3.219 \(\int \text {csch}^2(c+d x) (a+b \sinh ^4(c+d x))^3 \, dx\)

Optimal. Leaf size=181 \[ -\frac {a^3 \coth (c+d x)}{d}+\frac {b \left (384 a^2+528 a b+193 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{256 d}-\frac {3}{256} b x \left (128 a^2+80 a b+21 b^2\right )+\frac {b^2 (80 a+171 b) \sinh (c+d x) \cosh ^5(c+d x)}{160 d}-\frac {b^2 (208 a+149 b) \sinh (c+d x) \cosh ^3(c+d x)}{128 d}+\frac {b^3 \sinh (c+d x) \cosh ^9(c+d x)}{10 d}-\frac {41 b^3 \sinh (c+d x) \cosh ^7(c+d x)}{80 d} \]

[Out]

-3/256*b*(128*a^2+80*a*b+21*b^2)*x-a^3*coth(d*x+c)/d+1/256*b*(384*a^2+528*a*b+193*b^2)*cosh(d*x+c)*sinh(d*x+c)
/d-1/128*b^2*(208*a+149*b)*cosh(d*x+c)^3*sinh(d*x+c)/d+1/160*b^2*(80*a+171*b)*cosh(d*x+c)^5*sinh(d*x+c)/d-41/8
0*b^3*cosh(d*x+c)^7*sinh(d*x+c)/d+1/10*b^3*cosh(d*x+c)^9*sinh(d*x+c)/d

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Rubi [A]  time = 0.44, antiderivative size = 181, 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, 453, 206} \[ \frac {b \left (384 a^2+528 a b+193 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{256 d}-\frac {3}{256} b x \left (128 a^2+80 a b+21 b^2\right )-\frac {a^3 \coth (c+d x)}{d}+\frac {b^2 (80 a+171 b) \sinh (c+d x) \cosh ^5(c+d x)}{160 d}-\frac {b^2 (208 a+149 b) \sinh (c+d x) \cosh ^3(c+d x)}{128 d}+\frac {b^3 \sinh (c+d x) \cosh ^9(c+d x)}{10 d}-\frac {41 b^3 \sinh (c+d x) \cosh ^7(c+d x)}{80 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(128*a^2 + 80*a*b + 21*b^2)*x)/256 - (a^3*Coth[c + d*x])/d + (b*(384*a^2 + 528*a*b + 193*b^2)*Cosh[c + d
*x]*Sinh[c + d*x])/(256*d) - (b^2*(208*a + 149*b)*Cosh[c + d*x]^3*Sinh[c + d*x])/(128*d) + (b^2*(80*a + 171*b)
*Cosh[c + d*x]^5*Sinh[c + d*x])/(160*d) - (41*b^3*Cosh[c + d*x]^7*Sinh[c + d*x])/(80*d) + (b^3*Cosh[c + d*x]^9
*Sinh[c + d*x])/(10*d)

