Integrand size = 13, antiderivative size = 168 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=\frac {1}{16 (a+b) (1-\coth (x))^2}+\frac {3 a+5 b}{16 (a+b)^2 (1-\coth (x))}-\frac {1}{16 (a-b) (1+\coth (x))^2}-\frac {3 a-5 b}{16 (a-b)^2 (1+\coth (x))}-\frac {\left (3 a^2+9 a b+8 b^2\right ) \log (1-\coth (x))}{16 (a+b)^3}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \log (1+\coth (x))}{16 (a-b)^3}-\frac {b^5 \log (a+b \coth (x))}{\left (a^2-b^2\right )^3} \] Output:
1/16/(a+b)/(1-coth(x))^2+1/16*(3*a+5*b)/(a+b)^2/(1-coth(x))-1/16/(a-b)/(1+ coth(x))^2-1/16*(3*a-5*b)/(a-b)^2/(1+coth(x))-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
Time = 0.38 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.93 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=\frac {12 a^5 x-40 a^3 b^2 x+60 a b^4 x+4 b \left (a^4-4 a^2 b^2+3 b^4\right ) \cosh (2 x)-b \left (a^2-b^2\right )^2 \cosh (4 x)-32 b^5 \log (b \cosh (x)+a \sinh (x))-8 a^5 \sinh (2 x)+24 a^3 b^2 \sinh (2 x)-16 a b^4 \sinh (2 x)+a^5 \sinh (4 x)-2 a^3 b^2 \sinh (4 x)+a b^4 \sinh (4 x)}{32 (a-b)^3 (a+b)^3} \] Input:
Integrate[Sinh[x]^4/(a + b*Coth[x]),x]
Output:
(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)
Time = 0.56 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3987, 27, 477, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec \left (-\frac {\pi }{2}+i x\right )^4 \left (a-i b \tan \left (-\frac {\pi }{2}+i x\right )\right )}dx\) |
| \(\Big \downarrow \) 3987 |
| \(\displaystyle \frac {\int \frac {b^6}{(a+b \coth (x)) \left (b^2-b^2 \coth ^2(x)\right )^3}d(b \coth (x))}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle b^5 \int \frac {1}{(a+b \coth (x)) \left (b^2-b^2 \coth ^2(x)\right )^3}d(b \coth (x))\) |
| \(\Big \downarrow \) 477 |
| \(\displaystyle \frac {\int \left (-\frac {b^6}{\left (a^2-b^2\right )^3 (a+b \coth (x))}+\frac {b^3}{8 (a+b) (b-b \coth (x))^3}+\frac {b^3}{8 (a-b) (\coth (x) b+b)^3}+\frac {(3 a+5 b) b^2}{16 (a+b)^2 (b-b \coth (x))^2}+\frac {(3 a-5 b) b^2}{16 (a-b)^2 (\coth (x) b+b)^2}+\frac {\left (3 a^2+9 b a+8 b^2\right ) b}{16 (a+b)^3 (b-b \coth (x))}+\frac {\left (3 a^2-9 b a+8 b^2\right ) b}{16 (a-b)^3 (\coth (x) b+b)}\right )d(b \coth (x))}{b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {b \left (3 a^2+9 a b+8 b^2\right ) \log (b-b \coth (x))}{16 (a+b)^3}+\frac {b \left (3 a^2-9 a b+8 b^2\right ) \log (b \coth (x)+b)}{16 (a-b)^3}-\frac {b^6 \log (a+b \coth (x))}{\left (a^2-b^2\right )^3}+\frac {b^3}{16 (a+b) (b-b \coth (x))^2}-\frac {b^3}{16 (a-b) (b \coth (x)+b)^2}+\frac {b^2 (3 a+5 b)}{16 (a+b)^2 (b-b \coth (x))}-\frac {b^2 (3 a-5 b)}{16 (a-b)^2 (b \coth (x)+b)}}{b}\) |
Input:
Int[Sinh[x]^4/(a + b*Coth[x]),x]
Output:
(b^3/(16*(a + b)*(b - b*Coth[x])^2) + (b^2*(3*a + 5*b))/(16*(a + b)^2*(b - b*Coth[x])) - b^3/(16*(a - b)*(b + b*Coth[x])^2) - ((3*a - 5*b)*b^2)/(16* (a - b)^2*(b + b*Coth[x])) - (b*(3*a^2 + 9*a*b + 8*b^2)*Log[b - b*Coth[x]] )/(16*(a + b)^3) - (b^6*Log[a + b*Coth[x]])/(a^2 - b^2)^3 + (b*(3*a^2 - 9* a*b + 8*b^2)*Log[b + b*Coth[x]])/(16*(a - b)^3))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[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]
Time = 6.02 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.14
| method | result | size |
| risch | \(\frac {3 x \,a^{2}}{8 \left (a +b \right )^{3}}+\frac {9 x a b}{8 \left (a +b \right )^{3}}+\frac {x \,b^{2}}{\left (a +b \right )^{3}}+\frac {{\mathrm e}^{4 x}}{64 a +64 b}-\frac {{\mathrm e}^{2 x} a}{8 \left (a +b \right )^{2}}-\frac {3 \,{\mathrm e}^{2 x} b}{16 \left (a +b \right )^{2}}+\frac {{\mathrm e}^{-2 x} a}{8 \left (a -b \right )^{2}}-\frac {3 \,{\mathrm e}^{-2 x} b}{16 \left (a -b \right )^{2}}-\frac {{\mathrm e}^{-4 x}}{64 \left (a -b \right )}+\frac {2 b^{5} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}-\frac {b^{5} \ln \left ({\mathrm e}^{2 x}-\frac {a -b}{a +b}\right )}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}\) | \(192\) |
| default | \(-\frac {16}{\left (64 a -64 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {64}{\left (128 a -128 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {-a +3 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 a -5 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\left (3 a^{2}-9 a b +8 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 \left (a -b \right )^{3}}-\frac {b^{5} \ln \left (b \tanh \left (\frac {x}{2}\right )^{2}+2 a \tanh \left (\frac {x}{2}\right )+b \right )}{\left (a -b \right )^{3} \left (a +b \right )^{3}}+\frac {16}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {64}{\left (128 a +128 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a +3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {3 a +5 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-3 a^{2}-9 a b -8 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}\) | \(263\) |
Input:
int(sinh(x)^4/(a+b*coth(x)),x,method=_RETURNVERBOSE)
Output:
3/8*x/(a+b)^3*a^2+9/8*x/(a+b)^3*a*b+x/(a+b)^3*b^2+1/64/(a+b)*exp(4*x)-1/8/ (a+b)^2*exp(2*x)*a-3/16/(a+b)^2*exp(2*x)*b+1/8/(a-b)^2*exp(-2*x)*a-3/16/(a -b)^2*exp(-2*x)*b-1/64/(a-b)*exp(-4*x)+2*b^5/(a^6-3*a^4*b^2+3*a^2*b^4-b^6) *x-b^5/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*ln(exp(2*x)-(a-b)/(a+b))
Leaf count of result is larger than twice the leaf count of optimal. 1279 vs. \(2 (152) = 304\).
Time = 0.11 (sec) , antiderivative size = 1279, normalized size of antiderivative = 7.61 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=\text {Too large to display} \] Input:
integrate(sinh(x)^4/(a+b*coth(x)),x, algorithm="fricas")
Output:
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*co...
