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

Optimal. Leaf size=50 \[ -\frac{a^2 \coth (c+d x)}{d}+\frac{1}{2} b x (4 a-b)+\frac{b^2 \sinh (c+d x) \cosh (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0813633, antiderivative size = 64, normalized size of antiderivative = 1.28, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3187, 462, 385, 206} \[ \frac{\left (2 a^2+b^2\right ) \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{a^2 \cosh ^2(c+d x) \coth (c+d x)}{d}+\frac{1}{2} b x (4 a-b) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 385

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

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

Mathematica [A]  time = 0.162947, size = 56, normalized size = 1.12 \[ -\frac{a^2 \coth (c+d x)}{d}+2 a b x+\frac{b^2 (-c-d x)}{2 d}+\frac{b^2 \sinh (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 52, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( -{a}^{2}{\rm coth} \left (dx+c\right )+2\,ab \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)^2*(a+b*sinh(d*x+c)^2)^2,x)

[Out]

1/d*(-a^2*coth(d*x+c)+2*a*b*(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 [A]  time = 1.01429, size = 85, normalized size = 1.7 \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 )} + 2 \, a b x + \frac{2 \, a^{2}}{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)^2)^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) + 2*a*b*x + 2*a^2/(d*(e^(-2*d*x - 2*c) - 1))

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Fricas [A]  time = 1.87027, size = 215, normalized size = 4.3 \begin{align*} \frac{b^{2} \cosh \left (d x + c\right )^{3} + 3 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} -{\left (8 \, a^{2} + b^{2}\right )} \cosh \left (d x + c\right ) + 4 \,{\left ({\left (4 \, a b - b^{2}\right )} d x + 2 \, a^{2}\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)^2)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.22766, size = 189, normalized size = 3.78 \begin{align*} \frac{b^{2} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac{{\left (4 \, a b - b^{2}\right )}{\left (d x + c\right )}}{2 \, d} - \frac{4 \, a b e^{\left (4 \, d x + 4 \, c\right )} - b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 16 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}}{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)^2)^2,x, algorithm="giac")

[Out]

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