Rule 206

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

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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 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}^2(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^2 \left (1-x^2\right )^6} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b^3 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}-\frac {\operatorname {Subst}\left (\int \frac {-10 a^3+\left (50 a^3+b^3\right ) x^2-10 \left (10 a^3+3 a^2 b-b^3\right ) x^4+10 \left (10 a^3+9 a^2 b+b^3\right ) x^6-10 (5 a-b) (a+b)^2 x^8+10 (a+b)^3 x^{10}}{x^2 \left (1-x^2\right )^5} \, dx,x,\tanh (c+d x)\right )}{10 d}\\ &=-\frac {41 b^3 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac {b^3 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}+\frac {\operatorname {Subst}\left (\int \frac {80 a^3-\left (320 a^3-33 b^3\right ) x^2+240 \left (2 a^3+a^2 b+b^3\right ) x^4-160 (2 a-b) (a+b)^2 x^6+80 (a+b)^3 x^8}{x^2 \left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{80 d}\\ &=\frac {b^2 (80 a+171 b) \cosh ^5(c+d x) \sinh (c+d x)}{160 d}-\frac {41 b^3 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac {b^3 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}-\frac {\operatorname {Subst}\left (\int \frac {-480 a^3+15 \left (96 a^3+16 a b^2+21 b^3\right ) x^2-1440 (a-b) (a+b)^2 x^4+480 (a+b)^3 x^6}{x^2 \left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{480 d}\\ &=-\frac {b^2 (208 a+149 b) \cosh ^3(c+d x) \sinh (c+d x)}{128 d}+\frac {b^2 (80 a+171 b) \cosh ^5(c+d x) \sinh (c+d x)}{160 d}-\frac {41 b^3 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac {b^3 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}+\frac {\operatorname {Subst}\left (\int \frac {1920 a^3-15 \left (256 a^3-144 a b^2-65 b^3\right ) x^2+1920 (a+b)^3 x^4}{x^2 \left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{1920 d}\\ &=\frac {b \left (384 a^2+528 a b+193 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{256 d}-\frac {b^2 (208 a+149 b) \cosh ^3(c+d x) \sinh (c+d x)}{128 d}+\frac {b^2 (80 a+171 b) \cosh ^5(c+d x) \sinh (c+d x)}{160 d}-\frac {41 b^3 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac {b^3 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}-\frac {\operatorname {Subst}\left (\int \frac {-3840 a^3+15 \left (256 a^3+384 a^2 b+240 a b^2+63 b^3\right ) x^2}{x^2 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{3840 d}\\ &=-\frac {a^3 \coth (c+d x)}{d}+\frac {b \left (384 a^2+528 a b+193 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{256 d}-\frac {b^2 (208 a+149 b) \cosh ^3(c+d x) \sinh (c+d x)}{128 d}+\frac {b^2 (80 a+171 b) \cosh ^5(c+d x) \sinh (c+d x)}{160 d}-\frac {41 b^3 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac {b^3 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}-\frac {\left (3 b \left (128 a^2+80 a b+21 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{256 d}\\ &=-\frac {3}{256} b \left (128 a^2+80 a b+21 b^2\right ) x-\frac {a^3 \coth (c+d x)}{d}+\frac {b \left (384 a^2+528 a b+193 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{256 d}-\frac {b^2 (208 a+149 b) \cosh ^3(c+d x) \sinh (c+d x)}{128 d}+\frac {b^2 (80 a+171 b) \cosh ^5(c+d x) \sinh (c+d x)}{160 d}-\frac {41 b^3 \cosh ^7(c+d x) \sinh (c+d x)}{80 d}+\frac {b^3 \cosh ^9(c+d x) \sinh (c+d x)}{10 d}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 134, normalized size = 0.74 \[ \frac {-10240 a^3 \coth (c+d x)-120 b \left (128 a^2+80 a b+21 b^2\right ) (c+d x)+60 b \left (128 a^2+120 a b+35 b^2\right ) \sinh (2 (c+d x))-120 b^2 (12 a+5 b) \sinh (4 (c+d x))+10 b^2 (16 a+15 b) \sinh (6 (c+d x))-25 b^3 \sinh (8 (c+d x))+2 b^3 \sinh (10 (c+d x))}{10240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-120*b*(128*a^2 + 80*a*b + 21*b^2)*(c + d*x) - 10240*a^3*Coth[c + d*x] + 60*b*(128*a^2 + 120*a*b + 35*b^2)*Si
nh[2*(c + d*x)] - 120*b^2*(12*a + 5*b)*Sinh[4*(c + d*x)] + 10*b^2*(16*a + 15*b)*Sinh[6*(c + d*x)] - 25*b^3*Sin
h[8*(c + d*x)] + 2*b^3*Sinh[10*(c + d*x)])/(10240*d)

