### 3.645 $$\int (a \coth (x)+b \text{csch}(x))^4 \, dx$$

Optimal. Leaf size=101 $\frac{4}{3} a b \left (2 a^2-b^2\right ) \sinh (x)+\frac{1}{3} a^2 \left (3 a^2-2 b^2\right ) \sinh (x) \cosh (x)-\frac{1}{3} \text{csch}(x) (a \cosh (x)+b)^2 \left (\left (3 a^2-2 b^2\right ) \cosh (x)+a b\right )+a^4 x-\frac{1}{3} \text{csch}^3(x) (a \cosh (x)+b)^3 (a+b \cosh (x))$

[Out]

a^4*x - ((b + a*Cosh[x])^2*(a*b + (3*a^2 - 2*b^2)*Cosh[x])*Csch[x])/3 - ((b + a*Cosh[x])^3*(a + b*Cosh[x])*Csc
h[x]^3)/3 + (4*a*b*(2*a^2 - b^2)*Sinh[x])/3 + (a^2*(3*a^2 - 2*b^2)*Cosh[x]*Sinh[x])/3

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Rubi [A]  time = 0.238612, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.364, Rules used = {4392, 2691, 2861, 2734} $\frac{4}{3} a b \left (2 a^2-b^2\right ) \sinh (x)+\frac{1}{3} a^2 \left (3 a^2-2 b^2\right ) \sinh (x) \cosh (x)-\frac{1}{3} \text{csch}(x) (a \cosh (x)+b)^2 \left (\left (3 a^2-2 b^2\right ) \cosh (x)+a b\right )+a^4 x-\frac{1}{3} \text{csch}^3(x) (a \cosh (x)+b)^3 (a+b \cosh (x))$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*Coth[x] + b*Csch[x])^4,x]

[Out]

a^4*x - ((b + a*Cosh[x])^2*(a*b + (3*a^2 - 2*b^2)*Cosh[x])*Csch[x])/3 - ((b + a*Cosh[x])^3*(a + b*Cosh[x])*Csc
h[x]^3)/3 + (4*a*b*(2*a^2 - b^2)*Sinh[x])/3 + (a^2*(3*a^2 - 2*b^2)*Cosh[x]*Sinh[x])/3

Rule 4392

Int[(cot[(c_.) + (d_.)*(x_)]^(n_.)*(a_.) + csc[(c_.) + (d_.)*(x_)]^(n_.)*(b_.))^(p_)*(u_.), x_Symbol] :> Int[A
ctivateTrig[u]*Csc[c + d*x]^(n*p)*(b + a*Cos[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rule 2691

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[((g*C
os[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1)*(b + a*Sin[e + f*x]))/(f*g*(p + 1)), x] + Dist[1/(g^2*(p + 1
)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*Sin
[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[
2*m, 2*p] || IntegerQ[m])

Rule 2861

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*(d + c*Sin[e + f*x]))/(f*
g*(p + 1)), x] + Dist[1/(g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(p +
2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x
])

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.265863, size = 95, normalized size = 0.94 $-\frac{1}{12} \text{csch}^3(x) \left (6 a^2 b^2 \cosh (3 x)+6 b^2 \left (3 a^2+b^2\right ) \cosh (x)+24 a^3 b \cosh (2 x)-8 a^3 b+9 a^4 x \sinh (x)-3 a^4 x \sinh (3 x)+4 a^4 \cosh (3 x)+16 a b^3-2 b^4 \cosh (3 x)\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*Coth[x] + b*Csch[x])^4,x]

[Out]

-(Csch[x]^3*(-8*a^3*b + 16*a*b^3 + 6*b^2*(3*a^2 + b^2)*Cosh[x] + 24*a^3*b*Cosh[2*x] + 4*a^4*Cosh[3*x] + 6*a^2*
b^2*Cosh[3*x] - 2*b^4*Cosh[3*x] + 9*a^4*x*Sinh[x] - 3*a^4*x*Sinh[3*x]))/12

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Maple [A]  time = 0.02, size = 123, normalized size = 1.2 \begin{align*}{a}^{4} \left ( x-{\rm coth} \left (x\right )-{\frac{ \left ({\rm coth} \left (x\right ) \right ) ^{3}}{3}} \right ) +4\,{a}^{3}b \left ( -1/3\,{\frac{ \left ( \cosh \left ( x \right ) \right ) ^{2}}{ \left ( \sinh \left ( x \right ) \right ) ^{3}}}-2/3\,{\frac{ \left ( \cosh \left ( x \right ) \right ) ^{2}}{\sinh \left ( x \right ) }}+2/3\,\sinh \left ( x \right ) \right ) +6\,{a}^{2}{b}^{2} \left ( -1/2\,{\frac{\cosh \left ( x \right ) }{ \left ( \sinh \left ( x \right ) \right ) ^{3}}}-1/2\, \left ( 2/3-1/3\, \left ({\rm csch} \left (x\right ) \right ) ^{2} \right ){\rm coth} \left (x\right ) \right ) +4\,a{b}^{3} \left ( -1/3\,{\frac{ \left ( \cosh \left ( x \right ) \right ) ^{2}}{ \left ( \sinh \left ( x \right ) \right ) ^{3}}}+1/3\,{\frac{ \left ( \cosh \left ( x \right ) \right ) ^{2}}{\sinh \left ( x \right ) }}-1/3\,\sinh \left ( x \right ) \right ) +{b}^{4} \left ({\frac{2}{3}}-{\frac{ \left ({\rm csch} \left (x\right ) \right ) ^{2}}{3}} \right ){\rm coth} \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^4*(x-coth(x)-1/3*coth(x)^3)+4*a^3*b*(-1/3/sinh(x)^3*cosh(x)^2-2/3*cosh(x)^2/sinh(x)+2/3*sinh(x))+6*a^2*b^2*(
-1/2/sinh(x)^3*cosh(x)-1/2*(2/3-1/3*csch(x)^2)*coth(x))+4*a*b^3*(-1/3/sinh(x)^3*cosh(x)^2+1/3*cosh(x)^2/sinh(x
)-1/3*sinh(x))+b^4*(2/3-1/3*csch(x)^2)*coth(x)

