3.263 \(\int \frac{1}{(a-b \sinh ^4(c+d x))^3} \, dx\)

Optimal. Leaf size=320 \[ -\frac{b \tanh (c+d x) \left (\frac{17 a^2-40 a b+7 b^2}{(a-b)^3}-\frac{(33 a-13 b) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a^2 d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}+\frac{\left (-50 \sqrt{a} \sqrt{b}+32 a+21 b\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{11/4} d \left (\sqrt{a}-\sqrt{b}\right )^{5/2}}+\frac{\left (50 \sqrt{a} \sqrt{b}+32 a+21 b\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{11/4} d \left (\sqrt{a}+\sqrt{b}\right )^{5/2}}-\frac{b^2 \tanh (c+d x) \left (-4 (a+b) \tanh ^2(c+d x)+3 a+b\right )}{8 a d (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2} \]

[Out]

((32*a - 50*Sqrt[a]*Sqrt[b] + 21*b)*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(64*a^(11/4)*(Sq
rt[a] - Sqrt[b])^(5/2)*d) + ((32*a + 50*Sqrt[a]*Sqrt[b] + 21*b)*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x]
)/a^(1/4)])/(64*a^(11/4)*(Sqrt[a] + Sqrt[b])^(5/2)*d) - (b^2*Tanh[c + d*x]*(3*a + b - 4*(a + b)*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) - (b*Tanh[c + d*x]*((17*a^2 - 40*a
*b + 7*b^2)/(a - b)^3 - ((33*a - 13*b)*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 = 0.60877, antiderivative size = 320, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3209, 1205, 1678, 1166, 208} \[ -\frac{b \tanh (c+d x) \left (\frac{17 a^2-40 a b+7 b^2}{(a-b)^3}-\frac{(33 a-13 b) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a^2 d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}+\frac{\left (-50 \sqrt{a} \sqrt{b}+32 a+21 b\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{11/4} d \left (\sqrt{a}-\sqrt{b}\right )^{5/2}}+\frac{\left (50 \sqrt{a} \sqrt{b}+32 a+21 b\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{11/4} d \left (\sqrt{a}+\sqrt{b}\right )^{5/2}}-\frac{b^2 \tanh (c+d x) \left (-4 (a+b) \tanh ^2(c+d x)+3 a+b\right )}{8 a d (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*Sinh[c + d*x]^4)^(-3),x]

[Out]

((32*a - 50*Sqrt[a]*Sqrt[b] + 21*b)*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(64*a^(11/4)*(Sq
rt[a] - Sqrt[b])^(5/2)*d) + ((32*a + 50*Sqrt[a]*Sqrt[b] + 21*b)*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x]
)/a^(1/4)])/(64*a^(11/4)*(Sqrt[a] + Sqrt[b])^(5/2)*d) - (b^2*Tanh[c + d*x]*(3*a + b - 4*(a + b)*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) - (b*Tanh[c + d*x]*((17*a^2 - 40*a
*b + 7*b^2)/(a - b)^3 - ((33*a - 13*b)*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 3209

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

Rule 1205

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{f = Coeff[Polynom
ialRemainder[(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(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)*ExpandToS
um[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(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]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*
a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1] && LtQ[p, -1]

