3.97 \(\int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx\)

Optimal. Leaf size=155 \[ -\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\coth (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (\coth (x)+1)}{16 (a-b)^3}-\frac {\sinh ^4(x) (b-a \coth (x))}{4 \left (a^2-b^2\right )}-\frac {b^5 \log (a+b \coth (x))}{\left (a^2-b^2\right )^3}-\frac {\sinh ^2(x) \left (4 b^3-a b^2 \left (7-\frac {3 a^2}{b^2}\right ) \coth (x)\right )}{8 \left (a^2-b^2\right )^2} \]

[Out]

-1/16*(3*a^2+9*a*b+8*b^2)*ln(1-coth(x))/(a+b)^3+1/16*(3*a^2-9*a*b+8*b^2)*ln(1+coth(x))/(a-b)^3-b^5*ln(a+b*coth
(x))/(a^2-b^2)^3-1/8*(4*b^3-a*(7-3*a^2/b^2)*b^2*coth(x))*sinh(x)^2/(a^2-b^2)^2-1/4*(b-a*coth(x))*sinh(x)^4/(a^
2-b^2)

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Rubi [A]  time = 0.24, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3506, 741, 823, 801} \[ -\frac {b^5 \log (a+b \coth (x))}{\left (a^2-b^2\right )^3}-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\coth (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (\coth (x)+1)}{16 (a-b)^3}-\frac {\sinh ^4(x) (b-a \coth (x))}{4 \left (a^2-b^2\right )}-\frac {\sinh ^2(x) \left (4 b^3-a b^2 \left (7-\frac {3 a^2}{b^2}\right ) \coth (x)\right )}{8 \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4/(a + b*Coth[x]),x]

[Out]

-((3*a^2 + 9*a*b + 8*b^2)*Log[1 - Coth[x]])/(16*(a + b)^3) + ((3*a^2 - 9*a*b + 8*b^2)*Log[1 + Coth[x]])/(16*(a
- b)^3) - (b^5*Log[a + b*Coth[x]])/(a^2 - b^2)^3 - ((4*b^3 - a*(7 - (3*a^2)/b^2)*b^2*Coth[x])*Sinh[x]^2)/(8*(
a^2 - b^2)^2) - ((b - a*Coth[x])*Sinh[x]^4)/(4*(a^2 - b^2))

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{(a+x) \left (1-\frac {x^2}{b^2}\right )^3} \, dx,x,b \coth (x)\right )}{b}\\ &=-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )}+\frac {b \operatorname {Subst}\left (\int \frac {-4+\frac {3 a^2}{b^2}+\frac {3 a x}{b^2}}{(a+x) \left (1-\frac {x^2}{b^2}\right )^2} \, dx,x,b \coth (x)\right )}{4 \left (a^2-b^2\right )}\\ &=-\frac {\left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \coth (x)\right ) \sinh ^2(x)}{8 \left (a^2-b^2\right )^2}-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )}-\frac {b^5 \operatorname {Subst}\left (\int \frac {-\frac {3 a^4-7 a^2 b^2+8 b^4}{b^6}+\frac {a \left (7-\frac {3 a^2}{b^2}\right ) x}{b^4}}{(a+x) \left (1-\frac {x^2}{b^2}\right )} \, dx,x,b \coth (x)\right )}{8 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \coth (x)\right ) \sinh ^2(x)}{8 \left (a^2-b^2\right )^2}-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )}-\frac {b^5 \operatorname {Subst}\left (\int \left (-\frac {(a-b)^2 \left (3 a^2+9 a b+8 b^2\right )}{2 b^5 (a+b) (b-x)}+\frac {8}{(a-b) (a+b) (a+x)}-\frac {(a+b)^2 \left (3 a^2-9 a b+8 b^2\right )}{2 (a-b) b^5 (b+x)}\right ) \, dx,x,b \coth (x)\right )}{8 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\coth (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (1+\coth (x))}{16 (a-b)^3}-\frac {b^5 \log (a+b \coth (x))}{\left (a^2-b^2\right )^3}-\frac {\left (4 b^3-a \left (7-\frac {3 a^2}{b^2}\right ) b^2 \coth (x)\right ) \sinh ^2(x)}{8 \left (a^2-b^2\right )^2}-\frac {(b-a \coth (x)) \sinh ^4(x)}{4 \left (a^2-b^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 156, normalized size = 1.01 \[ \frac {12 a^5 x-8 a^5 \sinh (2 x)+a^5 \sinh (4 x)-40 a^3 b^2 x+24 a^3 b^2 \sinh (2 x)-2 a^3 b^2 \sinh (4 x)-b \left (a^2-b^2\right )^2 \cosh (4 x)+4 b \left (a^4-4 a^2 b^2+3 b^4\right ) \cosh (2 x)-32 b^5 \log (a \sinh (x)+b \cosh (x))+60 a b^4 x-16 a b^4 \sinh (2 x)+a b^4 \sinh (4 x)}{32 (a-b)^3 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a^5*x - 40*a^3*b^2*x + 60*a*b^4*x + 4*b*(a^4 - 4*a^2*b^2 + 3*b^4)*Cosh[2*x] - b*(a^2 - b^2)^2*Cosh[4*x] -
32*b^5*Log[b*Cosh[x] + a*Sinh[x]] - 8*a^5*Sinh[2*x] + 24*a^3*b^2*Sinh[2*x] - 16*a*b^4*Sinh[2*x] + a^5*Sinh[4*x
] - 2*a^3*b^2*Sinh[4*x] + a*b^4*Sinh[4*x])/(32*(a - b)^3*(a + b)^3)

