Integrand size = 12, antiderivative size = 46 \[ \int \frac {\sinh \left (a+\frac {b}{x}\right )}{x^4} \, dx=-\frac {2 \cosh \left (a+\frac {b}{x}\right )}{b^3}-\frac {\cosh \left (a+\frac {b}{x}\right )}{b x^2}+\frac {2 \sinh \left (a+\frac {b}{x}\right )}{b^2 x} \] Output:
-2*cosh(a+b/x)/b^3-cosh(a+b/x)/b/x^2+2*sinh(a+b/x)/b^2/x
Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {\sinh \left (a+\frac {b}{x}\right )}{x^4} \, dx=\frac {-\left (\left (b^2+2 x^2\right ) \cosh \left (a+\frac {b}{x}\right )\right )+2 b x \sinh \left (a+\frac {b}{x}\right )}{b^3 x^2} \] Input:
Integrate[Sinh[a + b/x]/x^4,x]
Output:
(-((b^2 + 2*x^2)*Cosh[a + b/x]) + 2*b*x*Sinh[a + b/x])/(b^3*x^2)
Result contains complex when optimal does not.
Time = 0.38 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.28, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {5843, 3042, 26, 3777, 3042, 3777, 26, 3042, 26, 3118}
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.
\(\displaystyle \int \frac {\sinh \left (a+\frac {b}{x}\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 5843 |
\(\displaystyle -\int \frac {\sinh \left (a+\frac {b}{x}\right )}{x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int -\frac {i \sin \left (i a+\frac {i b}{x}\right )}{x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {\sin \left (i a+\frac {i b}{x}\right )}{x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle i \left (\frac {i \cosh \left (a+\frac {b}{x}\right )}{b x^2}-\frac {2 i \int \frac {\cosh \left (a+\frac {b}{x}\right )}{x}d\frac {1}{x}}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle i \left (\frac {i \cosh \left (a+\frac {b}{x}\right )}{b x^2}-\frac {2 i \int \frac {\sin \left (i a+\frac {i b}{x}+\frac {\pi }{2}\right )}{x}d\frac {1}{x}}{b}\right )\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle i \left (\frac {i \cosh \left (a+\frac {b}{x}\right )}{b x^2}-\frac {2 i \left (\frac {\sinh \left (a+\frac {b}{x}\right )}{b x}-\frac {i \int -i \sinh \left (a+\frac {b}{x}\right )d\frac {1}{x}}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \left (\frac {i \cosh \left (a+\frac {b}{x}\right )}{b x^2}-\frac {2 i \left (\frac {\sinh \left (a+\frac {b}{x}\right )}{b x}-\frac {\int \sinh \left (a+\frac {b}{x}\right )d\frac {1}{x}}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle i \left (\frac {i \cosh \left (a+\frac {b}{x}\right )}{b x^2}-\frac {2 i \left (\frac {\sinh \left (a+\frac {b}{x}\right )}{b x}-\frac {\int -i \sin \left (i a+\frac {i b}{x}\right )d\frac {1}{x}}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \left (\frac {i \cosh \left (a+\frac {b}{x}\right )}{b x^2}-\frac {2 i \left (\frac {\sinh \left (a+\frac {b}{x}\right )}{b x}+\frac {i \int \sin \left (i a+\frac {i b}{x}\right )d\frac {1}{x}}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle i \left (\frac {i \cosh \left (a+\frac {b}{x}\right )}{b x^2}-\frac {2 i \left (\frac {\sinh \left (a+\frac {b}{x}\right )}{b x}-\frac {\cosh \left (a+\frac {b}{x}\right )}{b^2}\right )}{b}\right )\) |
Input:
Int[Sinh[a + b/x]/x^4,x]
Output:
I*((I*Cosh[a + b/x])/(b*x^2) - ((2*I)*(-(Cosh[a + b/x]/b^2) + Sinh[a + b/x ]/(b*x)))/b)
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[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo l] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sinh[c + d*x] )^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplif y[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplif y[(m + 1)/n], 0]))
Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \(\frac {\left (-b^{2}-2 x^{2}\right ) \cosh \left (\frac {a x +b}{x}\right )+2 x b \sinh \left (\frac {a x +b}{x}\right )-2 x^{2}}{b^{3} x^{2}}\) | \(50\) |
