\(\int \frac {\sinh (x)}{\sqrt {a+i a \sinh (x)}} \, dx\) [64]

Optimal result

Integrand size = 16, antiderivative size = 57 \[ \int \frac {\sinh (x)}{\sqrt {a+i a \sinh (x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{\sqrt {a}}+\frac {2 \cosh (x)}{\sqrt {a+i a \sinh (x)}} \]

[Out]

-arctanh(1/2*cosh(x)*a^(1/2)*2^(1/2)/(a+I*a*sinh(x))^(1/2))*2^(1/2)/a^(1/2)+2*cosh(x)/(a+I*a*sinh(x))^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2830, 2728, 212} \[ \int \frac {\sinh (x)}{\sqrt {a+i a \sinh (x)}} \, dx=\frac {2 \cosh (x)}{\sqrt {a+i a \sinh (x)}}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{\sqrt {a}} \]

[In]

Int[Sinh[x]/Sqrt[a + I*a*Sinh[x]],x]

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[a]*Cosh[x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[x]])])/Sqrt[a]) + (2*Cosh[x])/Sqrt[a + I*a*Sin
h[x]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2830

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \cosh (x)}{\sqrt {a+i a \sinh (x)}}+i \int \frac {1}{\sqrt {a+i a \sinh (x)}} \, dx \\ & = \frac {2 \cosh (x)}{\sqrt {a+i a \sinh (x)}}-2 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cosh (x)}{\sqrt {a+i a \sinh (x)}}\right ) \\ & = -\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{\sqrt {a}}+\frac {2 \cosh (x)}{\sqrt {a+i a \sinh (x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.32 \[ \int \frac {\sinh (x)}{\sqrt {a+i a \sinh (x)}} \, dx=\frac {2 \left ((1+i) \sqrt [4]{-1} \arctan \left (\frac {i+\tanh \left (\frac {x}{4}\right )}{\sqrt {2}}\right )+\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )\right ) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )}{\sqrt {a+i a \sinh (x)}} \]

[In]

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

[Out]

(2*((1 + I)*(-1)^(1/4)*ArcTan[(I + Tanh[x/4])/Sqrt[2]] + Cosh[x/2] - I*Sinh[x/2])*(Cosh[x/2] + I*Sinh[x/2]))/S
qrt[a + I*a*Sinh[x]]

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (44 ) = 88\).

Time = 4.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.89

method	result	size
risch	\(\frac {\left ({\mathrm e}^{x}-i\right )^{2} \sqrt {2}\, {\mathrm e}^{-x}}{\sqrt {a \left (i {\mathrm e}^{2 x}+2 \,{\mathrm e}^{x}-i\right ) {\mathrm e}^{-x}}}-\frac {2 i \left (-{\mathrm e}^{x}+i\right ) \left (a^{\frac {3}{2}}+\arctan \left (\frac {\sqrt {i a \,{\mathrm e}^{x}}}{\sqrt {a}}\right ) a \sqrt {i a \,{\mathrm e}^{x}}\right ) \sqrt {2}\, {\mathrm e}^{-x}}{a^{\frac {3}{2}} \sqrt {a \left (i {\mathrm e}^{2 x}+2 \,{\mathrm e}^{x}-i\right ) {\mathrm e}^{-x}}}\)	\(108\)

[In]

int(sinh(x)/(a+I*a*sinh(x))^(1/2),x,method=_RETURNVERBOSE)

[Out]

(exp(x)-I)^2*2^(1/2)/(a*(I*exp(x)^2+2*exp(x)-I)/exp(x))^(1/2)/exp(x)-2*I*(-exp(x)+I)*(a^(3/2)+arctan((I*a*exp(
x))^(1/2)/a^(1/2))*a*(I*a*exp(x))^(1/2))/a^(3/2)*2^(1/2)/(a*(I*exp(x)^2+2*exp(x)-I)/exp(x))^(1/2)/exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.33 \[ \int \frac {\sinh (x)}{\sqrt {a+i a \sinh (x)}} \, dx=-\frac {\sqrt {2} \sqrt {a} \log \left (\frac {1}{2} \, \sqrt {2} \sqrt {a} + \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}}\right ) - \sqrt {2} \sqrt {a} \log \left (-\frac {1}{2} \, \sqrt {2} \sqrt {a} + \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}}\right ) + 2 \, \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (i \, e^{x} - 1\right )}}{a} \]

[In]

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

[Out]

-(sqrt(2)*sqrt(a)*log(1/2*sqrt(2)*sqrt(a) + sqrt(1/2*I*a*e^(-x))) - sqrt(2)*sqrt(a)*log(-1/2*sqrt(2)*sqrt(a) +
sqrt(1/2*I*a*e^(-x))) + 2*sqrt(1/2*I*a*e^(-x))*(I*e^x - 1))/a

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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