\(\int \frac {\sqrt {a+i a \sinh (e+f x)}}{x^3} \, dx\) [124]

Optimal result

Integrand size = 21, antiderivative size = 204 \[ \int \frac {\sqrt {a+i a \sinh (e+f x)}}{x^3} \, dx=-\frac {\sqrt {a+i a \sinh (e+f x)}}{2 x^2}+\frac {1}{8} i f^2 \text {Chi}\left (\frac {f x}{2}\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sinh \left (\frac {1}{4} (2 e-i \pi )\right ) \sqrt {a+i a \sinh (e+f x)}+\frac {1}{8} i f^2 \cosh \left (\frac {1}{4} (2 e-i \pi )\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)} \text {Shi}\left (\frac {f x}{2}\right )-\frac {f \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{4 x} \]

[Out]

-1/2*(a+I*a*sinh(f*x+e))^(1/2)/x^2+1/8*f^2*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)+1/8*f^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)-1/4*f*(a+I*a*sinh(f*x+e))^(1/2)*tanh(1/2*e+1/4*I*Pi+1/2*f*x)/x

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3400, 3378, 3384, 3379, 3382} \[ \int \frac {\sqrt {a+i a \sinh (e+f x)}}{x^3} \, dx=\frac {1}{8} i f^2 \sinh \left (\frac {1}{4} (2 e-i \pi )\right ) \text {Chi}\left (\frac {f x}{2}\right ) \text {sech}\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}+\frac {1}{8} i f^2 \cosh \left (\frac {1}{4} (2 e-i \pi )\right ) \text {Shi}\left (\frac {f x}{2}\right ) \text {sech}\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}-\frac {\sqrt {a+i a \sinh (e+f x)}}{2 x^2}-\frac {f \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)}}{4 x} \]

[In]

Int[Sqrt[a + I*a*Sinh[e + f*x]]/x^3,x]

[Out]

-1/2*Sqrt[a + I*a*Sinh[e + f*x]]/x^2 + (I/8)*f^2*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/8)*f^2*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] - (f*Sqrt[a + I*a*Sinh[e + f*x]]*Tanh[e/2 + (I/4)*Pi + (f*x)/2])/
(4*x)

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 (\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^3} \, dx \\ & = -\frac {\sqrt {a+i a \sinh (e+f x)}}{2 x^2}+\frac {1}{4} \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^2} \, dx \\ & = -\frac {\sqrt {a+i a \sinh (e+f x)}}{2 x^2}-\frac {f \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{4 x}-\frac {1}{8} \left (i f^2 \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)}}{2 x^2}-\frac {f \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{4 x}+\frac {1}{8} \left (f^2 \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}{8} \left (f^2 \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)}}{2 x^2}+\frac {1}{8} i f^2 \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}{8} i f^2 \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 )-\frac {f \sqrt {a+i a \sinh (e+f x)} \tanh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{4 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {a+i a \sinh (e+f x)}}{x^3} \, dx=\frac {\sqrt {a+i a \sinh (e+f x)} \left (-4 \cosh \left (\frac {1}{2} (e+f x)\right )-2 i f x \cosh \left (\frac {1}{2} (e+f x)\right )+f^2 x^2 \text {Chi}\left (\frac {f x}{2}\right ) \left (\cosh \left (\frac {e}{2}\right )+i \sinh \left (\frac {e}{2}\right )\right )-4 i \sinh \left (\frac {1}{2} (e+f x)\right )-2 f x \sinh \left (\frac {1}{2} (e+f x)\right )+f^2 x^2 \left (i \cosh \left (\frac {e}{2}\right )+\sinh \left (\frac {e}{2}\right )\right ) \text {Shi}\left (\frac {f x}{2}\right )\right )}{8 x^2 \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[Sqrt[a + I*a*Sinh[e + f*x]]/x^3,x]

[Out]

(Sqrt[a + I*a*Sinh[e + f*x]]*(-4*Cosh[(e + f*x)/2] - (2*I)*f*x*Cosh[(e + f*x)/2] + f^2*x^2*CoshIntegral[(f*x)/
2]*(Cosh[e/2] + I*Sinh[e/2]) - (4*I)*Sinh[(e + f*x)/2] - 2*f*x*Sinh[(e + f*x)/2] + f^2*x^2*(I*Cosh[e/2] + Sinh
[e/2])*SinhIntegral[(f*x)/2]))/(8*x^2*(Cosh[(e + f*x)/2] + I*Sinh[(e + f*x)/2]))

Maple [F]

\[\int \frac {\sqrt {a +i a \sinh \left (f x +e \right )}}{x^{3}}d x\]

[In]

int((a+I*a*sinh(f*x+e))^(1/2)/x^3,x)

[Out]

int((a+I*a*sinh(f*x+e))^(1/2)/x^3,x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+i a \sinh (e+f x)}}{x^3} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((a+I*a*sinh(f*x+e))^(1/2)/x^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {a+i a \sinh (e+f x)}}{x^3} \, dx=\int \frac {\sqrt {i a \left (\sinh {\left (e + f x \right )} - i\right )}}{x^{3}}\, dx \]

[In]

integrate((a+I*a*sinh(f*x+e))**(1/2)/x**3,x)

[Out]

Integral(sqrt(I*a*(sinh(e + f*x) - I))/x**3, x)

Maxima [F]

\[ \int \frac {\sqrt {a+i a \sinh (e+f x)}}{x^3} \, dx=\int { \frac {\sqrt {i \, a \sinh \left (f x + e\right ) + a}}{x^{3}} \,d x } \]

[In]

integrate((a+I*a*sinh(f*x+e))^(1/2)/x^3,x, algorithm="maxima")

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)/x^3, x)

Giac [F]

\[ \int \frac {\sqrt {a+i a \sinh (e+f x)}}{x^3} \, dx=\int { \frac {\sqrt {i \, a \sinh \left (f x + e\right ) + a}}{x^{3}} \,d x } \]

[In]

integrate((a+I*a*sinh(f*x+e))^(1/2)/x^3,x, algorithm="giac")

[Out]

integrate(sqrt(I*a*sinh(f*x + e) + a)/x^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+i a \sinh (e+f x)}}{x^3} \, dx=\int \frac {\sqrt {a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}}}{x^3} \,d x \]

[In]

int((a + a*sinh(e + f*x)*1i)^(1/2)/x^3,x)

[Out]

int((a + a*sinh(e + f*x)*1i)^(1/2)/x^3, x)