\(\int \frac {\sqrt {a+a \cosh (c+d x)}}{x^2} \, dx\) [125]

Optimal result

Integrand size = 18, antiderivative size = 110 \[ \int \frac {\sqrt {a+a \cosh (c+d x)}}{x^2} \, dx=-\frac {\sqrt {a+a \cosh (c+d x)}}{x}+\frac {1}{2} d \sqrt {a+a \cosh (c+d x)} \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}\right )+\frac {1}{2} d \cosh \left (\frac {c}{2}\right ) \sqrt {a+a \cosh (c+d x)} \text {sech}\left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Shi}\left (\frac {d x}{2}\right ) \]

[Out]

-(a+a*cosh(d*x+c))^(1/2)/x+1/2*d*cosh(1/2*c)*sech(1/2*d*x+1/2*c)*Shi(1/2*d*x)*(a+a*cosh(d*x+c))^(1/2)+1/2*d*Ch
i(1/2*d*x)*sech(1/2*d*x+1/2*c)*sinh(1/2*c)*(a+a*cosh(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3400, 3378, 3384, 3379, 3382} \[ \int \frac {\sqrt {a+a \cosh (c+d x)}}{x^2} \, dx=\frac {1}{2} d \sinh \left (\frac {c}{2}\right ) \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cosh (c+d x)+a}+\frac {1}{2} d \cosh \left (\frac {c}{2}\right ) \text {Shi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cosh (c+d x)+a}-\frac {\sqrt {a \cosh (c+d x)+a}}{x} \]

[In]

Int[Sqrt[a + a*Cosh[c + d*x]]/x^2,x]

[Out]

-(Sqrt[a + a*Cosh[c + d*x]]/x) + (d*Sqrt[a + a*Cosh[c + d*x]]*CoshIntegral[(d*x)/2]*Sech[c/2 + (d*x)/2]*Sinh[c
/2])/2 + (d*Cosh[c/2]*Sqrt[a + a*Cosh[c + d*x]]*Sech[c/2 + (d*x)/2]*SinhIntegral[(d*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 3400

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(2*a)^IntPart[n]
*((a + b*Sin[e + f*x])^FracPart[n]/Sin[e/2 + a*(Pi/(4*b)) + f*(x/2)]^(2*FracPart[n])), Int[(c + d*x)^m*Sin[e/2
+ a*(Pi/(4*b)) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {a+a \cosh (c+d x)} \csc \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right )\right ) \int \frac {\sin \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right )}{x^2} \, dx \\ & = -\frac {\sqrt {a+a \cosh (c+d x)}}{x}+\frac {1}{2} \left (d \sqrt {a+a \cosh (c+d x)} \csc \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right )\right ) \int \frac {\sinh \left (\frac {c}{2}+\frac {d x}{2}\right )}{x} \, dx \\ & = -\frac {\sqrt {a+a \cosh (c+d x)}}{x}+\frac {1}{2} \left (d \cosh \left (\frac {c}{2}\right ) \sqrt {a+a \cosh (c+d x)} \csc \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right )\right ) \int \frac {\sinh \left (\frac {d x}{2}\right )}{x} \, dx+\frac {1}{2} \left (d \sqrt {a+a \cosh (c+d x)} \csc \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right ) \sinh \left (\frac {c}{2}\right )\right ) \int \frac {\cosh \left (\frac {d x}{2}\right )}{x} \, dx \\ & = -\frac {\sqrt {a+a \cosh (c+d x)}}{x}+\frac {1}{2} d \sqrt {a+a \cosh (c+d x)} \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}\right )+\frac {1}{2} d \cosh \left (\frac {c}{2}\right ) \sqrt {a+a \cosh (c+d x)} \text {sech}\left (\frac {c}{2}+\frac {d x}{2}\right ) \text {Shi}\left (\frac {d x}{2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {a+a \cosh (c+d x)}}{x^2} \, dx=\frac {\sqrt {a (1+\cosh (c+d x))} \left (-2+d x \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {1}{2} (c+d x)\right ) \sinh \left (\frac {c}{2}\right )+d x \cosh \left (\frac {c}{2}\right ) \text {sech}\left (\frac {1}{2} (c+d x)\right ) \text {Shi}\left (\frac {d x}{2}\right )\right )}{2 x} \]

[In]

Integrate[Sqrt[a + a*Cosh[c + d*x]]/x^2,x]

[Out]

(Sqrt[a*(1 + Cosh[c + d*x])]*(-2 + d*x*CoshIntegral[(d*x)/2]*Sech[(c + d*x)/2]*Sinh[c/2] + d*x*Cosh[c/2]*Sech[
(c + d*x)/2]*SinhIntegral[(d*x)/2]))/(2*x)

Maple [F]

\[\int \frac {\sqrt {a +a \cosh \left (d x +c \right )}}{x^{2}}d x\]

[In]

int((a+a*cosh(d*x+c))^(1/2)/x^2,x)

[Out]

int((a+a*cosh(d*x+c))^(1/2)/x^2,x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+a \cosh (c+d x)}}{x^2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((a+a*cosh(d*x+c))^(1/2)/x^2,x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [F]

\[ \int \frac {\sqrt {a+a \cosh (c+d x)}}{x^2} \, dx=\int \frac {\sqrt {a \left (\cosh {\left (c + d x \right )} + 1\right )}}{x^{2}}\, dx \]

[In]

integrate((a+a*cosh(d*x+c))**(1/2)/x**2,x)

[Out]

Integral(sqrt(a*(cosh(c + d*x) + 1))/x**2, x)

Maxima [F]

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

[In]

integrate((a+a*cosh(d*x+c))^(1/2)/x^2,x, algorithm="maxima")

[Out]

integrate(sqrt(a*cosh(d*x + c) + a)/x^2, x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {a+a \cosh (c+d x)}}{x^2} \, dx=\frac {\sqrt {2} {\left (\sqrt {a} d x {\rm Ei}\left (\frac {1}{2} \, d x\right ) e^{\left (\frac {1}{2} \, c\right )} - \sqrt {a} d x {\rm Ei}\left (-\frac {1}{2} \, d x\right ) e^{\left (-\frac {1}{2} \, c\right )} - 2 \, \sqrt {a} e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - 2 \, \sqrt {a} e^{\left (-\frac {1}{2} \, d x - \frac {1}{2} \, c\right )}\right )}}{4 \, x} \]

[In]

integrate((a+a*cosh(d*x+c))^(1/2)/x^2,x, algorithm="giac")

[Out]

1/4*sqrt(2)*(sqrt(a)*d*x*Ei(1/2*d*x)*e^(1/2*c) - sqrt(a)*d*x*Ei(-1/2*d*x)*e^(-1/2*c) - 2*sqrt(a)*e^(1/2*d*x +
1/2*c) - 2*sqrt(a)*e^(-1/2*d*x - 1/2*c))/x

Mupad [F(-1)]

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

[In]

int((a + a*cosh(c + d*x))^(1/2)/x^2,x)

[Out]

int((a + a*cosh(c + d*x))^(1/2)/x^2, x)