risch | \(-\frac {\left (b^{2}-2 b x +2 x^{2}\right ) {\mathrm e}^{\frac {a x +b}{x}}}{2 b^{3} x^{2}}-\frac {\left (b^{2}+2 b x +2 x^{2}\right ) {\mathrm e}^{-\frac {a x +b}{x}}}{2 b^{3} x^{2}}\) | \(65\) |
orering | \(\frac {2 \left (3 b^{2}+4 x^{2}\right ) \sinh \left (a +\frac {b}{x}\right )}{x \,b^{4}}+\frac {\left (b^{2}+2 x^{2}\right ) x^{4} \left (-\frac {b \cosh \left (a +\frac {b}{x}\right )}{x^{6}}-\frac {4 \sinh \left (a +\frac {b}{x}\right )}{x^{5}}\right )}{b^{4}}\) | \(73\) |
derivativedivides | \(-\frac {a^{2} \cosh \left (a +\frac {b}{x}\right )-2 a \left (\left (a +\frac {b}{x}\right ) \cosh \left (a +\frac {b}{x}\right )-\sinh \left (a +\frac {b}{x}\right )\right )+\left (a +\frac {b}{x}\right )^{2} \cosh \left (a +\frac {b}{x}\right )-2 \left (a +\frac {b}{x}\right ) \sinh \left (a +\frac {b}{x}\right )+2 \cosh \left (a +\frac {b}{x}\right )}{b^{3}}\) | \(94\) |
default | \(-\frac {a^{2} \cosh \left (a +\frac {b}{x}\right )-2 a \left (\left (a +\frac {b}{x}\right ) \cosh \left (a +\frac {b}{x}\right )-\sinh \left (a +\frac {b}{x}\right )\right )+\left (a +\frac {b}{x}\right )^{2} \cosh \left (a +\frac {b}{x}\right )-2 \left (a +\frac {b}{x}\right ) \sinh \left (a +\frac {b}{x}\right )+2 \cosh \left (a +\frac {b}{x}\right )}{b^{3}}\) | \(94\) |
meijerg | \(-\frac {4 \sqrt {\pi }\, \cosh \left (a \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (\frac {b^{2}}{2 x^{2}}+1\right ) \cosh \left (\frac {b}{x}\right )}{2 \sqrt {\pi }}-\frac {b \sinh \left (\frac {b}{x}\right )}{2 \sqrt {\pi }\, x}\right )}{b^{3}}-\frac {4 i \sqrt {\pi }\, \sinh \left (a \right ) \left (\frac {i b \cosh \left (\frac {b}{x}\right )}{2 \sqrt {\pi }\, x}-\frac {i \left (\frac {3 b^{2}}{2 x^{2}}+3\right ) \sinh \left (\frac {b}{x}\right )}{6 \sqrt {\pi }}\right )}{b^{3}}\) | \(104\) |
Input:
int(sinh(a+b/x)/x^4,x,method=_RETURNVERBOSE)
Output:
((-b^2-2*x^2)*cosh((a*x+b)/x)+2*x*b*sinh((a*x+b)/x)-2*x^2)/b^3/x^2
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \frac {\sinh \left (a+\frac {b}{x}\right )}{x^4} \, dx=\frac {2 \, b x \sinh \left (\frac {a x + b}{x}\right ) - {\left (b^{2} + 2 \, x^{2}\right )} \cosh \left (\frac {a x + b}{x}\right )}{b^{3} x^{2}} \] Input:
integrate(sinh(a+b/x)/x^4,x, algorithm="fricas")
Output:
(2*b*x*sinh((a*x + b)/x) - (b^2 + 2*x^2)*cosh((a*x + b)/x))/(b^3*x^2)
Time = 0.53 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {\sinh \left (a+\frac {b}{x}\right )}{x^4} \, dx=\begin {cases} - \frac {\cosh {\left (a + \frac {b}{x} \right )}}{b x^{2}} + \frac {2 \sinh {\left (a + \frac {b}{x} \right )}}{b^{2} x} - \frac {2 \cosh {\left (a + \frac {b}{x} \right )}}{b^{3}} & \text {for}\: b \neq 0 \\- \frac {\sinh {\left (a \right )}}{3 x^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(sinh(a+b/x)/x**4,x)
Output:
Piecewise((-cosh(a + b/x)/(b*x**2) + 2*sinh(a + b/x)/(b**2*x) - 2*cosh(a + b/x)/b**3, Ne(b, 0)), (-sinh(a)/(3*x**3), True))
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \frac {\sinh \left (a+\frac {b}{x}\right )}{x^4} \, dx=-\frac {1}{6} \, b {\left (\frac {e^{\left (-a\right )} \Gamma \left (4, \frac {b}{x}\right )}{b^{4}} + \frac {e^{a} \Gamma \left (4, -\frac {b}{x}\right )}{b^{4}}\right )} - \frac {\sinh \left (a + \frac {b}{x}\right )}{3 \, x^{3}} \] Input:
integrate(sinh(a+b/x)/x^4,x, algorithm="maxima")
Output:
-1/6*b*(e^(-a)*gamma(4, b/x)/b^4 + e^a*gamma(4, -b/x)/b^4) - 1/3*sinh(a + b/x)/x^3
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (46) = 92\).
