### 3.17 $$\int \sinh ^4(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx$$

Optimal. Leaf size=182 $-\frac{b \left (6 a^2+35 a b+21 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 (a+b) \left (a^2+14 a b+21 b^2\right ) \tanh (c+d x)}{8 d}+\frac{3}{8} x (a+b) \left (a^2+14 a b+21 b^2\right )-\frac{3 b^2 (5 a+21 b) \tanh ^5(c+d x)}{40 d}-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\sinh ^3(c+d x) \cosh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d}$

[Out]

(3*(a + b)*(a^2 + 14*a*b + 21*b^2)*x)/8 - (3*(a + b)*(a^2 + 14*a*b + 21*b^2)*Tanh[c + d*x])/(8*d) - (b*(6*a^2
+ 35*a*b + 21*b^2)*Tanh[c + d*x]^3)/(8*d) - (3*b^2*(5*a + 21*b)*Tanh[c + d*x]^5)/(40*d) - (3*(a + 3*b)*Sinh[c
+ d*x]^2*Tanh[c + d*x]*(a + b*Tanh[c + d*x]^2)^2)/(8*d) + (Cosh[c + d*x]*Sinh[c + d*x]^3*(a + b*Tanh[c + d*x]^
2)^3)/(4*d)

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Rubi [A]  time = 0.22168, antiderivative size = 182, normalized size of antiderivative = 1., 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, 577, 570, 206} $-\frac{b \left (6 a^2+35 a b+21 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 (a+b) \left (a^2+14 a b+21 b^2\right ) \tanh (c+d x)}{8 d}+\frac{3}{8} x (a+b) \left (a^2+14 a b+21 b^2\right )-\frac{3 b^2 (5 a+21 b) \tanh ^5(c+d x)}{40 d}-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\sinh ^3(c+d x) \cosh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*(a + b)*(a^2 + 14*a*b + 21*b^2)*x)/8 - (3*(a + b)*(a^2 + 14*a*b + 21*b^2)*Tanh[c + d*x])/(8*d) - (b*(6*a^2
+ 35*a*b + 21*b^2)*Tanh[c + d*x]^3)/(8*d) - (3*b^2*(5*a + 21*b)*Tanh[c + d*x]^5)/(40*d) - (3*(a + 3*b)*Sinh[c
+ d*x]^2*Tanh[c + d*x]*(a + b*Tanh[c + d*x]^2)^2)/(8*d) + (Cosh[c + d*x]*Sinh[c + d*x]^3*(a + b*Tanh[c + d*x]^
2)^3)/(4*d)

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]

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 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 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 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])

Rubi steps

\begin{align*} \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+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 \tanh ^2(c+d x)\right )^3}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b x^2\right )^2 \left (3 a+9 b x^2\right )}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b x^2\right ) \left (-3 a (a+9 b)-3 b (5 a+21 b) x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d}-\frac{\operatorname{Subst}\left (\int \left (3 (a+b) \left (a^2+14 a b+21 b^2\right )+3 b \left (6 a^2+35 a b+21 b^2\right ) x^2+3 b^2 (5 a+21 b) x^4-\frac{3 \left (a^3+15 a^2 b+35 a b^2+21 b^3\right )}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 (a+b) \left (a^2+14 a b+21 b^2\right ) \tanh (c+d x)}{8 d}-\frac{b \left (6 a^2+35 a b+21 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 b^2 (5 a+21 b) \tanh ^5(c+d x)}{40 d}-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d}+\frac{\left (3 (a+b) \left (a^2+14 a b+21 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+b) \left (a^2+14 a b+21 b^2\right ) x-\frac{3 (a+b) \left (a^2+14 a b+21 b^2\right ) \tanh (c+d x)}{8 d}-\frac{b \left (6 a^2+35 a b+21 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 b^2 (5 a+21 b) \tanh ^5(c+d x)}{40 d}-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d}\\ \end{align*}

Mathematica [A]  time = 3.97828, size = 125, normalized size = 0.69 $\frac{60 \left (15 a^2 b+a^3+35 a b^2+21 b^3\right ) (c+d x)-32 b \tanh (c+d x) \left (15 a^2-b (5 a+7 b) \text{sech}^2(c+d x)+50 a b+b^2 \text{sech}^4(c+d x)+36 b^2\right )+5 (a+b)^3 \sinh (4 (c+d x))-40 (a+4 b) (a+b)^2 \sinh (2 (c+d x))}{160 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(60*(a^3 + 15*a^2*b + 35*a*b^2 + 21*b^3)*(c + d*x) - 40*(a + b)^2*(a + 4*b)*Sinh[2*(c + d*x)] + 5*(a + b)^3*Si
nh[4*(c + d*x)] - 32*b*(15*a^2 + 50*a*b + 36*b^2 - b*(5*a + 7*b)*Sech[c + d*x]^2 + b^2*Sech[c + d*x]^4)*Tanh[c
+ d*x])/(160*d)

