### 3.65 $$\int \sinh ^4(c+d x) (a+b \tanh ^3(c+d x))^3 \, dx$$

Optimal. Leaf size=275 $-\frac{b \left (3 a^2+10 b^2\right ) \tanh ^2(c+d x)}{2 d}+\frac{3 b \left (3 a^2+5 b^2\right ) \log (\cosh (c+d x))}{d}+\frac{\sinh (c+d x) \cosh ^3(c+d x) \left (b \left (3 a^2+b^2\right ) \tanh (c+d x)+a \left (a^2+3 b^2\right )\right )}{4 d}-\frac{\sinh (c+d x) \cosh (c+d x) \left (2 b \left (15 a^2+11 b^2\right ) \tanh (c+d x)+a \left (5 a^2+51 b^2\right )\right )}{8 d}+\frac{3}{8} a x \left (a^2+63 b^2\right )-\frac{3 a b^2 \tanh ^5(c+d x)}{5 d}-\frac{3 a b^2 \tanh ^3(c+d x)}{d}-\frac{18 a b^2 \tanh (c+d x)}{d}-\frac{b^3 \tanh ^8(c+d x)}{8 d}-\frac{b^3 \tanh ^6(c+d x)}{2 d}-\frac{3 b^3 \tanh ^4(c+d x)}{2 d}$

[Out]

(3*a*(a^2 + 63*b^2)*x)/8 + (3*b*(3*a^2 + 5*b^2)*Log[Cosh[c + d*x]])/d - (18*a*b^2*Tanh[c + d*x])/d - (b*(3*a^2
+ 10*b^2)*Tanh[c + d*x]^2)/(2*d) - (3*a*b^2*Tanh[c + d*x]^3)/d - (3*b^3*Tanh[c + d*x]^4)/(2*d) - (3*a*b^2*Tan
h[c + d*x]^5)/(5*d) - (b^3*Tanh[c + d*x]^6)/(2*d) - (b^3*Tanh[c + d*x]^8)/(8*d) + (Cosh[c + d*x]^3*Sinh[c + d*
x]*(a*(a^2 + 3*b^2) + b*(3*a^2 + b^2)*Tanh[c + d*x]))/(4*d) - (Cosh[c + d*x]*Sinh[c + d*x]*(a*(5*a^2 + 51*b^2)
+ 2*b*(15*a^2 + 11*b^2)*Tanh[c + d*x]))/(8*d)

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Rubi [A]  time = 0.493673, antiderivative size = 306, normalized size of antiderivative = 1.11, 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 = {3663, 1804, 1802, 633, 31} $-\frac{3 b \left (3 a^2+5 b^2\right ) \tanh ^2(c+d x)}{2 d}-\frac{3 a \left (a^2+63 b^2\right ) \tanh (c+d x)}{8 d}-\frac{3 (a+b) \left (a^2+23 a b+40 b^2\right ) \log (1-\tanh (c+d x))}{16 d}+\frac{3 (a-b) \left (a^2-23 a b+40 b^2\right ) \log (\tanh (c+d x)+1)}{16 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)+a \left (a^2+39 b^2\right )\right )}{8 d}+\frac{\sinh ^4(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{4 d}-\frac{3 a b^2 \tanh ^5(c+d x)}{5 d}-\frac{3 a b^2 \tanh ^3(c+d x)}{d}-\frac{b^3 \tanh ^8(c+d x)}{8 d}-\frac{b^3 \tanh ^6(c+d x)}{2 d}-\frac{3 b^3 \tanh ^4(c+d x)}{2 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-3*(a + b)*(a^2 + 23*a*b + 40*b^2)*Log[1 - Tanh[c + d*x]])/(16*d) + (3*(a - b)*(a^2 - 23*a*b + 40*b^2)*Log[1
+ Tanh[c + d*x]])/(16*d) - (3*a*(a^2 + 63*b^2)*Tanh[c + d*x])/(8*d) - (3*b*(3*a^2 + 5*b^2)*Tanh[c + d*x]^2)/(2
*d) - (3*a*b^2*Tanh[c + d*x]^3)/d - (3*b^3*Tanh[c + d*x]^4)/(2*d) - (3*a*b^2*Tanh[c + d*x]^5)/(5*d) - (b^3*Tan
h[c + d*x]^6)/(2*d) - (b^3*Tanh[c + d*x]^8)/(8*d) + (Sinh[c + d*x]^4*(b*(3*a^2 + b^2) + a*(a^2 + 3*b^2)*Tanh[c
+ d*x]))/(4*d) - (Sinh[c + d*x]^2*Tanh[c + d*x]*(a*(a^2 + 39*b^2) + 4*b*(6*a^2 + 5*b^2)*Tanh[c + d*x]))/(8*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 1804

