Integrand size = 12, antiderivative size = 47 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{b^3}-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x^4}+\frac {\sinh \left (a+\frac {b}{x^2}\right )}{b^2 x^2} \] Output:
-cosh(a+b/x^2)/b^3-1/2*cosh(a+b/x^2)/b/x^4+sinh(a+b/x^2)/b^2/x^2
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=\frac {-\left (\left (b^2+2 x^4\right ) \cosh \left (a+\frac {b}{x^2}\right )\right )+2 b x^2 \sinh \left (a+\frac {b}{x^2}\right )}{2 b^3 x^4} \] Input:
Integrate[Sinh[a + b/x^2]/x^7,x]
Output:
(-((b^2 + 2*x^4)*Cosh[a + b/x^2]) + 2*b*x^2*Sinh[a + b/x^2])/(2*b^3*x^4)
Result contains complex when optimal does not.
Time = 0.39 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.30, 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.
\(\displaystyle \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx\) |
\(\Big \downarrow \) 5843 |
\(\displaystyle -\frac {1}{2} \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^4}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{2} \int -\frac {i \sin \left (i a+\frac {i b}{x^2}\right )}{x^4}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {1}{2} i \int \frac {\sin \left (i a+\frac {i b}{x^2}\right )}{x^4}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{2} i \left (\frac {i \cosh \left (a+\frac {b}{x^2}\right )}{b x^4}-\frac {2 i \int \frac {\cosh \left (a+\frac {b}{x^2}\right )}{x^2}d\frac {1}{x^2}}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} i \left (\frac {i \cosh \left (a+\frac {b}{x^2}\right )}{b x^4}-\frac {2 i \int \frac {\sin \left (i a+\frac {i b}{x^2}+\frac {\pi }{2}\right )}{x^2}d\frac {1}{x^2}}{b}\right )\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {1}{2} i \left (\frac {i \cosh \left (a+\frac {b}{x^2}\right )}{b x^4}-\frac {2 i \left (\frac {\sinh \left (a+\frac {b}{x^2}\right )}{b x^2}-\frac {i \int -i \sinh \left (a+\frac {b}{x^2}\right )d\frac {1}{x^2}}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {1}{2} i \left (\frac {i \cosh \left (a+\frac {b}{x^2}\right )}{b x^4}-\frac {2 i \left (\frac {\sinh \left (a+\frac {b}{x^2}\right )}{b x^2}-\frac {\int \sinh \left (a+\frac {b}{x^2}\right )d\frac {1}{x^2}}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} i \left (\frac {i \cosh \left (a+\frac {b}{x^2}\right )}{b x^4}-\frac {2 i \left (\frac {\sinh \left (a+\frac {b}{x^2}\right )}{b x^2}-\frac {\int -i \sin \left (i a+\frac {i b}{x^2}\right )d\frac {1}{x^2}}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {1}{2} i \left (\frac {i \cosh \left (a+\frac {b}{x^2}\right )}{b x^4}-\frac {2 i \left (\frac {\sinh \left (a+\frac {b}{x^2}\right )}{b x^2}+\frac {i \int \sin \left (i a+\frac {i b}{x^2}\right )d\frac {1}{x^2}}{b}\right )}{b}\right )\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle \frac {1}{2} i \left (\frac {i \cosh \left (a+\frac {b}{x^2}\right )}{b x^4}-\frac {2 i \left (\frac {\sinh \left (a+\frac {b}{x^2}\right )}{b x^2}-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{b^2}\right )}{b}\right )\) |
Input:
Int[Sinh[a + b/x^2]/x^7,x]
Output:
(I/2)*((I*Cosh[a + b/x^2])/(b*x^4) - ((2*I)*(-(Cosh[a + b/x^2]/b^2) + Sinh [a + b/x^2]/(b*x^2)))/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 = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(\frac {\left (-2 x^{4}-b^{2}\right ) \cosh \left (\frac {a \,x^{2}+b}{x^{2}}\right )+2 x^{2} b \sinh \left (\frac {a \,x^{2}+b}{x^{2}}\right )-2 x^{4}}{2 b^{3} x^{4}}\) | \(57\) |
risch | \(-\frac {\left (2 x^{4}-2 b \,x^{2}+b^{2}\right ) {\mathrm e}^{\frac {a \,x^{2}+b}{x^{2}}}}{4 b^{3} x^{4}}-\frac {\left (2 x^{4}+2 b \,x^{2}+b^{2}\right ) {\mathrm e}^{-\frac {a \,x^{2}+b}{x^{2}}}}{4 b^{3} x^{4}}\) | \(73\) |
