3.1.9 \(\int \sinh ^4(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\) [9]

Optimal. Leaf size=118 \[ \frac {1}{8} \left (3 a^2+30 a b+35 b^2\right ) x-\frac {(a+b) (a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}-\frac {\left (a^2+10 a b+13 b^2\right ) \tanh (c+d x)}{4 d}+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d} \]

[Out]

1/8*(3*a^2+30*a*b+35*b^2)*x-1/8*(a+b)*(a+9*b)*cosh(d*x+c)*sinh(d*x+c)/d-1/4*(a^2+10*a*b+13*b^2)*tanh(d*x+c)/d+
1/4*(a+b)^2*sinh(d*x+c)^4*tanh(d*x+c)/d-1/3*b^2*tanh(d*x+c)^3/d

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Rubi [A]
time = 0.10, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3744, 474, 466, 1167, 212} \begin {gather*} -\frac {\left (a^2+10 a b+13 b^2\right ) \tanh (c+d x)}{4 d}+\frac {1}{8} x \left (3 a^2+30 a b+35 b^2\right )+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {(a+b) (a+9 b) \sinh (c+d x) \cosh (c+d x)}{8 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2 + 30*a*b + 35*b^2)*x)/8 - ((a + b)*(a + 9*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) - ((a^2 + 10*a*b + 13*
b^2)*Tanh[c + d*x])/(4*d) + ((a + b)^2*Sinh[c + d*x]^4*Tanh[c + d*x])/(4*d) - (b^2*Tanh[c + d*x]^3)/(3*d)

Rule 212

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

Rule 466

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 474

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

Rule 1167

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, 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[p, 0] && IGtQ[q, -2]

Rule 3744

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff^(m + 1)/f), Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)
^(m/2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx &=\frac {\text {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^2}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {\text {Subst}\left (\int \frac {x^4 \left (a^2+10 a b+5 b^2+4 b^2 x^2\right )}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=-\frac {(a+b) (a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {\text {Subst}\left (\int \frac {-(a+b) (a+9 b)-2 (a+b) (a+9 b) x^2-8 b^2 x^4}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac {(a+b) (a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {\text {Subst}\left (\int \left (2 \left (a^2+10 a b+13 b^2\right )+8 b^2 x^2+\frac {-3 a^2-30 a b-35 b^2}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac {(a+b) (a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}-\frac {\left (a^2+10 a b+13 b^2\right ) \tanh (c+d x)}{4 d}+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d}+\frac {\left (3 a^2+30 a b+35 b^2\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac {1}{8} \left (3 a^2+30 a b+35 b^2\right ) x-\frac {(a+b) (a+9 b) \cosh (c+d x) \sinh (c+d x)}{8 d}-\frac {\left (a^2+10 a b+13 b^2\right ) \tanh (c+d x)}{4 d}+\frac {(a+b)^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac {b^2 \tanh ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 1.03, size = 94, normalized size = 0.80 \begin {gather*} \frac {12 \left (3 a^2+30 a b+35 b^2\right ) (c+d x)-24 \left (a^2+4 a b+3 b^2\right ) \sinh (2 (c+d x))+3 (a+b)^2 \sinh (4 (c+d x))+32 b \left (-6 a-10 b+b \text {sech}^2(c+d x)\right ) \tanh (c+d x)}{96 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(3*a^2 + 30*a*b + 35*b^2)*(c + d*x) - 24*(a^2 + 4*a*b + 3*b^2)*Sinh[2*(c + d*x)] + 3*(a + b)^2*Sinh[4*(c +
d*x)] + 32*b*(-6*a - 10*b + b*Sech[c + d*x]^2)*Tanh[c + d*x])/(96*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(292\) vs. \(2(108)=216\).
time = 2.04, size = 293, normalized size = 2.48

method	result	size
risch	\(\frac {3 a^{2} x}{8}+\frac {15 a b x}{4}+\frac {35 b^{2} x}{8}+\frac {{\mathrm e}^{4 d x +4 c} a^{2}}{64 d}+\frac {{\mathrm e}^{4 d x +4 c} a b}{32 d}+\frac {{\mathrm e}^{4 d x +4 c} b^{2}}{64 d}-\frac {{\mathrm e}^{2 d x +2 c} a^{2}}{8 d}-\frac {{\mathrm e}^{2 d x +2 c} a b}{2 d}-\frac {3 \,{\mathrm e}^{2 d x +2 c} b^{2}}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} a^{2}}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} a b}{2 d}+\frac {3 \,{\mathrm e}^{-2 d x -2 c} b^{2}}{8 d}-\frac {{\mathrm e}^{-4 d x -4 c} a^{2}}{64 d}-\frac {{\mathrm e}^{-4 d x -4 c} a b}{32 d}-\frac {{\mathrm e}^{-4 d x -4 c} b^{2}}{64 d}+\frac {4 b \left (3 a \,{\mathrm e}^{4 d x +4 c}+6 b \,{\mathrm e}^{4 d x +4 c}+6 a \,{\mathrm e}^{2 d x +2 c}+9 b \,{\mathrm e}^{2 d x +2 c}+3 a +5 b \right )}{3 d \left (1+{\mathrm e}^{2 d x +2 c}\right )^{3}}\)	\(293\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