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, Simp[((c*x)^m*(a + b*x^2)^(p + 1)*(a*g - b*f*x))/(2*a*b*(p + 1)), x] + Dist[c/(2*a*b*(p + 1)), Int[
(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+b x^3\right )^3}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (-4 b \left (3 a^2+b^2\right )-a \left (a^2+15 b^2\right ) x-4 b \left (3 a^2+b^2\right ) x^2-12 a b^2 x^3-4 b^3 x^4-12 a b^2 x^5-4 b^3 x^6-4 b^3 x^8\right )}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (a \left (a^2+39 b^2\right )+4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)\right )}{8 d}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a \left (a^2+39 b^2\right )+72 b \left (a^2+b^2\right ) x+48 a b^2 x^2+24 b^3 x^3+24 a b^2 x^4+16 b^3 x^5+8 b^3 x^7\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (a \left (a^2+39 b^2\right )+4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)\right )}{8 d}+\frac{\operatorname{Subst}\left (\int \left (-3 a \left (a^2+63 b^2\right )-24 b \left (3 a^2+5 b^2\right ) x-72 a b^2 x^2-48 b^3 x^3-24 a b^2 x^4-24 b^3 x^5-8 b^3 x^7+\frac{3 \left (a^3+63 a b^2+8 b \left (3 a^2+5 b^2\right ) x\right )}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 a \left (a^2+63 b^2\right ) \tanh (c+d x)}{8 d}-\frac{3 b \left (3 a^2+5 b^2\right ) \tanh ^2(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^3(c+d x)}{d}-\frac{3 b^3 \tanh ^4(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^5(c+d x)}{5 d}-\frac{b^3 \tanh ^6(c+d x)}{2 d}-\frac{b^3 \tanh ^8(c+d x)}{8 d}+\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (a \left (a^2+39 b^2\right )+4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)\right )}{8 d}+\frac{3 \operatorname{Subst}\left (\int \frac{a^3+63 a b^2+8 b \left (3 a^2+5 b^2\right ) x}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 a \left (a^2+63 b^2\right ) \tanh (c+d x)}{8 d}-\frac{3 b \left (3 a^2+5 b^2\right ) \tanh ^2(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^3(c+d x)}{d}-\frac{3 b^3 \tanh ^4(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^5(c+d x)}{5 d}-\frac{b^3 \tanh ^6(c+d x)}{2 d}-\frac{b^3 \tanh ^8(c+d x)}{8 d}+\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (a \left (a^2+39 b^2\right )+4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)\right )}{8 d}-\frac{\left (3 (a-b) \left (a^2-23 a b+40 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x} \, dx,x,\tanh (c+d x)\right )}{16 d}+\frac{\left (3 (a+b) \left (a^2+23 a b+40 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,\tanh (c+d x)\right )}{16 d}\\ &=-\frac{3 (a+b) \left (a^2+23 a b+40 b^2\right ) \log (1-\tanh (c+d x))}{16 d}+\frac{3 (a-b) \left (a^2-23 a b+40 b^2\right ) \log (1+\tanh (c+d x))}{16 d}-\frac{3 a \left (a^2+63 b^2\right ) \tanh (c+d x)}{8 d}-\frac{3 b \left (3 a^2+5 b^2\right ) \tanh ^2(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^3(c+d x)}{d}-\frac{3 b^3 \tanh ^4(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^5(c+d x)}{5 d}-\frac{b^3 \tanh ^6(c+d x)}{2 d}-\frac{b^3 \tanh ^8(c+d x)}{8 d}+\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (a \left (a^2+39 b^2\right )+4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)\right )}{8 d}\\ \end{align*}

