3.2.23 \(\int \frac {\sqrt {a+i a \sinh (e+f x)}}{x^2} \, dx\) [123]

Optimal. Leaf size=149 \[ -\frac {\sqrt {a+i a \sinh (e+f x)}}{x}+\frac {1}{2} f \text {Chi}\left (\frac {f x}{2}\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sinh \left (\frac {1}{4} (2 e+i \pi )\right ) \sqrt {a+i a \sinh (e+f x)}+\frac {1}{2} f \cosh \left (\frac {1}{4} (2 e+i \pi )\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)} \text {Shi}\left (\frac {f x}{2}\right ) \]

[Out]

-(a+I*a*sinh(f*x+e))^(1/2)/x+1/2*f*cosh(1/2*e+1/4*I*Pi)*sech(1/2*e+1/4*I*Pi+1/2*f*x)*Shi(1/2*f*x)*(a+I*a*sinh(
f*x+e))^(1/2)+1/2*f*Chi(1/2*f*x)*sech(1/2*e+1/4*I*Pi+1/2*f*x)*sinh(1/2*e+1/4*I*Pi)*(a+I*a*sinh(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3400, 3378, 3384, 3379, 3382} \begin {gather*} \frac {1}{2} f \sinh \left (\frac {1}{4} (2 e+i \pi )\right ) \text {Chi}\left (\frac {f x}{2}\right ) \text {sech}\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}+\frac {1}{2} f \cosh \left (\frac {1}{4} (2 e+i \pi )\right ) \text {Shi}\left (\frac {f x}{2}\right ) \text {sech}\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}-\frac {\sqrt {a+i a \sinh (e+f x)}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*a*Sinh[e + f*x]]/x^2,x]

[Out]

-(Sqrt[a + I*a*Sinh[e + f*x]]/x) + (f*CoshIntegral[(f*x)/2]*Sech[e/2 + (I/4)*Pi + (f*x)/2]*Sinh[(2*e + I*Pi)/4
]*Sqrt[a + I*a*Sinh[e + f*x]])/2 + (f*Cosh[(2*e + I*Pi)/4]*Sech[e/2 + (I/4)*Pi + (f*x)/2]*Sqrt[a + I*a*Sinh[e
+ f*x]]*SinhIntegral[(f*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*} \int \frac {\sqrt {a+i a \sinh (e+f x)}}{x^2} \, dx &=\left (\text {csch}\left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)}\right ) \int \frac {\sinh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )}{x^2} \, dx\\ &=-\frac {\sqrt {a+i a \sinh (e+f x)}}{x}+\frac {1}{2} \left (f \text {csch}\left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)}\right ) \int \frac {\cosh \left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right )}{x} \, dx\\ &=-\frac {\sqrt {a+i a \sinh (e+f x)}}{x}-\frac {1}{2} \left (i f \cosh \left (\frac {1}{4} (2 e+i \pi )\right ) \text {csch}\left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)}\right ) \int \frac {\sinh \left (\frac {f x}{2}\right )}{x} \, dx-\frac {1}{2} \left (i f \text {csch}\left (\frac {e}{2}-\frac {i \pi }{4}+\frac {f x}{2}\right ) \sinh \left (\frac {1}{4} (2 e+i \pi )\right ) \sqrt {a+i a \sinh (e+f x)}\right ) \int \frac {\cosh \left (\frac {f x}{2}\right )}{x} \, dx\\ &=-\frac {\sqrt {a+i a \sinh (e+f x)}}{x}+\frac {1}{2} f \text {Chi}\left (\frac {f x}{2}\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sinh \left (\frac {1}{4} (2 e+i \pi )\right ) \sqrt {a+i a \sinh (e+f x)}+\frac {1}{2} f \cosh \left (\frac {1}{4} (2 e+i \pi )\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)} \text {Shi}\left (\frac {f x}{2}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.15, size = 133, normalized size = 0.89 \begin {gather*} \frac {\sqrt {a+i a \sinh (e+f x)} \left (f x \text {Chi}\left (\frac {f x}{2}\right ) \left (i \cosh \left (\frac {e}{2}\right )+\sinh \left (\frac {e}{2}\right )\right )-2 \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )+f x \left (\cosh \left (\frac {e}{2}\right )+i \sinh \left (\frac {e}{2}\right )\right ) \text {Shi}\left (\frac {f x}{2}\right )\right )}{2 x \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + I*a*Sinh[e + f*x]]/x^2,x]

[Out]

(Sqrt[a + I*a*Sinh[e + f*x]]*(f*x*CoshIntegral[(f*x)/2]*(I*Cosh[e/2] + Sinh[e/2]) - 2*(Cosh[(e + f*x)/2] + I*S
inh[(e + f*x)/2]) + f*x*(Cosh[e/2] + I*Sinh[e/2])*SinhIntegral[(f*x)/2]))/(2*x*(Cosh[(e + f*x)/2] + I*Sinh[(e
+ f*x)/2]))

________________________________________________________________________________________

Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {a +i a \sinh \left (f x +e \right )}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*sinh(f*x+e))^(1/2)/x^2,x)

[Out]

int((a+I*a*sinh(f*x+e))^(1/2)/x^2,x)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(f*x+e))^(1/2)/x^2,x, algorithm="maxima")

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)/x^2, x)

________________________________________________________________________________________

Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(f*x+e))^(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {i a \left (\sinh {\left (e + f x \right )} - i\right )}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(f*x+e))**(1/2)/x**2,x)

[Out]

Integral(sqrt(I*a*(sinh(e + f*x) - I))/x**2, x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*sinh(f*x+e))^(1/2)/x^2,x, algorithm="giac")

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)/x^2, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sinh(e + f*x)*1i)^(1/2)/x^2,x)

[Out]

int((a + a*sinh(e + f*x)*1i)^(1/2)/x^2, x)

________________________________________________________________________________________