Integrand size = 16, antiderivative size = 101 \[ \int \frac {\sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx=\frac {a^2 b x}{\left (a^2-b^2\right )^2}+\frac {b x}{2 \left (a^2-b^2\right )}-\frac {a^3 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^2}-\frac {b \cosh (x) \sinh (x)}{2 \left (a^2-b^2\right )}+\frac {a \sinh ^2(x)}{2 \left (a^2-b^2\right )} \]
a^2*b*x/(a^2-b^2)^2+1/2*b*x/(a^2-b^2)-a^3*ln(a*cosh(x)+b*sinh(x))/(a^2-b^2 )^2-1/2*b*cosh(x)*sinh(x)/(a^2-b^2)+1/2*a*sinh(x)^2/(a^2-b^2)
Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {\sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx=\frac {6 a^2 b x-2 b^3 x+a \left (a^2-b^2\right ) \cosh (2 x)-4 a^3 \log (a \cosh (x)+b \sinh (x))+\left (-a^2 b+b^3\right ) \sinh (2 x)}{4 (a-b)^2 (a+b)^2} \]
(6*a^2*b*x - 2*b^3*x + a*(a^2 - b^2)*Cosh[2*x] - 4*a^3*Log[a*Cosh[x] + b*S inh[x]] + (-(a^2*b) + b^3)*Sinh[2*x])/(4*(a - b)^2*(a + b)^2)
Result contains complex when optimal does not.
Time = 0.55 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.15, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {3042, 26, 3578, 25, 26, 3042, 25, 26, 3115, 24, 3576, 26, 3042, 3612}
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.
| \(\displaystyle \int \frac {\sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {i \sin (i x)^3}{a \cos (i x)-i b \sin (i x)}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \frac {\sin (i x)^3}{a \cos (i x)-i b \sin (i x)}dx\) |
| \(\Big \downarrow \) 3578 |
| \(\displaystyle i \left (-\frac {i b \int -\sinh ^2(x)dx}{a^2-b^2}+\frac {a^2 \int \frac {i \sinh (x)}{a \cosh (x)+b \sinh (x)}dx}{a^2-b^2}-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle i \left (\frac {i b \int \sinh ^2(x)dx}{a^2-b^2}+\frac {a^2 \int \frac {i \sinh (x)}{a \cosh (x)+b \sinh (x)}dx}{a^2-b^2}-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \left (\frac {i b \int \sinh ^2(x)dx}{a^2-b^2}+\frac {i a^2 \int \frac {\sinh (x)}{a \cosh (x)+b \sinh (x)}dx}{a^2-b^2}-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \left (\frac {i b \int -\sin (i x)^2dx}{a^2-b^2}+\frac {i a^2 \int -\frac {i \sin (i x)}{a \cos (i x)-i b \sin (i x)}dx}{a^2-b^2}-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle i \left (-\frac {i b \int \sin (i x)^2dx}{a^2-b^2}+\frac {i a^2 \int -\frac {i \sin (i x)}{a \cos (i x)-i b \sin (i x)}dx}{a^2-b^2}-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \left (-\frac {i b \int \sin (i x)^2dx}{a^2-b^2}+\frac {a^2 \int \frac {\sin (i x)}{a \cos (i x)-i b \sin (i x)}dx}{a^2-b^2}-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle i \left (\frac {a^2 \int \frac {\sin (i x)}{a \cos (i x)-i b \sin (i x)}dx}{a^2-b^2}-\frac {i b \left (\frac {\int 1dx}{2}-\frac {1}{2} \sinh (x) \cosh (x)\right )}{a^2-b^2}-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle i \left (\frac {a^2 \int \frac {\sin (i x)}{a \cos (i x)-i b \sin (i x)}dx}{a^2-b^2}-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac {i b \left (\frac {x}{2}-\frac {1}{2} \sinh (x) \cosh (x)\right )}{a^2-b^2}\right )\) |
| \(\Big \downarrow \) 3576 |