Mathematica [A]  time = 6.23935, size = 294, normalized size = 1.07 $\frac{3 a \left (a^2+63 b^2\right ) (c+d x)}{8 d}-\frac{a \left (a^2+12 b^2\right ) \sinh (2 (c+d x))}{4 d}+\frac{a \left (a^2+3 b^2\right ) \sinh (4 (c+d x))}{32 d}-\frac{b \left (15 a^2+11 b^2\right ) \cosh (2 (c+d x))}{8 d}+\frac{b \left (3 a^2+b^2\right ) \cosh (4 (c+d x))}{32 d}+\frac{b \left (3 a^2+20 b^2\right ) \text{sech}^2(c+d x)}{2 d}+\frac{3 \left (3 a^2 b+5 b^3\right ) \log (\cosh (c+d x))}{d}-\frac{108 a b^2 \tanh (c+d x)}{5 d}-\frac{3 a b^2 \tanh (c+d x) \text{sech}^4(c+d x)}{5 d}+\frac{21 a b^2 \tanh (c+d x) \text{sech}^2(c+d x)}{5 d}-\frac{b^3 \text{sech}^8(c+d x)}{8 d}+\frac{b^3 \text{sech}^6(c+d x)}{d}-\frac{15 b^3 \text{sech}^4(c+d x)}{4 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*a*(a^2 + 63*b^2)*(c + d*x))/(8*d) - (b*(15*a^2 + 11*b^2)*Cosh[2*(c + d*x)])/(8*d) + (b*(3*a^2 + b^2)*Cosh[4
*(c + d*x)])/(32*d) + (3*(3*a^2*b + 5*b^3)*Log[Cosh[c + d*x]])/d + (b*(3*a^2 + 20*b^2)*Sech[c + d*x]^2)/(2*d)
- (15*b^3*Sech[c + d*x]^4)/(4*d) + (b^3*Sech[c + d*x]^6)/d - (b^3*Sech[c + d*x]^8)/(8*d) - (a*(a^2 + 12*b^2)*S
inh[2*(c + d*x)])/(4*d) + (a*(a^2 + 3*b^2)*Sinh[4*(c + d*x)])/(32*d) - (108*a*b^2*Tanh[c + d*x])/(5*d) + (21*a
*b^2*Sech[c + d*x]^2*Tanh[c + d*x])/(5*d) - (3*a*b^2*Sech[c + d*x]^4*Tanh[c + d*x])/(5*d)

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Maple [A]  time = 0.076, size = 385, normalized size = 1.4 \begin{align*}{\frac{{a}^{3}\cosh \left ( dx+c \right ) \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,{a}^{3}\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{3}x}{8}}+{\frac{3\,{a}^{3}c}{8\,d}}+{\frac{3\,{a}^{2}b \left ( \sinh \left ( dx+c \right ) \right ) ^{6}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-{\frac{9\,{a}^{2}b \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+9\,{\frac{{a}^{2}b\ln \left ( \cosh \left ( dx+c \right ) \right ) }{d}}-{\frac{9\,{a}^{2}b \left ( \tanh \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{3\,a{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{9}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{27\,a{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{189\,a{b}^{2}x}{8}}+{\frac{189\,a{b}^{2}c}{8\,d}}-{\frac{189\,a{b}^{2}\tanh \left ( dx+c \right ) }{8\,d}}-{\frac{63\,a{b}^{2} \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{189\,a{b}^{2} \left ( \tanh \left ( dx+c \right ) \right ) ^{5}}{40\,d}}+{\frac{{b}^{3} \left ( \sinh \left ( dx+c \right ) \right ) ^{12}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{8}}}-{\frac{3\,{b}^{3} \left ( \sinh \left ( dx+c \right ) \right ) ^{10}}{2\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{8}}}+15\,{\frac{{b}^{3}\ln \left ( \cosh \left ( dx+c \right ) \right ) }{d}}-{\frac{15\,{b}^{3} \left ( \tanh \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{15\,{b}^{3} \left ( \tanh \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{5\,{b}^{3} \left ( \tanh \left ( dx+c \right ) \right ) ^{6}}{2\,d}}-{\frac{15\,{b}^{3} \left ( \tanh \left ( dx+c \right ) \right ) ^{8}}{8\,d}} \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)^3)^3,x)

[Out]

1/4/d*a^3*cosh(d*x+c)*sinh(d*x+c)^3-3/8/d*a^3*cosh(d*x+c)*sinh(d*x+c)+3/8*a^3*x+3/8/d*a^3*c+3/4/d*a^2*b*sinh(d
*x+c)^6/cosh(d*x+c)^2-9/4/d*a^2*b*sinh(d*x+c)^4/cosh(d*x+c)^2+9/d*a^2*b*ln(cosh(d*x+c))-9/2*a^2*b*tanh(d*x+c)^
2/d+3/4/d*a*b^2*sinh(d*x+c)^9/cosh(d*x+c)^5-27/8/d*a*b^2*sinh(d*x+c)^7/cosh(d*x+c)^5+189/8*a*b^2*x+189/8/d*a*b
^2*c-189/8*a*b^2*tanh(d*x+c)/d-63/8*a*b^2*tanh(d*x+c)^3/d-189/40*a*b^2*tanh(d*x+c)^5/d+1/4/d*b^3*sinh(d*x+c)^1
2/cosh(d*x+c)^8-3/2/d*b^3*sinh(d*x+c)^10/cosh(d*x+c)^8+15/d*b^3*ln(cosh(d*x+c))-15/2/d*b^3*tanh(d*x+c)^2-15/4*
b^3*tanh(d*x+c)^4/d-5/2*b^3*tanh(d*x+c)^6/d-15/8*b^3*tanh(d*x+c)^8/d

