3.2.90 \(\int \text {csch}^2(c+d x) (a+b \sinh ^4(c+d x)) \, dx\) [190]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3296, 1273, 464, 212} \begin {gather*} -\frac {a \coth (c+d x)}{d}+\frac {b \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac {b x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*x) - (a*Coth[c + d*x])/d + (b*Cosh[c + d*x]*Sinh[c + d*x])/(2*d)

Rule 212

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

Rule 464

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 1273

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/(d + e*x^2))*(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))], 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 3296

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}^2(c+d x) \left (a+b \sinh ^4(c+d x)\right ) \, dx &=\frac {\text {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 {\text {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 \text {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*}

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Mathematica [A]
time = 0.09, size = 45, normalized size = 1.15 \begin {gather*} \frac {b (-c-d x)}{2 d}-\frac {a \coth (c+d x)}{d}+\frac {b \sinh (2 (c+d x))}{4 d} \end {gather*}

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 = 1.09, size = 55, normalized size = 1.41

method	result	size
risch	\(-\frac {b x}{2}+\frac {b \,{\mathrm e}^{2 d x +2 c}}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} b}{8 d}-\frac {2 a}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )}\)	\(55\)

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,method=_RETURNVERBOSE)

[Out]

-1/2*b*x+1/8*b/d*exp(2*d*x+2*c)-1/8/d*exp(-2*d*x-2*c)*b-2*a/d/(exp(2*d*x+2*c)-1)

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Maxima [A]
time = 0.27, size = 54, normalized size = 1.38 \begin {gather*} -\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 {gather*}

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 = 0.39, size = 70, normalized size = 1.79 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (35) = 70\).
time = 0.43, size = 88, normalized size = 2.26 \begin {gather*} -\frac {4 \, {\left (d x + c\right )} b - b e^{\left (2 \, d x + 2 \, c\right )} - \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}{e^{\left (4 \, d x + 4 \, c\right )} - e^{\left (2 \, d x + 2 \, c\right )}}}{8 \, d} \end {gather*}

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]

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

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Mupad [B]
time = 0.72, size = 54, normalized size = 1.38 \begin {gather*} \frac {b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {2\,a}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {b\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}-\frac {b\,x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sinh(c + d*x)^4)/sinh(c + d*x)^2,x)

[Out]

(b*exp(2*c + 2*d*x))/(8*d) - (2*a)/(d*(exp(2*c + 2*d*x) - 1)) - (b*exp(- 2*c - 2*d*x))/(8*d) - (b*x)/2

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