Integrand size = 12, antiderivative size = 62 \[ \int x^3 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {1}{4} b x^2 \cosh \left (a+\frac {b}{x^2}\right )-\frac {1}{4} b^2 \text {Chi}\left (\frac {b}{x^2}\right ) \sinh (a)+\frac {1}{4} x^4 \sinh \left (a+\frac {b}{x^2}\right )-\frac {1}{4} b^2 \cosh (a) \text {Shi}\left (\frac {b}{x^2}\right ) \] Output:
1/4*b*x^2*cosh(a+b/x^2)-1/4*b^2*Chi(b/x^2)*sinh(a)+1/4*x^4*sinh(a+b/x^2)-1 /4*b^2*cosh(a)*Shi(b/x^2)
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.90 \[ \int x^3 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {1}{4} \left (b x^2 \cosh \left (a+\frac {b}{x^2}\right )-b^2 \text {Chi}\left (\frac {b}{x^2}\right ) \sinh (a)+x^4 \sinh \left (a+\frac {b}{x^2}\right )-b^2 \cosh (a) \text {Shi}\left (\frac {b}{x^2}\right )\right ) \] Input:
Integrate[x^3*Sinh[a + b/x^2],x]
Output:
(b*x^2*Cosh[a + b/x^2] - b^2*CoshIntegral[b/x^2]*Sinh[a] + x^4*Sinh[a + b/ x^2] - b^2*Cosh[a]*SinhIntegral[b/x^2])/4
Result contains complex when optimal does not.
Time = 0.61 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.21, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {5843, 3042, 26, 3778, 3042, 3778, 26, 3042, 26, 3784, 26, 3042, 26, 3779, 3782}
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.
| \(\displaystyle \int x^3 \sinh \left (a+\frac {b}{x^2}\right ) \, dx\) |
| \(\Big \downarrow \) 5843 |
| \(\displaystyle -\frac {1}{2} \int x^6 \sinh \left (a+\frac {b}{x^2}\right )d\frac {1}{x^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {1}{2} \int -i x^6 \sin \left (i a+\frac {i b}{x^2}\right )d\frac {1}{x^2}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {1}{2} i \int x^6 \sin \left (i a+\frac {i b}{x^2}\right )d\frac {1}{x^2}\) |
| \(\Big \downarrow \) 3778 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \int x^4 \cosh \left (a+\frac {b}{x^2}\right )d\frac {1}{x^2}-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \int x^4 \sin \left (i a+\frac {i b}{x^2}+\frac {\pi }{2}\right )d\frac {1}{x^2}-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 3778 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \left (-x^2 \cosh \left (a+\frac {b}{x^2}\right )+i b \int -i x^2 \sinh \left (a+\frac {b}{x^2}\right )d\frac {1}{x^2}\right )-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \left (b \int x^2 \sinh \left (a+\frac {b}{x^2}\right )d\frac {1}{x^2}-x^2 \cosh \left (a+\frac {b}{x^2}\right )\right )-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \left (-x^2 \cosh \left (a+\frac {b}{x^2}\right )+b \int -i x^2 \sin \left (i a+\frac {i b}{x^2}\right )d\frac {1}{x^2}\right )-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \left (x^2 \left (-\cosh \left (a+\frac {b}{x^2}\right )\right )-i b \int x^2 \sin \left (i a+\frac {i b}{x^2}\right )d\frac {1}{x^2}\right )-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 3784 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \left (x^2 \left (-\cosh \left (a+\frac {b}{x^2}\right )\right )-i b \left (i \sinh (a) \int x^2 \cosh \left (\frac {b}{x^2}\right )d\frac {1}{x^2}+\cosh (a) \int i x^2 \sinh \left (\frac {b}{x^2}\right )d\frac {1}{x^2}\right )\right )-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \left (x^2 \left (-\cosh \left (a+\frac {b}{x^2}\right )\right )-i b \left (i \sinh (a) \int x^2 \cosh \left (\frac {b}{x^2}\right )d\frac {1}{x^2}+i \cosh (a) \int x^2 \sinh \left (\frac {b}{x^2}\right )d\frac {1}{x^2}\right )\right )-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \left (x^2 \left (-\cosh \left (a+\frac {b}{x^2}\right )\right )-i b \left (i \sinh (a) \int x^2 \sin \left (\frac {i b}{x^2}+\frac {\pi }{2}\right )d\frac {1}{x^2}+i \cosh (a) \int -i x^2 \sin \left (\frac {i b}{x^2}\right )d\frac {1}{x^2}\right )\right )-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \left (x^2 \left (-\cosh \left (a+\frac {b}{x^2}\right )\right )-i b \left (i \sinh (a) \int x^2 \sin \left (\frac {i b}{x^2}+\frac {\pi }{2}\right )d\frac {1}{x^2}+\cosh (a) \int x^2 \sin \left (\frac {i b}{x^2}\right )d\frac {1}{x^2}\right )\right )-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 3779 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \left (x^2 \left (-\cosh \left (a+\frac {b}{x^2}\right )\right )-i b \left (i \sinh (a) \int x^2 \sin \left (\frac {i b}{x^2}+\frac {\pi }{2}\right )d\frac {1}{x^2}+i \cosh (a) \text {Shi}\left (\frac {b}{x^2}\right )\right )\right )-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
| \(\Big \downarrow \) 3782 |
| \(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b \left (x^2 \left (-\cosh \left (a+\frac {b}{x^2}\right )\right )-i b \left (i \sinh (a) \text {Chi}\left (\frac {b}{x^2}\right )+i \cosh (a) \text {Shi}\left (\frac {b}{x^2}\right )\right )\right )-\frac {1}{2} i x^4 \sinh \left (a+\frac {b}{x^2}\right )\right )\) |
Input:
Int[x^3*Sinh[a + b/x^2],x]
Output:
(I/2)*((-1/2*I)*x^4*Sinh[a + b/x^2] + (I/2)*b*(-(x^2*Cosh[a + b/x^2]) - I* b*(I*CoshIntegral[b/x^2]*Sinh[a] + I*Cosh[a]*SinhIntegral[b/x^2])))
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 + 1)*(Sin[e + f*x]/(d*(m + 1))), x] - Simp[f/(d*(m + 1)) Int[( c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[m, - 1]
Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbo l] :> Simp[I*(SinhIntegral[c*f*(fz/d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f , fz}, x] && EqQ[d*e - c*f*fz*I, 0]
Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbo l] :> Simp[CoshIntegral[c*f*(fz/d) + f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz }, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* e - c*f)/d] Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[Sin[(d*e - c* f)/d] Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f}, x] && NeQ[d*e - c*f, 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]))
Leaf count of result is larger than twice the leaf count of optimal. \(116\) vs. \(2(54)=108\).
Time = 0.22 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.89
| method | result | size |