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fricas [B]  time = 0.46, size = 474, normalized size = 2.62 \[ \frac {2 \, b^{3} \cosh \left (d x + c\right )^{11} + 22 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{10} - 27 \, b^{3} \cosh \left (d x + c\right )^{9} + 3 \, {\left (110 \, b^{3} \cosh \left (d x + c\right )^{3} - 81 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{8} + 5 \, {\left (32 \, a b^{2} + 35 \, b^{3}\right )} \cosh \left (d x + c\right )^{7} + 7 \, {\left (132 \, b^{3} \cosh \left (d x + c\right )^{5} - 324 \, b^{3} \cosh \left (d x + c\right )^{3} + 5 \, {\left (32 \, a b^{2} + 35 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{6} - 50 \, {\left (32 \, a b^{2} + 15 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + {\left (660 \, b^{3} \cosh \left (d x + c\right )^{7} - 3402 \, b^{3} \cosh \left (d x + c\right )^{5} + 175 \, {\left (32 \, a b^{2} + 35 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - 250 \, {\left (32 \, a b^{2} + 15 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 60 \, {\left (128 \, a^{2} b + 144 \, a b^{2} + 45 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + {\left (110 \, b^{3} \cosh \left (d x + c\right )^{9} - 972 \, b^{3} \cosh \left (d x + c\right )^{7} + 105 \, {\left (32 \, a b^{2} + 35 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} - 500 \, {\left (32 \, a b^{2} + 15 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 180 \, {\left (128 \, a^{2} b + 144 \, a b^{2} + 45 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 20 \, {\left (1024 \, a^{3} + 384 \, a^{2} b + 360 \, a b^{2} + 105 \, b^{3}\right )} \cosh \left (d x + c\right ) + 80 \, {\left (256 \, a^{3} - 3 \, {\left (128 \, a^{2} b + 80 \, a b^{2} + 21 \, b^{3}\right )} d x\right )} \sinh \left (d x + c\right )}{20480 \, d \sinh \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2*(a+b*sinh(d*x+c)^4)^3,x, algorithm="fricas")

[Out]

1/20480*(2*b^3*cosh(d*x + c)^11 + 22*b^3*cosh(d*x + c)*sinh(d*x + c)^10 - 27*b^3*cosh(d*x + c)^9 + 3*(110*b^3*
cosh(d*x + c)^3 - 81*b^3*cosh(d*x + c))*sinh(d*x + c)^8 + 5*(32*a*b^2 + 35*b^3)*cosh(d*x + c)^7 + 7*(132*b^3*c
osh(d*x + c)^5 - 324*b^3*cosh(d*x + c)^3 + 5*(32*a*b^2 + 35*b^3)*cosh(d*x + c))*sinh(d*x + c)^6 - 50*(32*a*b^2
+ 15*b^3)*cosh(d*x + c)^5 + (660*b^3*cosh(d*x + c)^7 - 3402*b^3*cosh(d*x + c)^5 + 175*(32*a*b^2 + 35*b^3)*cos
h(d*x + c)^3 - 250*(32*a*b^2 + 15*b^3)*cosh(d*x + c))*sinh(d*x + c)^4 + 60*(128*a^2*b + 144*a*b^2 + 45*b^3)*co
sh(d*x + c)^3 + (110*b^3*cosh(d*x + c)^9 - 972*b^3*cosh(d*x + c)^7 + 105*(32*a*b^2 + 35*b^3)*cosh(d*x + c)^5 -
500*(32*a*b^2 + 15*b^3)*cosh(d*x + c)^3 + 180*(128*a^2*b + 144*a*b^2 + 45*b^3)*cosh(d*x + c))*sinh(d*x + c)^2
- 20*(1024*a^3 + 384*a^2*b + 360*a*b^2 + 105*b^3)*cosh(d*x + c) + 80*(256*a^3 - 3*(128*a^2*b + 80*a*b^2 + 21*
b^3)*d*x)*sinh(d*x + c))/(d*sinh(d*x + c))

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giac [B]  time = 0.40, size = 355, normalized size = 1.96 \[ \frac {2 \, b^{3} e^{\left (10 \, d x + 10 \, c\right )} - 25 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 160 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 150 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 1440 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 600 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 7680 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 7200 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2100 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - 240 \, {\left (128 \, a^{2} b + 80 \, a b^{2} + 21 \, b^{3}\right )} {\left (d x + c\right )} - \frac {40960 \, a^{3}}{e^{\left (2 \, d x + 2 \, c\right )} - 1} + {\left (35072 \, a^{2} b e^{\left (10 \, d x + 10 \, c\right )} + 21920 \, a b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 5754 \, b^{3} e^{\left (10 \, d x + 10 \, c\right )} - 7680 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} - 7200 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 2100 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 1440 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 600 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} - 160 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 150 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 25 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b^{3}\right )} e^{\left (-10 \, d x - 10 \, c\right )}}{20480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2*(a+b*sinh(d*x+c)^4)^3,x, algorithm="giac")

