3.64 \(\int \frac{\sinh (x)}{\sqrt{a+i a \sinh (x)}} \, dx\)

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

[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]]

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Rubi [A]  time = 0.0559671, antiderivative size = 57, normalized size of antiderivative = 1., 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 = {2751, 2649, 206} \[ \frac{2 \cosh (x)}{\sqrt{a+i a \sinh (x)}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (x)}{\sqrt{2} \sqrt{a+i a \sinh (x)}}\right )}{\sqrt{a}} \]

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 2751

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*Sin
[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)]

Rule 2649

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 206

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

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 \operatorname{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*}

Mathematica [A]  time = 0.073706, size = 75, normalized size = 1.32 \[ \frac{2 \left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right ) \left (-i \sinh \left (\frac{x}{2}\right )+\cosh \left (\frac{x}{2}\right )+(1+i) \sqrt [4]{-1} \tan ^{-1}\left (\frac{\tanh \left (\frac{x}{4}\right )+i}{\sqrt{2}}\right )\right )}{\sqrt{a+i a \sinh (x)}} \]

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 = 0.355, size = 0, normalized size = 0. \begin{align*} \int{\sinh \left ( x \right ){\frac{1}{\sqrt{a+ia\sinh \left ( x \right ) }}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (x\right )}{\sqrt{i \, a \sinh \left (x\right ) + a}}\,{d x} \end{align*}

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 [B]  time = 2.12969, size = 501, normalized size = 8.79 \begin{align*} -\frac{2 \, \sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a}{\left (i \, e^{x} - 1\right )} e^{\left (-\frac{1}{2} \, x\right )} + \frac{\sqrt{2}{\left (a e^{x} - i \, a\right )} \log \left (\frac{2 \, \sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a} e^{\left (-\frac{1}{2} \, x\right )} + \frac{\sqrt{2}{\left (a e^{x} - i \, a\right )}}{\sqrt{a}}}{2 \, e^{x} - 2 i}\right )}{\sqrt{a}} - \frac{\sqrt{2}{\left (a e^{x} - i \, a\right )} \log \left (\frac{2 \, \sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a} e^{\left (-\frac{1}{2} \, x\right )} - \frac{\sqrt{2}{\left (a e^{x} - i \, a\right )}}{\sqrt{a}}}{2 \, e^{x} - 2 i}\right )}{\sqrt{a}}}{a e^{x} - i \, a} \end{align*}

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]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh{\left (x \right )}}{\sqrt{a \left (i \sinh{\left (x \right )} + 1\right )}}\, dx \end{align*}

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(a*(I*sinh(x) + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (x\right )}{\sqrt{i \, a \sinh \left (x\right ) + a}}\,{d x} \end{align*}

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)