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Maple [A]  time = 0.053, size = 246, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{4}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8}} \right ) \cosh \left ( dx+c \right ) +{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +3\,{a}^{2}b \left ( 1/4\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{\cosh \left ( dx+c \right ) }}-5/8\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{\cosh \left ( dx+c \right ) }}+{\frac{15\,dx}{8}}+{\frac{15\,c}{8}}-{\frac{15\,\tanh \left ( dx+c \right ) }{8}} \right ) +3\,a{b}^{2} \left ( 1/4\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{7}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-{\frac{7\, \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{8\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+{\frac{35\,dx}{8}}+{\frac{35\,c}{8}}-{\frac{35\,\tanh \left ( dx+c \right ) }{8}}-{\frac{35\, \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{24}} \right ) +{b}^{3} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{9}}{4\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{9\, \left ( \sinh \left ( dx+c \right ) \right ) ^{7}}{8\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{63\,dx}{8}}+{\frac{63\,c}{8}}-{\frac{63\,\tanh \left ( dx+c \right ) }{8}}-{\frac{21\, \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{8}}-{\frac{63\, \left ( \tanh \left ( dx+c \right ) \right ) ^{5}}{40}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,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*(1/4*sinh(d*x+c)^5/cosh(d*x+c
)-5/8*sinh(d*x+c)^3/cosh(d*x+c)+15/8*d*x+15/8*c-15/8*tanh(d*x+c))+3*a*b^2*(1/4*sinh(d*x+c)^7/cosh(d*x+c)^3-7/8
*sinh(d*x+c)^5/cosh(d*x+c)^3+35/8*d*x+35/8*c-35/8*tanh(d*x+c)-35/24*tanh(d*x+c)^3)+b^3*(1/4*sinh(d*x+c)^9/cosh
(d*x+c)^5-9/8*sinh(d*x+c)^7/cosh(d*x+c)^5+63/8*d*x+63/8*c-63/8*tanh(d*x+c)-21/8*tanh(d*x+c)^3-63/40*tanh(d*x+c
)^5))

