frac {\left (\left (\sqrt {a}+\sqrt {b}\right ) \left (12 a-14 \sqrt {a} \sqrt {b}+5 b\right )\right ) \text {Subst}\left (\int \frac {1}{-a-\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{64 a^2 \left (\sqrt {a}-\sqrt {b}\right )^2 \sqrt {b} d}-\frac {\left (\left (\sqrt {a}-\sqrt {b}\right ) \left (12 a+14 \sqrt {a} \sqrt {b}+5 b\right )\right ) \text {Subst}\left (\int \frac {1}{-a+\sqrt {a} \sqrt {b}+(a-b) x^2} \, dx,x,\tanh (c+d x)\right )}{64 a^2 \left (\sqrt {a}+\sqrt {b}\right )^2 \sqrt {b} d}\\ &=-\frac {\left (12 a-14 \sqrt {a} \sqrt {b}+5 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} \sqrt {b} d}+\frac {\left (12 a+14 \sqrt {a} \sqrt {b}+5 b\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} \sqrt {b} d}+\frac {b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a (a-b)^3 d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )^2}+\frac {\tanh (c+d x) \left (\frac {2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}-\frac {5 \left (2 a^2+3 a b-b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a^2 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 = 3.40, size = 343, normalized size = 0.99 \begin {gather*} \frac {\frac {\left (\sqrt {a}+\sqrt {b}\right )^2 \left (12 a-14 \sqrt {a} \sqrt {b}+5 b\right ) \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 {b}} \sqrt {b}}+\frac {\left (\sqrt {a}-\sqrt {b}\right )^2 \left (12 a+14 \sqrt {a} \sqrt {b}+5 b\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 {b}} \sqrt {b}}+\frac {4 \left (12 a^2+11 a b-5 b^2+b (-11 a+5 b) \cosh (2 (c+d x))\right ) \sinh (2 (c+d x))}{8 a-3 b+4 b \cosh (2 (c+d x))-b \cosh (4 (c+d x))}+\frac {128 a (a-b) (2 a+b-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)))^2}}{64 a^2 (a-b)^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^2/(a - b*Sinh[c + d*x]^4)^3,x]

[Out]

(((Sqrt[a] + Sqrt[b])^2*(12*a - 14*Sqrt[a]*Sqrt[b] + 5*b)*ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt[-a +
Sqrt[a]*Sqrt[b]]])/(Sqrt[-a + Sqrt[a]*Sqrt[b]]*Sqrt[b]) + ((Sqrt[a] - Sqrt[b])^2*(12*a + 14*Sqrt[a]*Sqrt[b] +
5*b)*ArcTanh[((Sqrt[a] + Sqrt[b])*Tanh[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(Sqrt[a + Sqrt[a]*Sqrt[b]]*Sqrt[
b]) + (4*(12*a^2 + 11*a*b - 5*b^2 + b*(-11*a + 5*b)*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)])/(8*a - 3*b + 4*b*Cos
h[2*(c + d*x)] - b*Cosh[4*(c + d*x)]) + (128*a*(a - b)*(2*a + b - 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)])^2)/(64*a^2*(a - b)^2*d)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 2.98, size = 583, normalized size = 1.68