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Maxima [B]  time = 1.04159, size = 289, normalized size = 2.86 \begin{align*} -2 \, a^{2} b^{2} \coth \left (x\right )^{3} + \frac{1}{3} \, a^{4}{\left (3 \, x - \frac{4 \,{\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} - 2\right )}}{3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1}\right )} + \frac{8}{3} \, a^{3} b{\left (\frac{3 \, e^{\left (-x\right )}}{3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1} - \frac{2 \, e^{\left (-3 \, x\right )}}{3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1} + \frac{3 \, e^{\left (-5 \, x\right )}}{3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1}\right )} + \frac{4}{3} \, b^{4}{\left (\frac{3 \, e^{\left (-2 \, x\right )}}{3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1} - \frac{1}{3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1}\right )} + \frac{32 \, a b^{3}}{3 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*coth(x)+b*csch(x))^4,x, algorithm="maxima")

[Out]

-2*a^2*b^2*coth(x)^3 + 1/3*a^4*(3*x - 4*(3*e^(-2*x) - 3*e^(-4*x) - 2)/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6*x) - 1)
) + 8/3*a^3*b*(3*e^(-x)/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6*x) - 1) - 2*e^(-3*x)/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6
*x) - 1) + 3*e^(-5*x)/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6*x) - 1)) + 4/3*b^4*(3*e^(-2*x)/(3*e^(-2*x) - 3*e^(-4*x)
+ e^(-6*x) - 1) - 1/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6*x) - 1)) + 32/3*a*b^3/(e^(-x) - e^x)^3

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Fricas [B]  time = 2.05329, size = 497, normalized size = 4.92 \begin{align*} -\frac{24 \, a^{3} b \cosh \left (x\right )^{2} - 8 \, a^{3} b + 16 \, a b^{3} + 2 \,{\left (2 \, a^{4} + 3 \, a^{2} b^{2} - b^{4}\right )} \cosh \left (x\right )^{3} -{\left (3 \, a^{4} x + 4 \, a^{4} + 6 \, a^{2} b^{2} - 2 \, b^{4}\right )} \sinh \left (x\right )^{3} + 6 \,{\left (4 \, a^{3} b +{\left (2 \, a^{4} + 3 \, a^{2} b^{2} - b^{4}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + 6 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) + 3 \,{\left (3 \, a^{4} x + 4 \, a^{4} + 6 \, a^{2} b^{2} - 2 \, b^{4} -{\left (3 \, a^{4} x + 4 \, a^{4} + 6 \, a^{2} b^{2} - 2 \, b^{4}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )}{3 \,{\left (\sinh \left (x\right )^{3} + 3 \,{\left (\cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*coth(x)+b*csch(x))^4,x, algorithm="fricas")

[Out]

-1/3*(24*a^3*b*cosh(x)^2 - 8*a^3*b + 16*a*b^3 + 2*(2*a^4 + 3*a^2*b^2 - b^4)*cosh(x)^3 - (3*a^4*x + 4*a^4 + 6*a
^2*b^2 - 2*b^4)*sinh(x)^3 + 6*(4*a^3*b + (2*a^4 + 3*a^2*b^2 - b^4)*cosh(x))*sinh(x)^2 + 6*(3*a^2*b^2 + b^4)*co
sh(x) + 3*(3*a^4*x + 4*a^4 + 6*a^2*b^2 - 2*b^4 - (3*a^4*x + 4*a^4 + 6*a^2*b^2 - 2*b^4)*cosh(x)^2)*sinh(x))/(si
nh(x)^3 + 3*(cosh(x)^2 - 1)*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \coth{\left (x \right )} + b \operatorname{csch}{\left (x \right )}\right )^{4}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*coth(x)+b*csch(x))**4,x)

[Out]

Integral((a*coth(x) + b*csch(x))**4, x)

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Giac [A]  time = 1.22368, size = 151, normalized size = 1.5 \begin{align*} a^{4} x - \frac{4 \,{\left (6 \, a^{3} b e^{\left (5 \, x\right )} + 3 \, a^{4} e^{\left (4 \, x\right )} + 9 \, a^{2} b^{2} e^{\left (4 \, x\right )} - 4 \, a^{3} b e^{\left (3 \, x\right )} + 8 \, a b^{3} e^{\left (3 \, x\right )} - 3 \, a^{4} e^{\left (2 \, x\right )} + 3 \, b^{4} e^{\left (2 \, x\right )} + 6 \, a^{3} b e^{x} + 2 \, a^{4} + 3 \, a^{2} b^{2} - b^{4}\right )}}{3 \,{\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^4*x - 4/3*(6*a^3*b*e^(5*x) + 3*a^4*e^(4*x) + 9*a^2*b^2*e^(4*x) - 4*a^3*b*e^(3*x) + 8*a*b^3*e^(3*x) - 3*a^4*e
^(2*x) + 3*b^4*e^(2*x) + 6*a^3*b*e^x + 2*a^4 + 3*a^2*b^2 - b^4)/(e^(2*x) - 1)^3