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(d*x+c)/d+2/3*a*coth(d*x+c)^3/d-1/5*a*coth(d*x+c)^5/d

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Rubi [A]  time = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.00, 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 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]]

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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [B]  time = 0.79, size = 333, normalized size = 7.09 \[ -\frac {4 \, {\left ({\left (4 \, a + 15 \, b\right )} \cosh \left (d x + c\right )^{4} - 16 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (4 \, a + 15 \, b\right )} \sinh \left (d x + c\right )^{4} - 20 \, {\left (a + 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (4 \, a + 15 \, b\right )} \cosh \left (d x + c\right )^{2} - 10 \, a - 30 \, b\right )} \sinh \left (d x + c\right )^{2} - 8 \, {\left (2 \, a \cosh \left (d x + c\right )^{3} - 5 \, a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 40 \, a + 45 \, b\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} - 6 \, d \cosh \left (d x + c\right )^{4} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{2} - 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, d \cosh \left (d x + c\right )^{3} - 4 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 15 \, d \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{4} - 12 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{5} - 8 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 10 \, d\right )}} \]

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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giac [B]  time = 0.19, size = 97, normalized size = 2.06 \[ -\frac {2 \, {\left (15 \, b e^{\left (8 \, d x + 8 \, c\right )} - 60 \, b e^{\left (6 \, d x + 6 \, c\right )} + 80 \, a e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b e^{\left (4 \, d x + 4 \, c\right )} - 40 \, a e^{\left (2 \, d x + 2 \, c\right )} - 60 \, b e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a + 15 \, b\right )}}{15 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}} \]

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)

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maple [A]  time = 0.09, size = 45, normalized size = 0.96 \[ \frac {a \left (-\frac {8}{15}-\frac {\mathrm {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )-b \coth \left (d x +c \right )}{d} \]

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 = 0.33, size = 228, normalized size = 4.85 \[ -\frac {16}{15} \, a {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + \frac {2 \, b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \]

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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mupad [B]  time = 0.72, size = 337, normalized size = 7.17 \[ \frac {\frac {2\,b}{5\,d}+\frac {6\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}-\frac {2\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (8\,a+3\,b\right )}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,\left (8\,a+3\,b\right )}{15\,d}-\frac {4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}+\frac {2\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {\frac {2\,b}{5\,d}-\frac {8\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}-\frac {8\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}+\frac {2\,b\,{\mathrm {e}}^{8\,c+8\,d\,x}}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (8\,a+3\,b\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {4\,b}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*b)/(5*d) + (6*b*exp(4*c + 4*d*x))/(5*d) - (2*b*exp(6*c + 6*d*x))/(5*d) - (2*exp(2*c + 2*d*x)*(8*a + 3*b))/
(5*d))/(6*exp(4*c + 4*d*x) - 4*exp(2*c + 2*d*x) - 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) - ((2*(8*a + 3*b)
)/(15*d) - (4*b*exp(2*c + 2*d*x))/(5*d) + (2*b*exp(4*c + 4*d*x))/(5*d))/(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*
x) + exp(6*c + 6*d*x) - 1) - ((2*b)/(5*d) - (8*b*exp(2*c + 2*d*x))/(5*d) - (8*b*exp(6*c + 6*d*x))/(5*d) + (2*b
*exp(8*c + 8*d*x))/(5*d) + (4*exp(4*c + 4*d*x)*(8*a + 3*b))/(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) - (4*b)/(5*d*(exp(2*c + 2*d*x) - 1))

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

[Out]

Timed out

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