3.69 \(\int \frac{1}{\sqrt{a+i a \sinh (c+d x)}} \, dx\)

Optimal. Leaf size=52 \[ \frac{i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{2} \sqrt{a+i a \sinh (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

(I*Sqrt[2]*ArcTanh[(Sqrt[a]*Cosh[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[c + d*x]])])/(Sqrt[a]*d)

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Rubi [A]  time = 0.0243704, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2649, 206} \[ \frac{i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{2} \sqrt{a+i a \sinh (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + I*a*Sinh[c + d*x]],x]

[Out]

(I*Sqrt[2]*ArcTanh[(Sqrt[a]*Cosh[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[c + d*x]])])/(Sqrt[a]*d)

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{1}{\sqrt{a+i a \sinh (c+d x)}} \, dx &=\frac{(2 i) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cosh (c+d x)}{\sqrt{a+i a \sinh (c+d x)}}\right )}{d}\\ &=\frac{i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (c+d x)}{\sqrt{2} \sqrt{a+i a \sinh (c+d x)}}\right )}{\sqrt{a} d}\\ \end{align*}

Mathematica [A]  time = 0.0852399, size = 84, normalized size = 1.62 \[ \frac{(2+2 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (1-i \tanh \left (\frac{1}{4} (c+d x)\right )\right )\right ) \left (\sinh \left (\frac{1}{2} (c+d x)\right )-i \cosh \left (\frac{1}{2} (c+d x)\right )\right )}{d \sqrt{a+i a \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + I*a*Sinh[c + d*x]],x]

[Out]

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

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Maple [F]  time = 0.506, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt{a+ia\sinh \left ( dx+c \right ) }}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(I*a*sinh(d*x + c) + a), x)

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Fricas [B]  time = 2.11263, size = 518, normalized size = 9.96 \begin{align*} i \, \sqrt{2} \sqrt{\frac{1}{a d^{2}}} \log \left (\frac{\sqrt{2}{\left (a d e^{\left (d x + c\right )} - i \, a d\right )} \sqrt{\frac{1}{a d^{2}}} + 2 \, \sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - i \, a} e^{\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c\right )}}{2 \, e^{\left (d x + c\right )} - 2 i}\right ) - i \, \sqrt{2} \sqrt{\frac{1}{a d^{2}}} \log \left (-\frac{\sqrt{2}{\left (a d e^{\left (d x + c\right )} - i \, a d\right )} \sqrt{\frac{1}{a d^{2}}} - 2 \, \sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - i \, a} e^{\left (-\frac{1}{2} \, d x - \frac{1}{2} \, c\right )}}{2 \, e^{\left (d x + c\right )} - 2 i}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*sqrt(2)*sqrt(1/(a*d^2))*log((sqrt(2)*(a*d*e^(d*x + c) - I*a*d)*sqrt(1/(a*d^2)) + 2*sqrt(1/2)*sqrt(I*a*e^(2*d
*x + 2*c) + 2*a*e^(d*x + c) - I*a)*e^(-1/2*d*x - 1/2*c))/(2*e^(d*x + c) - 2*I)) - I*sqrt(2)*sqrt(1/(a*d^2))*lo
g(-(sqrt(2)*(a*d*e^(d*x + c) - I*a*d)*sqrt(1/(a*d^2)) - 2*sqrt(1/2)*sqrt(I*a*e^(2*d*x + 2*c) + 2*a*e^(d*x + c)
- I*a)*e^(-1/2*d*x - 1/2*c))/(2*e^(d*x + c) - 2*I))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(I*a*sinh(c + d*x) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(I*a*sinh(d*x + c) + a), x)