orering | \(\frac {\left (14 x^{4}+11 b^{2}\right ) \sinh \left (a +\frac {b}{x^{2}}\right )}{4 x^{2} b^{4}}+\frac {\left (2 x^{4}+b^{2}\right ) x^{6} \left (-\frac {2 b \cosh \left (a +\frac {b}{x^{2}}\right )}{x^{10}}-\frac {7 \sinh \left (a +\frac {b}{x^{2}}\right )}{x^{8}}\right )}{4 b^{4}}\) | \(74\) |
meijerg | \(-\frac {2 \sqrt {\pi }\, \cosh \left (a \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (\frac {b^{2}}{2 x^{4}}+1\right ) \cosh \left (\frac {b}{x^{2}}\right )}{2 \sqrt {\pi }}-\frac {b \sinh \left (\frac {b}{x^{2}}\right )}{2 \sqrt {\pi }\, x^{2}}\right )}{b^{3}}-\frac {2 i \sqrt {\pi }\, \sinh \left (a \right ) \left (\frac {i b \cosh \left (\frac {b}{x^{2}}\right )}{2 \sqrt {\pi }\, x^{2}}-\frac {i \left (\frac {3 b^{2}}{2 x^{4}}+3\right ) \sinh \left (\frac {b}{x^{2}}\right )}{6 \sqrt {\pi }}\right )}{b^{3}}\) | \(104\) |
Input:
int(sinh(a+b/x^2)/x^7,x,method=_RETURNVERBOSE)
Output:
1/2*((-2*x^4-b^2)*cosh((a*x^2+b)/x^2)+2*x^2*b*sinh((a*x^2+b)/x^2)-2*x^4)/b ^3/x^4
Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.06 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=\frac {2 \, b x^{2} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) - {\left (2 \, x^{4} + b^{2}\right )} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right )}{2 \, b^{3} x^{4}} \] Input:
integrate(sinh(a+b/x^2)/x^7,x, algorithm="fricas")
Output:
1/2*(2*b*x^2*sinh((a*x^2 + b)/x^2) - (2*x^4 + b^2)*cosh((a*x^2 + b)/x^2))/ (b^3*x^4)
Time = 1.83 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.09 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=\begin {cases} - \frac {\cosh {\left (a + \frac {b}{x^{2}} \right )}}{2 b x^{4}} + \frac {\sinh {\left (a + \frac {b}{x^{2}} \right )}}{b^{2} x^{2}} - \frac {\cosh {\left (a + \frac {b}{x^{2}} \right )}}{b^{3}} & \text {for}\: b \neq 0 \\- \frac {\sinh {\left (a \right )}}{6 x^{6}} & \text {otherwise} \end {cases} \] Input:
integrate(sinh(a+b/x**2)/x**7,x)
Output:
Piecewise((-cosh(a + b/x**2)/(2*b*x**4) + sinh(a + b/x**2)/(b**2*x**2) - c osh(a + b/x**2)/b**3, Ne(b, 0)), (-sinh(a)/(6*x**6), True))
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=-\frac {1}{12} \, b {\left (\frac {e^{\left (-a\right )} \Gamma \left (4, \frac {b}{x^{2}}\right )}{b^{4}} + \frac {e^{a} \Gamma \left (4, -\frac {b}{x^{2}}\right )}{b^{4}}\right )} - \frac {\sinh \left (a + \frac {b}{x^{2}}\right )}{6 \, x^{6}} \] Input:
integrate(sinh(a+b/x^2)/x^7,x, algorithm="maxima")
Output:
-1/12*b*(e^(-a)*gamma(4, b/x^2)/b^4 + e^a*gamma(4, -b/x^2)/b^4) - 1/6*sinh (a + b/x^2)/x^6
\[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=\int { \frac {\sinh \left (a + \frac {b}{x^{2}}\right )}{x^{7}} \,d x } \] Input:
integrate(sinh(a+b/x^2)/x^7,x, algorithm="giac")
Output:
integrate(sinh(a + b/x^2)/x^7, x)
Time = 1.39 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.57 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=-\frac {{\mathrm {e}}^{a+\frac {b}{x^2}}\,\left (\frac {1}{4\,b}-\frac {x^2}{2\,b^2}+\frac {x^4}{2\,b^3}\right )}{x^4}-\frac {{\mathrm {e}}^{-a-\frac {b}{x^2}}\,\left (\frac {1}{4\,b}+\frac {x^2}{2\,b^2}+\frac {x^4}{2\,b^3}\right )}{x^4} \] Input:
int(sinh(a + b/x^2)/x^7,x)
Output:
- (exp(a + b/x^2)*(1/(4*b) - x^2/(2*b^2) + x^4/(2*b^3)))/x^4 - (exp(- a - b/x^2)*(1/(4*b) + x^2/(2*b^2) + x^4/(2*b^3)))/x^4
Time = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.21 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=\frac {-e^{\frac {2 a \,x^{2}+2 b}{x^{2}}} b^{2}+2 e^{\frac {2 a \,x^{2}+2 b}{x^{2}}} b \,x^{2}-2 e^{\frac {2 a \,x^{2}+2 b}{x^{2}}} x^{4}-b^{2}-2 b \,x^{2}-2 x^{4}}{4 e^{\frac {a \,x^{2}+b}{x^{2}}} b^{3} x^{4}} \] Input:
int(sinh(a+b/x^2)/x^7,x)
Output:
( - e**((2*a*x**2 + 2*b)/x**2)*b**2 + 2*e**((2*a*x**2 + 2*b)/x**2)*b*x**2 - 2*e**((2*a*x**2 + 2*b)/x**2)*x**4 - b**2 - 2*b*x**2 - 2*x**4)/(4*e**((a* x**2 + b)/x**2)*b**3*x**4)