\(\int \frac {\cosh ^2(a+b x^2)}{x^3} \, dx\) [14]

Optimal result

Integrand size = 14, antiderivative size = 57 \[ \int \frac {\cosh ^2\left (a+b x^2\right )}{x^3} \, dx=-\frac {1}{4 x^2}-\frac {\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}+\frac {1}{2} b \text {Chi}\left (2 b x^2\right ) \sinh (2 a)+\frac {1}{2} b \cosh (2 a) \text {Shi}\left (2 b x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5449, 5429, 3378, 3384, 3379, 3382} \[ \int \frac {\cosh ^2\left (a+b x^2\right )}{x^3} \, dx=\frac {1}{2} b \sinh (2 a) \text {Chi}\left (2 b x^2\right )+\frac {1}{2} b \cosh (2 a) \text {Shi}\left (2 b x^2\right )-\frac {\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}-\frac {1}{4 x^2} \]

[In]

Int[Cosh[a + b*x^2]^2/x^3,x]

[Out]

-1/4*1/x^2 - Cosh[2*(a + b*x^2)]/(4*x^2) + (b*CoshIntegral[2*b*x^2]*Sinh[2*a])/2 + (b*Cosh[2*a]*SinhIntegral[2
*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]))

Rule 5449

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(
e*x)^m, (a + b*Cosh[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2 x^3}+\frac {\cosh \left (2 a+2 b x^2\right )}{2 x^3}\right ) \, dx \\ & = -\frac {1}{4 x^2}+\frac {1}{2} \int \frac {\cosh \left (2 a+2 b x^2\right )}{x^3} \, dx \\ & = -\frac {1}{4 x^2}+\frac {1}{4} \text {Subst}\left (\int \frac {\cosh (2 a+2 b x)}{x^2} \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x^2}-\frac {\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}+\frac {1}{2} b \text {Subst}\left (\int \frac {\sinh (2 a+2 b x)}{x} \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x^2}-\frac {\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}+\frac {1}{2} (b \cosh (2 a)) \text {Subst}\left (\int \frac {\sinh (2 b x)}{x} \, dx,x,x^2\right )+\frac {1}{2} (b \sinh (2 a)) \text {Subst}\left (\int \frac {\cosh (2 b x)}{x} \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x^2}-\frac {\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}+\frac {1}{2} b \text {Chi}\left (2 b x^2\right ) \sinh (2 a)+\frac {1}{2} b \cosh (2 a) \text {Shi}\left (2 b x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.81 \[ \int \frac {\cosh ^2\left (a+b x^2\right )}{x^3} \, dx=\frac {1}{2} \left (-\frac {\cosh ^2\left (a+b x^2\right )}{x^2}+b \text {Chi}\left (2 b x^2\right ) \sinh (2 a)+b \cosh (2 a) \text {Shi}\left (2 b x^2\right )\right ) \]

[In]

Integrate[Cosh[a + b*x^2]^2/x^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.16

method	result	size
risch	\(-\frac {-2 \,{\mathrm e}^{-2 a} \operatorname {Ei}_{1}\left (2 b \,x^{2}\right ) b \,x^{2}+2 \,{\mathrm e}^{2 a} \operatorname {Ei}_{1}\left (-2 b \,x^{2}\right ) b \,x^{2}+{\mathrm e}^{-2 b \,x^{2}-2 a}+{\mathrm e}^{2 b \,x^{2}+2 a}+2}{8 x^{2}}\)	\(66\)

[In]

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

[Out]

-1/8*(-2*exp(-2*a)*Ei(1,2*b*x^2)*b*x^2+2*exp(2*a)*Ei(1,-2*b*x^2)*b*x^2+exp(-2*b*x^2-2*a)+exp(2*b*x^2+2*a)+2)/x
^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.58 \[ \int \frac {\cosh ^2\left (a+b x^2\right )}{x^3} \, dx=-\frac {\cosh \left (b x^{2} + a\right )^{2} - {\left (b x^{2} {\rm Ei}\left (2 \, b x^{2}\right ) - b x^{2} {\rm Ei}\left (-2 \, b x^{2}\right )\right )} \cosh \left (2 \, a\right ) + \sinh \left (b x^{2} + a\right )^{2} - {\left (b x^{2} {\rm Ei}\left (2 \, b x^{2}\right ) + b x^{2} {\rm Ei}\left (-2 \, b x^{2}\right )\right )} \sinh \left (2 \, a\right ) + 1}{4 \, x^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(cosh(a + b*x**2)**2/x**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.63 \[ \int \frac {\cosh ^2\left (a+b x^2\right )}{x^3} \, dx=-\frac {1}{4} \, b e^{\left (-2 \, a\right )} \Gamma \left (-1, 2 \, b x^{2}\right ) + \frac {1}{4} \, b e^{\left (2 \, a\right )} \Gamma \left (-1, -2 \, b x^{2}\right ) - \frac {1}{4 \, x^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (50) = 100\).

Time = 0.26 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.21 \[ \int \frac {\cosh ^2\left (a+b x^2\right )}{x^3} \, dx=\frac {2 \, {\left (b x^{2} + a\right )} b^{2} {\rm Ei}\left (2 \, b x^{2}\right ) e^{\left (2 \, a\right )} - 2 \, a b^{2} {\rm Ei}\left (2 \, b x^{2}\right ) e^{\left (2 \, a\right )} - 2 \, {\left (b x^{2} + a\right )} b^{2} {\rm Ei}\left (-2 \, b x^{2}\right ) e^{\left (-2 \, a\right )} + 2 \, a b^{2} {\rm Ei}\left (-2 \, b x^{2}\right ) e^{\left (-2 \, a\right )} - b^{2} e^{\left (2 \, b x^{2} + 2 \, a\right )} - b^{2} e^{\left (-2 \, b x^{2} - 2 \, a\right )} - 2 \, b^{2}}{8 \, b^{2} x^{2}} \]

[In]

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

[Out]

1/8*(2*(b*x^2 + a)*b^2*Ei(2*b*x^2)*e^(2*a) - 2*a*b^2*Ei(2*b*x^2)*e^(2*a) - 2*(b*x^2 + a)*b^2*Ei(-2*b*x^2)*e^(-
2*a) + 2*a*b^2*Ei(-2*b*x^2)*e^(-2*a) - b^2*e^(2*b*x^2 + 2*a) - b^2*e^(-2*b*x^2 - 2*a) - 2*b^2)/(b^2*x^2)

Mupad [F(-1)]

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

[In]

int(cosh(a + b*x^2)^2/x^3,x)

[Out]

int(cosh(a + b*x^2)^2/x^3, x)