3.220 \(\int \text {csch}^4(c+d x) (a+b \sinh ^4(c+d x))^3 \, dx\)

Optimal. Leaf size=161 \[ -\frac {a^3 \coth ^3(c+d x)}{3 d}+\frac {a^3 \coth (c+d x)}{d}+\frac {1}{128} b x \left (384 a^2+144 a b+35 b^2\right )+\frac {b^2 (144 a+163 b) \sinh (c+d x) \cosh ^3(c+d x)}{192 d}-\frac {3 b^2 (80 a+31 b) \sinh (c+d x) \cosh (c+d x)}{128 d}+\frac {b^3 \sinh (c+d x) \cosh ^7(c+d x)}{8 d}-\frac {25 b^3 \sinh (c+d x) \cosh ^5(c+d x)}{48 d} \]

[Out]

1/128*b*(384*a^2+144*a*b+35*b^2)*x+a^3*coth(d*x+c)/d-1/3*a^3*coth(d*x+c)^3/d-3/128*b^2*(80*a+31*b)*cosh(d*x+c)
*sinh(d*x+c)/d+1/192*b^2*(144*a+163*b)*cosh(d*x+c)^3*sinh(d*x+c)/d-25/48*b^3*cosh(d*x+c)^5*sinh(d*x+c)/d+1/8*b
^3*cosh(d*x+c)^7*sinh(d*x+c)/d

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Rubi [A]  time = 0.38, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3217, 1259, 1805, 1261, 207} \[ \frac {1}{128} b x \left (384 a^2+144 a b+35 b^2\right )-\frac {a^3 \coth ^3(c+d x)}{3 d}+\frac {a^3 \coth (c+d x)}{d}+\frac {b^2 (144 a+163 b) \sinh (c+d x) \cosh ^3(c+d x)}{192 d}-\frac {3 b^2 (80 a+31 b) \sinh (c+d x) \cosh (c+d x)}{128 d}+\frac {b^3 \sinh (c+d x) \cosh ^7(c+d x)}{8 d}-\frac {25 b^3 \sinh (c+d x) \cosh ^5(c+d x)}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(384*a^2 + 144*a*b + 35*b^2)*x)/128 + (a^3*Coth[c + d*x])/d - (a^3*Coth[c + d*x]^3)/(3*d) - (3*b^2*(80*a +
31*b)*Cosh[c + d*x]*Sinh[c + d*x])/(128*d) + (b^2*(144*a + 163*b)*Cosh[c + d*x]^3*Sinh[c + d*x])/(192*d) - (25
*b^3*Cosh[c + d*x]^5*Sinh[c + d*x])/(48*d) + (b^3*Cosh[c + d*x]^7*Sinh[c + d*x])/(8*d)

