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

Optimal. Leaf size=314 \[ \frac {3 \left (2 \sqrt {a}-\sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{7/4} \sqrt {b} d \left (\sqrt {a}-\sqrt {b}\right )^{5/2}}-\frac {3 \left (2 \sqrt {a}+\sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{7/4} \sqrt {b} d \left (\sqrt {a}+\sqrt {b}\right )^{5/2}}-\frac {\tanh (c+d x) \left (\frac {9 a^2-24 a b-b^2}{(a-b)^3}-\frac {(17 a+3 b) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}-\frac {b \tanh (c+d x) \left (-4 (a+b) \tanh ^2(c+d x)+3 a+b\right )}{8 d (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2} \]

[Out]

3/64*arctanh((a^(1/2)-b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))*(2*a^(1/2)-b^(1/2))/a^(7/4)/d/(a^(1/2)-b^(1/2))^(5/2
)/b^(1/2)-3/64*arctanh((a^(1/2)+b^(1/2))^(1/2)*tanh(d*x+c)/a^(1/4))*(2*a^(1/2)+b^(1/2))/a^(7/4)/d/b^(1/2)/(a^(
1/2)+b^(1/2))^(5/2)-1/8*b*tanh(d*x+c)*(3*a+b-4*(a+b)*tanh(d*x+c)^2)/(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)*((9*a^2-24*a*b-b^2)/(a-b)^3-(17*a+3*b)*tanh(d*x+c)^2/(a-b)^2)/a/d/(a-2*a*tanh(d*x
+c)^2+(a-b)*tanh(d*x+c)^4)

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Rubi [A]  time = 0.65, antiderivative size = 314, 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 = {3217, 1333, 1678, 1166, 208} \[ -\frac {\tanh (c+d x) \left (\frac {9 a^2-24 a b-b^2}{(a-b)^3}-\frac {(17 a+3 b) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a d \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )}+\frac {3 \left (2 \sqrt {a}-\sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{7/4} \sqrt {b} d \left (\sqrt {a}-\sqrt {b}\right )^{5/2}}-\frac {3 \left (2 \sqrt {a}+\sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{7/4} \sqrt {b} d \left (\sqrt {a}+\sqrt {b}\right )^{5/2}}-\frac {b \tanh (c+d x) \left (-4 (a+b) \tanh ^2(c+d x)+3 a+b\right )}{8 d (a-b)^3 \left ((a-b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(2*Sqrt[a] - Sqrt[b])*ArcTanh[(Sqrt[Sqrt[a] - Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(64*a^(7/4)*(Sqrt[a] - Sqrt
[b])^(5/2)*Sqrt[b]*d) - (3*(2*Sqrt[a] + Sqrt[b])*ArcTanh[(Sqrt[Sqrt[a] + Sqrt[b]]*Tanh[c + d*x])/a^(1/4)])/(64
*a^(7/4)*(Sqrt[a] + Sqrt[b])^(5/2)*Sqrt[b]*d) - (b*Tanh[c + d*x]*(3*a + b - 4*(a + b)*Tanh[c + d*x]^2))/(8*(a
- b)^3*d*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4)^2) - (Tanh[c + d*x]*((9*a^2 - 24*a*b - b^2)/(a -
b)^3 - ((17*a + 3*b)*Tanh[c + d*x]^2)/(a - b)^2))/(32*a*d*(a - 2*a*Tanh[c + d*x]^2 + (a - b)*Tanh[c + d*x]^4))

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]

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 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 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]

