3.205 \(\int \text{csch}^6(c+d x) (a+b \sinh ^4(c+d x))^2 \, dx\)

Optimal. Leaf size=84 \[ -\frac{a^2 \coth ^5(c+d x)}{5 d}+\frac{2 a^2 \coth ^3(c+d x)}{3 d}-\frac{a (a+2 b) \coth (c+d x)}{d}+\frac{b^2 \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{b^2 x}{2} \]

[Out]

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

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Rubi [A]  time = 0.145736, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3217, 1259, 1802, 207} \[ -\frac{a^2 \coth ^5(c+d x)}{5 d}+\frac{2 a^2 \coth ^3(c+d x)}{3 d}-\frac{a (a+2 b) \coth (c+d x)}{d}+\frac{b^2 \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{b^2 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^6*(a + b*Sinh[c + d*x]^4)^2,x]

[Out]

-(b^2*x)/2 - (a*(a + 2*b)*Coth[c + d*x])/d + (2*a^2*Coth[c + d*x]^3)/(3*d) - (a^2*Coth[c + d*x]^5)/(5*d) + (b^
2*Cosh[c + d*x]*Sinh[c + d*x])/(2*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 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.635186, size = 67, normalized size = 0.8 \[ \frac{15 b^2 (\sinh (2 (c+d x))-2 (c+d x))-4 a \coth (c+d x) \left (3 a \text{csch}^4(c+d x)-4 a \text{csch}^2(c+d x)+8 a+30 b\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^6*(a + b*Sinh[c + d*x]^4)^2,x]

[Out]

(-4*a*Coth[c + d*x]*(8*a + 30*b - 4*a*Csch[c + d*x]^2 + 3*a*Csch[c + d*x]^4) + 15*b^2*(-2*(c + d*x) + Sinh[2*(
c + d*x)]))/(60*d)

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Maple [A]  time = 0.049, size = 74, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({a}^{2} \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 )-2\,ab{\rm coth} \left (dx+c\right )+{b}^{2} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \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)^2,x)

[Out]

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

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Maxima [B]  time = 1.06529, size = 360, normalized size = 4.29 \begin{align*} -\frac{1}{8} \, b^{2}{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac{16}{15} \, a^{2}{\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{4 \, a 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.71391, size = 1185, normalized size = 14.11 \begin{align*} \frac{15 \, b^{2} \cosh \left (d x + c\right )^{7} + 105 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} -{\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} - 4 \,{\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \sinh \left (d x + c\right )^{5} + 5 \,{\left (105 \, b^{2} \cosh \left (d x + c\right )^{3} -{\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 5 \,{\left (64 \, a^{2} + 144 \, a b + 27 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 20 \,{\left (15 \, b^{2} d x - 2 \,{\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{2} - 16 \, a^{2} - 60 \, a b\right )} \sinh \left (d x + c\right )^{3} + 5 \,{\left (63 \, b^{2} \cosh \left (d x + c\right )^{5} - 2 \,{\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (64 \, a^{2} + 144 \, a b + 27 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 5 \,{\left (128 \, a^{2} + 96 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right ) - 20 \,{\left ({\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{4} + 30 \, b^{2} d x - 3 \,{\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{2} - 32 \, a^{2} - 120 \, a b\right )} \sinh \left (d x + c\right )}{120 \,{\left (d \sinh \left (d x + c\right )^{5} + 5 \,{\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \,{\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\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)^2,x, algorithm="fricas")

[Out]

1/120*(15*b^2*cosh(d*x + c)^7 + 105*b^2*cosh(d*x + c)*sinh(d*x + c)^6 - (64*a^2 + 240*a*b + 75*b^2)*cosh(d*x +
c)^5 - 4*(15*b^2*d*x - 16*a^2 - 60*a*b)*sinh(d*x + c)^5 + 5*(105*b^2*cosh(d*x + c)^3 - (64*a^2 + 240*a*b + 75
*b^2)*cosh(d*x + c))*sinh(d*x + c)^4 + 5*(64*a^2 + 144*a*b + 27*b^2)*cosh(d*x + c)^3 + 20*(15*b^2*d*x - 2*(15*
b^2*d*x - 16*a^2 - 60*a*b)*cosh(d*x + c)^2 - 16*a^2 - 60*a*b)*sinh(d*x + c)^3 + 5*(63*b^2*cosh(d*x + c)^5 - 2*
(64*a^2 + 240*a*b + 75*b^2)*cosh(d*x + c)^3 + 3*(64*a^2 + 144*a*b + 27*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 - 5
*(128*a^2 + 96*a*b + 15*b^2)*cosh(d*x + c) - 20*((15*b^2*d*x - 16*a^2 - 60*a*b)*cosh(d*x + c)^4 + 30*b^2*d*x -
3*(15*b^2*d*x - 16*a^2 - 60*a*b)*cosh(d*x + c)^2 - 32*a^2 - 120*a*b)*sinh(d*x + c))/(d*sinh(d*x + c)^5 + 5*(2
*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^3 + 5*(d*cosh(d*x + c)^4 - 3*d*cosh(d*x + c)^2 + 2*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)**6*(a+b*sinh(d*x+c)**4)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.27883, size = 234, normalized size = 2.79 \begin{align*} -\frac{{\left (d x + c\right )} b^{2}}{2 \, d} + \frac{b^{2} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac{{\left (2 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, d} - \frac{4 \,{\left (15 \, a b e^{\left (8 \, d x + 8 \, c\right )} - 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 40 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 90 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 20 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 60 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{2} + 15 \, a 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)^2,x, algorithm="giac")

[Out]

-1/2*(d*x + c)*b^2/d + 1/8*b^2*e^(2*d*x + 2*c)/d + 1/8*(2*b^2*e^(2*d*x + 2*c) - b^2)*e^(-2*d*x - 2*c)/d - 4/15
*(15*a*b*e^(8*d*x + 8*c) - 60*a*b*e^(6*d*x + 6*c) + 40*a^2*e^(4*d*x + 4*c) + 90*a*b*e^(4*d*x + 4*c) - 20*a^2*e
^(2*d*x + 2*c) - 60*a*b*e^(2*d*x + 2*c) + 4*a^2 + 15*a*b)/(d*(e^(2*d*x + 2*c) - 1)^5)