3.165 \(\int \text{csch}^2(c+d x) (a+b \sinh ^3(c+d x))^3 \, dx\)

Optimal. Leaf size=152 \[ \frac{3 a^2 b \cosh (c+d x)}{d}-\frac{a^3 \coth (c+d x)}{d}+\frac{3 a b^2 \sinh ^3(c+d x) \cosh (c+d x)}{4 d}-\frac{9 a b^2 \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{9}{8} a b^2 x+\frac{b^3 \cosh ^7(c+d x)}{7 d}-\frac{3 b^3 \cosh ^5(c+d x)}{5 d}+\frac{b^3 \cosh ^3(c+d x)}{d}-\frac{b^3 \cosh (c+d x)}{d} \]

[Out]

(9*a*b^2*x)/8 + (3*a^2*b*Cosh[c + d*x])/d - (b^3*Cosh[c + d*x])/d + (b^3*Cosh[c + d*x]^3)/d - (3*b^3*Cosh[c +
d*x]^5)/(5*d) + (b^3*Cosh[c + d*x]^7)/(7*d) - (a^3*Coth[c + d*x])/d - (9*a*b^2*Cosh[c + d*x]*Sinh[c + d*x])/(8
*d) + (3*a*b^2*Cosh[c + d*x]*Sinh[c + d*x]^3)/(4*d)

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Rubi [A]  time = 0.138335, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3220, 3767, 8, 2638, 2635, 2633} \[ \frac{3 a^2 b \cosh (c+d x)}{d}-\frac{a^3 \coth (c+d x)}{d}+\frac{3 a b^2 \sinh ^3(c+d x) \cosh (c+d x)}{4 d}-\frac{9 a b^2 \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac{9}{8} a b^2 x+\frac{b^3 \cosh ^7(c+d x)}{7 d}-\frac{3 b^3 \cosh ^5(c+d x)}{5 d}+\frac{b^3 \cosh ^3(c+d x)}{d}-\frac{b^3 \cosh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*a*b^2*x)/8 + (3*a^2*b*Cosh[c + d*x])/d - (b^3*Cosh[c + d*x])/d + (b^3*Cosh[c + d*x]^3)/d - (3*b^3*Cosh[c +
d*x]^5)/(5*d) + (b^3*Cosh[c + d*x]^7)/(7*d) - (a^3*Coth[c + d*x])/d - (9*a*b^2*Cosh[c + d*x]*Sinh[c + d*x])/(8
*d) + (3*a*b^2*Cosh[c + d*x]*Sinh[c + d*x]^3)/(4*d)

Rule 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \text{csch}^2(c+d x) \left (a+b \sinh ^3(c+d x)\right )^3 \, dx &=-\int \left (-a^3 \text{csch}^2(c+d x)-3 a^2 b \sinh (c+d x)-3 a b^2 \sinh ^4(c+d x)-b^3 \sinh ^7(c+d x)\right ) \, dx\\ &=a^3 \int \text{csch}^2(c+d x) \, dx+\left (3 a^2 b\right ) \int \sinh (c+d x) \, dx+\left (3 a b^2\right ) \int \sinh ^4(c+d x) \, dx+b^3 \int \sinh ^7(c+d x) \, dx\\ &=\frac{3 a^2 b \cosh (c+d x)}{d}+\frac{3 a b^2 \cosh (c+d x) \sinh ^3(c+d x)}{4 d}-\frac{1}{4} \left (9 a b^2\right ) \int \sinh ^2(c+d x) \, dx-\frac{\left (i a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-i \coth (c+d x))}{d}-\frac{b^3 \operatorname{Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{3 a^2 b \cosh (c+d x)}{d}-\frac{b^3 \cosh (c+d x)}{d}+\frac{b^3 \cosh ^3(c+d x)}{d}-\frac{3 b^3 \cosh ^5(c+d x)}{5 d}+\frac{b^3 \cosh ^7(c+d x)}{7 d}-\frac{a^3 \coth (c+d x)}{d}-\frac{9 a b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{3 a b^2 \cosh (c+d x) \sinh ^3(c+d x)}{4 d}+\frac{1}{8} \left (9 a b^2\right ) \int 1 \, dx\\ &=\frac{9}{8} a b^2 x+\frac{3 a^2 b \cosh (c+d x)}{d}-\frac{b^3 \cosh (c+d x)}{d}+\frac{b^3 \cosh ^3(c+d x)}{d}-\frac{3 b^3 \cosh ^5(c+d x)}{5 d}+\frac{b^3 \cosh ^7(c+d x)}{7 d}-\frac{a^3 \coth (c+d x)}{d}-\frac{9 a b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{3 a b^2 \cosh (c+d x) \sinh ^3(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.14364, size = 140, normalized size = 0.92 \[ \frac{35 b \left (192 a^2-35 b^2\right ) \cosh (c+d x)-1120 a^3 \tanh \left (\frac{1}{2} (c+d x)\right )-1120 a^3 \coth \left (\frac{1}{2} (c+d x)\right )-1680 a b^2 \sinh (2 (c+d x))+210 a b^2 \sinh (4 (c+d x))+2520 a b^2 c+2520 a b^2 d x+245 b^3 \cosh (3 (c+d x))-49 b^3 \cosh (5 (c+d x))+5 b^3 \cosh (7 (c+d x))}{2240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2520*a*b^2*c + 2520*a*b^2*d*x + 35*b*(192*a^2 - 35*b^2)*Cosh[c + d*x] + 245*b^3*Cosh[3*(c + d*x)] - 49*b^3*Co
sh[5*(c + d*x)] + 5*b^3*Cosh[7*(c + d*x)] - 1120*a^3*Coth[(c + d*x)/2] - 1680*a*b^2*Sinh[2*(c + d*x)] + 210*a*
b^2*Sinh[4*(c + d*x)] - 1120*a^3*Tanh[(c + d*x)/2])/(2240*d)

