### 3.278 $$\int \frac{\cosh (x) \sinh (x)}{x^2} \, dx$$

Optimal. Leaf size=16 $\text{Chi}(2 x)-\frac{\sinh (2 x)}{2 x}$

[Out]

CoshIntegral[2*x] - Sinh[2*x]/(2*x)

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Rubi [A]  time = 0.0474439, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {5448, 12, 3297, 3301} $\text{Chi}(2 x)-\frac{\sinh (2 x)}{2 x}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Cosh[x]*Sinh[x])/x^2,x]

[Out]

CoshIntegral[2*x] - Sinh[2*x]/(2*x)

Rule 5448

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 (x) \sinh (x)}{x^2} \, dx &=\int \frac{\sinh (2 x)}{2 x^2} \, dx\\ &=\frac{1}{2} \int \frac{\sinh (2 x)}{x^2} \, dx\\ &=-\frac{\sinh (2 x)}{2 x}+\int \frac{\cosh (2 x)}{x} \, dx\\ &=\text{Chi}(2 x)-\frac{\sinh (2 x)}{2 x}\\ \end{align*}

Mathematica [A]  time = 0.0061389, size = 16, normalized size = 1. $\text{Chi}(2 x)-\frac{\sinh (2 x)}{2 x}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Cosh[x]*Sinh[x])/x^2,x]

[Out]

CoshIntegral[2*x] - Sinh[2*x]/(2*x)

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Maple [A]  time = 0.007, size = 15, normalized size = 0.9 \begin{align*}{\it Chi} \left ( 2\,x \right ) -{\frac{\sinh \left ( 2\,x \right ) }{2\,x}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)*sinh(x)/x^2,x)

[Out]

Chi(2*x)-1/2*sinh(2*x)/x

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Maxima [A]  time = 1.23678, size = 20, normalized size = 1.25 \begin{align*} \frac{1}{2} \, \Gamma \left (-1, 2 \, x\right ) + \frac{1}{2} \, \Gamma \left (-1, -2 \, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sinh(x)/x^2,x, algorithm="maxima")

[Out]

1/2*gamma(-1, 2*x) + 1/2*gamma(-1, -2*x)

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Fricas [A]  time = 2.02682, size = 70, normalized size = 4.38 \begin{align*} \frac{x{\rm Ei}\left (2 \, x\right ) + x{\rm Ei}\left (-2 \, x\right ) - 2 \, \cosh \left (x\right ) \sinh \left (x\right )}{2 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sinh(x)/x^2,x, algorithm="fricas")

[Out]

1/2*(x*Ei(2*x) + x*Ei(-2*x) - 2*cosh(x)*sinh(x))/x

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sinh(x)/x**2,x)

[Out]

Integral(sinh(x)*cosh(x)/x**2, x)

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Giac [B]  time = 1.12932, size = 41, normalized size = 2.56 \begin{align*} \frac{2 \, x{\rm Ei}\left (2 \, x\right ) + 2 \, x{\rm Ei}\left (-2 \, x\right ) - e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}}{4 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sinh(x)/x^2,x, algorithm="giac")

[Out]

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