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

Optimal. Leaf size=109 \[ -\frac{b^2 \left (17 a^2-2 b^2\right ) \coth (c+d x)}{3 d}-\frac{2 a b \left (2 a^2-b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{d}+a^4 x-\frac{4 a b^3 \coth (c+d x) \text{csch}(c+d x)}{3 d}-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))^2}{3 d} \]

[Out]

a^4*x - (2*a*b*(2*a^2 - b^2)*ArcTanh[Cosh[c + d*x]])/d - (b^2*(17*a^2 - 2*b^2)*Coth[c + d*x])/(3*d) - (4*a*b^3
*Coth[c + d*x]*Csch[c + d*x])/(3*d) - (b^2*Coth[c + d*x]*(a + b*Csch[c + d*x])^2)/(3*d)

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Rubi [A]  time = 0.127954, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3782, 4048, 3770, 3767, 8} \[ -\frac{b^2 \left (17 a^2-2 b^2\right ) \coth (c+d x)}{3 d}-\frac{2 a b \left (2 a^2-b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{d}+a^4 x-\frac{4 a b^3 \coth (c+d x) \text{csch}(c+d x)}{3 d}-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csch[c + d*x])^4,x]

[Out]

a^4*x - (2*a*b*(2*a^2 - b^2)*ArcTanh[Cosh[c + d*x]])/d - (b^2*(17*a^2 - 2*b^2)*Coth[c + d*x])/(3*d) - (4*a*b^3
*Coth[c + d*x]*Csch[c + d*x])/(3*d) - (b^2*Coth[c + d*x]*(a + b*Csch[c + d*x])^2)/(3*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 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

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))^4 \, dx &=-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))^2}{3 d}+\frac{1}{3} \int (a+b \text{csch}(c+d x)) \left (3 a^3+b \left (9 a^2-2 b^2\right ) \text{csch}(c+d x)+8 a b^2 \text{csch}^2(c+d x)\right ) \, dx\\ &=-\frac{4 a b^3 \coth (c+d x) \text{csch}(c+d x)}{3 d}-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))^2}{3 d}+\frac{1}{6} \int \left (6 a^4+12 a b \left (2 a^2-b^2\right ) \text{csch}(c+d x)+2 b^2 \left (17 a^2-2 b^2\right ) \text{csch}^2(c+d x)\right ) \, dx\\ &=a^4 x-\frac{4 a b^3 \coth (c+d x) \text{csch}(c+d x)}{3 d}-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))^2}{3 d}+\frac{1}{3} \left (b^2 \left (17 a^2-2 b^2\right )\right ) \int \text{csch}^2(c+d x) \, dx+\left (2 a b \left (2 a^2-b^2\right )\right ) \int \text{csch}(c+d x) \, dx\\ &=a^4 x-\frac{2 a b \left (2 a^2-b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{d}-\frac{4 a b^3 \coth (c+d x) \text{csch}(c+d x)}{3 d}-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))^2}{3 d}-\frac{\left (i b^2 \left (17 a^2-2 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-i \coth (c+d x))}{3 d}\\ &=a^4 x-\frac{2 a b \left (2 a^2-b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{d}-\frac{b^2 \left (17 a^2-2 b^2\right ) \coth (c+d x)}{3 d}-\frac{4 a b^3 \coth (c+d x) \text{csch}(c+d x)}{3 d}-\frac{b^2 \coth (c+d x) (a+b \text{csch}(c+d x))^2}{3 d}\\ \end{align*}

