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

Optimal. Leaf size=182 \[ \frac {b \left (6 a^2-23 a b-8 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac {3 a \left (a^2-12 a b+8 b^2\right ) \tanh (c+d x)}{8 d}+\frac {3}{8} a x \left (a^2-12 a b+8 b^2\right )-\frac {3 b^2 (5 a-16 b) \tanh ^5(c+d x)}{40 d}-\frac {3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )^2}{8 d}+\frac {\sinh ^3(c+d x) \cosh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )^3}{4 d} \]

[Out]

3/8*a*(a^2-12*a*b+8*b^2)*x-3/8*a*(a^2-12*a*b+8*b^2)*tanh(d*x+c)/d+1/8*b*(6*a^2-23*a*b-8*b^2)*tanh(d*x+c)^3/d-3
/40*(5*a-16*b)*b^2*tanh(d*x+c)^5/d-3/8*(a-2*b)*sinh(d*x+c)^2*tanh(d*x+c)*(a+b-b*tanh(d*x+c)^2)^2/d+1/4*cosh(d*
x+c)*sinh(d*x+c)^3*(a+b-b*tanh(d*x+c)^2)^3/d

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Rubi [A]  time = 0.23, antiderivative size = 182, 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 = {4132, 467, 577, 570, 206} \[ \frac {b \left (6 a^2-23 a b-8 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac {3 a \left (a^2-12 a b+8 b^2\right ) \tanh (c+d x)}{8 d}+\frac {3}{8} a x \left (a^2-12 a b+8 b^2\right )-\frac {3 b^2 (5 a-16 b) \tanh ^5(c+d x)}{40 d}-\frac {3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )^2}{8 d}+\frac {\sinh ^3(c+d x) \cosh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(a^2 - 12*a*b + 8*b^2)*x)/8 - (3*a*(a^2 - 12*a*b + 8*b^2)*Tanh[c + d*x])/(8*d) + (b*(6*a^2 - 23*a*b - 8*b
^2)*Tanh[c + d*x]^3)/(8*d) - (3*(5*a - 16*b)*b^2*Tanh[c + d*x]^5)/(40*d) - (3*(a - 2*b)*Sinh[c + d*x]^2*Tanh[c
+ d*x]*(a + b - b*Tanh[c + d*x]^2)^2)/(8*d) + (Cosh[c + d*x]*Sinh[c + d*x]^3*(a + b - b*Tanh[c + d*x]^2)^3)/(
4*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 467

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

Rule 570

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

Rule 577

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*b*g*n*(p + 1)), x] + Dist[
1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(m
+ 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] &&
IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rule 4132

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

Rubi steps

\begin {align*} \int \left (a+b \text {sech}^2(c+d x)\right )^3 \sinh ^4(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (a+b-b x^2\right )^3}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\cosh (c+d x) \sinh ^3(c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 (a+b)-9 b x^2\right ) \left (a+b-b x^2\right )^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=-\frac {3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{8 d}+\frac {\cosh (c+d x) \sinh ^3(c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (a+b-b x^2\right ) \left (-3 (a-8 b) (a+b)+3 (5 a-16 b) b x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac {3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{8 d}+\frac {\cosh (c+d x) \sinh ^3(c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{4 d}-\frac {\operatorname {Subst}\left (\int \left (3 a \left (a^2-12 a b+8 b^2\right )-3 b \left (6 a^2-23 a b-8 b^2\right ) x^2+3 (5 a-16 b) b^2 x^4-\frac {3 \left (a^3-12 a^2 b+8 a b^2\right )}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac {3 a \left (a^2-12 a b+8 b^2\right ) \tanh (c+d x)}{8 d}+\frac {b \left (6 a^2-23 a b-8 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac {3 (5 a-16 b) b^2 \tanh ^5(c+d x)}{40 d}-\frac {3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{8 d}+\frac {\cosh (c+d x) \sinh ^3(c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{4 d}+\frac {\left (3 a \left (a^2-12 a b+8 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac {3}{8} a \left (a^2-12 a b+8 b^2\right ) x-\frac {3 a \left (a^2-12 a b+8 b^2\right ) \tanh (c+d x)}{8 d}+\frac {b \left (6 a^2-23 a b-8 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac {3 (5 a-16 b) b^2 \tanh ^5(c+d x)}{40 d}-\frac {3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{8 d}+\frac {\cosh (c+d x) \sinh ^3(c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{4 d}\\ \end {align*}