| risch | \(\frac {{\mathrm e}^{-a} {\mathrm e}^{\frac {2 a \,x^{2}+b}{x^{2}}} x^{4}}{8}-\frac {{\mathrm e}^{-a} x^{4} {\mathrm e}^{-\frac {b}{x^{2}}}}{8}+\frac {{\mathrm e}^{-a} \operatorname {expIntegral}_{1}\left (-\frac {b}{x^{2}}\right ) {\mathrm e}^{2 a} b^{2}}{8}+\frac {{\mathrm e}^{-a} {\mathrm e}^{\frac {2 a \,x^{2}+b}{x^{2}}} b \,x^{2}}{8}+\frac {{\mathrm e}^{-a} b \,x^{2} {\mathrm e}^{-\frac {b}{x^{2}}}}{8}-\frac {{\mathrm e}^{-a} b^{2} \operatorname {expIntegral}_{1}\left (\frac {b}{x^{2}}\right )}{8}\) | \(117\) |
| meijerg | \(-\frac {i b^{2} \sqrt {\pi }\, \cosh \left (a \right ) \left (\frac {4 i x^{2} \cosh \left (\frac {b}{x^{2}}\right )}{b \sqrt {\pi }}+\frac {4 i x^{4} \sinh \left (\frac {b}{x^{2}}\right )}{b^{2} \sqrt {\pi }}-\frac {4 i \operatorname {Shi}\left (\frac {b}{x^{2}}\right )}{\sqrt {\pi }}\right )}{16}+\frac {b^{2} \sqrt {\pi }\, \sinh \left (a \right ) \left (\frac {4 x^{4}}{\sqrt {\pi }\, b^{2}}-\frac {2 \left (2 \gamma -3-4 \ln \left (x \right )+2 \ln \left (i b \right )\right )}{\sqrt {\pi }}-\frac {4 x^{4} \left (\frac {9 b^{2}}{2 x^{4}}+3\right )}{3 \sqrt {\pi }\, b^{2}}+\frac {4 x^{4} \cosh \left (\frac {b}{x^{2}}\right )}{\sqrt {\pi }\, b^{2}}+\frac {4 x^{2} \sinh \left (\frac {b}{x^{2}}\right )}{\sqrt {\pi }\, b}-\frac {4 \left (\operatorname {Chi}\left (\frac {b}{x^{2}}\right )-\ln \left (\frac {b}{x^{2}}\right )-\gamma \right )}{\sqrt {\pi }}\right )}{16}\) | \(183\) |
Input:
int(x^3*sinh(a+b/x^2),x,method=_RETURNVERBOSE)
Output:
1/8*exp(-a)*exp((2*a*x^2+b)/x^2)*x^4-1/8*exp(-a)*x^4*exp(-b/x^2)+1/8*exp(- a)*Ei(1,-b/x^2)*exp(2*a)*b^2+1/8*exp(-a)*exp((2*a*x^2+b)/x^2)*b*x^2+1/8*ex p(-a)*b*x^2*exp(-b/x^2)-1/8*exp(-a)*b^2*Ei(1,b/x^2)
Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.44 \[ \int x^3 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {1}{4} \, x^{4} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) + \frac {1}{4} \, b x^{2} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) - \frac {1}{8} \, {\left (b^{2} {\rm Ei}\left (\frac {b}{x^{2}}\right ) - b^{2} {\rm Ei}\left (-\frac {b}{x^{2}}\right )\right )} \cosh \left (a\right ) - \frac {1}{8} \, {\left (b^{2} {\rm Ei}\left (\frac {b}{x^{2}}\right ) + b^{2} {\rm Ei}\left (-\frac {b}{x^{2}}\right )\right )} \sinh \left (a\right ) \] Input:
integrate(x^3*sinh(a+b/x^2),x, algorithm="fricas")
Output:
1/4*x^4*sinh((a*x^2 + b)/x^2) + 1/4*b*x^2*cosh((a*x^2 + b)/x^2) - 1/8*(b^2 *Ei(b/x^2) - b^2*Ei(-b/x^2))*cosh(a) - 1/8*(b^2*Ei(b/x^2) + b^2*Ei(-b/x^2) )*sinh(a)
\[ \int x^3 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\int x^{3} \sinh {\left (a + \frac {b}{x^{2}} \right )}\, dx \] Input:
integrate(x**3*sinh(a+b/x**2),x)
Output:
Integral(x**3*sinh(a + b/x**2), x)
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.71 \[ \int x^3 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {1}{4} \, x^{4} \sinh \left (a + \frac {b}{x^{2}}\right ) + \frac {1}{8} \, {\left (b e^{\left (-a\right )} \Gamma \left (-1, \frac {b}{x^{2}}\right ) - b e^{a} \Gamma \left (-1, -\frac {b}{x^{2}}\right )\right )} b \] Input:
integrate(x^3*sinh(a+b/x^2),x, algorithm="maxima")
Output:
1/4*x^4*sinh(a + b/x^2) + 1/8*(b*e^(-a)*gamma(-1, b/x^2) - b*e^a*gamma(-1, -b/x^2))*b
Leaf count of result is larger than twice the leaf count of optimal. 353 vs. \(2 (54) = 108\).
