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]

3/8*a*(a^2+63*b^2)*x+3*b*(3*a^2+5*b^2)*ln(cosh(d*x+c))/d-18*a*b^2*tanh(d*x+c)/d-1/2*b*(3*a^2+10*b^2)*tanh(d*x+
c)^2/d-3*a*b^2*tanh(d*x+c)^3/d-3/2*b^3*tanh(d*x+c)^4/d-3/5*a*b^2*tanh(d*x+c)^5/d-1/2*b^3*tanh(d*x+c)^6/d-1/8*b
^3*tanh(d*x+c)^8/d+1/4*cosh(d*x+c)^3*sinh(d*x+c)*(a*(a^2+3*b^2)+b*(3*a^2+b^2)*tanh(d*x+c))/d-1/8*cosh(d*x+c)*s
inh(d*x+c)*(a*(5*a^2+51*b^2)+2*b*(15*a^2+11*b^2)*tanh(d*x+c))/d

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Rubi [A]  time = 0.49, 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 31

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

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 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 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 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 ^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*}

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Mathematica [A]  time = 6.27, size = 294, normalized size = 1.07 \[ \frac {3 \left (3 a^2 b+5 b^3\right ) \log (\cosh (c+d x))}{d}+\frac {3 a \left (a^2+63 b^2\right ) (c+d x)}{8 d}-\frac {a \left (a^2+12 b^2\right ) \sinh (2 (c+d x))}{4 d}+\frac {a \left (a^2+3 b^2\right ) \sinh (4 (c+d x))}{32 d}-\frac {b \left (15 a^2+11 b^2\right ) \cosh (2 (c+d x))}{8 d}+\frac {b \left (3 a^2+b^2\right ) \cosh (4 (c+d x))}{32 d}+\frac {b \left (3 a^2+20 b^2\right ) \text {sech}^2(c+d x)}{2 d}-\frac {108 a b^2 \tanh (c+d x)}{5 d}-\frac {3 a b^2 \tanh (c+d x) \text {sech}^4(c+d x)}{5 d}+\frac {21 a b^2 \tanh (c+d x) \text {sech}^2(c+d x)}{5 d}-\frac {b^3 \text {sech}^8(c+d x)}{8 d}+\frac {b^3 \text {sech}^6(c+d x)}{d}-\frac {15 b^3 \text {sech}^4(c+d x)}{4 d} \]

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

[Out]

Timed out

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giac [B]  time = 1.63, size = 708, normalized size = 2.57 \[ \frac {840 \, {\left (a^{3} - 24 \, a^{2} b + 63 \, a b^{2} - 40 \, b^{3}\right )} d x + 6720 \, {\left (3 \, a^{2} b e^{\left (2 \, c\right )} + 5 \, b^{3} e^{\left (2 \, c\right )}\right )} e^{\left (-2 \, c\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) - 35 \, {\left (18 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - 432 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 1134 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 720 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 60 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 96 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 44 \, 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 )} + 35 \, {\left (a^{3} e^{\left (4 \, d x + 48 \, c\right )} + 3 \, a^{2} b e^{\left (4 \, d x + 48 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, d x + 48 \, c\right )} + b^{3} e^{\left (4 \, d x + 48 \, c\right )} - 8 \, a^{3} e^{\left (2 \, d x + 46 \, c\right )} - 60 \, a^{2} b e^{\left (2 \, d x + 46 \, c\right )} - 96 \, a b^{2} e^{\left (2 \, d x + 46 \, c\right )} - 44 \, b^{3} e^{\left (2 \, d x + 46 \, c\right )}\right )} e^{\left (-44 \, c\right )} - \frac {8 \, {\left (6849 \, a^{2} b e^{\left (16 \, d x + 16 \, c\right )} + 11415 \, b^{3} e^{\left (16 \, d x + 16 \, c\right )} + 53112 \, a^{2} b e^{\left (14 \, d x + 14 \, c\right )} - 16800 \, a b^{2} e^{\left (14 \, d x + 14 \, c\right )} + 80120 \, b^{3} e^{\left (14 \, d x + 14 \, c\right )} + 181692 \, a^{2} b e^{\left (12 \, d x + 12 \, c\right )} - 100800 \, a b^{2} e^{\left (12 \, d x + 12 \, c\right )} + 269220 \, b^{3} e^{\left (12 \, d x + 12 \, c\right )} + 358344 \, a^{2} b e^{\left (10 \, d x + 10 \, c\right )} - 272160 \, a b^{2} e^{\left (10 \, d x + 10 \, c\right )} + 520520 \, b^{3} e^{\left (10 \, d x + 10 \, c\right )} + 445830 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} - 423360 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 648970 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 358344 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 405216 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 520520 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 181692 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 237888 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 269220 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 53112 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 79968 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 80120 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 6849 \, a^{2} b - 12096 \, a b^{2} + 11415 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{8}}}{2240 \, d} \]