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Maxima [B]  time = 1.07719, size = 648, normalized size = 3.56 \begin{align*} \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 )} + \frac{1}{320} \, b^{3}{\left (\frac{2520 \,{\left (d x + c\right )}}{d} + \frac{5 \,{\left (32 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}\right )}}{d} - \frac{135 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5358 \, e^{\left (-4 \, d x - 4 \, c\right )} + 18190 \, e^{\left (-6 \, d x - 6 \, c\right )} + 28455 \, e^{\left (-8 \, d x - 8 \, c\right )} + 19995 \, e^{\left (-10 \, d x - 10 \, c\right )} + 6560 \, e^{\left (-12 \, d x - 12 \, c\right )} - 5}{d{\left (e^{\left (-4 \, d x - 4 \, c\right )} + 5 \, e^{\left (-6 \, d x - 6 \, c\right )} + 10 \, e^{\left (-8 \, d x - 8 \, c\right )} + 10 \, e^{\left (-10 \, d x - 10 \, c\right )} + 5 \, e^{\left (-12 \, d x - 12 \, c\right )} + e^{\left (-14 \, d x - 14 \, c\right )}\right )}}\right )} + \frac{1}{64} \, a 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{3}{64} \, a^{2} 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{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,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) + 1/320*
b^3*(2520*(d*x + c)/d + 5*(32*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c))/d - (135*e^(-2*d*x - 2*c) + 5358*e^(-4*d*x
- 4*c) + 18190*e^(-6*d*x - 6*c) + 28455*e^(-8*d*x - 8*c) + 19995*e^(-10*d*x - 10*c) + 6560*e^(-12*d*x - 12*c)
- 5)/(d*(e^(-4*d*x - 4*c) + 5*e^(-6*d*x - 6*c) + 10*e^(-8*d*x - 8*c) + 10*e^(-10*d*x - 10*c) + 5*e^(-12*d*x -
12*c) + e^(-14*d*x - 14*c)))) + 1/64*a*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)))) + 3/64*a^2*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]  time = 2.07457, size = 2273, normalized size = 12.49 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*(5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^9 - 15*(a^3 + 11*a^2*b + 19*a*b^2 + 9*b^3 - 12*(a^3 + 3
*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^7 + 8*(120*a^2*b + 400*a*b^2 + 288*b^3 + 15*(a^3 + 15*a
^2*b + 35*a*b^2 + 21*b^3)*d*x)*cosh(d*x + c)^5 + 40*(120*a^2*b + 400*a*b^2 + 288*b^3 + 15*(a^3 + 15*a^2*b + 35
*a*b^2 + 21*b^3)*d*x)*cosh(d*x + c)*sinh(d*x + c)^4 + (630*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 - 1
50*a^3 - 2010*a^2*b - 4850*a*b^2 - 3054*b^3 - 315*(a^3 + 11*a^2*b + 19*a*b^2 + 9*b^3)*cosh(d*x + c)^2)*sinh(d*
x + c)^5 + 40*(120*a^2*b + 400*a*b^2 + 288*b^3 + 15*(a^3 + 15*a^2*b + 35*a*b^2 + 21*b^3)*d*x)*cosh(d*x + c)^3
+ 5*(84*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^6 - 105*(a^3 + 11*a^2*b + 19*a*b^2 + 9*b^3)*cosh(d*x + c
)^4 - 62*a^3 - 978*a^2*b - 2282*a*b^2 - 1302*b^3 - 4*(75*a^3 + 1005*a^2*b + 2425*a*b^2 + 1527*b^3)*cosh(d*x +
c)^2)*sinh(d*x + c)^3 + 40*(2*(120*a^2*b + 400*a*b^2 + 288*b^3 + 15*(a^3 + 15*a^2*b + 35*a*b^2 + 21*b^3)*d*x)*
cosh(d*x + c)^3 + 3*(120*a^2*b + 400*a*b^2 + 288*b^3 + 15*(a^3 + 15*a^2*b + 35*a*b^2 + 21*b^3)*d*x)*cosh(d*x +
c))*sinh(d*x + c)^2 + 80*(120*a^2*b + 400*a*b^2 + 288*b^3 + 15*(a^3 + 15*a^2*b + 35*a*b^2 + 21*b^3)*d*x)*cosh
(d*x + c) + 5*(9*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^8 - 21*(a^3 + 11*a^2*b + 19*a*b^2 + 9*b^3)*cosh
(d*x + c)^6 - 2*(75*a^3 + 1005*a^2*b + 2425*a*b^2 + 1527*b^3)*cosh(d*x + c)^4 - 36*a^3 - 612*a^2*b - 1372*a*b^
2 - 924*b^3 - 6*(31*a^3 + 489*a^2*b + 1141*a*b^2 + 651*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**4*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 2.74754, size = 684, normalized size = 3.76 \begin{align*} \frac{120 \,{\left (a^{3} + 15 \, a^{2} b + 35 \, a b^{2} + 21 \, b^{3}\right )} d x - 5 \,{\left (18 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 630 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 378 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 48 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 72 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 32 \, 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 (-4 \, d x - 4 \, c\right )} + 5 \,{\left (a^{3} e^{\left (4 \, d x + 36 \, c\right )} + 3 \, a^{2} b e^{\left (4 \, d x + 36 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, d x + 36 \, c\right )} + b^{3} e^{\left (4 \, d x + 36 \, c\right )} - 8 \, a^{3} e^{\left (2 \, d x + 34 \, c\right )} - 48 \, a^{2} b e^{\left (2 \, d x + 34 \, c\right )} - 72 \, a b^{2} e^{\left (2 \, d x + 34 \, c\right )} - 32 \, b^{3} e^{\left (2 \, d x + 34 \, c\right )}\right )} e^{\left (-32 \, c\right )} + \frac{128 \,{\left (15 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 50 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 210 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 150 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 290 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 210 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 190 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 130 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} b + 50 \, a b^{2} + 36 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{320 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*(120*(a^3 + 15*a^2*b + 35*a*b^2 + 21*b^3)*d*x - 5*(18*a^3*e^(4*d*x + 4*c) + 270*a^2*b*e^(4*d*x + 4*c) +
630*a*b^2*e^(4*d*x + 4*c) + 378*b^3*e^(4*d*x + 4*c) - 8*a^3*e^(2*d*x + 2*c) - 48*a^2*b*e^(2*d*x + 2*c) - 72*a*
b^2*e^(2*d*x + 2*c) - 32*b^3*e^(2*d*x + 2*c) + a^3 + 3*a^2*b + 3*a*b^2 + b^3)*e^(-4*d*x - 4*c) + 5*(a^3*e^(4*d
*x + 36*c) + 3*a^2*b*e^(4*d*x + 36*c) + 3*a*b^2*e^(4*d*x + 36*c) + b^3*e^(4*d*x + 36*c) - 8*a^3*e^(2*d*x + 34*
c) - 48*a^2*b*e^(2*d*x + 34*c) - 72*a*b^2*e^(2*d*x + 34*c) - 32*b^3*e^(2*d*x + 34*c))*e^(-32*c) + 128*(15*a^2*
b*e^(8*d*x + 8*c) + 60*a*b^2*e^(8*d*x + 8*c) + 50*b^3*e^(8*d*x + 8*c) + 60*a^2*b*e^(6*d*x + 6*c) + 210*a*b^2*e
^(6*d*x + 6*c) + 150*b^3*e^(6*d*x + 6*c) + 90*a^2*b*e^(4*d*x + 4*c) + 290*a*b^2*e^(4*d*x + 4*c) + 210*b^3*e^(4
*d*x + 4*c) + 60*a^2*b*e^(2*d*x + 2*c) + 190*a*b^2*e^(2*d*x + 2*c) + 130*b^3*e^(2*d*x + 2*c) + 15*a^2*b + 50*a
*b^2 + 36*b^3)/(e^(2*d*x + 2*c) + 1)^5)/d