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{1}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^5}{\left (a-2 a x^2+(a-b) x^4\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac{b^2 \tanh (c+d x) \left (3 a+b-4 (a+b) \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 b \left (8 a^3-24 a^2 b+27 a b^2-7 b^3\right )}{(a-b)^3}+\frac{8 a b \left (6 a^3-18 a^2 b+15 a b^2-5 b^3\right ) x^2}{(a-b)^3}-\frac{16 a^2 (3 a-5 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^2 \tanh (c+d x) \left (3 a+b-4 (a+b) \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{b \tanh (c+d x) \left (\frac{17 a^2-40 a b+7 b^2}{(a-b)^3}-\frac{(33 a-13 b) \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{4 a^2 b^2 \left (32 a^2-47 a b+21 b^2\right )}{(a-b)^2}-\frac{4 a^2 b^2 \left (32 a^2-33 a b+13 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^2 \tanh (c+d x) \left (3 a+b-4 (a+b) \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{b \tanh (c+d x) \left (\frac{17 a^2-40 a b+7 b^2}{(a-b)^3}-\frac{(33 a-13 b) \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 (32 a-50 \sqrt{a} \sqrt{b}+21 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^{5/2} \left (\sqrt{a}-\sqrt{b}\right )^2 d}-\frac{\left (\left (\sqrt{a}-\sqrt{b}\right ) \left (32 a+50 \sqrt{a} \sqrt{b}+21 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^{5/2} \left (\sqrt{a}+\sqrt{b}\right )^2 d}\\ &=\frac{\left (32 a-50 \sqrt{a} \sqrt{b}+21 b\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{11/4} \left (\sqrt{a}-\sqrt{b}\right )^{5/2} d}+\frac{\left (32 a+50 \sqrt{a} \sqrt{b}+21 b\right ) \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{11/4} \left (\sqrt{a}+\sqrt{b}\right )^{5/2} d}-\frac{b^2 \tanh (c+d x) \left (3 a+b-4 (a+b) \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{b \tanh (c+d x) \left (\frac{17 a^2-40 a b+7 b^2}{(a-b)^3}-\frac{(33 a-13 b) \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 = 2.98798, size = 333, normalized size = 1.04 \[ \frac{\frac{64 a^{3/2} b (a-b) (\sinh (4 (c+d x))-6 \sinh (2 (c+d x)))}{(-8 a-4 b \cosh (2 (c+d x))+b \cosh (4 (c+d x))+3 b)^2}+\frac{\left (50 \sqrt{a} \sqrt{b}+32 a+21 b\right ) \left (\sqrt{a}-\sqrt{b}\right )^2 \tanh ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}+a}}-\frac{\left (\sqrt{a}+\sqrt{b}\right )^2 \left (-50 \sqrt{a} \sqrt{b}+32 a+21 b\right ) \tan ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tanh (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\sqrt{\sqrt{a} \sqrt{b}-a}}+\frac{8 \sqrt{a} b \sinh (2 (c+d x)) ((6 a-3 b) \cosh (2 (c+d x))-19 a+10 b)}{8 a+4 b \cosh (2 (c+d x))-b \cosh (4 (c+d x))-3 b}}{64 a^{5/2} d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(((Sqrt[a] + Sqrt[b])^2*(32*a - 50*Sqrt[a]*Sqrt[b] + 21*b)*ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt[-
a + Sqrt[a]*Sqrt[b]]])/Sqrt[-a + Sqrt[a]*Sqrt[b]]) + ((Sqrt[a] - Sqrt[b])^2*(32*a + 50*Sqrt[a]*Sqrt[b] + 21*b)
*ArcTanh[((Sqrt[a] + Sqrt[b])*Tanh[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/Sqrt[a + Sqrt[a]*Sqrt[b]] + (8*Sqrt[a
]*b*(-19*a + 10*b + (6*a - 3*b)*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)])/(8*a - 3*b + 4*b*Cosh[2*(c + d*x)] - b*C
osh[4*(c + d*x)]) + (64*a^(3/2)*(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^(5/2)*(a - b)^2*d)