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Mathematica [B]  time = 2.51, size = 651, normalized size = 3.58 \[ \frac {\text {sech}(c) \text {sech}^5(c+d x) \left (a \cosh ^2(c+d x)+b\right )^3 \left (-180 a^3 \sinh (2 c+d x)-310 a^3 \sinh (2 c+3 d x)-310 a^3 \sinh (4 c+3 d x)-150 a^3 \sinh (4 c+5 d x)-150 a^3 \sinh (6 c+5 d x)-15 a^3 \sinh (6 c+7 d x)-15 a^3 \sinh (8 c+7 d x)+5 a^3 \sinh (8 c+9 d x)+5 a^3 \sinh (10 c+9 d x)+600 a^3 d x \cosh (2 c+3 d x)+600 a^3 d x \cosh (4 c+3 d x)+120 a^3 d x \cosh (4 c+5 d x)+120 a^3 d x \cosh (6 c+5 d x)-180 a^3 \sinh (d x)+1200 a d x \left (a^2-12 a b+8 b^2\right ) \cosh (2 c+d x)+1200 a d x \left (a^2-12 a b+8 b^2\right ) \cosh (d x)-7080 a^2 b \sinh (2 c+d x)+8760 a^2 b \sinh (2 c+3 d x)-840 a^2 b \sinh (4 c+3 d x)+2520 a^2 b \sinh (4 c+5 d x)+600 a^2 b \sinh (6 c+5 d x)+120 a^2 b \sinh (6 c+7 d x)+120 a^2 b \sinh (8 c+7 d x)-7200 a^2 b d x \cosh (2 c+3 d x)-7200 a^2 b d x \cosh (4 c+3 d x)-1440 a^2 b d x \cosh (4 c+5 d x)-1440 a^2 b d x \cosh (6 c+5 d x)+12120 a^2 b \sinh (d x)+11520 a b^2 \sinh (2 c+d x)-8960 a b^2 \sinh (2 c+3 d x)+3840 a b^2 \sinh (4 c+3 d x)-2560 a b^2 \sinh (4 c+5 d x)+4800 a b^2 d x \cosh (2 c+3 d x)+4800 a b^2 d x \cosh (4 c+3 d x)+960 a b^2 d x \cosh (4 c+5 d x)+960 a b^2 d x \cosh (6 c+5 d x)-14080 a b^2 \sinh (d x)-640 b^3 \sinh (4 c+3 d x)+128 b^3 \sinh (4 c+5 d x)+1280 b^3 \sinh (d x)\right )}{1280 d (a \cosh (2 (c+d x))+a+2 b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cosh[c + d*x]^2)^3*Sech[c]*Sech[c + d*x]^5*(1200*a*(a^2 - 12*a*b + 8*b^2)*d*x*Cosh[d*x] + 1200*a*(a^2
- 12*a*b + 8*b^2)*d*x*Cosh[2*c + d*x] + 600*a^3*d*x*Cosh[2*c + 3*d*x] - 7200*a^2*b*d*x*Cosh[2*c + 3*d*x] + 480
0*a*b^2*d*x*Cosh[2*c + 3*d*x] + 600*a^3*d*x*Cosh[4*c + 3*d*x] - 7200*a^2*b*d*x*Cosh[4*c + 3*d*x] + 4800*a*b^2*
d*x*Cosh[4*c + 3*d*x] + 120*a^3*d*x*Cosh[4*c + 5*d*x] - 1440*a^2*b*d*x*Cosh[4*c + 5*d*x] + 960*a*b^2*d*x*Cosh[
4*c + 5*d*x] + 120*a^3*d*x*Cosh[6*c + 5*d*x] - 1440*a^2*b*d*x*Cosh[6*c + 5*d*x] + 960*a*b^2*d*x*Cosh[6*c + 5*d
*x] - 180*a^3*Sinh[d*x] + 12120*a^2*b*Sinh[d*x] - 14080*a*b^2*Sinh[d*x] + 1280*b^3*Sinh[d*x] - 180*a^3*Sinh[2*
c + d*x] - 7080*a^2*b*Sinh[2*c + d*x] + 11520*a*b^2*Sinh[2*c + d*x] - 310*a^3*Sinh[2*c + 3*d*x] + 8760*a^2*b*S
inh[2*c + 3*d*x] - 8960*a*b^2*Sinh[2*c + 3*d*x] - 310*a^3*Sinh[4*c + 3*d*x] - 840*a^2*b*Sinh[4*c + 3*d*x] + 38
40*a*b^2*Sinh[4*c + 3*d*x] - 640*b^3*Sinh[4*c + 3*d*x] - 150*a^3*Sinh[4*c + 5*d*x] + 2520*a^2*b*Sinh[4*c + 5*d
*x] - 2560*a*b^2*Sinh[4*c + 5*d*x] + 128*b^3*Sinh[4*c + 5*d*x] - 150*a^3*Sinh[6*c + 5*d*x] + 600*a^2*b*Sinh[6*
c + 5*d*x] - 15*a^3*Sinh[6*c + 7*d*x] + 120*a^2*b*Sinh[6*c + 7*d*x] - 15*a^3*Sinh[8*c + 7*d*x] + 120*a^2*b*Sin
h[8*c + 7*d*x] + 5*a^3*Sinh[8*c + 9*d*x] + 5*a^3*Sinh[10*c + 9*d*x]))/(1280*d*(a + 2*b + a*Cosh[2*(c + d*x)])^
3)