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fricas [B]  time = 0.43, size = 1279, normalized size = 8.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^8 + 8*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 +
a*b^4 - b^5)*cosh(x)*sinh(x)^7 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^8 - 4*(2*a^5 - a
^4*b - 6*a^3*b^2 + 4*a^2*b^3 + 4*a*b^4 - 3*b^5)*cosh(x)^6 - 4*(2*a^5 - a^4*b - 6*a^3*b^2 + 4*a^2*b^3 + 4*a*b^4
- 3*b^5 - 7*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^6 + 8*(3*a^5 - 10*a^3*b^2
+ 15*a*b^4 + 8*b^5)*x*cosh(x)^4 + 8*(7*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 - 3*(2*a^
5 - a^4*b - 6*a^3*b^2 + 4*a^2*b^3 + 4*a*b^4 - 3*b^5)*cosh(x))*sinh(x)^5 - a^5 - a^4*b + 2*a^3*b^2 + 2*a^2*b^3
- a*b^4 - b^5 + 2*(35*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 - 30*(2*a^5 - a^4*b - 6*a^
3*b^2 + 4*a^2*b^3 + 4*a*b^4 - 3*b^5)*cosh(x)^2 + 4*(3*a^5 - 10*a^3*b^2 + 15*a*b^4 + 8*b^5)*x)*sinh(x)^4 + 8*(7
*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^5 - 10*(2*a^5 - a^4*b - 6*a^3*b^2 + 4*a^2*b^3 + 4
*a*b^4 - 3*b^5)*cosh(x)^3 + 4*(3*a^5 - 10*a^3*b^2 + 15*a*b^4 + 8*b^5)*x*cosh(x))*sinh(x)^3 + 4*(2*a^5 + a^4*b
- 6*a^3*b^2 - 4*a^2*b^3 + 4*a*b^4 + 3*b^5)*cosh(x)^2 + 4*(7*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5
)*cosh(x)^6 + 2*a^5 + a^4*b - 6*a^3*b^2 - 4*a^2*b^3 + 4*a*b^4 + 3*b^5 - 15*(2*a^5 - a^4*b - 6*a^3*b^2 + 4*a^2*
b^3 + 4*a*b^4 - 3*b^5)*cosh(x)^4 + 12*(3*a^5 - 10*a^3*b^2 + 15*a*b^4 + 8*b^5)*x*cosh(x)^2)*sinh(x)^2 - 64*(b^5
*cosh(x)^4 + 4*b^5*cosh(x)^3*sinh(x) + 6*b^5*cosh(x)^2*sinh(x)^2 + 4*b^5*cosh(x)*sinh(x)^3 + b^5*sinh(x)^4)*lo
g(2*(b*cosh(x) + a*sinh(x))/(cosh(x) - sinh(x))) + 8*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh
(x)^7 - 3*(2*a^5 - a^4*b - 6*a^3*b^2 + 4*a^2*b^3 + 4*a*b^4 - 3*b^5)*cosh(x)^5 + 4*(3*a^5 - 10*a^3*b^2 + 15*a*b
^4 + 8*b^5)*x*cosh(x)^3 + (2*a^5 + a^4*b - 6*a^3*b^2 - 4*a^2*b^3 + 4*a*b^4 + 3*b^5)*cosh(x))*sinh(x))/((a^6 -
3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^4 + 4*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3*sinh(x) + 6*(a^6 - 3*
a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x)^2 + 4*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^3 + (a^
6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^4)

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giac [A]  time = 0.13, size = 229, normalized size = 1.48 \[ -\frac {b^{5} \log \left ({\left | -a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} - 9 \, a b + 8 \, b^{2}\right )} x}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac {{\left (18 \, a^{2} e^{\left (4 \, x\right )} - 54 \, a b e^{\left (4 \, x\right )} + 48 \, b^{2} e^{\left (4 \, x\right )} - 8 \, a^{2} e^{\left (2 \, x\right )} + 20 \, a b e^{\left (2 \, x\right )} - 12 \, b^{2} e^{\left (2 \, x\right )} + a^{2} - 2 \, a b + b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} - 8 \, a e^{\left (2 \, x\right )} - 12 \, b e^{\left (2 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^5*log(abs(-a*e^(2*x) - b*e^(2*x) + a - b))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + 1/8*(3*a^2 - 9*a*b + 8*b^2
)*x/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 1/64*(18*a^2*e^(4*x) - 54*a*b*e^(4*x) + 48*b^2*e^(4*x) - 8*a^2*e^(2*x) +
20*a*b*e^(2*x) - 12*b^2*e^(2*x) + a^2 - 2*a*b + b^2)*e^(-4*x)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 1/64*(a*e^(4*
x) + b*e^(4*x) - 8*a*e^(2*x) - 12*b*e^(2*x))/(a^2 + 2*a*b + b^2)