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Maple [C]  time = 0.103, size = 2290, normalized size = 7.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-17/16/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+11/16/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tan
h(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^2/a
/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)+149/16/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
-95/16/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^2-345/16/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+427/16/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^2-7/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*b^3/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^5+213/16/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)^7-1111/16/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)^7+31/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+213/16/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)^9-1111/16/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+31/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/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^9*b^3-345/16/
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*ta
nh(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+427/16/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^2-7/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*b^3/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^
11+149/16/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)^13-95/16/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*b^2-17/16/d/(tanh(1/2*d*x+1/2*c)^8*a-4*tanh(1/2*d*x+1/2*c)^6*a+6*tan
h(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+11/16/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^2/a/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^15-1/128/d/(a^2-2*a*b+b^2
)/a^2*sum(((32*a^2-47*a*b+21*b^2)*_R^6+(-96*a^2+85*a*b-31*b^2)*_R^4+(96*a^2-85*a*b+31*b^2)*_R^2-32*a^2+47*a*b-
21*b^2)/(_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., 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(1/(a-b*sinh(d*x+c)^4)^3,x, algorithm="maxima")

[Out]

1/8*(6*a*b^2 - 3*b^3 + (7*a*b^2*e^(14*c) - 4*b^3*e^(14*c))*e^(14*d*x) - (32*a^2*b*e^(12*c) + 2*a*b^2*e^(12*c)
- 7*b^3*e^(12*c))*e^(12*d*x) - (16*a^2*b*e^(10*c) - 3*a*b^2*e^(10*c) - 28*b^3*e^(10*c))*e^(10*d*x) + 3*(256*a^
3*e^(8*c) - 320*a^2*b*e^(8*c) + 166*a*b^2*e^(8*c) - 35*b^3*e^(8*c))*e^(8*d*x) + (784*a^2*b*e^(6*c) - 723*a*b^2
*e^(6*c) + 140*b^3*e^(6*c))*e^(6*d*x) - (160*a^2*b*e^(4*c) - 266*a*b^2*e^(4*c) + 91*b^3*e^(4*c))*e^(4*d*x) - (
55*a*b^2*e^(2*c) - 28*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)) + integrate(1/4
*((7*a*b*e^(6*c) - 4*b^2*e^(6*c))*e^(6*d*x) - 2*(32*a^2*e^(4*c) - 40*a*b*e^(4*c) + 17*b^2*e^(4*c))*e^(4*d*x) +
(7*a*b*e^(2*c) - 4*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(1/(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(1/(a-b*sinh(d*x+c)**4)**3,x)

[Out]

Timed out

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Giac [A]  time = 17.8292, size = 529, normalized size = 1.65 \begin{align*} \frac{7 \, 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 )} - 2 \, a b^{2} e^{\left (12 \, d x + 12 \, c\right )} + 7 \, b^{3} e^{\left (12 \, d x + 12 \, c\right )} - 16 \, a^{2} b e^{\left (10 \, d x + 10 \, c\right )} + 3 \, a b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 28 \, b^{3} e^{\left (10 \, d x + 10 \, c\right )} + 768 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} - 960 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 498 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 105 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 784 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 723 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 140 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 160 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 266 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 91 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 55 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 28 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 6 \, a b^{2} - 3 \, b^{3}}{8 \,{\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(1/(a-b*sinh(d*x+c)^4)^3,x, algorithm="giac")

[Out]

1/8*(7*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) - 2*a*b^2*e^(12*d*x + 12
*c) + 7*b^3*e^(12*d*x + 12*c) - 16*a^2*b*e^(10*d*x + 10*c) + 3*a*b^2*e^(10*d*x + 10*c) + 28*b^3*e^(10*d*x + 10
*c) + 768*a^3*e^(8*d*x + 8*c) - 960*a^2*b*e^(8*d*x + 8*c) + 498*a*b^2*e^(8*d*x + 8*c) - 105*b^3*e^(8*d*x + 8*c
) + 784*a^2*b*e^(6*d*x + 6*c) - 723*a*b^2*e^(6*d*x + 6*c) + 140*b^3*e^(6*d*x + 6*c) - 160*a^2*b*e^(4*d*x + 4*c
) + 266*a*b^2*e^(4*d*x + 4*c) - 91*b^3*e^(4*d*x + 4*c) - 55*a*b^2*e^(2*d*x + 2*c) + 28*b^3*e^(2*d*x + 2*c) + 6
*a*b^2 - 3*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)