3.17 \(\int \text{csch}^3(c+d x) (a+b \sinh ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=56 \[ -\frac{a^2 \coth (c+d x) \text{csch}(c+d x)}{2 d}+\frac{a (a-4 b) \tanh ^{-1}(\cosh (c+d x))}{2 d}+\frac{b^2 \cosh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0880729, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3186, 390, 385, 206} \[ -\frac{a^2 \coth (c+d x) \text{csch}(c+d x)}{2 d}+\frac{a (a-4 b) \tanh ^{-1}(\cosh (c+d x))}{2 d}+\frac{b^2 \cosh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^3*(a + b*Sinh[c + d*x]^2)^2,x]

[Out]

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

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

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 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}^3(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-b+b x^2\right )^2}{\left (1-x^2\right )^2} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (b^2+\frac{a (a-2 b)+2 a b x^2}{\left (1-x^2\right )^2}\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{b^2 \cosh (c+d x)}{d}+\frac{\operatorname{Subst}\left (\int \frac{a (a-2 b)+2 a b x^2}{\left (1-x^2\right )^2} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{b^2 \cosh (c+d x)}{d}-\frac{a^2 \coth (c+d x) \text{csch}(c+d x)}{2 d}+\frac{(a (a-4 b)) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cosh (c+d x)\right )}{2 d}\\ &=\frac{a (a-4 b) \tanh ^{-1}(\cosh (c+d x))}{2 d}+\frac{b^2 \cosh (c+d x)}{d}-\frac{a^2 \coth (c+d x) \text{csch}(c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 0.0639887, size = 134, normalized size = 2.39 \[ -\frac{a^2 \text{csch}^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{a^2 \text{sech}^2\left (\frac{1}{2} (c+d x)\right )}{8 d}-\frac{a^2 \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}+\frac{2 a b \log \left (\sinh \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d}-\frac{2 a b \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{d}+\frac{b^2 \sinh (c) \sinh (d x)}{d}+\frac{b^2 \cosh (c) \cosh (d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^3*(a + b*Sinh[c + d*x]^2)^2,x]

[Out]

(b^2*Cosh[c]*Cosh[d*x])/d - (a^2*Csch[(c + d*x)/2]^2)/(8*d) - (2*a*b*Log[Cosh[c/2 + (d*x)/2]])/d + (2*a*b*Log[
Sinh[c/2 + (d*x)/2]])/d - (a^2*Log[Tanh[(c + d*x)/2]])/(2*d) - (a^2*Sech[(c + d*x)/2]^2)/(8*d) + (b^2*Sinh[c]*
Sinh[d*x])/d

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Maple [A]  time = 0.041, size = 53, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{{\rm csch} \left (dx+c\right ){\rm coth} \left (dx+c\right )}{2}}+{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) \right ) -4\,ab{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) +{b}^{2}\cosh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^3*(a+b*sinh(d*x+c)^2)^2,x)

[Out]

1/d*(a^2*(-1/2*csch(d*x+c)*coth(d*x+c)+arctanh(exp(d*x+c)))-4*a*b*arctanh(exp(d*x+c))+b^2*cosh(d*x+c))

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Maxima [B]  time = 1.07468, size = 212, normalized size = 3.79 \begin{align*} \frac{1}{2} \, b^{2}{\left (\frac{e^{\left (d x + c\right )}}{d} + \frac{e^{\left (-d x - c\right )}}{d}\right )} + \frac{1}{2} \, a^{2}{\left (\frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac{2 \,{\left (e^{\left (-d x - c\right )} + e^{\left (-3 \, d x - 3 \, c\right )}\right )}}{d{\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} - 2 \, a b{\left (\frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^3*(a+b*sinh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.95247, size = 2310, normalized size = 41.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^3*(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/2*(b^2*cosh(d*x + c)^6 + 6*b^2*cosh(d*x + c)*sinh(d*x + c)^5 + b^2*sinh(d*x + c)^6 - (2*a^2 + b^2)*cosh(d*x
+ c)^4 + (15*b^2*cosh(d*x + c)^2 - 2*a^2 - b^2)*sinh(d*x + c)^4 + 4*(5*b^2*cosh(d*x + c)^3 - (2*a^2 + b^2)*cos
h(d*x + c))*sinh(d*x + c)^3 - (2*a^2 + b^2)*cosh(d*x + c)^2 + (15*b^2*cosh(d*x + c)^4 - 6*(2*a^2 + b^2)*cosh(d
*x + c)^2 - 2*a^2 - b^2)*sinh(d*x + c)^2 + b^2 + ((a^2 - 4*a*b)*cosh(d*x + c)^5 + 5*(a^2 - 4*a*b)*cosh(d*x + c
)*sinh(d*x + c)^4 + (a^2 - 4*a*b)*sinh(d*x + c)^5 - 2*(a^2 - 4*a*b)*cosh(d*x + c)^3 + 2*(5*(a^2 - 4*a*b)*cosh(
d*x + c)^2 - a^2 + 4*a*b)*sinh(d*x + c)^3 + 2*(5*(a^2 - 4*a*b)*cosh(d*x + c)^3 - 3*(a^2 - 4*a*b)*cosh(d*x + c)
)*sinh(d*x + c)^2 + (a^2 - 4*a*b)*cosh(d*x + c) + (5*(a^2 - 4*a*b)*cosh(d*x + c)^4 - 6*(a^2 - 4*a*b)*cosh(d*x
+ c)^2 + a^2 - 4*a*b)*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) + 1) - ((a^2 - 4*a*b)*cosh(d*x + c)^5 +
5*(a^2 - 4*a*b)*cosh(d*x + c)*sinh(d*x + c)^4 + (a^2 - 4*a*b)*sinh(d*x + c)^5 - 2*(a^2 - 4*a*b)*cosh(d*x + c)
^3 + 2*(5*(a^2 - 4*a*b)*cosh(d*x + c)^2 - a^2 + 4*a*b)*sinh(d*x + c)^3 + 2*(5*(a^2 - 4*a*b)*cosh(d*x + c)^3 -
3*(a^2 - 4*a*b)*cosh(d*x + c))*sinh(d*x + c)^2 + (a^2 - 4*a*b)*cosh(d*x + c) + (5*(a^2 - 4*a*b)*cosh(d*x + c)^
4 - 6*(a^2 - 4*a*b)*cosh(d*x + c)^2 + a^2 - 4*a*b)*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 2*(
3*b^2*cosh(d*x + c)^5 - 2*(2*a^2 + b^2)*cosh(d*x + c)^3 - (2*a^2 + b^2)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(
d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + d*sinh(d*x + c)^5 - 2*d*cosh(d*x + c)^3 + 2*(5*d*cosh(d*x + c
)^2 - d)*sinh(d*x + c)^3 + 2*(5*d*cosh(d*x + c)^3 - 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + d*cosh(d*x + c) + (5*
d*cosh(d*x + c)^4 - 6*d*cosh(d*x + c)^2 + 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)**3*(a+b*sinh(d*x+c)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.35923, size = 180, normalized size = 3.21 \begin{align*} \frac{b^{2}{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}{2 \, d} - \frac{a^{2}{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}{{\left ({\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} - 4\right )} d} + \frac{{\left (a^{2} - 4 \, a b\right )} \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )} + 2\right )}{4 \, d} - \frac{{\left (a^{2} - 4 \, a b\right )} \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )} - 2\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^3*(a+b*sinh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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