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

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maple [A]  time = 0.41, size = 385, normalized size = 1.40 \[ \frac {a^{3} \cosh \left (d x +c \right ) \left (\sinh ^{3}\left (d x +c \right )\right )}{4 d}-\frac {3 a^{3} \cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{8 d}+\frac {3 a^{3} x}{8}+\frac {3 a^{3} c}{8 d}+\frac {3 a^{2} b \left (\sinh ^{6}\left (d x +c \right )\right )}{4 d \cosh \left (d x +c \right )^{2}}-\frac {9 a^{2} b \left (\sinh ^{4}\left (d x +c \right )\right )}{4 d \cosh \left (d x +c \right )^{2}}+\frac {9 a^{2} b \ln \left (\cosh \left (d x +c \right )\right )}{d}-\frac {9 a^{2} b \left (\tanh ^{2}\left (d x +c \right )\right )}{2 d}+\frac {3 a \,b^{2} \left (\sinh ^{9}\left (d x +c \right )\right )}{4 d \cosh \left (d x +c \right )^{5}}-\frac {27 a \,b^{2} \left (\sinh ^{7}\left (d x +c \right )\right )}{8 d \cosh \left (d x +c \right )^{5}}+\frac {189 a \,b^{2} x}{8}+\frac {189 a \,b^{2} c}{8 d}-\frac {189 a \,b^{2} \tanh \left (d x +c \right )}{8 d}-\frac {63 a \,b^{2} \left (\tanh ^{3}\left (d x +c \right )\right )}{8 d}-\frac {189 a \,b^{2} \left (\tanh ^{5}\left (d x +c \right )\right )}{40 d}+\frac {b^{3} \left (\sinh ^{12}\left (d x +c \right )\right )}{4 d \cosh \left (d x +c \right )^{8}}-\frac {3 b^{3} \left (\sinh ^{10}\left (d x +c \right )\right )}{2 d \cosh \left (d x +c \right )^{8}}+\frac {15 b^{3} \ln \left (\cosh \left (d x +c \right )\right )}{d}-\frac {15 b^{3} \left (\tanh ^{2}\left (d x +c \right )\right )}{2 d}-\frac {15 b^{3} \left (\tanh ^{4}\left (d x +c \right )\right )}{4 d}-\frac {5 b^{3} \left (\tanh ^{6}\left (d x +c \right )\right )}{2 d}-\frac {15 b^{3} \left (\tanh ^{8}\left (d x +c \right )\right )}{8 d} \]

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*a^3*cosh(d*x+c)*sinh(d*x+c)/d+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*b^3*tanh(d*x+c)^2/d-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 = 0.43, size = 647, normalized size = 2.35 \[ \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 {3}{320} \, a b^{2} {\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} \, b^{3} {\left (\frac {960 \, {\left (d x + c\right )}}{d} - \frac {44 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} + \frac {960 \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} - \frac {36 \, e^{\left (-2 \, d x - 2 \, c\right )} + 324 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1384 \, e^{\left (-6 \, d x - 6 \, c\right )} - 9126 \, e^{\left (-8 \, d x - 8 \, c\right )} - 24112 \, e^{\left (-10 \, d x - 10 \, c\right )} - 31868 \, e^{\left (-12 \, d x - 12 \, c\right )} - 25912 \, e^{\left (-14 \, d x - 14 \, c\right )} - 11169 \, e^{\left (-16 \, d x - 16 \, c\right )} - 2516 \, e^{\left (-18 \, d x - 18 \, c\right )} - 1}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + 8 \, e^{\left (-6 \, d x - 6 \, c\right )} + 28 \, e^{\left (-8 \, d x - 8 \, c\right )} + 56 \, e^{\left (-10 \, d x - 10 \, c\right )} + 70 \, e^{\left (-12 \, d x - 12 \, c\right )} + 56 \, e^{\left (-14 \, d x - 14 \, c\right )} + 28 \, e^{\left (-16 \, d x - 16 \, c\right )} + 8 \, e^{\left (-18 \, d x - 18 \, c\right )} + e^{\left (-20 \, d x - 20 \, c\right )}\right )}}\right )} + \frac {3}{64} \, a^{2} b {\left (\frac {192 \, {\left (d x + c\right )}}{d} - \frac {20 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} + \frac {192 \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} - \frac {18 \, e^{\left (-2 \, d x - 2 \, c\right )} + 39 \, e^{\left (-4 \, d x - 4 \, c\right )} - 108 \, e^{\left (-6 \, d x - 6 \, c\right )} - 1}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} + 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} \]

