\(\int \frac {\cosh (a+b x^2)}{x^3} \, dx\) [7]

Optimal result

Integrand size = 12, antiderivative size = 42 \[ \int \frac {\cosh \left (a+b x^2\right )}{x^3} \, dx=-\frac {\cosh \left (a+b x^2\right )}{2 x^2}+\frac {1}{2} b \text {Chi}\left (b x^2\right ) \sinh (a)+\frac {1}{2} b \cosh (a) \text {Shi}\left (b x^2\right ) \]

[Out]

-1/2*cosh(b*x^2+a)/x^2+1/2*b*cosh(a)*Shi(b*x^2)+1/2*b*Chi(b*x^2)*sinh(a)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5429, 3378, 3384, 3379, 3382} \[ \int \frac {\cosh \left (a+b x^2\right )}{x^3} \, dx=\frac {1}{2} b \sinh (a) \text {Chi}\left (b x^2\right )+\frac {1}{2} b \cosh (a) \text {Shi}\left (b x^2\right )-\frac {\cosh \left (a+b x^2\right )}{2 x^2} \]

[In]

Int[Cosh[a + b*x^2]/x^3,x]

[Out]

-1/2*Cosh[a + b*x^2]/x^2 + (b*CoshIntegral[b*x^2]*Sinh[a])/2 + (b*Cosh[a]*SinhIntegral[b*x^2])/2

Rule 3378

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] - Dist[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]

Rule 3379

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> 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]

Rule 3382

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> 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]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[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]

Rule 5429

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Cosh[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Sim
plify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\cosh (a+b x)}{x^2} \, dx,x,x^2\right ) \\ & = -\frac {\cosh \left (a+b x^2\right )}{2 x^2}+\frac {1}{2} b \text {Subst}\left (\int \frac {\sinh (a+b x)}{x} \, dx,x,x^2\right ) \\ & = -\frac {\cosh \left (a+b x^2\right )}{2 x^2}+\frac {1}{2} (b \cosh (a)) \text {Subst}\left (\int \frac {\sinh (b x)}{x} \, dx,x,x^2\right )+\frac {1}{2} (b \sinh (a)) \text {Subst}\left (\int \frac {\cosh (b x)}{x} \, dx,x,x^2\right ) \\ & = -\frac {\cosh \left (a+b x^2\right )}{2 x^2}+\frac {1}{2} b \text {Chi}\left (b x^2\right ) \sinh (a)+\frac {1}{2} b \cosh (a) \text {Shi}\left (b x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int \frac {\cosh \left (a+b x^2\right )}{x^3} \, dx=\frac {1}{2} \left (-\frac {\cosh \left (a+b x^2\right )}{x^2}+b \text {Chi}\left (b x^2\right ) \sinh (a)+b \cosh (a) \text {Shi}\left (b x^2\right )\right ) \]

[In]

Integrate[Cosh[a + b*x^2]/x^3,x]

[Out]

(-(Cosh[a + b*x^2]/x^2) + b*CoshIntegral[b*x^2]*Sinh[a] + b*Cosh[a]*SinhIntegral[b*x^2])/2

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.38

method	result	size
risch	\(-\frac {-{\mathrm e}^{-a} \operatorname {Ei}_{1}\left (b \,x^{2}\right ) b \,x^{2}+\operatorname {Ei}_{1}\left (-b \,x^{2}\right ) {\mathrm e}^{a} b \,x^{2}+{\mathrm e}^{-b \,x^{2}-a}+{\mathrm e}^{b \,x^{2}+a}}{4 x^{2}}\)	\(58\)
meijerg	\(\frac {i \cosh \left (a \right ) \sqrt {\pi }\, b \left (\frac {4 i \cosh \left (b \,x^{2}\right )}{b \,x^{2} \sqrt {\pi }}-\frac {4 i \operatorname {Shi}\left (b \,x^{2}\right )}{\sqrt {\pi }}\right )}{8}+\frac {\sinh \left (a \right ) \sqrt {\pi }\, b \left (\frac {4 \gamma -4+8 \ln \left (x \right )+4 \ln \left (i b \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \sinh \left (b \,x^{2}\right )}{\sqrt {\pi }\, b \,x^{2}}+\frac {4 \,\operatorname {Chi}\left (b \,x^{2}\right )-4 \ln \left (b \,x^{2}\right )-4 \gamma }{\sqrt {\pi }}\right )}{8}\)	\(117\)