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Maple [A]  time = 0.06, size = 111, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( -{a}^{3}{\rm coth} \left (dx+c\right )+3\,{a}^{2}b\cosh \left ( dx+c \right ) +3\,a{b}^{2} \left ( \left ( 1/4\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}-3/8\,\sinh \left ( dx+c \right ) \right ) \cosh \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{b}^{3} \left ( -{\frac{16}{35}}+{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{35}}+{\frac{8\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{35}} \right ) \cosh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-a^3*coth(d*x+c)+3*a^2*b*cosh(d*x+c)+3*a*b^2*((1/4*sinh(d*x+c)^3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8
*c)+b^3*(-16/35+1/7*sinh(d*x+c)^6-6/35*sinh(d*x+c)^4+8/35*sinh(d*x+c)^2)*cosh(d*x+c))

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Maxima [A]  time = 1.20085, size = 297, normalized size = 1.95 \begin{align*} \frac{3}{64} \, a b^{2}{\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}{4480} \, b^{3}{\left (\frac{{\left (49 \, e^{\left (-2 \, d x - 2 \, c\right )} - 245 \, e^{\left (-4 \, d x - 4 \, c\right )} + 1225 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5\right )} e^{\left (7 \, d x + 7 \, c\right )}}{d} + \frac{1225 \, e^{\left (-d x - c\right )} - 245 \, e^{\left (-3 \, d x - 3 \, c\right )} + 49 \, e^{\left (-5 \, d x - 5 \, c\right )} - 5 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d}\right )} + \frac{3}{2} \, a^{2} b{\left (\frac{e^{\left (d x + c\right )}}{d} + \frac{e^{\left (-d x - c\right )}}{d}\right )} + \frac{2 \, a^{3}}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/64*a*b^2*(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) - 1/44
80*b^3*((49*e^(-2*d*x - 2*c) - 245*e^(-4*d*x - 4*c) + 1225*e^(-6*d*x - 6*c) - 5)*e^(7*d*x + 7*c)/d + (1225*e^(
-d*x - c) - 245*e^(-3*d*x - 3*c) + 49*e^(-5*d*x - 5*c) - 5*e^(-7*d*x - 7*c))/d) + 3/2*a^2*b*(e^(d*x + c)/d + e
^(-d*x - c)/d) + 2*a^3/(d*(e^(-2*d*x - 2*c) - 1))