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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mupad [B]  time = 1.70, size = 682, normalized size = 2.48 \[ x\,\left (\frac {3\,a^3}{8}-9\,a^2\,b+\frac {189\,a\,b^2}{8}-15\,b^3\right )-\frac {4\,\left (71\,b^3+12\,a\,b^2\right )}{d\,\left (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\right )}-\frac {256\,b^3}{d\,\left (6\,{\mathrm {e}}^{2\,c+2\,d\,x}+15\,{\mathrm {e}}^{4\,c+4\,d\,x}+20\,{\mathrm {e}}^{6\,c+6\,d\,x}+15\,{\mathrm {e}}^{8\,c+8\,d\,x}+6\,{\mathrm {e}}^{10\,c+10\,d\,x}+{\mathrm {e}}^{12\,c+12\,d\,x}+1\right )}+\frac {\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+1\right )\,\left (9\,a^2\,b+15\,b^3\right )}{d}+\frac {2\,\left (3\,a^2\,b+30\,a\,b^2+20\,b^3\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {32\,\left (50\,b^3+3\,a\,b^2\right )}{5\,d\,\left (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\right )}+\frac {128\,b^3}{d\,\left (7\,{\mathrm {e}}^{2\,c+2\,d\,x}+21\,{\mathrm {e}}^{4\,c+4\,d\,x}+35\,{\mathrm {e}}^{6\,c+6\,d\,x}+35\,{\mathrm {e}}^{8\,c+8\,d\,x}+21\,{\mathrm {e}}^{10\,c+10\,d\,x}+7\,{\mathrm {e}}^{12\,c+12\,d\,x}+{\mathrm {e}}^{14\,c+14\,d\,x}+1\right )}-\frac {2\,\left (3\,a^2\,b+30\,a\,b^2+50\,b^3\right )}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {32\,b^3}{d\,\left (8\,{\mathrm {e}}^{2\,c+2\,d\,x}+28\,{\mathrm {e}}^{4\,c+4\,d\,x}+56\,{\mathrm {e}}^{6\,c+6\,d\,x}+70\,{\mathrm {e}}^{8\,c+8\,d\,x}+56\,{\mathrm {e}}^{10\,c+10\,d\,x}+28\,{\mathrm {e}}^{12\,c+12\,d\,x}+8\,{\mathrm {e}}^{14\,c+14\,d\,x}+{\mathrm {e}}^{16\,c+16\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,{\left (a+b\right )}^3}{64\,d}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,{\left (a-b\right )}^3}{64\,d}+\frac {8\,\left (23\,b^3+9\,a\,b^2\right )}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2\,\left (2\,a+11\,b\right )}{16\,d}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a-b\right )}^2\,\left (2\,a-11\,b\right )}{16\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((189*a*b^2)/8 - 9*a^2*b + (3*a^3)/8 - 15*b^3) - (4*(12*a*b^2 + 71*b^3))/(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)) - (256*b^3)/(d*(6*exp(2*c + 2*d*x) + 15*exp(4*c + 4*d*x
) + 20*exp(6*c + 6*d*x) + 15*exp(8*c + 8*d*x) + 6*exp(10*c + 10*d*x) + exp(12*c + 12*d*x) + 1)) + (log(exp(2*c
)*exp(2*d*x) + 1)*(9*a^2*b + 15*b^3))/d + (2*(30*a*b^2 + 3*a^2*b + 20*b^3))/(d*(exp(2*c + 2*d*x) + 1)) + (32*(
3*a*b^2 + 50*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)) + (128*b^3)/(d*(7*exp(2*c + 2*d*x) + 21*exp(4*c + 4*d*x) + 35*exp(6*c + 6*d*x) + 35*
exp(8*c + 8*d*x) + 21*exp(10*c + 10*d*x) + 7*exp(12*c + 12*d*x) + exp(14*c + 14*d*x) + 1)) - (2*(30*a*b^2 + 3*
a^2*b + 50*b^3))/(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) - (32*b^3)/(d*(8*exp(2*c + 2*d*x) + 28*exp(4*
c + 4*d*x) + 56*exp(6*c + 6*d*x) + 70*exp(8*c + 8*d*x) + 56*exp(10*c + 10*d*x) + 28*exp(12*c + 12*d*x) + 8*exp
(14*c + 14*d*x) + exp(16*c + 16*d*x) + 1)) + (exp(4*c + 4*d*x)*(a + b)^3)/(64*d) - (exp(- 4*c - 4*d*x)*(a - b)
^3)/(64*d) + (8*(9*a*b^2 + 23*b^3))/(d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) - (ex
p(2*c + 2*d*x)*(a + b)^2*(2*a + 11*b))/(16*d) + (exp(- 2*c - 2*d*x)*(a - b)^2*(2*a - 11*b))/(16*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(sinh(d*x+c)**4*(a+b*tanh(d*x+c)**3)**3,x)

[Out]

Timed out

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