3.3.78 \(\int \frac {\cosh (x) \sinh (x)}{x^2} \, dx\) [278]

3.3.78.1 Optimal result

Integrand size = 8, antiderivative size = 16 \[ \int \frac {\cosh (x) \sinh (x)}{x^2} \, dx=\text {Chi}(2 x)-\frac {\sinh (2 x)}{2 x} \]

output

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

3.3.78.2 Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\cosh (x) \sinh (x)}{x^2} \, dx=\text {Chi}(2 x)-\frac {\sinh (2 x)}{2 x} \]

input

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

output

CoshIntegral[2*x] - Sinh[2*x]/(2*x)
 

3.3.78.3 Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.62, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5971, 27, 3042, 26, 3778, 3042, 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.

\(\displaystyle \int \frac {\sinh (x) \cosh (x)}{x^2} \, dx\)

\(\Big \downarrow \) 5971

\(\displaystyle \int \frac {\sinh (2 x)}{2 x^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int \frac {\sinh (2 x)}{x^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int -\frac {i \sin (2 i x)}{x^2}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {1}{2} i \int \frac {\sin (2 i x)}{x^2}dx\)

\(\Big \downarrow \) 3778

\(\displaystyle -\frac {1}{2} i \left (2 i \int \frac {\cosh (2 x)}{x}dx-\frac {i \sinh (2 x)}{x}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {1}{2} i \left (2 i \int \frac {\sin \left (2 i x+\frac {\pi }{2}\right )}{x}dx-\frac {i \sinh (2 x)}{x}\right )\)

\(\Big \downarrow \) 3782

\(\displaystyle -\frac {1}{2} i \left (2 i \text {Chi}(2 x)-\frac {i \sinh (2 x)}{x}\right )\)

input

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

output

(-1/2*I)*((2*I)*CoshIntegral[2*x] - (I*Sinh[2*x])/x)
 

3.3.78.3.1 Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3778

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]
 

rule 3782

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]
 

rule 5971

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]
 

3.3.78.4 Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
default	\(\operatorname {Chi}\left (2 x \right )-\frac {\sinh \left (2 x \right )}{2 x}\)	\(15\)
risch	\(\frac {-2 \,\operatorname {Ei}_{1}\left (2 x \right ) x -2 \,\operatorname {Ei}_{1}\left (-2 x \right ) x +{\mathrm e}^{-2 x}-{\mathrm e}^{2 x}}{4 x}\)	\(33\)
meijerg	\(\frac {\sqrt {\pi }\, \left (\frac {4 \gamma -4+4 \ln \left (2\right )+4 \ln \left (x \right )+2 i \pi }{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {2 \sinh \left (2 x \right )}{\sqrt {\pi }\, x}+\frac {4 \,\operatorname {Chi}\left (2 x \right )-4 \ln \left (2 x \right )-4 \gamma }{\sqrt {\pi }}\right )}{4}\)	\(65\)

input

int(cosh(x)*sinh(x)/x^2,x,method=_RETURNVERBOSE)
 

output

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

3.3.78.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\cosh (x) \sinh (x)}{x^2} \, dx=\frac {x {\rm Ei}\left (2 \, x\right ) + x {\rm Ei}\left (-2 \, x\right ) - 2 \, \cosh \left (x\right ) \sinh \left (x\right )}{2 \, x} \]

input

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

output

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

3.3.78.6 Sympy [F]

\[ \int \frac {\cosh (x) \sinh (x)}{x^2} \, dx=\int \frac {\sinh {\left (x \right )} \cosh {\left (x \right )}}{x^{2}}\, dx \]

input

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

output

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

3.3.78.7 Maxima [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {\cosh (x) \sinh (x)}{x^2} \, dx=\frac {1}{2} \, \Gamma \left (-1, 2 \, x\right ) + \frac {1}{2} \, \Gamma \left (-1, -2 \, x\right ) \]

input

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

output

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

3.3.78.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.88 \[ \int \frac {\cosh (x) \sinh (x)}{x^2} \, dx=\frac {2 \, x {\rm Ei}\left (2 \, x\right ) + 2 \, x {\rm Ei}\left (-2 \, x\right ) - e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}}{4 \, x} \]

input

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

output

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

3.3.78.9 Mupad [F(-1)]

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

input

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

output

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