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

Optimal. Leaf size=57 \[ \frac{1}{2} b \sinh (2 a) \text{Chi}\left (2 b x^2\right )+\frac{1}{2} b \cosh (2 a) \text{Shi}\left (2 b x^2\right )-\frac{\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}-\frac{1}{4 x^2} \]

[Out]

-1/(4*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

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Rubi [A]  time = 0.125971, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {5341, 5321, 3297, 3303, 3298, 3301} \[ \frac{1}{2} b \sinh (2 a) \text{Chi}\left (2 b x^2\right )+\frac{1}{2} b \cosh (2 a) \text{Shi}\left (2 b x^2\right )-\frac{\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}-\frac{1}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x^2]^2/x^3,x]

[Out]

-1/(4*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 5341

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]

Rule 5321

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 3297

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 3303

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 3298

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 3301

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]

Rubi steps

\begin{align*} \int \frac{\cosh ^2\left (a+b x^2\right )}{x^3} \, dx &=\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} \operatorname{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 \operatorname{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)) \operatorname{Subst}\left (\int \frac{\sinh (2 b x)}{x} \, dx,x,x^2\right )+\frac{1}{2} (b \sinh (2 a)) \operatorname{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]  time = 0.088959, size = 46, normalized size = 0.81 \[ \frac{1}{2} \left (b \sinh (2 a) \text{Chi}\left (2 b x^2\right )+b \cosh (2 a) \text{Shi}\left (2 b x^2\right )-\frac{\cosh ^2\left (a+b x^2\right )}{x^2}\right ) \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.035, size = 69, normalized size = 1.2 \begin{align*} -{\frac{1}{4\,{x}^{2}}}-{\frac{{{\rm e}^{-2\,a}}{{\rm e}^{-2\,b{x}^{2}}}}{8\,{x}^{2}}}+{\frac{{{\rm e}^{-2\,a}}b{\it Ei} \left ( 1,2\,b{x}^{2} \right ) }{4}}-{\frac{{{\rm e}^{2\,a}}{{\rm e}^{2\,b{x}^{2}}}}{8\,{x}^{2}}}-{\frac{{{\rm e}^{2\,a}}b{\it Ei} \left ( 1,-2\,b{x}^{2} \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/x^2-1/8*exp(-2*a)/x^2*exp(-2*b*x^2)+1/4*exp(-2*a)*b*Ei(1,2*b*x^2)-1/8*exp(2*a)/x^2*exp(2*b*x^2)-1/4*exp(2
*a)*b*Ei(1,-2*b*x^2)

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Maxima [A]  time = 1.10873, size = 49, normalized size = 0.86 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 2.00239, size = 216, normalized size = 3.79 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh ^{2}{\left (a + b x^{2} \right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cosh(a + b*x**2)**2/x**3, x)

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Giac [B]  time = 1.29606, size = 170, normalized size = 2.98 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)