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

Optimal. Leaf size=43 \[ a^2 x-\frac {b (2 a-b) \coth (c+d x)}{d}-\frac {b^2 \coth ^3(c+d x)}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4128, 390, 206} \[ a^2 x-\frac {b (2 a-b) \coth (c+d x)}{d}-\frac {b^2 \coth ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csch[c + d*x]^2)^2,x]

[Out]

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

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])

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.64, size = 84, normalized size = 1.95 \[ \frac {4 \sinh ^4(c+d x) \left (a+b \text {csch}^2(c+d x)\right )^2 \left (3 a^2 (c+d x)-b \coth (c+d x) \left (6 a+b \text {csch}^2(c+d x)-2 b\right )\right )}{3 d (a (-\cosh (2 (c+d x)))+a-2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 180, normalized size = 4.19 \[ -\frac {2 \, {\left (3 \, a b - b^{2}\right )} \cosh \left (d x + c\right )^{3} + 6 \, {\left (3 \, a b - b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} - {\left (3 \, a^{2} d x + 6 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} - 6 \, {\left (a b - b^{2}\right )} \cosh \left (d x + c\right ) + 3 \, {\left (3 \, a^{2} d x - {\left (3 \, a^{2} d x + 6 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 6 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )}{3 \, {\left (d \sinh \left (d x + c\right )^{3} + 3 \, {\left (d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.13, size = 81, normalized size = 1.88 \[ \frac {3 \, {\left (d x + c\right )} a^{2} - \frac {4 \, {\left (3 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a b - b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.42, size = 47, normalized size = 1.09 \[ \frac {a^{2} \left (d x +c \right )-2 a b \coth \left (d x +c \right )+b^{2} \left (\frac {2}{3}-\frac {\mathrm {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.49, size = 121, normalized size = 2.81 \[ a^{2} x + \frac {4}{3} \, b^{2} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + \frac {4 \, a b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x + 4/3*b^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))

________________________________________________________________________________________

mupad [B]  time = 1.50, size = 166, normalized size = 3.86 \[ a^2\,x-\frac {\frac {4\,a\,b}{3\,d}-\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b-b^2\right )}{3\,d}+\frac {4\,a\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}+\frac {\frac {4\,\left (a\,b-b^2\right )}{3\,d}-\frac {4\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {4\,a\,b}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/sinh(c + d*x)^2)^2,x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {csch}^{2}{\left (c + d x \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(d*x+c)**2)**2,x)

[Out]

Integral((a + b*csch(c + d*x)**2)**2, x)

________________________________________________________________________________________