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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 \text {csch}^4(c+d x) \left (a+b \sinh ^4(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a-2 a x^2+(a+b) x^4\right )^3}{x^4 \left (1-x^2\right )^5} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}+\frac {\operatorname {Subst}\left (\int \frac {8 a^3-40 a^3 x^2+\left (80 a^3+24 a^2 b-b^3\right ) x^4-8 \left (10 a^3+9 a^2 b+b^3\right ) x^6+8 (5 a-b) (a+b)^2 x^8-8 (a+b)^3 x^{10}}{x^4 \left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac {25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac {b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}-\frac {\operatorname {Subst}\left (\int \frac {-48 a^3+192 a^3 x^2-\left (288 a^3+144 a^2 b+19 b^3\right ) x^4+96 (2 a-b) (a+b)^2 x^6-48 (a+b)^3 x^8}{x^4 \left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{48 d}\\ &=\frac {b^2 (144 a+163 b) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}-\frac {25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac {b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}+\frac {\operatorname {Subst}\left (\int \frac {192 a^3-576 a^3 x^2+3 \left (192 a^3+192 a^2 b-48 a b^2-29 b^3\right ) x^4-192 (a+b)^3 x^6}{x^4 \left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{192 d}\\ &=-\frac {3 b^2 (80 a+31 b) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac {b^2 (144 a+163 b) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}-\frac {25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac {b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}-\frac {\operatorname {Subst}\left (\int \frac {-384 a^3+768 a^3 x^2-3 \left (128 a^3+384 a^2 b+144 a b^2+35 b^3\right ) x^4}{x^4 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{384 d}\\ &=-\frac {3 b^2 (80 a+31 b) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac {b^2 (144 a+163 b) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}-\frac {25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac {b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}-\frac {\operatorname {Subst}\left (\int \left (-\frac {384 a^3}{x^4}+\frac {384 a^3}{x^2}+\frac {3 b \left (384 a^2+144 a b+35 b^2\right )}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{384 d}\\ &=\frac {a^3 \coth (c+d x)}{d}-\frac {a^3 \coth ^3(c+d x)}{3 d}-\frac {3 b^2 (80 a+31 b) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac {b^2 (144 a+163 b) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}-\frac {25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac {b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}-\frac {\left (b \left (384 a^2+144 a b+35 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{128 d}\\ &=\frac {1}{128} b \left (384 a^2+144 a b+35 b^2\right ) x+\frac {a^3 \coth (c+d x)}{d}-\frac {a^3 \coth ^3(c+d x)}{3 d}-\frac {3 b^2 (80 a+31 b) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac {b^2 (144 a+163 b) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}-\frac {25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac {b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 131, normalized size = 0.81 \[ \frac {b \left (9216 a^2 c+9216 a^2 d x-96 b (24 a+7 b) \sinh (2 (c+d x))+24 b (12 a+7 b) \sinh (4 (c+d x))+3456 a b c+3456 a b d x-32 b^2 \sinh (6 (c+d x))+3 b^2 \sinh (8 (c+d x))+840 b^2 c+840 b^2 d x\right )-1024 a^3 \coth (c+d x) \left (\text {csch}^2(c+d x)-2\right )}{3072 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1024*a^3*Coth[c + d*x]*(-2 + Csch[c + d*x]^2) + b*(9216*a^2*c + 3456*a*b*c + 840*b^2*c + 9216*a^2*d*x + 3456
*a*b*d*x + 840*b^2*d*x - 96*b*(24*a + 7*b)*Sinh[2*(c + d*x)] + 24*b*(12*a + 7*b)*Sinh[4*(c + d*x)] - 32*b^2*Si
nh[6*(c + d*x)] + 3*b^2*Sinh[8*(c + d*x)]))/(3072*d)

