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

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

[Out]

-((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)*(Sqr
t[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))

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Rubi [A]  time = 0.664354, antiderivative size = 348, normalized size of antiderivative = 1., 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 = {3217, 1333, 1678, 1166, 208} \[ \frac{\tanh (c+d x) \left (\frac{2 a \left (5 a^2-9 a b-4 b^2\right )}{(a-b)^3}-\frac{5 \left (2 a^2+3 a b-b^2\right ) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a^2 d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}+\frac{b \tanh (c+d x) \left (a (a+3 b)-\left (a^2+6 a b+b^2\right ) \tanh ^2(c+d x)\right )}{8 a d (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}-\frac{\left (-14 \sqrt{a} \sqrt{b}+12 a+5 b\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \sqrt{b} d \left (\sqrt{a}-\sqrt{b}\right )^{5/2}}+\frac{\left (14 \sqrt{a} \sqrt{b}+12 a+5 b\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{9/4} \sqrt{b} d \left (\sqrt{a}+\sqrt{b}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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)*(Sqr
t[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 3217

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]

Rule 1333

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 1678

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 1166

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 208

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

Rubi steps

\begin{align*} \int \frac{\sinh ^2(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx &=\frac{\operatorname{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{\operatorname{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{\operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 4.87274, size = 343, normalized size = 0.99 \[ \frac{\frac{4 \sinh (2 (c+d x)) \left (12 a^2+b (5 b-11 a) \cosh (2 (c+d x))+11 a b-5 b^2\right )}{8 a+4 b \cosh (2 (c+d x))-b \cosh (4 (c+d x))-3 b}+\frac{\left (14 \sqrt{a} \sqrt{b}+12 a+5 b\right ) \left (\sqrt{a}-\sqrt{b}\right )^2 \tanh ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\sqrt{b} \sqrt{\sqrt{a} \sqrt{b}+a}}+\frac{\left (\sqrt{a}+\sqrt{b}\right )^2 \left (-14 \sqrt{a} \sqrt{b}+12 a+5 b\right ) \tan ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\sqrt{b} \sqrt{\sqrt{a} \sqrt{b}-a}}+\frac{128 a (a-b) \sinh (2 (c+d x)) (2 a-b \cosh (2 (c+d x))+b)}{(-8 a-4 b \cosh (2 (c+d x))+b \cosh (4 (c+d x))+3 b)^2}}{64 a^2 d (a-b)^2} \]

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]  time = 0.095, size = 2670, normalized size = 7.7 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

-1/64/d/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=RootOf(a*_Z^8-4*a*_Z^6+(6*a-
16*b)*_Z^4-4*a*_Z^2+a))+97/2/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16
*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/a*b^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^9+97/2/d/(ta
nh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/
2*d*x+1/2*c)^2*a+a)^2/a*b^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^7-20/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x
+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/a^2/(a^2-2*a*b
+b^2)*tanh(1/2*d*x+1/2*c)^7*b^3+9/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)
^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3*b^2+9/4
/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*t
anh(1/2*d*x+1/2*c)^2*a+a)^2/a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^13*b^2-20/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(
1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/a^2/(a^
2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^9*b^3-27/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*
x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/(a^2-2*a*b+b^2)/a*tanh(1/2*d*x+1/2*c)^5
*b^2-27/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2
*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/(a^2-2*a*b+b^2)/a*tanh(1/2*d*x+1/2*c)^11*b^2-5/2/d/(tanh(1/2*d*x+1/2*c)^8
*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)
^2*b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^13-5/2/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/
2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*
c)^3-1/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*
c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^15*b-1/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*
tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/(a
^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)*b+45/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1
/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^11*a+3
/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*t
anh(1/2*d*x+1/2*c)^2*a+a)^2*b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^11-25/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/
2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*a/(a^2-2*
a*b+b^2)*tanh(1/2*d*x+1/2*c)^13-25/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c
)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^3+45/8/d
/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tan
h(1/2*d*x+1/2*c)^2*a+a)^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^5*a+3/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+
1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*b/(a^2-2*a*b+b^
2)*tanh(1/2*d*x+1/2*c)^5-25/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-1
6*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^7*a-1/4/d/(tanh(1
/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*
x+1/2*c)^2*a+a)^2*b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^7-25/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c
)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2/(a^2-2*a*b+b^2)*tanh
(1/2*d*x+1/2*c)^9*a-1/4/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*ta
nh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*b/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^9+5/8/d/(tanh(1/2*d*x
+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*
c)^2*a+a)^2*a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^15+5/8/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+
6*tanh(1/2*d*x+1/2*c)^4*a-16*b*tanh(1/2*d*x+1/2*c)^4-4*tanh(1/2*d*x+1/2*c)^2*a+a)^2*a/(a^2-2*a*b+b^2)*tanh(1/2
*d*x+1/2*c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

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

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]

Timed out

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

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 = 17.2049, size = 608, normalized size = 1.75 \begin{align*} -\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} d - 2 \, a^{3} b d + a^{2} b^{2} d\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}} \end{align*}

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*d - 2*a^3*b*d + a^2*b^2*d)*(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)