Rubi steps

\begin {align*} \int \frac {\sinh ^4(c+d x)}{\left (a-b \sinh ^4(c+d x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (1-x^2\right )^3}{\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 (3 a+b-4 (a+b) \tanh ^2(c+d x)\right )}{8 (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 (3 a+b)}{(a-b)^3}-\frac {8 a^2 (3 a-b) b^2 x^2}{(a-b)^3}-\frac {16 a^2 (a-3 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 (3 a+b-4 (a+b) \tanh ^2(c+d x)\right )}{8 (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 {9 a^2-24 a b-b^2}{(a-b)^3}-\frac {(17 a+3 b) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a 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 {12 a^3 (3 a-b) b^2}{(a-b)^2}-\frac {12 a^3 (5 a-b) b^2 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 (3 a+b-4 (a+b) \tanh ^2(c+d x)\right )}{8 (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 {9 a^2-24 a b-b^2}{(a-b)^3}-\frac {(17 a+3 b) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a d \left (a-2 a \tanh ^2(c+d x)+(a-b) \tanh ^4(c+d x)\right )}+\frac {\left (3 \left (2 a-\sqrt {a} \sqrt {b}-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^{3/2} \left (\sqrt {a}+\sqrt {b}\right )^2 \sqrt {b} d}-\frac {\left (3 \left (2 a+\sqrt {a} \sqrt {b}-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^{3/2} \left (\sqrt {a}-\sqrt {b}\right )^2 \sqrt {b} d}\\ &=\frac {3 \left (2 \sqrt {a}-\sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{7/4} \left (\sqrt {a}-\sqrt {b}\right )^{5/2} \sqrt {b} d}-\frac {3 \left (2 \sqrt {a}+\sqrt {b}\right ) \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tanh (c+d x)}{\sqrt [4]{a}}\right )}{64 a^{7/4} \left (\sqrt {a}+\sqrt {b}\right )^{5/2} \sqrt {b} d}-\frac {b \tanh (c+d x) \left (3 a+b-4 (a+b) \tanh ^2(c+d x)\right )}{8 (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 {9 a^2-24 a b-b^2}{(a-b)^3}-\frac {(17 a+3 b) \tanh ^2(c+d x)}{(a-b)^2}\right )}{32 a 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 = 4.88, size = 316, normalized size = 1.01 \[ \frac {-\frac {3 \left (2 a^{3/2}-3 a \sqrt {b}+b^{3/2}\right ) \tanh ^{-1}\left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {\sqrt {a} \sqrt {b}+a}}\right )}{a^{3/2} \sqrt {b} \sqrt {\sqrt {a} \sqrt {b}+a}}-\frac {3 \left (2 a^{3/2}+3 a \sqrt {b}-b^{3/2}\right ) \tan ^{-1}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tanh (c+d x)}{\sqrt {\sqrt {a} \sqrt {b}-a}}\right )}{a^{3/2} \sqrt {b} \sqrt {\sqrt {a} \sqrt {b}-a}}+\frac {8 \sinh (2 (c+d x)) ((2 a+b) \cosh (2 (c+d x))-7 a-2 b)}{a (8 a+4 b \cosh (2 (c+d x))-b \cosh (4 (c+d x))-3 b)}+\frac {64 (a-b) (\sinh (4 (c+d x))-6 \sinh (2 (c+d x)))}{(-8 a-4 b \cosh (2 (c+d x))+b \cosh (4 (c+d x))+3 b)^2}}{64 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*(2*a^(3/2) + 3*a*Sqrt[b] - b^(3/2))*ArcTan[((Sqrt[a] - Sqrt[b])*Tanh[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]
])/(a^(3/2)*Sqrt[-a + Sqrt[a]*Sqrt[b]]*Sqrt[b]) - (3*(2*a^(3/2) - 3*a*Sqrt[b] + b^(3/2))*ArcTanh[((Sqrt[a] + S
qrt[b])*Tanh[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/(a^(3/2)*Sqrt[a + Sqrt[a]*Sqrt[b]]*Sqrt[b]) + (8*(-7*a - 2*
b + (2*a + b)*Cosh[2*(c + d*x)])*Sinh[2*(c + d*x)])/(a*(8*a - 3*b + 4*b*Cosh[2*(c + d*x)] - b*Cosh[4*(c + d*x)
])) + (64*(a - 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*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(sinh(d*x+c)^4/(a-b*sinh(d*x+c)^4)^3,x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 1.71, size = 362, normalized size = 1.15 \[ \frac {3 \, a b^{2} e^{\left (14 \, d x + 14 \, c\right )} - 30 \, a b^{2} e^{\left (12 \, d x + 12 \, c\right )} + 3 \, b^{3} e^{\left (12 \, d x + 12 \, c\right )} - 80 \, a^{2} b e^{\left (10 \, d x + 10 \, c\right )} + 111 \, a b^{2} e^{\left (10 \, d x + 10 \, c\right )} - 16 \, b^{3} e^{\left (10 \, d x + 10 \, c\right )} + 256 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} - 64 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} - 26 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 35 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 336 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 95 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 40 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 64 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 54 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 25 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 19 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 8 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b^{2} + b^{3}}{8 \, {\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} {\left (b e^{\left (8 \, d x + 8 \, c\right )} - 4 \, b e^{\left (6 \, d x + 6 \, c\right )} - 16 \, a e^{\left (4 \, d x + 4 \, c\right )} + 6 \, b e^{\left (4 \, d x + 4 \, c\right )} - 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + b\right )}^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*a*b^2*e^(14*d*x + 14*c) - 30*a*b^2*e^(12*d*x + 12*c) + 3*b^3*e^(12*d*x + 12*c) - 80*a^2*b*e^(10*d*x + 1