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fricas [B]  time = 0.99, size = 567, normalized size = 3.52 \[ \frac {3 \, b^{3} \cosh \left (d x + c\right )^{11} + 33 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{10} - 41 \, b^{3} \cosh \left (d x + c\right )^{9} + 9 \, {\left (55 \, b^{3} \cosh \left (d x + c\right )^{3} - 41 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{8} + 3 \, {\left (96 \, a b^{2} + 91 \, b^{3}\right )} \cosh \left (d x + c\right )^{7} + 21 \, {\left (66 \, b^{3} \cosh \left (d x + c\right )^{5} - 164 \, b^{3} \cosh \left (d x + c\right )^{3} + {\left (96 \, a b^{2} + 91 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{6} - 3 \, {\left (1056 \, a b^{2} + 425 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 3 \, {\left (330 \, b^{3} \cosh \left (d x + c\right )^{7} - 1722 \, b^{3} \cosh \left (d x + c\right )^{5} + 35 \, {\left (96 \, a b^{2} + 91 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - 5 \, {\left (1056 \, a b^{2} + 425 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 8 \, {\left (512 \, a^{3} + 972 \, a b^{2} + 319 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - 16 \, {\left (256 \, a^{3} - 3 \, {\left (384 \, a^{2} b + 144 \, a b^{2} + 35 \, b^{3}\right )} d x\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (55 \, b^{3} \cosh \left (d x + c\right )^{9} - 492 \, b^{3} \cosh \left (d x + c\right )^{7} + 21 \, {\left (96 \, a b^{2} + 91 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} - 10 \, {\left (1056 \, a b^{2} + 425 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 8 \, {\left (512 \, a^{3} + 972 \, a b^{2} + 319 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 24 \, {\left (512 \, a^{3} + 204 \, a b^{2} + 63 \, b^{3}\right )} \cosh \left (d x + c\right ) + 48 \, {\left (256 \, a^{3} - 3 \, {\left (384 \, a^{2} b + 144 \, a b^{2} + 35 \, b^{3}\right )} d x - {\left (256 \, a^{3} - 3 \, {\left (384 \, a^{2} b + 144 \, a b^{2} + 35 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{6144 \, {\left (d \sinh \left (d x + c\right )^{3} + 3 \, {\left (d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6144*(3*b^3*cosh(d*x + c)^11 + 33*b^3*cosh(d*x + c)*sinh(d*x + c)^10 - 41*b^3*cosh(d*x + c)^9 + 9*(55*b^3*co
sh(d*x + c)^3 - 41*b^3*cosh(d*x + c))*sinh(d*x + c)^8 + 3*(96*a*b^2 + 91*b^3)*cosh(d*x + c)^7 + 21*(66*b^3*cos
h(d*x + c)^5 - 164*b^3*cosh(d*x + c)^3 + (96*a*b^2 + 91*b^3)*cosh(d*x + c))*sinh(d*x + c)^6 - 3*(1056*a*b^2 +
425*b^3)*cosh(d*x + c)^5 + 3*(330*b^3*cosh(d*x + c)^7 - 1722*b^3*cosh(d*x + c)^5 + 35*(96*a*b^2 + 91*b^3)*cosh
(d*x + c)^3 - 5*(1056*a*b^2 + 425*b^3)*cosh(d*x + c))*sinh(d*x + c)^4 + 8*(512*a^3 + 972*a*b^2 + 319*b^3)*cosh
(d*x + c)^3 - 16*(256*a^3 - 3*(384*a^2*b + 144*a*b^2 + 35*b^3)*d*x)*sinh(d*x + c)^3 + 3*(55*b^3*cosh(d*x + c)^
9 - 492*b^3*cosh(d*x + c)^7 + 21*(96*a*b^2 + 91*b^3)*cosh(d*x + c)^5 - 10*(1056*a*b^2 + 425*b^3)*cosh(d*x + c)
^3 + 8*(512*a^3 + 972*a*b^2 + 319*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 - 24*(512*a^3 + 204*a*b^2 + 63*b^3)*cosh
(d*x + c) + 48*(256*a^3 - 3*(384*a^2*b + 144*a*b^2 + 35*b^3)*d*x - (256*a^3 - 3*(384*a^2*b + 144*a*b^2 + 35*b^
3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*sinh(d*x + c)^3 + 3*(d*cosh(d*x + c)^2 - d)*sinh(d*x + c))

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giac [A]  time = 0.45, size = 285, normalized size = 1.77 \[ \frac {3 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} - 32 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 288 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 168 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 2304 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 672 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 48 \, {\left (384 \, a^{2} b + 144 \, a b^{2} + 35 \, b^{3}\right )} {\left (d x + c\right )} - {\left (19200 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 7200 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 1750 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} - 2304 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 672 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 288 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 168 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 32 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{3}\right )} e^{\left (-8 \, d x - 8 \, c\right )} - \frac {8192 \, {\left (3 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - a^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{6144 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6144*(3*b^3*e^(8*d*x + 8*c) - 32*b^3*e^(6*d*x + 6*c) + 288*a*b^2*e^(4*d*x + 4*c) + 168*b^3*e^(4*d*x + 4*c) -
2304*a*b^2*e^(2*d*x + 2*c) - 672*b^3*e^(2*d*x + 2*c) + 48*(384*a^2*b + 144*a*b^2 + 35*b^3)*(d*x + c) - (19200
*a^2*b*e^(8*d*x + 8*c) + 7200*a*b^2*e^(8*d*x + 8*c) + 1750*b^3*e^(8*d*x + 8*c) - 2304*a*b^2*e^(6*d*x + 6*c) -
672*b^3*e^(6*d*x + 6*c) + 288*a*b^2*e^(4*d*x + 4*c) + 168*b^3*e^(4*d*x + 4*c) - 32*b^3*e^(2*d*x + 2*c) + 3*b^3
)*e^(-8*d*x - 8*c) - 8192*(3*a^3*e^(2*d*x + 2*c) - a^3)/(e^(2*d*x + 2*c) - 1)^3)/d

