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

Optimal. Leaf size=40 \[ -\frac {a^2 \coth ^3(c+d x)}{3 d}+\frac {a (a-2 b) \coth (c+d x)}{d}+b^2 x \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3187, 461, 207} \[ -\frac {a^2 \coth ^3(c+d x)}{3 d}+\frac {a (a-2 b) \coth (c+d x)}{d}+b^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 3187

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + (a + b)*ff^2*x^2)^p)/(1 + ff^2*x^2)^(m/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}^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a-(a-b) x^2\right )^2}{x^4 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^2}{x^4}-\frac {a (a-2 b)}{x^2}-\frac {b^2}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {a (a-2 b) \coth (c+d x)}{d}-\frac {a^2 \coth ^3(c+d x)}{3 d}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=b^2 x+\frac {a (a-2 b) \coth (c+d x)}{d}-\frac {a^2 \coth ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [B]  time = 0.72, size = 85, normalized size = 2.12 \[ \frac {4 \sinh ^4(c+d x) \left (a \text {csch}^2(c+d x)+b\right )^2 \left (3 b^2 (c+d x)-a \coth (c+d x) \left (a \text {csch}^2(c+d x)-2 a+6 b\right )\right )}{3 d (2 a+b \cosh (2 (c+d x))-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(b + a*Csch[c + d*x]^2)^2*(3*b^2*(c + d*x) - a*Coth[c + d*x]*(-2*a + 6*b + a*Csch[c + d*x]^2))*Sinh[c + d*x
]^4)/(3*d*(2*a - b + b*Cosh[2*(c + d*x)])^2)

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fricas [B]  time = 0.39, size = 174, normalized size = 4.35 \[ \frac {2 \, {\left (a^{2} - 3 \, a b\right )} \cosh \left (d x + c\right )^{3} + 6 \, {\left (a^{2} - 3 \, a b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (3 \, b^{2} d x - 2 \, a^{2} + 6 \, a b\right )} \sinh \left (d x + c\right )^{3} - 6 \, {\left (a^{2} - a b\right )} \cosh \left (d x + c\right ) - 3 \, {\left (3 \, b^{2} d x - {\left (3 \, b^{2} d x - 2 \, a^{2} + 6 \, a b\right )} \cosh \left (d x + c\right )^{2} - 2 \, a^{2} + 6 \, a b\right )} \sinh \left (d x + c\right )}{3 \, {\left (d \sinh \left (d x + c\right )^{3} + 3 \, {\left (d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4*(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/3*(2*(a^2 - 3*a*b)*cosh(d*x + c)^3 + 6*(a^2 - 3*a*b)*cosh(d*x + c)*sinh(d*x + c)^2 + (3*b^2*d*x - 2*a^2 + 6*
a*b)*sinh(d*x + c)^3 - 6*(a^2 - a*b)*cosh(d*x + c) - 3*(3*b^2*d*x - (3*b^2*d*x - 2*a^2 + 6*a*b)*cosh(d*x + c)^
2 - 2*a^2 + 6*a*b)*sinh(d*x + c))/(d*sinh(d*x + c)^3 + 3*(d*cosh(d*x + c)^2 - d)*sinh(d*x + c))

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giac [B]  time = 0.18, size = 81, normalized size = 2.02 \[ \frac {3 \, {\left (d x + c\right )} b^{2} - \frac {4 \, {\left (3 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 6 \, a b e^{\left (2 \, d x + 2 \, c\right )} - a^{2} + 3 \, a b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4*(a+b*sinh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)*b^2 - 4*(3*a*b*e^(4*d*x + 4*c) + 3*a^2*e^(2*d*x + 2*c) - 6*a*b*e^(2*d*x + 2*c) - a^2 + 3*a*b)
/(e^(2*d*x + 2*c) - 1)^3)/d

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maple [A]  time = 0.09, size = 47, normalized size = 1.18 \[ \frac {a^{2} \left (\frac {2}{3}-\frac {\mathrm {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )-2 a b \coth \left (d x +c \right )+b^{2} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.47, size = 121, normalized size = 3.02 \[ b^{2} x + \frac {4}{3} \, a^{2} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + \frac {4 \, a b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4*(a+b*sinh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

b^2*x + 4/3*a^2*(3*e^(-2*d*x - 2*c)/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1)) - 1/(
d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) - 1))) + 4*a*b/(d*(e^(-2*d*x - 2*c) - 1))

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mupad [B]  time = 0.62, size = 166, normalized size = 4.15 \[ b^2\,x-\frac {\frac {4\,a\,b}{3\,d}-\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b-a^2\right )}{3\,d}+\frac {4\,a\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}+\frac {\frac {4\,\left (a\,b-a^2\right )}{3\,d}-\frac {4\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {4\,a\,b}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x - ((4*a*b)/(3*d) - (8*exp(2*c + 2*d*x)*(a*b - a^2))/(3*d) + (4*a*b*exp(4*c + 4*d*x))/(3*d))/(3*exp(2*c +
2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1) + ((4*(a*b - a^2))/(3*d) - (4*a*b*exp(2*c + 2*d*x))/(3*d)
)/(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1) - (4*a*b)/(3*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)**4*(a+b*sinh(d*x+c)**2)**2,x)

[Out]

Timed out

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