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

Optimal. Leaf size=39 \[ -\frac{a \coth (c+d x)}{d}+\frac{b \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{b x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0522995, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3217, 1259, 453, 206} \[ -\frac{a \coth (c+d x)}{d}+\frac{b \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{b x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x)/2 - (a*Coth[c + d*x])/d + (b*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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.119255, size = 45, normalized size = 1.15 \[ -\frac{a \coth (c+d x)}{d}+\frac{b (-c-d x)}{2 d}+\frac{b \sinh (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(-c - d*x))/(2*d) - (a*Coth[c + d*x])/d + (b*Sinh[2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.03, size = 39, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( -{\rm coth} \left (dx+c\right )a+b \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)^2*(a+b*sinh(d*x+c)^4),x)

[Out]

1/d*(-coth(d*x+c)*a+b*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c))

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Maxima [A]  time = 1.1604, size = 73, normalized size = 1.87 \begin{align*} -\frac{1}{8} \, b{\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{2 \, a}{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)^4),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.6723, size = 185, normalized size = 4.74 \begin{align*} \frac{b \cosh \left (d x + c\right )^{3} + 3 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} -{\left (8 \, a + b\right )} \cosh \left (d x + c\right ) - 4 \,{\left (b d x - 2 \, a\right )} \sinh \left (d x + c\right )}{8 \, 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)^4),x, algorithm="fricas")

[Out]

1/8*(b*cosh(d*x + c)^3 + 3*b*cosh(d*x + c)*sinh(d*x + c)^2 - (8*a + b)*cosh(d*x + c) - 4*(b*d*x - 2*a)*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)**4),x)

[Out]

Timed out

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Giac [B]  time = 1.15407, size = 124, normalized size = 3.18 \begin{align*} -\frac{{\left (d x + c\right )} b}{2 \, d} + \frac{b e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac{b e^{\left (4 \, d x + 4 \, c\right )} - 16 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + b}{8 \, d{\left (e^{\left (4 \, d x + 4 \, c\right )} - e^{\left (2 \, d x + 2 \, c\right )}\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)^4),x, algorithm="giac")

[Out]

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