3.19 \(\int \sinh ^2(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=122 \[ \frac {b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac {3 b (a+b)^2 \tanh (c+d x)}{d}+\frac {(a+b)^3}{4 d (1-\tanh (c+d x))}-\frac {(a+b)^3}{4 d (\tanh (c+d x)+1)}-\frac {1}{2} x (a+b)^2 (a+7 b)+\frac {b^3 \tanh ^5(c+d x)}{5 d} \]

[Out]

-1/2*(a+b)^2*(a+7*b)*x+1/4*(a+b)^3/d/(1-tanh(d*x+c))+3*b*(a+b)^2*tanh(d*x+c)/d+1/3*b^2*(3*a+2*b)*tanh(d*x+c)^3
/d+1/5*b^3*tanh(d*x+c)^5/d-1/4*(a+b)^3/d/(1+tanh(d*x+c))

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Rubi [A]  time = 0.19, antiderivative size = 139, normalized size of antiderivative = 1.14, 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 = {3663, 467, 528, 388, 206} \[ \frac {b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac {7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac {\sinh (c+d x) \cosh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac {1}{2} x (a+b)^2 (a+7 b) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b)^2*(a + 7*b)*x)/2 + (b*(81*a^2 + 190*a*b + 105*b^2)*Tanh[c + d*x])/(30*d) + (b*(33*a + 35*b)*Tanh[c +
d*x]*(a + b*Tanh[c + d*x]^2))/(30*d) + (7*b*Tanh[c + d*x]*(a + b*Tanh[c + d*x]^2)^2)/(10*d) + (Cosh[c + d*x]*
Sinh[c + d*x]*(a + b*Tanh[c + d*x]^2)^3)/(2*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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 528

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

Rule 3663

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 ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (a+b x^2\right )^3}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^2 \left (a+7 b x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac {7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}+\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right ) \left (-a (5 a+7 b)-b (33 a+35 b) x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{10 d}\\ &=\frac {b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac {7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {a \left (15 a^2+54 a b+35 b^2\right )+b \left (81 a^2+190 a b+105 b^2\right ) x^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{30 d}\\ &=\frac {b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac {b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac {7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac {\left ((a+b)^2 (a+7 b)\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=-\frac {1}{2} (a+b)^2 (a+7 b) x+\frac {b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac {b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac {7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}\\ \end {align*}

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Mathematica [A]  time = 2.20, size = 95, normalized size = 0.78 \[ \frac {4 b \tanh (c+d x) \left (45 a^2-b (15 a+16 b) \text {sech}^2(c+d x)+105 a b+3 b^2 \text {sech}^4(c+d x)+58 b^2\right )-30 (a+7 b) (a+b)^2 (c+d x)+15 (a+b)^3 \sinh (2 (c+d x))}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-30*(a + b)^2*(a + 7*b)*(c + d*x) + 15*(a + b)^3*Sinh[2*(c + d*x)] + 4*b*(45*a^2 + 105*a*b + 58*b^2 - b*(15*a
+ 16*b)*Sech[c + d*x]^2 + 3*b^2*Sech[c + d*x]^4)*Tanh[c + d*x])/(60*d)

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fricas [B]  time = 0.87, size = 725, normalized size = 5.94 \[ \frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sinh \left (d x + c\right )^{7} - 4 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{5} - 20 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + {\left (75 \, a^{3} + 585 \, a^{2} b + 1065 \, a b^{2} + 539 \, b^{3} + 315 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{5} - 20 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (105 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{4} + 27 \, a^{3} + 297 \, a^{2} b + 489 \, a b^{2} + 203 \, b^{3} + 2 \, {\left (75 \, a^{3} + 585 \, a^{2} b + 1065 \, a b^{2} + 539 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} - 20 \, {\left (2 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 40 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) + 5 \, {\left (21 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{6} + {\left (75 \, a^{3} + 585 \, a^{2} b + 1065 \, a b^{2} + 539 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 15 \, a^{3} + 189 \, a^{2} b + 285 \, a b^{2} + 175 \, b^{3} + 3 \, {\left (27 \, a^{3} + 297 \, a^{2} b + 489 \, a b^{2} + 203 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{120 \, {\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(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/120*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^7 - 4*(90*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^
2*b + 15*a*b^2 + 7*b^3)*d*x)*cosh(d*x + c)^5 - 20*(90*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^2*b + 15*a*b
^2 + 7*b^3)*d*x)*cosh(d*x + c)*sinh(d*x + c)^4 + (75*a^3 + 585*a^2*b + 1065*a*b^2 + 539*b^3 + 315*(a^3 + 3*a^2
*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 - 20*(90*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^2*b
+ 15*a*b^2 + 7*b^3)*d*x)*cosh(d*x + c)^3 + 5*(105*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 27*a^3 + 2
97*a^2*b + 489*a*b^2 + 203*b^3 + 2*(75*a^3 + 585*a^2*b + 1065*a*b^2 + 539*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^
3 - 20*(2*(90*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^2*b + 15*a*b^2 + 7*b^3)*d*x)*cosh(d*x + c)^3 + 3*(90
*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^2*b + 15*a*b^2 + 7*b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c)^2 - 40*
(90*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^2*b + 15*a*b^2 + 7*b^3)*d*x)*cosh(d*x + c) + 5*(21*(a^3 + 3*a^
2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^6 + (75*a^3 + 585*a^2*b + 1065*a*b^2 + 539*b^3)*cosh(d*x + c)^4 + 15*a^3 +
189*a^2*b + 285*a*b^2 + 175*b^3 + 3*(27*a^3 + 297*a^2*b + 489*a*b^2 + 203*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*c
osh(d*x + c))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))

