3.194 \(\int \text{csch}^6(c+d x) (a+b \sinh ^4(c+d x)) \, dx\)

Optimal. Leaf size=47 \[ -\frac{(a+b) \coth (c+d x)}{d}-\frac{a \coth ^5(c+d x)}{5 d}+\frac{2 a \coth ^3(c+d x)}{3 d} \]

[Out]

-(((a + b)*Coth[c + d*x])/d) + (2*a*Coth[c + d*x]^3)/(3*d) - (a*Coth[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.0434493, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3217, 14} \[ -\frac{(a+b) \coth (c+d x)}{d}-\frac{a \coth ^5(c+d x)}{5 d}+\frac{2 a \coth ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^6*(a + b*Sinh[c + d*x]^4),x]

[Out]

-(((a + b)*Coth[c + d*x])/d) + (2*a*Coth[c + d*x]^3)/(3*d) - (a*Coth[c + d*x]^5)/(5*d)

Rule 3217

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p)/(1 + ff^2
*x^2)^(m/2 + 2*p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0342195, size = 71, normalized size = 1.51 \[ -\frac{8 a \coth (c+d x)}{15 d}-\frac{a \coth (c+d x) \text{csch}^4(c+d x)}{5 d}+\frac{4 a \coth (c+d x) \text{csch}^2(c+d x)}{15 d}-\frac{b \coth (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^6*(a + b*Sinh[c + d*x]^4),x]

[Out]

(-8*a*Coth[c + d*x])/(15*d) - (b*Coth[c + d*x])/d + (4*a*Coth[c + d*x]*Csch[c + d*x]^2)/(15*d) - (a*Coth[c + d
*x]*Csch[c + d*x]^4)/(5*d)

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Maple [A]  time = 0.036, size = 45, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( a \left ( -{\frac{8}{15}}-{\frac{ \left ({\rm csch} \left (dx+c\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm csch} \left (dx+c\right ) \right ) ^{2}}{15}} \right ){\rm coth} \left (dx+c\right )-b{\rm coth} \left (dx+c\right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^6*(a+b*sinh(d*x+c)^4),x)

[Out]

1/d*(a*(-8/15-1/5*csch(d*x+c)^4+4/15*csch(d*x+c)^2)*coth(d*x+c)-b*coth(d*x+c))

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Maxima [B]  time = 1.04262, size = 308, normalized size = 6.55 \begin{align*} -\frac{16}{15} \, a{\left (\frac{5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac{10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac{1}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + \frac{2 \, b}{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)^6*(a+b*sinh(d*x+c)^4),x, algorithm="maxima")

[Out]

-16/15*a*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x
- 8*c) + e^(-10*d*x - 10*c) - 1)) - 10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-
6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) - 1/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c)
+ 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1))) + 2*b/(d*(e^(-2*d*x - 2*c) - 1))

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Fricas [B]  time = 1.66127, size = 903, normalized size = 19.21 \begin{align*} -\frac{4 \,{\left ({\left (4 \, a + 15 \, b\right )} \cosh \left (d x + c\right )^{4} - 16 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (4 \, a + 15 \, b\right )} \sinh \left (d x + c\right )^{4} - 20 \,{\left (a + 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \,{\left (4 \, a + 15 \, b\right )} \cosh \left (d x + c\right )^{2} - 10 \, a - 30 \, b\right )} \sinh \left (d x + c\right )^{2} - 8 \,{\left (2 \, a \cosh \left (d x + c\right )^{3} - 5 \, a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 40 \, a + 45 \, b\right )}}{15 \,{\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} - 6 \, d \cosh \left (d x + c\right )^{4} + 3 \,{\left (5 \, d \cosh \left (d x + c\right )^{2} - 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \,{\left (5 \, d \cosh \left (d x + c\right )^{3} - 4 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 15 \, d \cosh \left (d x + c\right )^{2} + 3 \,{\left (5 \, d \cosh \left (d x + c\right )^{4} - 12 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{2} + 2 \,{\left (3 \, d \cosh \left (d x + c\right )^{5} - 8 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 10 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^6*(a+b*sinh(d*x+c)^4),x, algorithm="fricas")

[Out]

-4/15*((4*a + 15*b)*cosh(d*x + c)^4 - 16*a*cosh(d*x + c)*sinh(d*x + c)^3 + (4*a + 15*b)*sinh(d*x + c)^4 - 20*(
a + 3*b)*cosh(d*x + c)^2 + 2*(3*(4*a + 15*b)*cosh(d*x + c)^2 - 10*a - 30*b)*sinh(d*x + c)^2 - 8*(2*a*cosh(d*x
+ c)^3 - 5*a*cosh(d*x + c))*sinh(d*x + c) + 40*a + 45*b)/(d*cosh(d*x + c)^6 + 6*d*cosh(d*x + c)*sinh(d*x + c)^
5 + d*sinh(d*x + c)^6 - 6*d*cosh(d*x + c)^4 + 3*(5*d*cosh(d*x + c)^2 - 2*d)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*x
+ c)^3 - 4*d*cosh(d*x + c))*sinh(d*x + c)^3 + 15*d*cosh(d*x + c)^2 + 3*(5*d*cosh(d*x + c)^4 - 12*d*cosh(d*x +
c)^2 + 5*d)*sinh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^5 - 8*d*cosh(d*x + c)^3 + 5*d*cosh(d*x + c))*sinh(d*x + c)
- 10*d)

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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)**6*(a+b*sinh(d*x+c)**4),x)

[Out]

Timed out

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Giac [B]  time = 1.14916, size = 131, normalized size = 2.79 \begin{align*} -\frac{2 \,{\left (15 \, b e^{\left (8 \, d x + 8 \, c\right )} - 60 \, b e^{\left (6 \, d x + 6 \, c\right )} + 80 \, a e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b e^{\left (4 \, d x + 4 \, c\right )} - 40 \, a e^{\left (2 \, d x + 2 \, c\right )} - 60 \, b e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a + 15 \, b\right )}}{15 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^6*(a+b*sinh(d*x+c)^4),x, algorithm="giac")

[Out]

-2/15*(15*b*e^(8*d*x + 8*c) - 60*b*e^(6*d*x + 6*c) + 80*a*e^(4*d*x + 4*c) + 90*b*e^(4*d*x + 4*c) - 40*a*e^(2*d
*x + 2*c) - 60*b*e^(2*d*x + 2*c) + 8*a + 15*b)/(d*(e^(2*d*x + 2*c) - 1)^5)