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Fricas [B]  time = 1.87147, size = 809, normalized size = 5.32 \begin{align*} \frac{20 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{7} + 105 \, a b^{2} \cosh \left (d x + c\right )^{5} + 525 \, a b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - 945 \, a b^{2} \cosh \left (d x + c\right )^{3} + 2 \,{\left (70 \, b^{3} \cosh \left (d x + c\right )^{3} - 81 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{5} + 4 \,{\left (35 \, b^{3} \cosh \left (d x + c\right )^{5} - 135 \, b^{3} \cosh \left (d x + c\right )^{3} + 147 \, b^{3} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 105 \,{\left (10 \, a b^{2} \cosh \left (d x + c\right )^{3} - 27 \, a b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 280 \,{\left (8 \, a^{3} - 3 \, a b^{2}\right )} \cosh \left (d x + c\right ) + 2 \,{\left (10 \, b^{3} \cosh \left (d x + c\right )^{7} - 81 \, b^{3} \cosh \left (d x + c\right )^{5} + 294 \, b^{3} \cosh \left (d x + c\right )^{3} + 1260 \, a b^{2} d x + 1120 \, a^{3} + 105 \,{\left (32 \, a^{2} b - 7 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2240 \, d \sinh \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2240*(20*b^3*cosh(d*x + c)*sinh(d*x + c)^7 + 105*a*b^2*cosh(d*x + c)^5 + 525*a*b^2*cosh(d*x + c)*sinh(d*x +
c)^4 - 945*a*b^2*cosh(d*x + c)^3 + 2*(70*b^3*cosh(d*x + c)^3 - 81*b^3*cosh(d*x + c))*sinh(d*x + c)^5 + 4*(35*b
^3*cosh(d*x + c)^5 - 135*b^3*cosh(d*x + c)^3 + 147*b^3*cosh(d*x + c))*sinh(d*x + c)^3 + 105*(10*a*b^2*cosh(d*x
+ c)^3 - 27*a*b^2*cosh(d*x + c))*sinh(d*x + c)^2 - 280*(8*a^3 - 3*a*b^2)*cosh(d*x + c) + 2*(10*b^3*cosh(d*x +
c)^7 - 81*b^3*cosh(d*x + c)^5 + 294*b^3*cosh(d*x + c)^3 + 1260*a*b^2*d*x + 1120*a^3 + 105*(32*a^2*b - 7*b^3)*
cosh(d*x + c))*sinh(d*x + c))/(d*sinh(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(csch(d*x+c)**2*(a+b*sinh(d*x+c)**3)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.44803, size = 410, normalized size = 2.7 \begin{align*} \frac{9 \,{\left (d x + c\right )} a b^{2}}{8 \, d} - \frac{{\left (1890 \, a b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 294 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 210 \, a b^{2} e^{\left (3 \, d x + 3 \, c\right )} - 54 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 5 \, b^{3} - 35 \,{\left (192 \, a^{2} b - 35 \, b^{3}\right )} e^{\left (8 \, d x + 8 \, c\right )} + 560 \,{\left (16 \, a^{3} - 3 \, a b^{2}\right )} e^{\left (7 \, d x + 7 \, c\right )} + 210 \,{\left (32 \, a^{2} b - 7 \, b^{3}\right )} e^{\left (6 \, d x + 6 \, c\right )}\right )} e^{\left (-7 \, d x - 7 \, c\right )}}{4480 \, d{\left (e^{\left (d x + c\right )} + 1\right )}{\left (e^{\left (d x + c\right )} - 1\right )}} + \frac{5 \, b^{3} d^{6} e^{\left (7 \, d x + 7 \, c\right )} - 49 \, b^{3} d^{6} e^{\left (5 \, d x + 5 \, c\right )} + 210 \, a b^{2} d^{6} e^{\left (4 \, d x + 4 \, c\right )} + 245 \, b^{3} d^{6} e^{\left (3 \, d x + 3 \, c\right )} - 1680 \, a b^{2} d^{6} e^{\left (2 \, d x + 2 \, c\right )} + 6720 \, a^{2} b d^{6} e^{\left (d x + c\right )} - 1225 \, b^{3} d^{6} e^{\left (d x + c\right )}}{4480 \, d^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/8*(d*x + c)*a*b^2/d - 1/4480*(1890*a*b^2*e^(5*d*x + 5*c) + 294*b^3*e^(4*d*x + 4*c) - 210*a*b^2*e^(3*d*x + 3*
c) - 54*b^3*e^(2*d*x + 2*c) + 5*b^3 - 35*(192*a^2*b - 35*b^3)*e^(8*d*x + 8*c) + 560*(16*a^3 - 3*a*b^2)*e^(7*d*
x + 7*c) + 210*(32*a^2*b - 7*b^3)*e^(6*d*x + 6*c))*e^(-7*d*x - 7*c)/(d*(e^(d*x + c) + 1)*(e^(d*x + c) - 1)) +
1/4480*(5*b^3*d^6*e^(7*d*x + 7*c) - 49*b^3*d^6*e^(5*d*x + 5*c) + 210*a*b^2*d^6*e^(4*d*x + 4*c) + 245*b^3*d^6*e
^(3*d*x + 3*c) - 1680*a*b^2*d^6*e^(2*d*x + 2*c) + 6720*a^2*b*d^6*e^(d*x + c) - 1225*b^3*d^6*e^(d*x + c))/d^7