Time = 0.12 (sec) , antiderivative size = 163, normalized size of antiderivative = 3.54 \[ \int \frac {\sinh \left (a+\frac {b}{x}\right )}{x^4} \, dx=-\frac {{\left (a^{2} e^{\left (\frac {2 \, {\left (a x + b\right )}}{x}\right )} + a^{2} + 2 \, a e^{\left (\frac {2 \, {\left (a x + b\right )}}{x}\right )} - \frac {2 \, {\left (a x + b\right )} a e^{\left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}}{x} - 2 \, a - \frac {2 \, {\left (a x + b\right )} a}{x} + \frac {{\left (a x + b\right )}^{2} e^{\left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}}{x^{2}} - \frac {2 \, {\left (a x + b\right )} e^{\left (\frac {2 \, {\left (a x + b\right )}}{x}\right )}}{x} + \frac {{\left (a x + b\right )}^{2}}{x^{2}} + \frac {2 \, {\left (a x + b\right )}}{x} + 2 \, e^{\left (\frac {2 \, {\left (a x + b\right )}}{x}\right )} + 2\right )} e^{\left (-\frac {a x + b}{x}\right )}}{2 \, b^{3}} \] Input:
integrate(sinh(a+b/x)/x^4,x, algorithm="giac")
Output:
-1/2*(a^2*e^(2*(a*x + b)/x) + a^2 + 2*a*e^(2*(a*x + b)/x) - 2*(a*x + b)*a* e^(2*(a*x + b)/x)/x - 2*a - 2*(a*x + b)*a/x + (a*x + b)^2*e^(2*(a*x + b)/x )/x^2 - 2*(a*x + b)*e^(2*(a*x + b)/x)/x + (a*x + b)^2/x^2 + 2*(a*x + b)/x + 2*e^(2*(a*x + b)/x) + 2)*e^(-(a*x + b)/x)/b^3
Time = 1.76 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.46 \[ \int \frac {\sinh \left (a+\frac {b}{x}\right )}{x^4} \, dx=-\frac {{\mathrm {e}}^{a+\frac {b}{x}}\,\left (\frac {1}{2\,b}-\frac {x}{b^2}+\frac {x^2}{b^3}\right )}{x^2}-\frac {{\mathrm {e}}^{-a-\frac {b}{x}}\,\left (\frac {x}{b^2}+\frac {1}{2\,b}+\frac {x^2}{b^3}\right )}{x^2} \] Input:
int(sinh(a + b/x)/x^4,x)
Output:
- (exp(a + b/x)*(1/(2*b) - x/b^2 + x^2/b^3))/x^2 - (exp(- a - b/x)*(x/b^2 + 1/(2*b) + x^2/b^3))/x^2
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.00 \[ \int \frac {\sinh \left (a+\frac {b}{x}\right )}{x^4} \, dx=\frac {-e^{\frac {2 a x +2 b}{x}} b^{2}+2 e^{\frac {2 a x +2 b}{x}} b x -2 e^{\frac {2 a x +2 b}{x}} x^{2}-b^{2}-2 b x -2 x^{2}}{2 e^{\frac {a x +b}{x}} b^{3} x^{2}} \] Input:
int(sinh(a+b/x)/x^4,x)
Output:
( - e**((2*a*x + 2*b)/x)*b**2 + 2*e**((2*a*x + 2*b)/x)*b*x - 2*e**((2*a*x + 2*b)/x)*x**2 - b**2 - 2*b*x - 2*x**2)/(2*e**((a*x + b)/x)*b**3*x**2)