method	result	size
derivativedivides	\(\frac {-\frac {8 \left (-\frac {\left (5 a -2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{2}+20 a b -18 b^{2}\right ) \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (15 a^{2}+8 a b -18 b^{2}\right ) \left (\tanh ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{3}+2 a^{2} b -388 a \,b^{2}+160 b^{3}\right ) \left (\tanh ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{3}+2 a^{2} b -388 a \,b^{2}+160 b^{3}\right ) \left (\tanh ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (15 a^{2}+8 a b -18 b^{2}\right ) \left (\tanh ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{2}+20 a b -18 b^{2}\right ) \left (\tanh ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a -2 b \right ) \left (\tanh ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 \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 (a \left (-5 a +2 b \right ) \textit {\_R}^{6}+\left (39 a^{2}-28 a b +10 b^{2}\right ) \textit {\_R}^{4}+\left (-39 a^{2}+28 a b -10 b^{2}\right ) \textit {\_R}^{2}+5 a^{2}-2 a 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}}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}}{d}\)	\(583\)
default	\(\frac {-\frac {8 \left (-\frac {\left (5 a -2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{2}+20 a b -18 b^{2}\right ) \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (15 a^{2}+8 a b -18 b^{2}\right ) \left (\tanh ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{3}+2 a^{2} b -388 a \,b^{2}+160 b^{3}\right ) \left (\tanh ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{3}+2 a^{2} b -388 a \,b^{2}+160 b^{3}\right ) \left (\tanh ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}-\frac {3 \left (15 a^{2}+8 a b -18 b^{2}\right ) \left (\tanh ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a \left (a^{2}-2 a b +b^{2}\right )}+\frac {\left (25 a^{2}+20 a b -18 b^{2}\right ) \left (\tanh ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 a \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (5 a -2 b \right ) \left (\tanh ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{64 \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 (a \left (-5 a +2 b \right ) \textit {\_R}^{6}+\left (39 a^{2}-28 a b +10 b^{2}\right ) \textit {\_R}^{4}+\left (-39 a^{2}+28 a b -10 b^{2}\right ) \textit {\_R}^{2}+5 a^{2}-2 a 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}}{64 a^{2} \left (a^{2}-2 a b +b^{2}\right )}}{d}\)	\(583\)
risch	\(\text {Expression too large to display}\)	\(2536\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^2/(a-b*sinh(d*x+c)^4)^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-8*(-1/64*(5*a-2*b)/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)+1/64*(25*a^2+20*a*b-18*b^2)/a/(a^2-2*a*b+b^2)*tan
h(1/2*d*x+1/2*c)^3-3/64/a*(15*a^2+8*a*b-18*b^2)/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^5+1/64*(25*a^3+2*a^2*b-388
*a*b^2+160*b^3)/a^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^7+1/64*(25*a^3+2*a^2*b-388*a*b^2+160*b^3)/a^2/(a^2-2*a
*b+b^2)*tanh(1/2*d*x+1/2*c)^9-3/64/a*(15*a^2+8*a*b-18*b^2)/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^11+1/64*(25*a^2
+20*a*b-18*b^2)/a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^13-1/64*(5*a-2*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)^2-1/64/a^2/(a^2-2*a*b+b^2)*sum((a*(-5*a+2*b)*_R^6+(39*a^2-28*a*b+10*b^2)*_R^4+(-39*a^
2+28*a*b-10*b^2)*_R^2+5*a^2-2*a*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=Root
Of(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)^2/(a-b*sinh(d*x+c)^4)^3,x, algorithm="maxima")

[Out]