| \(\displaystyle i \left (\frac {a^2 \left (-\frac {a \int -\frac {i (b \cosh (x)+a \sinh (x))}{a \cosh (x)+b \sinh (x)}dx}{a^2-b^2}-\frac {i b x}{a^2-b^2}\right )}{a^2-b^2}-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac {i b \left (\frac {x}{2}-\frac {1}{2} \sinh (x) \cosh (x)\right )}{a^2-b^2}\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \left (\frac {a^2 \left (\frac {i a \int \frac {b \cosh (x)+a \sinh (x)}{a \cosh (x)+b \sinh (x)}dx}{a^2-b^2}-\frac {i b x}{a^2-b^2}\right )}{a^2-b^2}-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac {i b \left (\frac {x}{2}-\frac {1}{2} \sinh (x) \cosh (x)\right )}{a^2-b^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \left (\frac {a^2 \left (\frac {i a \int \frac {b \cos (i x)-i a \sin (i x)}{a \cos (i x)-i b \sin (i x)}dx}{a^2-b^2}-\frac {i b x}{a^2-b^2}\right )}{a^2-b^2}-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac {i b \left (\frac {x}{2}-\frac {1}{2} \sinh (x) \cosh (x)\right )}{a^2-b^2}\right )\) |
| \(\Big \downarrow \) 3612 |
| \(\displaystyle i \left (-\frac {i a \sinh ^2(x)}{2 \left (a^2-b^2\right )}-\frac {i b \left (\frac {x}{2}-\frac {1}{2} \sinh (x) \cosh (x)\right )}{a^2-b^2}+\frac {a^2 \left (\frac {i a \log (a \cosh (x)+b \sinh (x))}{a^2-b^2}-\frac {i b x}{a^2-b^2}\right )}{a^2-b^2}\right )\) |
I*((a^2*(((-I)*b*x)/(a^2 - b^2) + (I*a*Log[a*Cosh[x] + b*Sinh[x]])/(a^2 - b^2)))/(a^2 - b^2) - ((I/2)*a*Sinh[x]^2)/(a^2 - b^2) - (I*b*(x/2 - (Cosh[x ]*Sinh[x])/2))/(a^2 - b^2))
3.7.90.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[sin[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_. ) + (d_.)*(x_)]), x_Symbol] :> Simp[b*(x/(a^2 + b^2)), x] - Simp[a/(a^2 + b ^2) Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c + d*x ]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]
Int[sin[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin [(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[(-a)*(Sin[c + d*x]^(m - 1)/(d*(a^2 + b^2)*(m - 1))), x] + (Simp[a^2/(a^2 + b^2) Int[Sin[c + d*x]^(m - 2)/(a *Cos[c + d*x] + b*Sin[c + d*x]), x], x] + Simp[b/(a^2 + b^2) Int[Sin[c + d*x]^(m - 1), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ [m, 1]
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]) /((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x _Symbol] :> Simp[(b*B + c*C)*(x/(b^2 + c^2)), x] + Simp[(c*B - b*C)*(Log[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/(e*(b^2 + c^2))), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2 + c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C ), 0]
Time = 1.25 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.05
| method | result | size |
| risch | \(-\frac {a x}{\left (a +b \right )^{2}}-\frac {x b}{2 \left (a +b \right )^{2}}+\frac {{\mathrm e}^{2 x}}{8 a +8 b}+\frac {{\mathrm e}^{-2 x}}{8 a -8 b}+\frac {2 a^{3} x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {a^{3} \ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}\) | \(106\) |
| default | \(\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {16}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (2 a +b \right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {16}{\left (32 a -32 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {8}{\left (16 a -16 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\left (2 a -b \right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 \left (a -b \right )^{2}}-\frac {a^{3} \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a +2 b \tanh \left (\frac {x}{2}\right )+a \right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}\) | \(155\) |
-a*x/(a+b)^2-1/2*x/(a+b)^2*b+1/8/(a+b)*exp(2*x)+1/8/(a-b)*exp(-2*x)+2*a^3/ (a^4-2*a^2*b^2+b^4)*x-a^3/(a^4-2*a^2*b^2+b^4)*ln(exp(2*x)+(a-b)/(a+b))
Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (95) = 190\).