[Out]

1/20480*(2*b^3*e^(10*d*x + 10*c) - 25*b^3*e^(8*d*x + 8*c) + 160*a*b^2*e^(6*d*x + 6*c) + 150*b^3*e^(6*d*x + 6*c
) - 1440*a*b^2*e^(4*d*x + 4*c) - 600*b^3*e^(4*d*x + 4*c) + 7680*a^2*b*e^(2*d*x + 2*c) + 7200*a*b^2*e^(2*d*x +
2*c) + 2100*b^3*e^(2*d*x + 2*c) - 240*(128*a^2*b + 80*a*b^2 + 21*b^3)*(d*x + c) - 40960*a^3/(e^(2*d*x + 2*c) -
1) + (35072*a^2*b*e^(10*d*x + 10*c) + 21920*a*b^2*e^(10*d*x + 10*c) + 5754*b^3*e^(10*d*x + 10*c) - 7680*a^2*b
*e^(8*d*x + 8*c) - 7200*a*b^2*e^(8*d*x + 8*c) - 2100*b^3*e^(8*d*x + 8*c) + 1440*a*b^2*e^(6*d*x + 6*c) + 600*b^
3*e^(6*d*x + 6*c) - 160*a*b^2*e^(4*d*x + 4*c) - 150*b^3*e^(4*d*x + 4*c) + 25*b^3*e^(2*d*x + 2*c) - 2*b^3)*e^(-
10*d*x - 10*c))/d

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maple [A]  time = 0.08, size = 163, normalized size = 0.90 \[ \frac {-a^{3} \coth \left (d x +c \right )+3 a^{2} b \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+3 a \,b^{2} \left (\left (\frac {\left (\sinh ^{5}\left (d x +c \right )\right )}{6}-\frac {5 \left (\sinh ^{3}\left (d x +c \right )\right )}{24}+\frac {5 \sinh \left (d x +c \right )}{16}\right ) \cosh \left (d x +c \right )-\frac {5 d x}{16}-\frac {5 c}{16}\right )+b^{3} \left (\left (\frac {\left (\sinh ^{9}\left (d x +c \right )\right )}{10}-\frac {9 \left (\sinh ^{7}\left (d x +c \right )\right )}{80}+\frac {21 \left (\sinh ^{5}\left (d x +c \right )\right )}{160}-\frac {21 \left (\sinh ^{3}\left (d x +c \right )\right )}{128}+\frac {63 \sinh \left (d x +c \right )}{256}\right ) \cosh \left (d x +c \right )-\frac {63 d x}{256}-\frac {63 c}{256}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-a^3*coth(d*x+c)+3*a^2*b*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+3*a*b^2*((1/6*sinh(d*x+c)^5-5/24*sin
h(d*x+c)^3+5/16*sinh(d*x+c))*cosh(d*x+c)-5/16*d*x-5/16*c)+b^3*((1/10*sinh(d*x+c)^9-9/80*sinh(d*x+c)^7+21/160*s
inh(d*x+c)^5-21/128*sinh(d*x+c)^3+63/256*sinh(d*x+c))*cosh(d*x+c)-63/256*d*x-63/256*c))

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maxima [A]  time = 0.33, size = 284, normalized size = 1.57 \[ -\frac {3}{8} \, a^{2} b {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{20480} \, b^{3} {\left (\frac {{\left (25 \, e^{\left (-2 \, d x - 2 \, c\right )} - 150 \, e^{\left (-4 \, d x - 4 \, c\right )} + 600 \, e^{\left (-6 \, d x - 6 \, c\right )} - 2100 \, e^{\left (-8 \, d x - 8 \, c\right )} - 2\right )} e^{\left (10 \, d x + 10 \, c\right )}}{d} + \frac {5040 \, {\left (d x + c\right )}}{d} + \frac {2100 \, e^{\left (-2 \, d x - 2 \, c\right )} - 600 \, e^{\left (-4 \, d x - 4 \, c\right )} + 150 \, e^{\left (-6 \, d x - 6 \, c\right )} - 25 \, e^{\left (-8 \, d x - 8 \, c\right )} + 2 \, e^{\left (-10 \, d x - 10 \, c\right )}}{d}\right )} - \frac {1}{128} \, a b^{2} {\left (\frac {{\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 45 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac {120 \, {\left (d x + c\right )}}{d} + \frac {45 \, e^{\left (-2 \, d x - 2 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} + \frac {2 \, a^{3}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2*(a+b*sinh(d*x+c)^4)^3,x, algorithm="maxima")