-1/16*(11*a*b^2 - 5*b^3 + (12*a^2*b*e^(14*c) - 11*a*b^2*e^(14*c) + 5*b^3*e^(14*c))*e^(14*d*x) - (104*a^2*b*e^(
12*c) - 85*a*b^2*e^(12*c) + 35*b^3*e^(12*c))*e^(12*d*x) - (320*a^3*e^(10*c) - 652*a^2*b*e^(10*c) + 407*a*b^2*e
^(10*c) - 105*b^3*e^(10*c))*e^(10*d*x) + (1408*a^3*e^(8*c) - 1696*a^2*b*e^(8*c) + 865*a*b^2*e^(8*c) - 175*b^3*
e^(8*c))*e^(8*d*x) + (320*a^3*e^(6*c) + 756*a^2*b*e^(6*c) - 849*a*b^2*e^(6*c) + 175*b^3*e^(6*c))*e^(6*d*x) - (
248*a^2*b*e^(4*c) - 383*a*b^2*e^(4*c) + 105*b^3*e^(4*c))*e^(4*d*x) - (12*a^2*b*e^(2*c) + 77*a*b^2*e^(2*c) - 35
*b^3*e^(2*c))*e^(2*d*x))/(a^4*b^2*d - 2*a^3*b^3*d + a^2*b^4*d + (a^4*b^2*d*e^(16*c) - 2*a^3*b^3*d*e^(16*c) + a
^2*b^4*d*e^(16*c))*e^(16*d*x) - 8*(a^4*b^2*d*e^(14*c) - 2*a^3*b^3*d*e^(14*c) + a^2*b^4*d*e^(14*c))*e^(14*d*x)
- 4*(8*a^5*b*d*e^(12*c) - 23*a^4*b^2*d*e^(12*c) + 22*a^3*b^3*d*e^(12*c) - 7*a^2*b^4*d*e^(12*c))*e^(12*d*x) + 8
*(16*a^5*b*d*e^(10*c) - 39*a^4*b^2*d*e^(10*c) + 30*a^3*b^3*d*e^(10*c) - 7*a^2*b^4*d*e^(10*c))*e^(10*d*x) + 2*(
128*a^6*d*e^(8*c) - 352*a^5*b*d*e^(8*c) + 355*a^4*b^2*d*e^(8*c) - 166*a^3*b^3*d*e^(8*c) + 35*a^2*b^4*d*e^(8*c)
)*e^(8*d*x) + 8*(16*a^5*b*d*e^(6*c) - 39*a^4*b^2*d*e^(6*c) + 30*a^3*b^3*d*e^(6*c) - 7*a^2*b^4*d*e^(6*c))*e^(6*
d*x) - 4*(8*a^5*b*d*e^(4*c) - 23*a^4*b^2*d*e^(4*c) + 22*a^3*b^3*d*e^(4*c) - 7*a^2*b^4*d*e^(4*c))*e^(4*d*x) - 8
*(a^4*b^2*d*e^(2*c) - 2*a^3*b^3*d*e^(2*c) + a^2*b^4*d*e^(2*c))*e^(2*d*x)) - 1/4*integrate(1/2*((12*a^2*e^(6*c)
- 11*a*b*e^(6*c) + 5*b^2*e^(6*c))*e^(6*d*x) - 2*(32*a^2*e^(4*c) - 19*a*b*e^(4*c) + 5*b^2*e^(4*c))*e^(4*d*x) +
(12*a^2*e^(2*c) - 11*a*b*e^(2*c) + 5*b^2*e^(2*c))*e^(2*d*x))/(a^4*b - 2*a^3*b^2 + a^2*b^3 + (a^4*b*e^(8*c) -
2*a^3*b^2*e^(8*c) + a^2*b^3*e^(8*c))*e^(8*d*x) - 4*(a^4*b*e^(6*c) - 2*a^3*b^2*e^(6*c) + a^2*b^3*e^(6*c))*e^(6*
d*x) - 2*(8*a^5*e^(4*c) - 19*a^4*b*e^(4*c) + 14*a^3*b^2*e^(4*c) - 3*a^2*b^3*e^(4*c))*e^(4*d*x) - 4*(a^4*b*e^(2
*c) - 2*a^3*b^2*e^(2*c) + a^2*b^3*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. 23355 vs. \(2 (296) = 592\).
time = 1.33, size = 23355, normalized size = 67.11 \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)^2/(a-b*sinh(d*x+c)^4)^3,x, algorithm="fricas")

[Out]

-1/128*(8*(12*a^2*b - 11*a*b^2 + 5*b^3)*cosh(d*x + c)^14 + 112*(12*a^2*b - 11*a*b^2 + 5*b^3)*cosh(d*x + c)*sin
h(d*x + c)^13 + 8*(12*a^2*b - 11*a*b^2 + 5*b^3)*sinh(d*x + c)^14 - 8*(104*a^2*b - 85*a*b^2 + 35*b^3)*cosh(d*x
+ c)^12 - 8*(104*a^2*b - 85*a*b^2 + 35*b^3 - 91*(12*a^2*b - 11*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^1
2 + 32*(91*(12*a^2*b - 11*a*b^2 + 5*b^3)*cosh(d*x + c)^3 - 3*(104*a^2*b - 85*a*b^2 + 35*b^3)*cosh(d*x + c))*si
nh(d*x + c)^11 - 8*(320*a^3 - 652*a^2*b + 407*a*b^2 - 105*b^3)*cosh(d*x + c)^10 + 8*(1001*(12*a^2*b - 11*a*b^2
+ 5*b^3)*cosh(d*x + c)^4 - 320*a^3 + 652*a^2*b - 407*a*b^2 + 105*b^3 - 66*(104*a^2*b - 85*a*b^2 + 35*b^3)*cos
h(d*x + c)^2)*sinh(d*x + c)^10 + 16*(1001*(12*a^2*b - 11*a*b^2 + 5*b^3)*cosh(d*x + c)^5 - 110*(104*a^2*b - 85*
a*b^2 + 35*b^3)*cosh(d*x + c)^3 - 5*(320*a^3 - 652*a^2*b + 407*a*b^2 - 105*b^3)*cosh(d*x + c))*sinh(d*x + c)^9
+ 8*(1408*a^3 - 1696*a^2*b + 865*a*b^2 - 175*b^3)*cosh(d*x + c)^8 + 8*(3003*(12*a^2*b - 11*a*b^2 + 5*b^3)*cos
h(d*x + c) ...