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fricas [B]  time = 0.45, size = 727, normalized size = 3.99 \[ \frac {5 \, a^{3} \sinh \left (d x + c\right )^{9} + 15 \, {\left (12 \, a^{3} \cosh \left (d x + c\right )^{2} - a^{3} + 8 \, a^{2} b\right )} \sinh \left (d x + c\right )^{7} - 8 \, {\left (120 \, a^{2} b - 160 \, a b^{2} + 8 \, b^{3} - 15 \, {\left (a^{3} - 12 \, a^{2} b + 8 \, a b^{2}\right )} d x\right )} \cosh \left (d x + c\right )^{5} - 40 \, {\left (120 \, a^{2} b - 160 \, a b^{2} + 8 \, b^{3} - 15 \, {\left (a^{3} - 12 \, a^{2} b + 8 \, a b^{2}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + {\left (630 \, a^{3} \cosh \left (d x + c\right )^{4} - 150 \, a^{3} + 1560 \, a^{2} b - 1280 \, a b^{2} + 64 \, b^{3} - 315 \, {\left (a^{3} - 8 \, a^{2} b\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{5} - 40 \, {\left (120 \, a^{2} b - 160 \, a b^{2} + 8 \, b^{3} - 15 \, {\left (a^{3} - 12 \, a^{2} b + 8 \, a b^{2}\right )} d x\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (84 \, a^{3} \cosh \left (d x + c\right )^{6} - 105 \, {\left (a^{3} - 8 \, a^{2} b\right )} \cosh \left (d x + c\right )^{4} - 62 \, a^{3} + 792 \, a^{2} b - 512 \, a b^{2} - 64 \, b^{3} - 4 \, {\left (75 \, a^{3} - 780 \, a^{2} b + 640 \, a b^{2} - 32 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} - 40 \, {\left (2 \, {\left (120 \, a^{2} b - 160 \, a b^{2} + 8 \, b^{3} - 15 \, {\left (a^{3} - 12 \, a^{2} b + 8 \, a b^{2}\right )} d x\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (120 \, a^{2} b - 160 \, a b^{2} + 8 \, b^{3} - 15 \, {\left (a^{3} - 12 \, a^{2} b + 8 \, a b^{2}\right )} d x\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 80 \, {\left (120 \, a^{2} b - 160 \, a b^{2} + 8 \, b^{3} - 15 \, {\left (a^{3} - 12 \, a^{2} b + 8 \, a b^{2}\right )} d x\right )} \cosh \left (d x + c\right ) + 5 \, {\left (9 \, a^{3} \cosh \left (d x + c\right )^{8} - 21 \, {\left (a^{3} - 8 \, a^{2} b\right )} \cosh \left (d x + c\right )^{6} - 2 \, {\left (75 \, a^{3} - 780 \, a^{2} b + 640 \, a b^{2} - 32 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} - 36 \, a^{3} + 504 \, a^{2} b - 256 \, a b^{2} + 128 \, b^{3} - 6 \, {\left (31 \, a^{3} - 396 \, a^{2} b + 256 \, a b^{2} + 32 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{320 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*(5*a^3*sinh(d*x + c)^9 + 15*(12*a^3*cosh(d*x + c)^2 - a^3 + 8*a^2*b)*sinh(d*x + c)^7 - 8*(120*a^2*b - 16
0*a*b^2 + 8*b^3 - 15*(a^3 - 12*a^2*b + 8*a*b^2)*d*x)*cosh(d*x + c)^5 - 40*(120*a^2*b - 160*a*b^2 + 8*b^3 - 15*
(a^3 - 12*a^2*b + 8*a*b^2)*d*x)*cosh(d*x + c)*sinh(d*x + c)^4 + (630*a^3*cosh(d*x + c)^4 - 150*a^3 + 1560*a^2*
b - 1280*a*b^2 + 64*b^3 - 315*(a^3 - 8*a^2*b)*cosh(d*x + c)^2)*sinh(d*x + c)^5 - 40*(120*a^2*b - 160*a*b^2 + 8
*b^3 - 15*(a^3 - 12*a^2*b + 8*a*b^2)*d*x)*cosh(d*x + c)^3 + 5*(84*a^3*cosh(d*x + c)^6 - 105*(a^3 - 8*a^2*b)*co
sh(d*x + c)^4 - 62*a^3 + 792*a^2*b - 512*a*b^2 - 64*b^3 - 4*(75*a^3 - 780*a^2*b + 640*a*b^2 - 32*b^3)*cosh(d*x
+ c)^2)*sinh(d*x + c)^3 - 40*(2*(120*a^2*b - 160*a*b^2 + 8*b^3 - 15*(a^3 - 12*a^2*b + 8*a*b^2)*d*x)*cosh(d*x
+ c)^3 + 3*(120*a^2*b - 160*a*b^2 + 8*b^3 - 15*(a^3 - 12*a^2*b + 8*a*b^2)*d*x)*cosh(d*x + c))*sinh(d*x + c)^2
- 80*(120*a^2*b - 160*a*b^2 + 8*b^3 - 15*(a^3 - 12*a^2*b + 8*a*b^2)*d*x)*cosh(d*x + c) + 5*(9*a^3*cosh(d*x + c
)^8 - 21*(a^3 - 8*a^2*b)*cosh(d*x + c)^6 - 2*(75*a^3 - 780*a^2*b + 640*a*b^2 - 32*b^3)*cosh(d*x + c)^4 - 36*a^
3 + 504*a^2*b - 256*a*b^2 + 128*b^3 - 6*(31*a^3 - 396*a^2*b + 256*a*b^2 + 32*b^3)*cosh(d*x + c)^2)*sinh(d*x +
c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*
d*cosh(d*x + c))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))