[In]

int(cosh(b*x^2+a)/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/4*(-exp(-a)*Ei(1,b*x^2)*b*x^2+Ei(1,-b*x^2)*exp(a)*b*x^2+exp(-b*x^2-a)+exp(b*x^2+a))/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.69 \[ \int \frac {\cosh \left (a+b x^2\right )}{x^3} \, dx=\frac {{\left (b x^{2} {\rm Ei}\left (b x^{2}\right ) - b x^{2} {\rm Ei}\left (-b x^{2}\right )\right )} \cosh \left (a\right ) + {\left (b x^{2} {\rm Ei}\left (b x^{2}\right ) + b x^{2} {\rm Ei}\left (-b x^{2}\right )\right )} \sinh \left (a\right ) - 2 \, \cosh \left (b x^{2} + a\right )}{4 \, x^{2}} \]

[In]

integrate(cosh(b*x^2+a)/x^3,x, algorithm="fricas")

[Out]

1/4*((b*x^2*Ei(b*x^2) - b*x^2*Ei(-b*x^2))*cosh(a) + (b*x^2*Ei(b*x^2) + b*x^2*Ei(-b*x^2))*sinh(a) - 2*cosh(b*x^
2 + a))/x^2

Sympy [F]

\[ \int \frac {\cosh \left (a+b x^2\right )}{x^3} \, dx=\int \frac {\cosh {\left (a + b x^{2} \right )}}{x^{3}}\, dx \]

[In]

integrate(cosh(b*x**2+a)/x**3,x)

[Out]

Integral(cosh(a + b*x**2)/x**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {\cosh \left (a+b x^2\right )}{x^3} \, dx=-\frac {1}{4} \, {\left ({\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} - {\rm Ei}\left (b x^{2}\right ) e^{a}\right )} b - \frac {\cosh \left (b x^{2} + a\right )}{2 \, x^{2}} \]

[In]

integrate(cosh(b*x^2+a)/x^3,x, algorithm="maxima")

[Out]

-1/4*(Ei(-b*x^2)*e^(-a) - Ei(b*x^2)*e^a)*b - 1/2*cosh(b*x^2 + a)/x^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (36) = 72\).

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.57 \[ \int \frac {\cosh \left (a+b x^2\right )}{x^3} \, dx=-\frac {{\left (b x^{2} + a\right )} b^{2} {\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} - a b^{2} {\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} - {\left (b x^{2} + a\right )} b^{2} {\rm Ei}\left (b x^{2}\right ) e^{a} + a b^{2} {\rm Ei}\left (b x^{2}\right ) e^{a} + b^{2} e^{\left (b x^{2} + a\right )} + b^{2} e^{\left (-b x^{2} - a\right )}}{4 \, b^{2} x^{2}} \]

[In]

integrate(cosh(b*x^2+a)/x^3,x, algorithm="giac")

[Out]

-1/4*((b*x^2 + a)*b^2*Ei(-b*x^2)*e^(-a) - a*b^2*Ei(-b*x^2)*e^(-a) - (b*x^2 + a)*b^2*Ei(b*x^2)*e^a + a*b^2*Ei(b
*x^2)*e^a + b^2*e^(b*x^2 + a) + b^2*e^(-b*x^2 - a))/(b^2*x^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cosh \left (a+b x^2\right )}{x^3} \, dx=\int \frac {\mathrm {cosh}\left (b\,x^2+a\right )}{x^3} \,d x \]

[In]

int(cosh(a + b*x^2)/x^3,x)

[Out]

int(cosh(a + b*x^2)/x^3, x)