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

[Out]

Timed out

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Giac [A]
time = 0.71, size = 449, normalized size = 1.29 \begin {gather*} -\frac {12 \, a^{2} b e^{\left (14 \, d x + 14 \, c\right )} - 11 \, a b^{2} e^{\left (14 \, d x + 14 \, c\right )} + 5 \, b^{3} e^{\left (14 \, d x + 14 \, c\right )} - 104 \, a^{2} b e^{\left (12 \, d x + 12 \, c\right )} + 85 \, a b^{2} e^{\left (12 \, d x + 12 \, c\right )} - 35 \, b^{3} e^{\left (12 \, d x + 12 \, c\right )} - 320 \, a^{3} e^{\left (10 \, d x + 10 \, c\right )} + 652 \, a^{2} b e^{\left (10 \, d x + 10 \, c\right )} - 407 \, a b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 105 \, b^{3} e^{\left (10 \, d x + 10 \, c\right )} + 1408 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} - 1696 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 865 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 175 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 320 \, a^{3} e^{\left (6 \, d x + 6 \, c\right )} + 756 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 849 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 175 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 248 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 383 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 105 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 12 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 77 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 35 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 11 \, a b^{2} - 5 \, b^{3}}{16 \, {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\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)^2/(a-b*sinh(d*x+c)^4)^3,x, algorithm="giac")

[Out]

-1/16*(12*a^2*b*e^(14*d*x + 14*c) - 11*a*b^2*e^(14*d*x + 14*c) + 5*b^3*e^(14*d*x + 14*c) - 104*a^2*b*e^(12*d*x
+ 12*c) + 85*a*b^2*e^(12*d*x + 12*c) - 35*b^3*e^(12*d*x + 12*c) - 320*a^3*e^(10*d*x + 10*c) + 652*a^2*b*e^(10
*d*x + 10*c) - 407*a*b^2*e^(10*d*x + 10*c) + 105*b^3*e^(10*d*x + 10*c) + 1408*a^3*e^(8*d*x + 8*c) - 1696*a^2*b
*e^(8*d*x + 8*c) + 865*a*b^2*e^(8*d*x + 8*c) - 175*b^3*e^(8*d*x + 8*c) + 320*a^3*e^(6*d*x + 6*c) + 756*a^2*b*e
^(6*d*x + 6*c) - 849*a*b^2*e^(6*d*x + 6*c) + 175*b^3*e^(6*d*x + 6*c) - 248*a^2*b*e^(4*d*x + 4*c) + 383*a*b^2*e
^(4*d*x + 4*c) - 105*b^3*e^(4*d*x + 4*c) - 12*a^2*b*e^(2*d*x + 2*c) - 77*a*b^2*e^(2*d*x + 2*c) + 35*b^3*e^(2*d
*x + 2*c) + 11*a*b^2 - 5*b^3)/((a^4 - 2*a^3*b + a^2*b^2)*(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 )}^2}{{\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)^2/(a - b*sinh(c + d*x)^4)^3,x)

[Out]

int(sinh(c + d*x)^2/(a - b*sinh(c + d*x)^4)^3, x)

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