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giac [A]  time = 0.22, size = 339, normalized size = 1.86 \[ \frac {5 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - 40 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 120 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 120 \, {\left (a^{3} - 12 \, a^{2} b + 8 \, a b^{2}\right )} {\left (d x + c\right )} - 5 \, {\left (18 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - 216 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 144 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + a^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - \frac {128 \, {\left (15 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} - 30 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 5 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 90 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 110 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 70 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} b - 20 \, a b^{2} + b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{320 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*(5*a^3*e^(4*d*x + 4*c) - 40*a^3*e^(2*d*x + 2*c) + 120*a^2*b*e^(2*d*x + 2*c) + 120*(a^3 - 12*a^2*b + 8*a*
b^2)*(d*x + c) - 5*(18*a^3*e^(4*d*x + 4*c) - 216*a^2*b*e^(4*d*x + 4*c) + 144*a*b^2*e^(4*d*x + 4*c) - 8*a^3*e^(
2*d*x + 2*c) + 24*a^2*b*e^(2*d*x + 2*c) + a^3)*e^(-4*d*x - 4*c) - 128*(15*a^2*b*e^(8*d*x + 8*c) - 30*a*b^2*e^(
8*d*x + 8*c) + 5*b^3*e^(8*d*x + 8*c) + 60*a^2*b*e^(6*d*x + 6*c) - 90*a*b^2*e^(6*d*x + 6*c) + 90*a^2*b*e^(4*d*x
+ 4*c) - 110*a*b^2*e^(4*d*x + 4*c) + 10*b^3*e^(4*d*x + 4*c) + 60*a^2*b*e^(2*d*x + 2*c) - 70*a*b^2*e^(2*d*x +
2*c) + 15*a^2*b - 20*a*b^2 + b^3)/(e^(2*d*x + 2*c) + 1)^5)/d

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maple [A]  time = 0.44, size = 182, normalized size = 1.00 \[ \frac {a^{3} \left (\left (\frac {\left (\sinh ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )+3 a \,b^{2} \left (d x +c -\tanh \left (d x +c \right )-\frac {\left (\tanh ^{3}\left (d x +c \right )\right )}{3}\right )+b^{3} \left (-\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\mathrm {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*((1/4*sinh(d*x+c)^3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8*c)+3*a^2*b*(1/2*sinh(d*x+c)^3/cosh(d*x+c
)-3/2*d*x-3/2*c+3/2*tanh(d*x+c))+3*a*b^2*(d*x+c-tanh(d*x+c)-1/3*tanh(d*x+c)^3)+b^3*(-1/2*sinh(d*x+c)^3/cosh(d*
x+c)^5-3/8*sinh(d*x+c)/cosh(d*x+c)^5+3/8*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)^2)*tanh(d*x+c)))

