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

Optimal. Leaf size=57 \[ -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{\sqrt {a}}+\frac {2 \cosh (x)}{\sqrt {a+i a \sinh (x)}} \]

[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)

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Rubi [A]
time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2830, 2728, 212} \begin {gather*} \frac {2 \cosh (x)}{\sqrt {a+i a \sinh (x)}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {\sinh (x)}{\sqrt {a+i a \sinh (x)}} \, dx &=\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} \tanh ^{-1}\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*}

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Mathematica [A]
time = 0.06, size = 75, normalized size = 1.32 \begin {gather*} \frac {2 \left ((1+i) \sqrt [4]{-1} \text {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)}} \end {gather*}

Antiderivative was successfully verified.

[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]]

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Maple [F]
time = 0.64, size = 0, normalized size = 0.00 \[\int \frac {\sinh \left (x \right )}{\sqrt {a +i a \sinh \left (x \right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 76, normalized size = 1.33 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sinh {\left (x \right )}}{\sqrt {i a \left (\sinh {\left (x \right )} - i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\mathrm {sinh}\left (x\right )}{\sqrt {a+a\,\mathrm {sinh}\left (x\right )\,1{}\mathrm {i}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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