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

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Rubi [A]  time = 0.02, antiderivative size = 52, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.09, 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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fricas [B]  time = 0.67, size = 93, normalized size = 1.79 \[ i \, \sqrt {2} \sqrt {\frac {1}{a d^{2}}} \log \left (\frac {1}{2} \, \sqrt {2} a d \sqrt {\frac {1}{a d^{2}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}}\right ) - i \, \sqrt {2} \sqrt {\frac {1}{a d^{2}}} \log \left (-\frac {1}{2} \, \sqrt {2} a d \sqrt {\frac {1}{a d^{2}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}}\right ) \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {i \, a \sinh \left (d x + c\right ) + a}}\,{d x} \]

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)

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maple [F]  time = 0.54, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a +i a \sinh \left (d x +c \right )}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {i \, a \sinh \left (d x + c\right ) + a}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {i a \sinh {\left (c + d x \right )} + a}}\, dx \]

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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