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maple [B]  time = 0.13, size = 354, normalized size = 2.28 \[ -\frac {b^{5} \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b +2 a \tanh \left (\frac {x}{2}\right )+b \right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}+\frac {16}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {64}{\left (128 a +128 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {3 a}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {5 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a^{2}}{8 \left (a +b \right )^{3}}-\frac {9 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a b}{8 \left (a +b \right )^{3}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) b^{2}}{\left (a +b \right )^{3}}-\frac {16}{\left (64 a -64 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {64}{\left (128 a -128 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {a}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 a}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {5 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a^{2}}{8 \left (a -b \right )^{3}}-\frac {9 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a b}{8 \left (a -b \right )^{3}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) b^{2}}{\left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^5/(a-b)^3/(a+b)^3*ln(tanh(1/2*x)^2*b+2*a*tanh(1/2*x)+b)+16/(64*a+64*b)/(tanh(1/2*x)-1)^4+64/(128*a+128*b)/(
tanh(1/2*x)-1)^3-1/8/(a+b)^2/(tanh(1/2*x)-1)^2*a-3/8/(a+b)^2/(tanh(1/2*x)-1)^2*b-3/8/(a+b)^2/(tanh(1/2*x)-1)*a
-5/8/(a+b)^2/(tanh(1/2*x)-1)*b-3/8/(a+b)^3*ln(tanh(1/2*x)-1)*a^2-9/8/(a+b)^3*ln(tanh(1/2*x)-1)*a*b-1/(a+b)^3*l
n(tanh(1/2*x)-1)*b^2-16/(64*a-64*b)/(tanh(1/2*x)+1)^4+64/(128*a-128*b)/(tanh(1/2*x)+1)^3+1/8/(a-b)^2/(tanh(1/2
*x)+1)^2*a-3/8/(a-b)^2/(tanh(1/2*x)+1)^2*b-3/8/(a-b)^2/(tanh(1/2*x)+1)*a+5/8/(a-b)^2/(tanh(1/2*x)+1)*b+3/8/(a-
b)^3*ln(tanh(1/2*x)+1)*a^2-9/8/(a-b)^3*ln(tanh(1/2*x)+1)*a*b+1/(a-b)^3*ln(tanh(1/2*x)+1)*b^2

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maxima [A]  time = 0.32, size = 166, normalized size = 1.07 \[ -\frac {b^{5} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} + 9 \, a b + 8 \, b^{2}\right )} x}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (4 \, {\left (2 \, a + 3 \, b\right )} e^{\left (-2 \, x\right )} - a - b\right )} e^{\left (4 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {4 \, {\left (2 \, a - 3 \, b\right )} e^{\left (-2 \, x\right )} - {\left (a - b\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^5*log(-(a - b)*e^(-2*x) + a + b)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + 1/8*(3*a^2 + 9*a*b + 8*b^2)*x/(a^3 +
3*a^2*b + 3*a*b^2 + b^3) - 1/64*(4*(2*a + 3*b)*e^(-2*x) - a - b)*e^(4*x)/(a^2 + 2*a*b + b^2) + 1/64*(4*(2*a -
3*b)*e^(-2*x) - (a - b)*e^(-4*x))/(a^2 - 2*a*b + b^2)

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mupad [B]  time = 1.63, size = 143, normalized size = 0.92 \[ \frac {{\mathrm {e}}^{4\,x}}{64\,a+64\,b}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a-64\,b}+\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a-3\,b\right )}{16\,{\left (a-b\right )}^2}-\frac {b^5\,\ln \left (b-a+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {x\,\left (3\,a^2-9\,a\,b+8\,b^2\right )}{8\,{\left (a-b\right )}^3}-\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a+3\,b\right )}{16\,{\left (a+b\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(a + b*coth(x)),x)

[Out]

exp(4*x)/(64*a + 64*b) - exp(-4*x)/(64*a - 64*b) + (exp(-2*x)*(2*a - 3*b))/(16*(a - b)^2) - (b^5*log(b - a + a
*exp(2*x) + b*exp(2*x)))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (x*(3*a^2 - 9*a*b + 8*b^2))/(8*(a - b)^3) - (ex
p(2*x)*(2*a + 3*b))/(16*(a + b)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{4}{\relax (x )}}{a + b \coth {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**4/(a+b*coth(x)),x)

[Out]

Integral(sinh(x)**4/(a + b*coth(x)), x)

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