Mathematica [B]  time = 6.22972, size = 508, normalized size = 4.66 \[ \frac{\sinh ^4(c+d x) \text{csch}\left (\frac{1}{2} (c+d x)\right ) \left (b^4 \cosh \left (\frac{1}{2} (c+d x)\right )-9 a^2 b^2 \cosh \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \text{csch}(c+d x))^4}{3 d (a \sinh (c+d x)+b)^4}+\frac{\sinh ^4(c+d x) \text{sech}\left (\frac{1}{2} (c+d x)\right ) \left (b^4 \sinh \left (\frac{1}{2} (c+d x)\right )-9 a^2 b^2 \sinh \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \text{csch}(c+d x))^4}{3 d (a \sinh (c+d x)+b)^4}+\frac{2 a b \left (2 a^2-b^2\right ) \sinh ^4(c+d x) \log \left (\tanh \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \text{csch}(c+d x))^4}{d (a \sinh (c+d x)+b)^4}+\frac{a^4 (c+d x) \sinh ^4(c+d x) (a+b \text{csch}(c+d x))^4}{d (a \sinh (c+d x)+b)^4}-\frac{a b^3 \sinh ^4(c+d x) \text{csch}^2\left (\frac{1}{2} (c+d x)\right ) (a+b \text{csch}(c+d x))^4}{2 d (a \sinh (c+d x)+b)^4}-\frac{b^4 \sinh ^4(c+d x) \coth \left (\frac{1}{2} (c+d x)\right ) \text{csch}^2\left (\frac{1}{2} (c+d x)\right ) (a+b \text{csch}(c+d x))^4}{24 d (a \sinh (c+d x)+b)^4}-\frac{a b^3 \sinh ^4(c+d x) \text{sech}^2\left (\frac{1}{2} (c+d x)\right ) (a+b \text{csch}(c+d x))^4}{2 d (a \sinh (c+d x)+b)^4}+\frac{b^4 \sinh ^4(c+d x) \tanh \left (\frac{1}{2} (c+d x)\right ) \text{sech}^2\left (\frac{1}{2} (c+d x)\right ) (a+b \text{csch}(c+d x))^4}{24 d (a \sinh (c+d x)+b)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csch[c + d*x])^4,x]

[Out]

(a^4*(c + d*x)*(a + b*Csch[c + d*x])^4*Sinh[c + d*x]^4)/(d*(b + a*Sinh[c + d*x])^4) + ((-9*a^2*b^2*Cosh[(c + d
*x)/2] + b^4*Cosh[(c + d*x)/2])*Csch[(c + d*x)/2]*(a + b*Csch[c + d*x])^4*Sinh[c + d*x]^4)/(3*d*(b + a*Sinh[c
+ d*x])^4) - (a*b^3*Csch[(c + d*x)/2]^2*(a + b*Csch[c + d*x])^4*Sinh[c + d*x]^4)/(2*d*(b + a*Sinh[c + d*x])^4)
- (b^4*Coth[(c + d*x)/2]*Csch[(c + d*x)/2]^2*(a + b*Csch[c + d*x])^4*Sinh[c + d*x]^4)/(24*d*(b + a*Sinh[c + d
*x])^4) + (2*a*b*(2*a^2 - b^2)*(a + b*Csch[c + d*x])^4*Log[Tanh[(c + d*x)/2]]*Sinh[c + d*x]^4)/(d*(b + a*Sinh[
c + d*x])^4) - (a*b^3*(a + b*Csch[c + d*x])^4*Sech[(c + d*x)/2]^2*Sinh[c + d*x]^4)/(2*d*(b + a*Sinh[c + d*x])^
4) + ((a + b*Csch[c + d*x])^4*Sech[(c + d*x)/2]*(-9*a^2*b^2*Sinh[(c + d*x)/2] + b^4*Sinh[(c + d*x)/2])*Sinh[c
+ d*x]^4)/(3*d*(b + a*Sinh[c + d*x])^4) + (b^4*(a + b*Csch[c + d*x])^4*Sech[(c + d*x)/2]^2*Sinh[c + d*x]^4*Tan
h[(c + d*x)/2])/(24*d*(b + a*Sinh[c + d*x])^4)