\[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=\int \frac {\sinh ^{4}{\left (x \right )}}{a + b \coth {\left (x \right )}}\, dx \] Input:
integrate(sinh(x)**4/(a+b*coth(x)),x)
Output:
Integral(sinh(x)**4/(a + b*coth(x)), x)
Time = 0.04 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.99 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=-\frac {b^{5} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} + 9 \, a b + 8 \, b^{2}\right )} x}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (4 \, {\left (2 \, a + 3 \, b\right )} e^{\left (-2 \, x\right )} - a - b\right )} e^{\left (4 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {4 \, {\left (2 \, a - 3 \, b\right )} e^{\left (-2 \, x\right )} - {\left (a - b\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} \] Input:
integrate(sinh(x)^4/(a+b*coth(x)),x, algorithm="maxima")
Output:
-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)
Time = 0.14 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.36 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=-\frac {b^{5} \log \left ({\left | -a e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (3 \, a^{2} - 9 \, a b + 8 \, b^{2}\right )} x}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac {{\left (18 \, a^{2} e^{\left (4 \, x\right )} - 54 \, a b e^{\left (4 \, x\right )} + 48 \, b^{2} e^{\left (4 \, x\right )} - 8 \, a^{2} e^{\left (2 \, x\right )} + 20 \, a b e^{\left (2 \, x\right )} - 12 \, b^{2} e^{\left (2 \, x\right )} + a^{2} - 2 \, a b + b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} - 8 \, a e^{\left (2 \, x\right )} - 12 \, b e^{\left (2 \, x\right )}}{64 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} \] Input:
integrate(sinh(x)^4/(a+b*coth(x)),x, algorithm="giac")
Output:
-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) + 2 0*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)
Time = 2.74 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.85 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=\frac {{\mathrm {e}}^{4\,x}}{64\,a+64\,b}-\frac {{\mathrm {e}}^{-4\,x}}{64\,a-64\,b}+\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a-3\,b\right )}{16\,{\left (a-b\right )}^2}-\frac {b^5\,\ln \left (b-a+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6}+\frac {x\,\left (3\,a^2-9\,a\,b+8\,b^2\right )}{8\,{\left (a-b\right )}^3}-\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a+3\,b\right )}{16\,{\left (a+b\right )}^2} \] Input:
int(sinh(x)^4/(a + b*coth(x)),x)
Output:
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) - (exp( 2*x)*(2*a + 3*b))/(16*(a + b)^2)
Time = 0.20 (sec) , antiderivative size = 355, normalized size of antiderivative = 2.11 \[ \int \frac {\sinh ^4(x)}{a+b \coth (x)} \, dx=\frac {-2 e^{8 x} a^{3} b^{2}+2 e^{8 x} a^{2} b^{3}+4 e^{6 x} a^{4} b +24 e^{6 x} a^{3} b^{2}-16 e^{6 x} a^{2} b^{3}-16 e^{6 x} a \,b^{4}-64 e^{4 x} \mathrm {log}\left (e^{2 x} a +e^{2 x} b -a +b \right ) b^{5}+24 e^{4 x} a^{5} x +64 e^{4 x} b^{5} x +4 e^{2 x} a^{4} b -24 e^{2 x} a^{3} b^{2}-16 e^{2 x} a^{2} b^{3}+16 e^{2 x} a \,b^{4}-e^{8 x} a^{4} b +e^{8 x} a \,b^{4}-a^{5}+2 a^{3} b^{2}-b^{5}+2 a^{2} b^{3}-80 e^{4 x} a^{3} b^{2} x +120 e^{4 x} a \,b^{4} x -a^{4} b +e^{8 x} a^{5}-e^{8 x} b^{5}-8 e^{6 x} a^{5}+12 e^{6 x} b^{5}+8 e^{2 x} a^{5}+12 e^{2 x} b^{5}-a \,b^{4}}{64 e^{4 x} \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )} \] Input:
int(sinh(x)^4/(a+b*coth(x)),x)
Output:
(e**(8*x)*a**5 - e**(8*x)*a**4*b - 2*e**(8*x)*a**3*b**2 + 2*e**(8*x)*a**2* b**3 + e**(8*x)*a*b**4 - e**(8*x)*b**5 - 8*e**(6*x)*a**5 + 4*e**(6*x)*a**4 *b + 24*e**(6*x)*a**3*b**2 - 16*e**(6*x)*a**2*b**3 - 16*e**(6*x)*a*b**4 + 12*e**(6*x)*b**5 - 64*e**(4*x)*log(e**(2*x)*a + e**(2*x)*b - a + b)*b**5 + 24*e**(4*x)*a**5*x - 80*e**(4*x)*a**3*b**2*x + 120*e**(4*x)*a*b**4*x + 64 *e**(4*x)*b**5*x + 8*e**(2*x)*a**5 + 4*e**(2*x)*a**4*b - 24*e**(2*x)*a**3* b**2 - 16*e**(2*x)*a**2*b**3 + 16*e**(2*x)*a*b**4 + 12*e**(2*x)*b**5 - a** 5 - a**4*b + 2*a**3*b**2 + 2*a**2*b**3 - a*b**4 - b**5)/(64*e**(4*x)*(a**6 - 3*a**4*b**2 + 3*a**2*b**4 - b**6))