0*c) + 111*a*b^2*e^(10*d*x + 10*c) - 16*b^3*e^(10*d*x + 10*c) + 256*a^3*e^(8*d*x + 8*c) - 64*a^2*b*e^(8*d*x +
8*c) - 26*a*b^2*e^(8*d*x + 8*c) + 35*b^3*e^(8*d*x + 8*c) + 336*a^2*b*e^(6*d*x + 6*c) - 95*a*b^2*e^(6*d*x + 6*c
) - 40*b^3*e^(6*d*x + 6*c) - 64*a^2*b*e^(4*d*x + 4*c) + 54*a*b^2*e^(4*d*x + 4*c) + 25*b^3*e^(4*d*x + 4*c) - 19
*a*b^2*e^(2*d*x + 2*c) - 8*b^3*e^(2*d*x + 2*c) + 2*a*b^2 + b^3)/((a^3*b - 2*a^2*b^2 + a*b^3)*(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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maple [C]  time = 0.17, size = 2214, normalized size = 7.05 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/16/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)+3/16/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+77/16/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)^3-23/16/
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)^3-177/16/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)^5*a+131/16/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+1/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+109/16/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)^7*a-367/16/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)^7-9/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+109/16/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)^9*a-367/16/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)^9-9/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*b^2/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c)^9-177/16/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)^11*a+131/16/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)^11+
1/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+77/16/d/(tanh(1/2*d*x+1/2*c)^8*a-4*t
anh(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)^13-23/16/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)^1
3-9/16/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)^15+3/16/d/(tanh(1/2*d*x+1/2*c)^8*a-4*t
anh(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-3/128/d/(a^2-2*a*b+b^2)/a*sum(((3*a-b)*_R^6+(-17*a+3*b)*_R^4+(17*a-3*b)*
_R^2-3*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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*a*b^2*e^(14*d*x + 14*c) + 2*a*b^2 + b^3 - 3*(10*a*b^2*e^(12*c) - b^3*e^(12*c))*e^(12*d*x) - (80*a^2*b*e
^(10*c) - 111*a*b^2*e^(10*c) + 16*b^3*e^(10*c))*e^(10*d*x) + (256*a^3*e^(8*c) - 64*a^2*b*e^(8*c) - 26*a*b^2*e^
(8*c) + 35*b^3*e^(8*c))*e^(8*d*x) + (336*a^2*b*e^(6*c) - 95*a*b^2*e^(6*c) - 40*b^3*e^(6*c))*e^(6*d*x) - (64*a^
2*b*e^(4*c) - 54*a*b^2*e^(4*c) - 25*b^3*e^(4*c))*e^(4*d*x) - (19*a*b^2*e^(2*c) + 8*b^3*e^(2*c))*e^(2*d*x))/(a^
3*b^3*d - 2*a^2*b^4*d + a*b^5*d + (a^3*b^3*d*e^(16*c) - 2*a^2*b^4*d*e^(16*c) + a*b^5*d*e^(16*c))*e^(16*d*x) -
8*(a^3*b^3*d*e^(14*c) - 2*a^2*b^4*d*e^(14*c) + a*b^5*d*e^(14*c))*e^(14*d*x) - 4*(8*a^4*b^2*d*e^(12*c) - 23*a^3
*b^3*d*e^(12*c) + 22*a^2*b^4*d*e^(12*c) - 7*a*b^5*d*e^(12*c))*e^(12*d*x) + 8*(16*a^4*b^2*d*e^(10*c) - 39*a^3*b
^3*d*e^(10*c) + 30*a^2*b^4*d*e^(10*c) - 7*a*b^5*d*e^(10*c))*e^(10*d*x) + 2*(128*a^5*b*d*e^(8*c) - 352*a^4*b^2*
d*e^(8*c) + 355*a^3*b^3*d*e^(8*c) - 166*a^2*b^4*d*e^(8*c) + 35*a*b^5*d*e^(8*c))*e^(8*d*x) + 8*(16*a^4*b^2*d*e^
(6*c) - 39*a^3*b^3*d*e^(6*c) + 30*a^2*b^4*d*e^(6*c) - 7*a*b^5*d*e^(6*c))*e^(6*d*x) - 4*(8*a^4*b^2*d*e^(4*c) -
23*a^3*b^3*d*e^(4*c) + 22*a^2*b^4*d*e^(4*c) - 7*a*b^5*d*e^(4*c))*e^(4*d*x) - 8*(a^3*b^3*d*e^(2*c) - 2*a^2*b^4*
d*e^(2*c) + a*b^5*d*e^(2*c))*e^(2*d*x)) + 1/16*integrate(-12*(2*(4*a*e^(4*c) - b*e^(4*c))*e^(4*d*x) - a*e^(6*d
*x + 6*c) - a*e^(2*d*x + 2*c))/(a^3*b - 2*a^2*b^2 + a*b^3 + (a^3*b*e^(8*c) - 2*a^2*b^2*e^(8*c) + a*b^3*e^(8*c)
)*e^(8*d*x) - 4*(a^3*b*e^(6*c) - 2*a^2*b^2*e^(6*c) + a*b^3*e^(6*c))*e^(6*d*x) - 2*(8*a^4*e^(4*c) - 19*a^3*b*e^
(4*c) + 14*a^2*b^2*e^(4*c) - 3*a*b^3*e^(4*c))*e^(4*d*x) - 4*(a^3*b*e^(2*c) - 2*a^2*b^2*e^(2*c) + a*b^3*e^(2*c)
)*e^(2*d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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