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Maple [A]  time = 0.039, size = 92, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({a}^{4} \left ( dx+c \right ) -8\,{a}^{3}b{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) -6\,{a}^{2}{b}^{2}{\rm coth} \left (dx+c\right )+4\,a{b}^{3} \left ( -1/2\,{\rm csch} \left (dx+c\right ){\rm coth} \left (dx+c\right )+{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) \right ) +{b}^{4} \left ({\frac{2}{3}}-{\frac{ \left ({\rm csch} \left (dx+c\right ) \right ) ^{2}}{3}} \right ){\rm coth} \left (dx+c\right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^4*(d*x+c)-8*a^3*b*arctanh(exp(d*x+c))-6*a^2*b^2*coth(d*x+c)+4*a*b^3*(-1/2*csch(d*x+c)*coth(d*x+c)+arcta
nh(exp(d*x+c)))+b^4*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c))

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Maxima [B]  time = 1.02551, size = 316, normalized size = 2.9 \begin{align*} a^{4} x + 2 \, a 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{4}{3} \, b^{4}{\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^{3} b \log \left (\tanh \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{d} + \frac{12 \, a^{2} 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))^4,x, algorithm="maxima")

[Out]

a^4*x + 2*a*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))) + 4/3*b^4*(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^3*b*log(tanh(1/2*d*x + 1/2*c))/d + 12*a^2*b^2/(d*(e^(-2*d*x - 2*c) - 1))

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Fricas [B]  time = 1.73342, size = 3453, normalized size = 31.68 \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))^4,x, algorithm="fricas")

[Out]

1/3*(3*a^4*d*x*cosh(d*x + c)^6 + 3*a^4*d*x*sinh(d*x + c)^6 - 12*a*b^3*cosh(d*x + c)^5 - 3*a^4*d*x + 6*(3*a^4*d
*x*cosh(d*x + c) - 2*a*b^3)*sinh(d*x + c)^5 + 12*a*b^3*cosh(d*x + c) - 9*(a^4*d*x + 4*a^2*b^2)*cosh(d*x + c)^4
+ 3*(15*a^4*d*x*cosh(d*x + c)^2 - 3*a^4*d*x - 20*a*b^3*cosh(d*x + c) - 12*a^2*b^2)*sinh(d*x + c)^4 - 36*a^2*b
^2 + 4*b^4 + 12*(5*a^4*d*x*cosh(d*x + c)^3 - 10*a*b^3*cosh(d*x + c)^2 - 3*(a^4*d*x + 4*a^2*b^2)*cosh(d*x + c))
*sinh(d*x + c)^3 + 3*(3*a^4*d*x + 24*a^2*b^2 - 4*b^4)*cosh(d*x + c)^2 + 3*(15*a^4*d*x*cosh(d*x + c)^4 - 40*a*b
^3*cosh(d*x + c)^3 + 3*a^4*d*x + 24*a^2*b^2 - 4*b^4 - 18*(a^4*d*x + 4*a^2*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^
2 - 6*((2*a^3*b - a*b^3)*cosh(d*x + c)^6 + 6*(2*a^3*b - a*b^3)*cosh(d*x + c)*sinh(d*x + c)^5 + (2*a^3*b - a*b^
3)*sinh(d*x + c)^6 - 3*(2*a^3*b - a*b^3)*cosh(d*x + c)^4 - 3*(2*a^3*b - a*b^3 - 5*(2*a^3*b - a*b^3)*cosh(d*x +
c)^2)*sinh(d*x + c)^4 - 2*a^3*b + a*b^3 + 4*(5*(2*a^3*b - a*b^3)*cosh(d*x + c)^3 - 3*(2*a^3*b - a*b^3)*cosh(d
*x + c))*sinh(d*x + c)^3 + 3*(2*a^3*b - a*b^3)*cosh(d*x + c)^2 + 3*(5*(2*a^3*b - a*b^3)*cosh(d*x + c)^4 + 2*a^
3*b - a*b^3 - 6*(2*a^3*b - a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 6*((2*a^3*b - a*b^3)*cosh(d*x + c)^5 - 2*
(2*a^3*b - a*b^3)*cosh(d*x + c)^3 + (2*a^3*b - a*b^3)*cosh(d*x + c))*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d
*x + c) + 1) + 6*((2*a^3*b - a*b^3)*cosh(d*x + c)^6 + 6*(2*a^3*b - a*b^3)*cosh(d*x + c)*sinh(d*x + c)^5 + (2*a
^3*b - a*b^3)*sinh(d*x + c)^6 - 3*(2*a^3*b - a*b^3)*cosh(d*x + c)^4 - 3*(2*a^3*b - a*b^3 - 5*(2*a^3*b - a*b^3)
*cosh(d*x + c)^2)*sinh(d*x + c)^4 - 2*a^3*b + a*b^3 + 4*(5*(2*a^3*b - a*b^3)*cosh(d*x + c)^3 - 3*(2*a^3*b - a*
b^3)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(2*a^3*b - a*b^3)*cosh(d*x + c)^2 + 3*(5*(2*a^3*b - a*b^3)*cosh(d*x +
c)^4 + 2*a^3*b - a*b^3 - 6*(2*a^3*b - a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 6*((2*a^3*b - a*b^3)*cosh(d*x
+ c)^5 - 2*(2*a^3*b - a*b^3)*cosh(d*x + c)^3 + (2*a^3*b - a*b^3)*cosh(d*x + c))*sinh(d*x + c))*log(cosh(d*x +
c) + sinh(d*x + c) - 1) + 6*(3*a^4*d*x*cosh(d*x + c)^5 - 10*a*b^3*cosh(d*x + c)^4 + 2*a*b^3 - 6*(a^4*d*x + 4*a
^2*b^2)*cosh(d*x + c)^3 + (3*a^4*d*x + 24*a^2*b^2 - 4*b^4)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^6 +
6*d*cosh(d*x + c)*sinh(d*x + c)^5 + d*sinh(d*x + c)^6 - 3*d*cosh(d*x + c)^4 + 3*(5*d*cosh(d*x + c)^2 - d)*sinh
(d*x + c)^4 + 4*(5*d*cosh(d*x + c)^3 - 3*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*d*cosh(d*x + c)^2 + 3*(5*d*cosh(
d*x + c)^4 - 6*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^2 + 6*(d*cosh(d*x + c)^5 - 2*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 )^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csch(d*x+c))**4,x)