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maple [A]  time = 0.10, size = 137, normalized size = 0.85 \[ \frac {a^{3} \left (\frac {2}{3}-\frac {\mathrm {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+3 a^{2} b \left (d x +c \right )+3 a \,b^{2} \left (\left (\frac {\left (\sinh ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+b^{3} \left (\left (\frac {\left (\sinh ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sinh ^{5}\left (d x +c \right )\right )}{48}+\frac {35 \left (\sinh ^{3}\left (d x +c \right )\right )}{192}-\frac {35 \sinh \left (d x +c \right )}{128}\right ) \cosh \left (d x +c \right )+\frac {35 d x}{128}+\frac {35 c}{128}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c)+3*a^2*b*(d*x+c)+3*a*b^2*((1/4*sinh(d*x+c)^3-3/8*sinh(d*x+c))*cosh
(d*x+c)+3/8*d*x+3/8*c)+b^3*((1/8*sinh(d*x+c)^7-7/48*sinh(d*x+c)^5+35/192*sinh(d*x+c)^3-35/128*sinh(d*x+c))*cos
h(d*x+c)+35/128*d*x+35/128*c))

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maxima [A]  time = 0.34, size = 282, normalized size = 1.75 \[ \frac {3}{64} \, a b^{2} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + 3 \, a^{2} b x - \frac {1}{6144} \, b^{3} {\left (\frac {{\left (32 \, e^{\left (-2 \, d x - 2 \, c\right )} - 168 \, e^{\left (-4 \, d x - 4 \, c\right )} + 672 \, e^{\left (-6 \, d x - 6 \, c\right )} - 3\right )} e^{\left (8 \, d x + 8 \, c\right )}}{d} - \frac {1680 \, {\left (d x + c\right )}}{d} - \frac {672 \, e^{\left (-2 \, d x - 2 \, c\right )} - 168 \, e^{\left (-4 \, d x - 4 \, c\right )} + 32 \, e^{\left (-6 \, d x - 6 \, c\right )} - 3 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d}\right )} + \frac {4}{3} \, a^{3} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/64*a*b^2*(24*x + e^(4*d*x + 4*c)/d - 8*e^(2*d*x + 2*c)/d + 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) + 3*a^
2*b*x - 1/6144*b^3*((32*e^(-2*d*x - 2*c) - 168*e^(-4*d*x - 4*c) + 672*e^(-6*d*x - 6*c) - 3)*e^(8*d*x + 8*c)/d
- 1680*(d*x + c)/d - (672*e^(-2*d*x - 2*c) - 168*e^(-4*d*x - 4*c) + 32*e^(-6*d*x - 6*c) - 3*e^(-8*d*x - 8*c))/
d) + 4/3*a^3*(3*e^(-2*d*x - 2*c)/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1)) - 1/(d*(
3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1)))

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mupad [B]  time = 1.07, size = 269, normalized size = 1.67 \[ x\,\left (3\,a^2\,b+\frac {9\,a\,b^2}{8}+\frac {35\,b^3}{128}\right )-\frac {4\,a^3}{3\,d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {b^3\,{\mathrm {e}}^{-6\,c-6\,d\,x}}{192\,d}-\frac {b^3\,{\mathrm {e}}^{6\,c+6\,d\,x}}{192\,d}-\frac {b^3\,{\mathrm {e}}^{-8\,c-8\,d\,x}}{2048\,d}+\frac {b^3\,{\mathrm {e}}^{8\,c+8\,d\,x}}{2048\,d}-\frac {8\,a^3\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )}-\frac {b^2\,{\mathrm {e}}^{-4\,c-4\,d\,x}\,\left (12\,a+7\,b\right )}{256\,d}+\frac {b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (12\,a+7\,b\right )}{256\,d}+\frac {b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (24\,a+7\,b\right )}{64\,d}-\frac {b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (24\,a+7\,b\right )}{64\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((9*a*b^2)/8 + 3*a^2*b + (35*b^3)/128) - (4*a^3)/(3*d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) + (b^3*ex
p(- 6*c - 6*d*x))/(192*d) - (b^3*exp(6*c + 6*d*x))/(192*d) - (b^3*exp(- 8*c - 8*d*x))/(2048*d) + (b^3*exp(8*c
+ 8*d*x))/(2048*d) - (8*a^3*exp(2*c + 2*d*x))/(3*d*(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x)
- 1)) - (b^2*exp(- 4*c - 4*d*x)*(12*a + 7*b))/(256*d) + (b^2*exp(4*c + 4*d*x)*(12*a + 7*b))/(256*d) + (b^2*ex
p(- 2*c - 2*d*x)*(24*a + 7*b))/(64*d) - (b^2*exp(2*c + 2*d*x)*(24*a + 7*b))/(64*d)

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

[Out]

Timed out

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