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giac [B]  time = 0.64, size = 395, normalized size = 3.24 \[ -\frac {60 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x - 15 \, {\left (2 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 18 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 30 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 14 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - 15 \, {\left (a^{3} e^{\left (2 \, d x + 16 \, c\right )} + 3 \, a^{2} b e^{\left (2 \, d x + 16 \, c\right )} + 3 \, a b^{2} e^{\left (2 \, d x + 16 \, c\right )} + b^{3} e^{\left (2 \, d x + 16 \, c\right )}\right )} e^{\left (-14 \, c\right )} + \frac {16 \, {\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 135 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 90 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 450 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 240 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 600 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 340 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 390 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 200 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b + 105 \, a b^{2} + 58 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(60*(a^3 + 9*a^2*b + 15*a*b^2 + 7*b^3)*d*x - 15*(2*a^3*e^(2*d*x + 2*c) + 18*a^2*b*e^(2*d*x + 2*c) + 30*
a*b^2*e^(2*d*x + 2*c) + 14*b^3*e^(2*d*x + 2*c) - a^3 - 3*a^2*b - 3*a*b^2 - b^3)*e^(-2*d*x - 2*c) - 15*(a^3*e^(
2*d*x + 16*c) + 3*a^2*b*e^(2*d*x + 16*c) + 3*a*b^2*e^(2*d*x + 16*c) + b^3*e^(2*d*x + 16*c))*e^(-14*c) + 16*(45
*a^2*b*e^(8*d*x + 8*c) + 135*a*b^2*e^(8*d*x + 8*c) + 90*b^3*e^(8*d*x + 8*c) + 180*a^2*b*e^(6*d*x + 6*c) + 450*
a*b^2*e^(6*d*x + 6*c) + 240*b^3*e^(6*d*x + 6*c) + 270*a^2*b*e^(4*d*x + 4*c) + 600*a*b^2*e^(4*d*x + 4*c) + 340*
b^3*e^(4*d*x + 4*c) + 180*a^2*b*e^(2*d*x + 2*c) + 390*a*b^2*e^(2*d*x + 2*c) + 200*b^3*e^(2*d*x + 2*c) + 45*a^2
*b + 105*a*b^2 + 58*b^3)/(e^(2*d*x + 2*c) + 1)^5)/d

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maple [A]  time = 0.23, size = 180, normalized size = 1.48 \[ \frac {a^{3} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+3 a^{2} b \left (\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )+3 a \,b^{2} \left (\frac {\sinh ^{5}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}-\frac {5 d x}{2}-\frac {5 c}{2}+\frac {5 \tanh \left (d x +c \right )}{2}+\frac {5 \left (\tanh ^{3}\left (d x +c \right )\right )}{6}\right )+b^{3} \left (\frac {\sinh ^{7}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{5}}-\frac {7 d x}{2}-\frac {7 c}{2}+\frac {7 \tanh \left (d x +c \right )}{2}+\frac {7 \left (\tanh ^{3}\left (d x +c \right )\right )}{6}+\frac {7 \left (\tanh ^{5}\left (d x +c \right )\right )}{10}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*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*(1/2*sinh(d*x+c)^5/cosh(d*x+c)^3-5/2*d*x-5/2*c+5/2*tanh(d*x+c)+5/6*tanh(d*x+c)^3)+b^3*(1/
2*sinh(d*x+c)^7/cosh(d*x+c)^5-7/2*d*x-7/2*c+7/2*tanh(d*x+c)+7/6*tanh(d*x+c)^3+7/10*tanh(d*x+c)^5))