3/8*a^2*x+15/4*a*b*x+35/8*b^2*x+1/64/d*exp(4*d*x+4*c)*a^2+1/32/d*exp(4*d*x+4*c)*a*b+1/64/d*exp(4*d*x+4*c)*b^2-
1/8/d*exp(2*d*x+2*c)*a^2-1/2/d*exp(2*d*x+2*c)*a*b-3/8/d*exp(2*d*x+2*c)*b^2+1/8/d*exp(-2*d*x-2*c)*a^2+1/2/d*exp
(-2*d*x-2*c)*a*b+3/8/d*exp(-2*d*x-2*c)*b^2-1/64/d*exp(-4*d*x-4*c)*a^2-1/32/d*exp(-4*d*x-4*c)*a*b-1/64/d*exp(-4
*d*x-4*c)*b^2+4/3*b*(3*a*exp(4*d*x+4*c)+6*b*exp(4*d*x+4*c)+6*a*exp(2*d*x+2*c)+9*b*exp(2*d*x+2*c)+3*a+5*b)/d/(1
+exp(2*d*x+2*c))^3

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (108) = 216\).
time = 0.28, size = 295, normalized size = 2.50 \begin {gather*} \frac {1}{64} \, a^{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 )} + \frac {1}{192} \, b^{2} {\left (\frac {840 \, {\left (d x + c\right )}}{d} + \frac {3 \, {\left (24 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}\right )}}{d} - \frac {63 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1487 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2517 \, e^{\left (-6 \, d x - 6 \, c\right )} + 1608 \, e^{\left (-8 \, d x - 8 \, c\right )} - 3}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )}\right )}}\right )} + \frac {1}{32} \, a b {\left (\frac {120 \, {\left (d x + c\right )}}{d} + \frac {16 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} - \frac {15 \, e^{\left (-2 \, d x - 2 \, c\right )} + 144 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*a^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) + 1/192*
b^2*(840*(d*x + c)/d + 3*(24*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c))/d - (63*e^(-2*d*x - 2*c) + 1487*e^(-4*d*x -
4*c) + 2517*e^(-6*d*x - 6*c) + 1608*e^(-8*d*x - 8*c) - 3)/(d*(e^(-4*d*x - 4*c) + 3*e^(-6*d*x - 6*c) + 3*e^(-8*
d*x - 8*c) + e^(-10*d*x - 10*c)))) + 1/32*a*b*(120*(d*x + c)/d + (16*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c))/d -
(15*e^(-2*d*x - 2*c) + 144*e^(-4*d*x - 4*c) - 1)/(d*(e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (108) = 216\).
time = 0.33, size = 394, normalized size = 3.34 \begin {gather*} \frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{7} + 3 \, {\left (21 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} - 5 \, a^{2} - 26 \, a b - 21 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} + 8 \, {\left (3 \, {\left (3 \, a^{2} + 30 \, a b + 35 \, b^{2}\right )} d x + 48 \, a b + 80 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 24 \, {\left (3 \, {\left (3 \, a^{2} + 30 \, a b + 35 \, b^{2}\right )} d x + 48 \, a b + 80 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (105 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} - 30 \, {\left (5 \, a^{2} + 26 \, a b + 21 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} - 63 \, a^{2} - 654 \, a b - 847 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 24 \, {\left (3 \, {\left (3 \, a^{2} + 30 \, a b + 35 \, b^{2}\right )} d x + 48 \, a b + 80 \, b^{2}\right )} \cosh \left (d x + c\right ) + 3 \, {\left (7 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{6} - 5 \, {\left (5 \, a^{2} + 26 \, a b + 21 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} - {\left (63 \, a^{2} + 654 \, a b + 847 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} - 15 \, a^{2} - 190 \, a b - 175 \, b^{2}\right )} \sinh \left (d x + c\right )}{192 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*(a^2 + 2*a*b + b^2)*sinh(d*x + c)^7 + 3*(21*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 - 5*a^2 - 26*a*b - 21
*b^2)*sinh(d*x + c)^5 + 8*(3*(3*a^2 + 30*a*b + 35*b^2)*d*x + 48*a*b + 80*b^2)*cosh(d*x + c)^3 + 24*(3*(3*a^2 +
30*a*b + 35*b^2)*d*x + 48*a*b + 80*b^2)*cosh(d*x + c)*sinh(d*x + c)^2 + (105*(a^2 + 2*a*b + b^2)*cosh(d*x + c
)^4 - 30*(5*a^2 + 26*a*b + 21*b^2)*cosh(d*x + c)^2 - 63*a^2 - 654*a*b - 847*b^2)*sinh(d*x + c)^3 + 24*(3*(3*a^
2 + 30*a*b + 35*b^2)*d*x + 48*a*b + 80*b^2)*cosh(d*x + c) + 3*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 - 5*(5*a^
2 + 26*a*b + 21*b^2)*cosh(d*x + c)^4 - (63*a^2 + 654*a*b + 847*b^2)*cosh(d*x + c)^2 - 15*a^2 - 190*a*b - 175*b
^2)*sinh(d*x + c))/(d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)*sinh(d*x + c)^2 + 3*d*cosh(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \sinh ^{4}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**4*(a+b*tanh(d*x+c)**2)**2,x)

