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

Optimal. Leaf size=75 \[ -\frac{b \left (6 a^2-b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 d}+a^3 x-\frac{5 a b^2 \coth (c+d x)}{2 d}-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))}{2 d} \]

[Out]

a^3*x - (b*(6*a^2 - b^2)*ArcTanh[Cosh[c + d*x]])/(2*d) - (5*a*b^2*Coth[c + d*x])/(2*d) - (b^2*Coth[c + d*x]*(a
+ b*Csch[c + d*x]))/(2*d)

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Rubi [A]  time = 0.051388, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3782, 3770, 3767, 8} \[ -\frac{b \left (6 a^2-b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 d}+a^3 x-\frac{5 a b^2 \coth (c+d x)}{2 d}-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csch[c + d*x])^3,x]

[Out]

a^3*x - (b*(6*a^2 - b^2)*ArcTanh[Cosh[c + d*x]])/(2*d) - (5*a*b^2*Coth[c + d*x])/(2*d) - (b^2*Coth[c + d*x]*(a
+ b*Csch[c + d*x]))/(2*d)

Rule 3782

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) +
3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (a+b \text{csch}(c+d x))^3 \, dx &=-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))}{2 d}+\frac{1}{2} \int \left (2 a^3+b \left (6 a^2-b^2\right ) \text{csch}(c+d x)+5 a b^2 \text{csch}^2(c+d x)\right ) \, dx\\ &=a^3 x-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))}{2 d}+\frac{1}{2} \left (5 a b^2\right ) \int \text{csch}^2(c+d x) \, dx+\frac{1}{2} \left (b \left (6 a^2-b^2\right )\right ) \int \text{csch}(c+d x) \, dx\\ &=a^3 x-\frac{b \left (6 a^2-b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 d}-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))}{2 d}-\frac{\left (5 i a b^2\right ) \operatorname{Subst}(\int 1 \, dx,x,-i \coth (c+d x))}{2 d}\\ &=a^3 x-\frac{b \left (6 a^2-b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 d}-\frac{5 a b^2 \coth (c+d x)}{2 d}-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.880053, size = 118, normalized size = 1.57 \[ -\frac{-24 a^2 b \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )-8 a^3 c-8 a^3 d x+12 a b^2 \tanh \left (\frac{1}{2} (c+d x)\right )+12 a b^2 \coth \left (\frac{1}{2} (c+d x)\right )+b^3 \text{csch}^2\left (\frac{1}{2} (c+d x)\right )+b^3 \text{sech}^2\left (\frac{1}{2} (c+d x)\right )+4 b^3 \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csch[c + d*x])^3,x]

[Out]

-(-8*a^3*c - 8*a^3*d*x + 12*a*b^2*Coth[(c + d*x)/2] + b^3*Csch[(c + d*x)/2]^2 - 24*a^2*b*Log[Tanh[(c + d*x)/2]
] + 4*b^3*Log[Tanh[(c + d*x)/2]] + b^3*Sech[(c + d*x)/2]^2 + 12*a*b^2*Tanh[(c + d*x)/2])/(8*d)

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Maple [A]  time = 0.028, size = 66, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( dx+c \right ) -6\,{a}^{2}b{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) -3\,a{b}^{2}{\rm coth} \left (dx+c\right )+{b}^{3} \left ( -{\frac{{\rm csch} \left (dx+c\right ){\rm coth} \left (dx+c\right )}{2}}+{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(d*x+c)-6*a^2*b*arctanh(exp(d*x+c))-3*a*b^2*coth(d*x+c)+b^3*(-1/2*csch(d*x+c)*coth(d*x+c)+arctanh(exp
(d*x+c))))

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Maxima [A]  time = 0.994928, size = 184, normalized size = 2.45 \begin{align*} a^{3} x + \frac{1}{2} \, b^{3}{\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 )} + \frac{3 \, a^{2} b \log \left (\tanh \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{d} + \frac{6 \, a b^{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((a+b*csch(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.66081, size = 1844, normalized size = 24.59 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a^3*d*x*cosh(d*x + c)^4 + 2*a^3*d*x*sinh(d*x + c)^4 - 2*b^3*cosh(d*x + c)^3 + 2*a^3*d*x - 2*b^3*cosh(d*
x + c) + 2*(4*a^3*d*x*cosh(d*x + c) - b^3)*sinh(d*x + c)^3 + 12*a*b^2 - 4*(a^3*d*x + 3*a*b^2)*cosh(d*x + c)^2
+ 2*(6*a^3*d*x*cosh(d*x + c)^2 - 2*a^3*d*x - 3*b^3*cosh(d*x + c) - 6*a*b^2)*sinh(d*x + c)^2 - ((6*a^2*b - b^3)
*cosh(d*x + c)^4 + 4*(6*a^2*b - b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (6*a^2*b - b^3)*sinh(d*x + c)^4 + 6*a^2*b
- b^3 - 2*(6*a^2*b - b^3)*cosh(d*x + c)^2 - 2*(6*a^2*b - b^3 - 3*(6*a^2*b - b^3)*cosh(d*x + c)^2)*sinh(d*x +
c)^2 + 4*((6*a^2*b - b^3)*cosh(d*x + c)^3 - (6*a^2*b - b^3)*cosh(d*x + c))*sinh(d*x + c))*log(cosh(d*x + c) +
sinh(d*x + c) + 1) + ((6*a^2*b - b^3)*cosh(d*x + c)^4 + 4*(6*a^2*b - b^3)*cosh(d*x + c)*sinh(d*x + c)^3 + (6*a
^2*b - b^3)*sinh(d*x + c)^4 + 6*a^2*b - b^3 - 2*(6*a^2*b - b^3)*cosh(d*x + c)^2 - 2*(6*a^2*b - b^3 - 3*(6*a^2*
b - b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((6*a^2*b - b^3)*cosh(d*x + c)^3 - (6*a^2*b - b^3)*cosh(d*x + c)
)*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 2*(4*a^3*d*x*cosh(d*x + c)^3 - 3*b^3*cosh(d*x + c)^2
- b^3 - 4*(a^3*d*x + 3*a*b^2)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^4 + 4*d*cosh(d*x + c)*sinh(d*x +
c)^3 + d*sinh(d*x + c)^4 - 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^2 + 4*(d*cosh(d*x
+ c)^3 - d*cosh(d*x + c))*sinh(d*x + c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{csch}{\left (c + d x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(d*x+c))**3,x)

[Out]

Integral((a + b*csch(c + d*x))**3, x)

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Giac [A]  time = 1.18527, size = 174, normalized size = 2.32 \begin{align*} \frac{{\left (d x + c\right )} a^{3}}{d} - \frac{{\left (6 \, a^{2} b - b^{3}\right )} \log \left (e^{\left (d x + c\right )} + 1\right )}{2 \, d} + \frac{{\left (6 \, a^{2} b - b^{3}\right )} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{2 \, d} - \frac{b^{3} e^{\left (3 \, d x + 3 \, c\right )} + 6 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{3} e^{\left (d x + c\right )} - 6 \, a b^{2}}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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