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maxima [B]  time = 0.34, size = 422, normalized size = 2.32 \[ \frac {1}{64} \, a^{3} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + a b^{2} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} - \frac {3}{8} \, a^{2} b {\left (\frac {12 \, {\left (d x + c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} + \frac {2}{5} \, b^{3} {\left (\frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {5 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*a^3*(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) + a*b^2*
(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 2)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e
^(-6*d*x - 6*c) + 1))) - 3/8*a^2*b*(12*(d*x + c)/d + e^(-2*d*x - 2*c)/d - (17*e^(-2*d*x - 2*c) + 1)/(d*(e^(-2*
d*x - 2*c) + e^(-4*d*x - 4*c)))) + 2/5*b^3*(10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) +
10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 5*e^(-8*d*x - 8*c)/(d*(5*e^(-2*d*x - 2*
c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2
*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)))

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mupad [B]  time = 0.37, size = 686, normalized size = 3.77 \[ \frac {\frac {2\,\left (-3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b-2\,a\,b^2+b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (-3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2\,b-6\,a\,b^2+b^3\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2\,b-6\,a\,b^2+b^3\right )}{5\,d}-\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (-3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}+\frac {12\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b-2\,a\,b^2+b^3\right )}{5\,d}-\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (-3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (3\,a^2\,b-6\,a\,b^2+b^3\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}+\frac {\frac {2\,\left (-3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b-6\,a\,b^2+b^3\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2\,b-2\,a\,b^2+b^3\right )}{5\,d}-\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (-3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b-6\,a\,b^2+b^3\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {3\,a\,x\,\left (a^2-12\,a\,b+8\,b^2\right )}{8}-\frac {a^3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}+\frac {a^3\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}-\frac {2\,\left (3\,a^2\,b-6\,a\,b^2+b^3\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {a^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a-3\,b\right )}{8\,d}-\frac {a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a-3\,b\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*(3*a*b^2 - 3*a^2*b + b^3))/(5*d) - (6*exp(2*c + 2*d*x)*(3*a^2*b - 2*a*b^2 + b^3))/(5*d) + (6*exp(4*c + 4*d
*x)*(3*a*b^2 - 3*a^2*b + b^3))/(5*d) - (2*exp(6*c + 6*d*x)*(3*a^2*b - 6*a*b^2 + b^3))/(5*d))/(4*exp(2*c + 2*d*
x) + 6*exp(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) - ((2*(3*a^2*b - 6*a*b^2 + b^3))/(5*d) -
(8*exp(2*c + 2*d*x)*(3*a*b^2 - 3*a^2*b + b^3))/(5*d) + (12*exp(4*c + 4*d*x)*(3*a^2*b - 2*a*b^2 + b^3))/(5*d) -
(8*exp(6*c + 6*d*x)*(3*a*b^2 - 3*a^2*b + b^3))/(5*d) + (2*exp(8*c + 8*d*x)*(3*a^2*b - 6*a*b^2 + b^3))/(5*d))/
(5*exp(2*c + 2*d*x) + 10*exp(4*c + 4*d*x) + 10*exp(6*c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1)
+ ((2*(3*a*b^2 - 3*a^2*b + b^3))/(5*d) - (2*exp(2*c + 2*d*x)*(3*a^2*b - 6*a*b^2 + b^3))/(5*d))/(2*exp(2*c + 2
*d*x) + exp(4*c + 4*d*x) + 1) - ((2*(3*a^2*b - 2*a*b^2 + b^3))/(5*d) - (4*exp(2*c + 2*d*x)*(3*a*b^2 - 3*a^2*b
+ b^3))/(5*d) + (2*exp(4*c + 4*d*x)*(3*a^2*b - 6*a*b^2 + b^3))/(5*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x)
+ exp(6*c + 6*d*x) + 1) + (3*a*x*(a^2 - 12*a*b + 8*b^2))/8 - (a^3*exp(- 4*c - 4*d*x))/(64*d) + (a^3*exp(4*c +
4*d*x))/(64*d) - (2*(3*a^2*b - 6*a*b^2 + b^3))/(5*d*(exp(2*c + 2*d*x) + 1)) + (a^2*exp(- 2*c - 2*d*x)*(a - 3*
b))/(8*d) - (a^2*exp(2*c + 2*d*x)*(a - 3*b))/(8*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((a+b*sech(d*x+c)**2)**3*sinh(d*x+c)**4,x)

[Out]

Timed out

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