[Out]

-3/8*a^2*b*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/20480*b^3*((25*e^(-2*d*x - 2*c) - 150*e^(-4*d*x
- 4*c) + 600*e^(-6*d*x - 6*c) - 2100*e^(-8*d*x - 8*c) - 2)*e^(10*d*x + 10*c)/d + 5040*(d*x + c)/d + (2100*e^(-
2*d*x - 2*c) - 600*e^(-4*d*x - 4*c) + 150*e^(-6*d*x - 6*c) - 25*e^(-8*d*x - 8*c) + 2*e^(-10*d*x - 10*c))/d) -
1/128*a*b^2*((9*e^(-2*d*x - 2*c) - 45*e^(-4*d*x - 4*c) - 1)*e^(6*d*x + 6*c)/d + 120*(d*x + c)/d + (45*e^(-2*d*
x - 2*c) - 9*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/d) + 2*a^3/(d*(e^(-2*d*x - 2*c) - 1))

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mupad [B]  time = 1.09, size = 265, normalized size = 1.46 \[ \frac {5\,b^3\,{\mathrm {e}}^{-8\,c-8\,d\,x}}{4096\,d}-\frac {2\,a^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {5\,b^3\,{\mathrm {e}}^{8\,c+8\,d\,x}}{4096\,d}-\frac {b^3\,{\mathrm {e}}^{-10\,c-10\,d\,x}}{10240\,d}+\frac {b^3\,{\mathrm {e}}^{10\,c+10\,d\,x}}{10240\,d}-\frac {3\,b\,x\,\left (128\,a^2+80\,a\,b+21\,b^2\right )}{256}-\frac {3\,b\,{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (128\,a^2+120\,a\,b+35\,b^2\right )}{1024\,d}+\frac {3\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (128\,a^2+120\,a\,b+35\,b^2\right )}{1024\,d}+\frac {3\,b^2\,{\mathrm {e}}^{-4\,c-4\,d\,x}\,\left (12\,a+5\,b\right )}{512\,d}-\frac {3\,b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (12\,a+5\,b\right )}{512\,d}-\frac {b^2\,{\mathrm {e}}^{-6\,c-6\,d\,x}\,\left (16\,a+15\,b\right )}{2048\,d}+\frac {b^2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (16\,a+15\,b\right )}{2048\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*b^3*exp(- 8*c - 8*d*x))/(4096*d) - (2*a^3)/(d*(exp(2*c + 2*d*x) - 1)) - (5*b^3*exp(8*c + 8*d*x))/(4096*d) -
(b^3*exp(- 10*c - 10*d*x))/(10240*d) + (b^3*exp(10*c + 10*d*x))/(10240*d) - (3*b*x*(80*a*b + 128*a^2 + 21*b^2
))/256 - (3*b*exp(- 2*c - 2*d*x)*(120*a*b + 128*a^2 + 35*b^2))/(1024*d) + (3*b*exp(2*c + 2*d*x)*(120*a*b + 128
*a^2 + 35*b^2))/(1024*d) + (3*b^2*exp(- 4*c - 4*d*x)*(12*a + 5*b))/(512*d) - (3*b^2*exp(4*c + 4*d*x)*(12*a + 5
*b))/(512*d) - (b^2*exp(- 6*c - 6*d*x)*(16*a + 15*b))/(2048*d) + (b^2*exp(6*c + 6*d*x)*(16*a + 15*b))/(2048*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)**2*(a+b*sinh(d*x+c)**4)**3,x)

[Out]

Timed out

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