Time = 0.13 (sec) , antiderivative size = 353, normalized size of antiderivative = 5.69 \[ \int x^3 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {a^{2} b^{3} {\rm Ei}\left (a - \frac {a x^{2} + b}{x^{2}}\right ) e^{\left (-a\right )} - a^{2} b^{3} {\rm Ei}\left (-a + \frac {a x^{2} + b}{x^{2}}\right ) e^{a} - \frac {2 \, {\left (a x^{2} + b\right )} a b^{3} {\rm Ei}\left (a - \frac {a x^{2} + b}{x^{2}}\right ) e^{\left (-a\right )}}{x^{2}} + \frac {2 \, {\left (a x^{2} + b\right )} a b^{3} {\rm Ei}\left (-a + \frac {a x^{2} + b}{x^{2}}\right ) e^{a}}{x^{2}} - a b^{3} e^{\left (\frac {a x^{2} + b}{x^{2}}\right )} - a b^{3} e^{\left (-\frac {a x^{2} + b}{x^{2}}\right )} + b^{3} e^{\left (\frac {a x^{2} + b}{x^{2}}\right )} - b^{3} e^{\left (-\frac {a x^{2} + b}{x^{2}}\right )} + \frac {{\left (a x^{2} + b\right )}^{2} b^{3} {\rm Ei}\left (a - \frac {a x^{2} + b}{x^{2}}\right ) e^{\left (-a\right )}}{x^{4}} - \frac {{\left (a x^{2} + b\right )}^{2} b^{3} {\rm Ei}\left (-a + \frac {a x^{2} + b}{x^{2}}\right ) e^{a}}{x^{4}} + \frac {{\left (a x^{2} + b\right )} b^{3} e^{\left (\frac {a x^{2} + b}{x^{2}}\right )}}{x^{2}} + \frac {{\left (a x^{2} + b\right )} b^{3} e^{\left (-\frac {a x^{2} + b}{x^{2}}\right )}}{x^{2}}}{8 \, {\left (a^{2} - \frac {2 \, {\left (a x^{2} + b\right )} a}{x^{2}} + \frac {{\left (a x^{2} + b\right )}^{2}}{x^{4}}\right )} b} \] Input:
integrate(x^3*sinh(a+b/x^2),x, algorithm="giac")
Output:
1/8*(a^2*b^3*Ei(a - (a*x^2 + b)/x^2)*e^(-a) - a^2*b^3*Ei(-a + (a*x^2 + b)/ x^2)*e^a - 2*(a*x^2 + b)*a*b^3*Ei(a - (a*x^2 + b)/x^2)*e^(-a)/x^2 + 2*(a*x ^2 + b)*a*b^3*Ei(-a + (a*x^2 + b)/x^2)*e^a/x^2 - a*b^3*e^((a*x^2 + b)/x^2) - a*b^3*e^(-(a*x^2 + b)/x^2) + b^3*e^((a*x^2 + b)/x^2) - b^3*e^(-(a*x^2 + b)/x^2) + (a*x^2 + b)^2*b^3*Ei(a - (a*x^2 + b)/x^2)*e^(-a)/x^4 - (a*x^2 + b)^2*b^3*Ei(-a + (a*x^2 + b)/x^2)*e^a/x^4 + (a*x^2 + b)*b^3*e^((a*x^2 + b )/x^2)/x^2 + (a*x^2 + b)*b^3*e^(-(a*x^2 + b)/x^2)/x^2)/((a^2 - 2*(a*x^2 + b)*a/x^2 + (a*x^2 + b)^2/x^4)*b)
Timed out. \[ \int x^3 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\int x^3\,\mathrm {sinh}\left (a+\frac {b}{x^2}\right ) \,d x \] Input:
int(x^3*sinh(a + b/x^2),x)
Output:
int(x^3*sinh(a + b/x^2), x)
\[ \int x^3 \sinh \left (a+\frac {b}{x^2}\right ) \, dx=\frac {\cosh \left (\frac {a \,x^{2}+b}{x^{2}}\right ) b \,x^{2}}{4}+\frac {\left (\int \frac {\sinh \left (\frac {a \,x^{2}+b}{x^{2}}\right )}{x}d x \right ) b^{2}}{2}+\frac {\sinh \left (\frac {a \,x^{2}+b}{x^{2}}\right ) x^{4}}{4} \] Input:
int(x^3*sinh(a+b/x^2),x)
Output:
(cosh((a*x**2 + b)/x**2)*b*x**2 + 2*int(sinh((a*x**2 + b)/x**2)/x,x)*b**2 + sinh((a*x**2 + b)/x**2)*x**4)/4