[Out]

Integral((a + b*tanh(c + d*x)**2)**2*sinh(c + d*x)**4, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (108) = 216\).
time = 0.50, size = 293, normalized size = 2.48 \begin {gather*} \frac {3 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 24 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 96 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 72 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, {\left (3 \, a^{2} + 30 \, a b + 35 \, b^{2}\right )} {\left (d x + c\right )} - 3 \, {\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 210 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 32 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 24 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + \frac {256 \, {\left (3 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 9 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a b + 5 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{192 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*a^2*e^(4*d*x + 4*c) + 6*a*b*e^(4*d*x + 4*c) + 3*b^2*e^(4*d*x + 4*c) - 24*a^2*e^(2*d*x + 2*c) - 96*a*b
*e^(2*d*x + 2*c) - 72*b^2*e^(2*d*x + 2*c) + 24*(3*a^2 + 30*a*b + 35*b^2)*(d*x + c) - 3*(18*a^2*e^(4*d*x + 4*c)
+ 180*a*b*e^(4*d*x + 4*c) + 210*b^2*e^(4*d*x + 4*c) - 8*a^2*e^(2*d*x + 2*c) - 32*a*b*e^(2*d*x + 2*c) - 24*b^2
*e^(2*d*x + 2*c) + a^2 + 2*a*b + b^2)*e^(-4*d*x - 4*c) + 256*(3*a*b*e^(4*d*x + 4*c) + 6*b^2*e^(4*d*x + 4*c) +
6*a*b*e^(2*d*x + 2*c) + 9*b^2*e^(2*d*x + 2*c) + 3*a*b + 5*b^2)/(e^(2*d*x + 2*c) + 1)^3)/d

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Mupad [B]
time = 0.31, size = 293, normalized size = 2.48 \begin {gather*} \frac {\frac {4\,\left (b^2+a\,b\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,b^2+a\,b\right )}{3\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+x\,\left (\frac {3\,a^2}{8}+\frac {15\,a\,b}{4}+\frac {35\,b^2}{8}\right )+\frac {\frac {4\,\left (2\,b^2+a\,b\right )}{3\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+a\,b\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,b^2+a\,b\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {4\,\left (2\,b^2+a\,b\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a^2+4\,a\,b+3\,b^2\right )}{8\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2+4\,a\,b+3\,b^2\right )}{8\,d}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,{\left (a+b\right )}^2}{64\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,{\left (a+b\right )}^2}{64\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((4*(a*b + b^2))/(3*d) + (4*exp(2*c + 2*d*x)*(a*b + 2*b^2))/(3*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)
+ x*((15*a*b)/4 + (3*a^2)/8 + (35*b^2)/8) + ((4*(a*b + 2*b^2))/(3*d) + (8*exp(2*c + 2*d*x)*(a*b + b^2))/(3*d)
+ (4*exp(4*c + 4*d*x)*(a*b + 2*b^2))/(3*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)
+ (4*(a*b + 2*b^2))/(3*d*(exp(2*c + 2*d*x) + 1)) + (exp(- 2*c - 2*d*x)*(4*a*b + a^2 + 3*b^2))/(8*d) - (exp(2*c
+ 2*d*x)*(4*a*b + a^2 + 3*b^2))/(8*d) - (exp(- 4*c - 4*d*x)*(a + b)^2)/(64*d) + (exp(4*c + 4*d*x)*(a + b)^2)/
(64*d)

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