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Maxima [B]  time = 1.80729, size = 873, normalized size = 3.17 \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)^3)^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) + 3/320*
a*b^2*(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*b^3*(960*(d*x + c)/d - (44*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c))/d + 960
*log(e^(-2*d*x - 2*c) + 1)/d - (36*e^(-2*d*x - 2*c) + 324*e^(-4*d*x - 4*c) - 1384*e^(-6*d*x - 6*c) - 9126*e^(-
8*d*x - 8*c) - 24112*e^(-10*d*x - 10*c) - 31868*e^(-12*d*x - 12*c) - 25912*e^(-14*d*x - 14*c) - 11169*e^(-16*d
*x - 16*c) - 2516*e^(-18*d*x - 18*c) - 1)/(d*(e^(-4*d*x - 4*c) + 8*e^(-6*d*x - 6*c) + 28*e^(-8*d*x - 8*c) + 56
*e^(-10*d*x - 10*c) + 70*e^(-12*d*x - 12*c) + 56*e^(-14*d*x - 14*c) + 28*e^(-16*d*x - 16*c) + 8*e^(-18*d*x - 1
8*c) + e^(-20*d*x - 20*c)))) + 3/64*a^2*b*(192*(d*x + c)/d - (20*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c))/d + 192*
log(e^(-2*d*x - 2*c) + 1)/d - (18*e^(-2*d*x - 2*c) + 39*e^(-4*d*x - 4*c) - 108*e^(-6*d*x - 6*c) - 1)/(d*(e^(-4
*d*x - 4*c) + 2*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c))))

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Fricas [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)^3)^3,x, algorithm="fricas")

[Out]

Timed out

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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)**3)**3,x)

[Out]

Timed out

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Giac [B]  time = 4.87444, size = 956, normalized size = 3.48 \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)^3)^3,x, algorithm="giac")

[Out]

1/2240*(840*(a^3 - 24*a^2*b + 63*a*b^2 - 40*b^3)*d*x + 6720*(3*a^2*b*e^(2*c) + 5*b^3*e^(2*c))*e^(-2*c)*log(e^(
2*d*x + 2*c) + 1) - 35*(18*a^3*e^(4*d*x + 4*c) - 432*a^2*b*e^(4*d*x + 4*c) + 1134*a*b^2*e^(4*d*x + 4*c) - 720*
b^3*e^(4*d*x + 4*c) - 8*a^3*e^(2*d*x + 2*c) + 60*a^2*b*e^(2*d*x + 2*c) - 96*a*b^2*e^(2*d*x + 2*c) + 44*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) + 35*(a^3*e^(4*d*x + 48*c) + 3*a^2*b*e^(4*d*x +
48*c) + 3*a*b^2*e^(4*d*x + 48*c) + b^3*e^(4*d*x + 48*c) - 8*a^3*e^(2*d*x + 46*c) - 60*a^2*b*e^(2*d*x + 46*c)
- 96*a*b^2*e^(2*d*x + 46*c) - 44*b^3*e^(2*d*x + 46*c))*e^(-44*c) - 8*(6849*a^2*b*e^(16*d*x + 16*c) + 11415*b^3
*e^(16*d*x + 16*c) + 53112*a^2*b*e^(14*d*x + 14*c) - 16800*a*b^2*e^(14*d*x + 14*c) + 80120*b^3*e^(14*d*x + 14*
c) + 181692*a^2*b*e^(12*d*x + 12*c) - 100800*a*b^2*e^(12*d*x + 12*c) + 269220*b^3*e^(12*d*x + 12*c) + 358344*a
^2*b*e^(10*d*x + 10*c) - 272160*a*b^2*e^(10*d*x + 10*c) + 520520*b^3*e^(10*d*x + 10*c) + 445830*a^2*b*e^(8*d*x
+ 8*c) - 423360*a*b^2*e^(8*d*x + 8*c) + 648970*b^3*e^(8*d*x + 8*c) + 358344*a^2*b*e^(6*d*x + 6*c) - 405216*a*
b^2*e^(6*d*x + 6*c) + 520520*b^3*e^(6*d*x + 6*c) + 181692*a^2*b*e^(4*d*x + 4*c) - 237888*a*b^2*e^(4*d*x + 4*c)
+ 269220*b^3*e^(4*d*x + 4*c) + 53112*a^2*b*e^(2*d*x + 2*c) - 79968*a*b^2*e^(2*d*x + 2*c) + 80120*b^3*e^(2*d*x
+ 2*c) + 6849*a^2*b - 12096*a*b^2 + 11415*b^3)/(e^(2*d*x + 2*c) + 1)^8)/d