Time = 0.25 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.34 \[ \int \frac {\sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx=\frac {{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{4} + 4 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \sinh \left (x\right )^{4} + 4 \, {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} x \cosh \left (x\right )^{2} + a^{3} + a^{2} b - a b^{2} - b^{3} + 2 \, {\left (3 \, {\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} x\right )} \sinh \left (x\right )^{2} - 8 \, {\left (a^{3} \cosh \left (x\right )^{2} + 2 \, a^{3} \cosh \left (x\right ) \sinh \left (x\right ) + a^{3} \sinh \left (x\right )^{2}\right )} \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 4 \, {\left ({\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} \cosh \left (x\right )^{3} + 2 \, {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} x \cosh \left (x\right )\right )} \sinh \left (x\right )}{8 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}\right )}} \]
1/8*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^4 + 4*(a^3 - a^2*b - a*b^2 + b^3) *cosh(x)*sinh(x)^3 + (a^3 - a^2*b - a*b^2 + b^3)*sinh(x)^4 + 4*(2*a^3 + 3* a^2*b - b^3)*x*cosh(x)^2 + a^3 + a^2*b - a*b^2 - b^3 + 2*(3*(a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^2 + 2*(2*a^3 + 3*a^2*b - b^3)*x)*sinh(x)^2 - 8*(a^3* cosh(x)^2 + 2*a^3*cosh(x)*sinh(x) + a^3*sinh(x)^2)*log(2*(a*cosh(x) + b*si nh(x))/(cosh(x) - sinh(x))) + 4*((a^3 - a^2*b - a*b^2 + b^3)*cosh(x)^3 + 2 *(2*a^3 + 3*a^2*b - b^3)*x*cosh(x))*sinh(x))/((a^4 - 2*a^2*b^2 + b^4)*cosh (x)^2 + 2*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sinh(x) + (a^4 - 2*a^2*b^2 + b^4 )*sinh(x)^2)
Timed out. \[ \int \frac {\sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.86 \[ \int \frac {\sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx=-\frac {a^{3} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (2 \, a + b\right )} x}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {e^{\left (2 \, x\right )}}{8 \, {\left (a + b\right )}} + \frac {e^{\left (-2 \, x\right )}}{8 \, {\left (a - b\right )}} \]
-a^3*log(-(a - b)*e^(-2*x) - a - b)/(a^4 - 2*a^2*b^2 + b^4) - 1/2*(2*a + b )*x/(a^2 + 2*a*b + b^2) + 1/8*e^(2*x)/(a + b) + 1/8*e^(-2*x)/(a - b)
Time = 0.26 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.13 \[ \int \frac {\sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx=-\frac {a^{3} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (2 \, a - b\right )} x}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (4 \, a e^{\left (2 \, x\right )} - 2 \, b e^{\left (2 \, x\right )} - a + b\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {e^{\left (2 \, x\right )}}{8 \, {\left (a + b\right )}} \]
-a^3*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^4 - 2*a^2*b^2 + b^4) + 1/2 *(2*a - b)*x/(a^2 - 2*a*b + b^2) - 1/8*(4*a*e^(2*x) - 2*b*e^(2*x) - a + b) *e^(-2*x)/(a^2 - 2*a*b + b^2) + 1/8*e^(2*x)/(a + b)
Time = 2.42 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.85 \[ \int \frac {\sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx=\frac {{\mathrm {e}}^{-2\,x}}{8\,a-8\,b}+\frac {{\mathrm {e}}^{2\,x}}{8\,a+8\,b}-\frac {a^3\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {x\,\left (2\,a-b\right )}{2\,{\left (a-b\right )}^2} \]