3.4.16 \(\int \frac {e^{-2 i \text {ArcTan}(a x)}}{\sqrt {c+a^2 c x^2}} \, dx\) [316]

Optimal. Leaf size=63 \[ \frac {2 i (1-i a x)}{a \sqrt {c+a^2 c x^2}}-\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a \sqrt {c}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {5182, 667, 223, 212} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a \sqrt {c}}+\frac {2 i (1-i a x)}{a \sqrt {a^2 c x^2+c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^((2*I)*ArcTan[a*x])*Sqrt[c + a^2*c*x^2]),x]

[Out]

((2*I)*(1 - I*a*x))/(a*Sqrt[c + a^2*c*x^2]) - ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]]/(a*Sqrt[c])

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 667

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)*((a + c*x^2)^(p + 1)/(c*(p
+ 1))), x] - Dist[e^2*((p + 2)/(c*(p + 1))), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 5182

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[c^(I*(n/2)), Int[(c + d*x^2)^(p
- I*(n/2))*(1 - I*a*x)^(I*n), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[d, a^2*c] &&  !(IntegerQ[p] || GtQ[c, 0]
) && IGtQ[I*(n/2), 0]

Rubi steps

\begin {align*} \int \frac {e^{-2 i \tan ^{-1}(a x)}}{\sqrt {c+a^2 c x^2}} \, dx &=c \int \frac {(1-i a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac {2 i (1-i a x)}{a \sqrt {c+a^2 c x^2}}-\int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx\\ &=\frac {2 i (1-i a x)}{a \sqrt {c+a^2 c x^2}}-\text {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )\\ &=\frac {2 i (1-i a x)}{a \sqrt {c+a^2 c x^2}}-\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a \sqrt {c}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 117, normalized size = 1.86 \begin {gather*} \frac {2 \sqrt {1+a^2 x^2} \left ((1-i a x) \sqrt {1+i a x}-i \sqrt {1-i a x} (-i+a x) \text {ArcSin}\left (\frac {\sqrt {1-i a x}}{\sqrt {2}}\right )\right )}{a \sqrt {1-i a x} (-i+a x) \sqrt {c+a^2 c x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^((2*I)*ArcTan[a*x])*Sqrt[c + a^2*c*x^2]),x]

[Out]

(2*Sqrt[1 + a^2*x^2]*((1 - I*a*x)*Sqrt[1 + I*a*x] - I*Sqrt[1 - I*a*x]*(-I + a*x)*ArcSin[Sqrt[1 - I*a*x]/Sqrt[2
]]))/(a*Sqrt[1 - I*a*x]*(-I + a*x)*Sqrt[c + a^2*c*x^2])

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Maple [A]
time = 0.08, size = 87, normalized size = 1.38

method	result	size
default	\(-\frac {\ln \left (\frac {a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}+c}\right )}{\sqrt {a^{2} c}}+\frac {2 \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}{a^{2} c \left (x -\frac {i}{a}\right )}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(a^2*c*x/(a^2*c)^(1/2)+(a^2*c*x^2+c)^(1/2))/(a^2*c)^(1/2)+2/a^2/c/(x-I/a)*((x-I/a)^2*a^2*c+2*I*a*c*(x-I/a))
^(1/2)

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Maxima [A]
time = 0.46, size = 40, normalized size = 0.63 \begin {gather*} \frac {2 i \, \sqrt {a^{2} c x^{2} + c}}{i \, a^{2} c x + a c} - \frac {\operatorname {arsinh}\left (a x\right )}{a \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*sqrt(a^2*c*x^2 + c)/(I*a^2*c*x + a*c) - arcsinh(a*x)/(a*sqrt(c))

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (50) = 100\).
time = 3.01, size = 152, normalized size = 2.41 \begin {gather*} -\frac {{\left (a^{2} c x - i \, a c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {2 \, {\left (a^{2} c x + \sqrt {a^{2} c x^{2} + c} a^{2} c \sqrt {\frac {1}{a^{2} c}}\right )}}{x}\right ) - {\left (a^{2} c x - i \, a c\right )} \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {2 \, {\left (a^{2} c x - \sqrt {a^{2} c x^{2} + c} a^{2} c \sqrt {\frac {1}{a^{2} c}}\right )}}{x}\right ) - 4 \, \sqrt {a^{2} c x^{2} + c}}{2 \, {\left (a^{2} c x - i \, a c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a^2*c*x - I*a*c)*sqrt(1/(a^2*c))*log(2*(a^2*c*x + sqrt(a^2*c*x^2 + c)*a^2*c*sqrt(1/(a^2*c)))/x) - (a^2*
c*x - I*a*c)*sqrt(1/(a^2*c))*log(2*(a^2*c*x - sqrt(a^2*c*x^2 + c)*a^2*c*sqrt(1/(a^2*c)))/x) - 4*sqrt(a^2*c*x^2
+ c))/(a^2*c*x - I*a*c)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {a^{2} x^{2}}{a^{2} x^{2} \sqrt {a^{2} c x^{2} + c} - 2 i a x \sqrt {a^{2} c x^{2} + c} - \sqrt {a^{2} c x^{2} + c}}\, dx - \int \frac {1}{a^{2} x^{2} \sqrt {a^{2} c x^{2} + c} - 2 i a x \sqrt {a^{2} c x^{2} + c} - \sqrt {a^{2} c x^{2} + c}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a**2*x**2/(a**2*x**2*sqrt(a**2*c*x**2 + c) - 2*I*a*x*sqrt(a**2*c*x**2 + c) - sqrt(a**2*c*x**2 + c)),
x) - Integral(1/(a**2*x**2*sqrt(a**2*c*x**2 + c) - 2*I*a*x*sqrt(a**2*c*x**2 + c) - sqrt(a**2*c*x**2 + c)), x)

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Giac [A]
time = 0.44, size = 70, normalized size = 1.11 \begin {gather*} \frac {\log \left ({\left | -\sqrt {a^{2} c} x + \sqrt {a^{2} c x^{2} + c} \right |}\right )}{a \sqrt {c}} - \frac {4}{{\left (-i \, \sqrt {a^{2} c} x + i \, \sqrt {a^{2} c x^{2} + c} - \sqrt {c}\right )} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 + c)))/(a*sqrt(c)) - 4/((-I*sqrt(a^2*c)*x + I*sqrt(a^2*c*x^2 + c) - sq
rt(c))*a)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {a^2\,x^2+1}{\sqrt {c\,a^2\,x^2+c}\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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