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maxima [B]  time = 0.33, size = 377, normalized size = 3.09 \[ -\frac {1}{8} \, a^{3} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{120} \, b^{3} {\left (\frac {420 \, {\left (d x + c\right )}}{d} + \frac {15 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {1003 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3350 \, e^{\left (-4 \, d x - 4 \, c\right )} + 5590 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3915 \, e^{\left (-8 \, d x - 8 \, c\right )} + 1455 \, e^{\left (-10 \, d x - 10 \, c\right )} + 15}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 10 \, e^{\left (-8 \, d x - 8 \, c\right )} + 5 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )}\right )}}\right )} - \frac {1}{8} \, a b^{2} {\left (\frac {60 \, {\left (d x + c\right )}}{d} + \frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {121 \, e^{\left (-2 \, d x - 2 \, c\right )} + 201 \, e^{\left (-4 \, d x - 4 \, c\right )} + 147 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} - \frac {3}{8} \, a^{2} b {\left (\frac {12 \, {\left (d x + c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a^3*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/120*b^3*(420*(d*x + c)/d + 15*e^(-2*d*x - 2*c)/d -
(1003*e^(-2*d*x - 2*c) + 3350*e^(-4*d*x - 4*c) + 5590*e^(-6*d*x - 6*c) + 3915*e^(-8*d*x - 8*c) + 1455*e^(-10*
d*x - 10*c) + 15)/(d*(e^(-2*d*x - 2*c) + 5*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 10*e^(-8*d*x - 8*c) + 5*e^
(-10*d*x - 10*c) + e^(-12*d*x - 12*c)))) - 1/8*a*b^2*(60*(d*x + c)/d + 3*e^(-2*d*x - 2*c)/d - (121*e^(-2*d*x -
2*c) + 201*e^(-4*d*x - 4*c) + 147*e^(-6*d*x - 6*c) + 3)/(d*(e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 3*e^(-6*d
*x - 6*c) + e^(-8*d*x - 8*c)))) - 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))))

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mupad [B]  time = 0.32, size = 668, normalized size = 5.48 \[ \frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^3}{8\,d}-\frac {\frac {2\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (9\,a^2\,b+15\,a\,b^2+10\,b^3\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,\left (9\,a^2\,b+15\,a\,b^2+10\,b^3\right )}{15\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {6\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a+b\right )}^3}{8\,d}-\frac {\frac {6\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (9\,a^2\,b+15\,a\,b^2+10\,b^3\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {x\,{\left (a+b\right )}^2\,\left (a+7\,b\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(2*c + 2*d*x)*(a + b)^3)/(8*d) - ((2*(6*a*b^2 + 3*a^2*b + 2*b^3))/(5*d) + (6*exp(2*c + 2*d*x)*(3*a*b^2 + a
^2*b + 2*b^3))/(5*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - ((2*(6*a*b^2 + 3*a^2*b + 2*b^3))/(5*d) + (
6*exp(6*c + 6*d*x)*(3*a*b^2 + a^2*b + 2*b^3))/(5*d) + (6*exp(4*c + 4*d*x)*(6*a*b^2 + 3*a^2*b + 2*b^3))/(5*d) +
(2*exp(2*c + 2*d*x)*(15*a*b^2 + 9*a^2*b + 10*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*(15*a*b^2 + 9*a^2*b + 10*b^3))/(15*d) + (6*exp(4*c + 4*d*x)*(3*a*b^2
+ a^2*b + 2*b^3))/(5*d) + (4*exp(2*c + 2*d*x)*(6*a*b^2 + 3*a^2*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) - (6*(3*a*b^2 + a^2*b + 2*b^3))/(5*d*(exp(2*c + 2*d*x) + 1)) - (exp(- 2*c
- 2*d*x)*(a + b)^3)/(8*d) - ((6*(3*a*b^2 + a^2*b + 2*b^3))/(5*d) + (8*exp(2*c + 2*d*x)*(6*a*b^2 + 3*a^2*b + 2
*b^3))/(5*d) + (6*exp(8*c + 8*d*x)*(3*a*b^2 + a^2*b + 2*b^3))/(5*d) + (8*exp(6*c + 6*d*x)*(6*a*b^2 + 3*a^2*b +
2*b^3))/(5*d) + (4*exp(4*c + 4*d*x)*(15*a*b^2 + 9*a^2*b + 10*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) - (x*(a + b)^2*(a + 7*b))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \sinh ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**2*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Integral((a + b*tanh(c + d*x)**2)**3*sinh(c + d*x)**2, x)

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