[Out]

Integral((a + b*csch(c + d*x))**4, x)

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Giac [A]  time = 1.13597, size = 236, normalized size = 2.17 \begin{align*} \frac{{\left (d x + c\right )} a^{4}}{d} - \frac{2 \,{\left (2 \, a^{3} b - a b^{3}\right )} \log \left (e^{\left (d x + c\right )} + 1\right )}{d} + \frac{2 \,{\left (2 \, a^{3} b - a b^{3}\right )} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{d} - \frac{4 \,{\left (3 \, a b^{3} e^{\left (5 \, d x + 5 \, c\right )} + 9 \, a^{2} b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 18 \, a^{2} b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{4} e^{\left (2 \, d x + 2 \, c\right )} - 3 \, a b^{3} e^{\left (d x + c\right )} + 9 \, a^{2} b^{2} - b^{4}\right )}}{3 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x + c)*a^4/d - 2*(2*a^3*b - a*b^3)*log(e^(d*x + c) + 1)/d + 2*(2*a^3*b - a*b^3)*log(abs(e^(d*x + c) - 1))/d
- 4/3*(3*a*b^3*e^(5*d*x + 5*c) + 9*a^2*b^2*e^(4*d*x + 4*c) - 18*a^2*b^2*e^(2*d*x + 2*c) + 3*b^4*e^(2*d*x + 2*
c) - 3*a*b^3*e^(d*x + c) + 9*a^2*b^2 - b^4)/(d*(e^(2*d*x + 2*c) - 1)^3)