3.2.22 \(\int \frac {\sqrt {a+i a \sinh (e+f x)}}{x} \, dx\) [122]

Optimal. Leaf size=125 \[ i \text {Chi}\left (\frac {f x}{2}\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sinh \left (\frac {1}{4} (2 e-i \pi )\right ) \sqrt {a+i a \sinh (e+f x)}+i \cosh \left (\frac {1}{4} (2 e-i \pi )\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)} \text {Shi}\left (\frac {f x}{2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3400, 3384, 3379, 3382} \begin {gather*} i \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)}+i \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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*a*Sinh[e + f*x]]/x,x]

[Out]

I*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]] + I*Co
sh[(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]

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} \, 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} \, dx\\ &=\left (\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+\left (\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\\ &=i \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)}+i \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*}

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Mathematica [A]
time = 0.09, size = 96, normalized size = 0.77 \begin {gather*} \frac {\sqrt {a+i a \sinh (e+f x)} \left (\text {Chi}\left (\frac {f x}{2}\right ) \left (\cosh \left (\frac {e}{2}\right )+i \sinh \left (\frac {e}{2}\right )\right )+\left (i \cosh \left (\frac {e}{2}\right )+\sinh \left (\frac {e}{2}\right )\right ) \text {Shi}\left (\frac {f x}{2}\right )\right )}{\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + I*a*Sinh[e + f*x]]*(CoshIntegral[(f*x)/2]*(Cosh[e/2] + I*Sinh[e/2]) + (I*Cosh[e/2] + Sinh[e/2])*Sinh
Integral[(f*x)/2]))/(Cosh[(e + f*x)/2] + I*Sinh[(e + f*x)/2])

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Maple [F]
time = 0.31, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {a +i a \sinh \left (f x +e \right )}}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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,x, algorithm="maxima")

[Out]

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

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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,x, algorithm="fricas")

[Out]

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

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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}\, 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,x)

[Out]

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

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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,x, algorithm="giac")

[Out]

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

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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} \,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,x)

[Out]

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

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