3.315 \(\int \frac{e^{-i \tan ^{-1}(a x)}}{\sqrt{c+a^2 c x^2}} \, dx\)

Optimal. Leaf size=43 \[ -\frac{i \sqrt{a^2 x^2+1} \log (-a x+i)}{a \sqrt{a^2 c x^2+c}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0658911, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {5076, 5073, 31} \[ -\frac{i \sqrt{a^2 x^2+1} \log (-a x+i)}{a \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5076

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

Rule 5073

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{e^{-i \tan ^{-1}(a x)}}{\sqrt{c+a^2 c x^2}} \, dx &=\frac{\sqrt{1+a^2 x^2} \int \frac{e^{-i \tan ^{-1}(a x)}}{\sqrt{1+a^2 x^2}} \, dx}{\sqrt{c+a^2 c x^2}}\\ &=\frac{\sqrt{1+a^2 x^2} \int \frac{1}{1+i a x} \, dx}{\sqrt{c+a^2 c x^2}}\\ &=-\frac{i \sqrt{1+a^2 x^2} \log (i-a x)}{a \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0157528, size = 43, normalized size = 1. \[ -\frac{i \sqrt{a^2 x^2+1} \log (-a x+i)}{a \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.109, size = 42, normalized size = 1. \begin{align*}{\frac{-i\ln \left ( 1+iax \right ) }{ac}\sqrt{c \left ({a}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.998187, size = 20, normalized size = 0.47 \begin{align*} -\frac{i \, \log \left (i \, a x + 1\right )}{a \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*log(I*a*x + 1)/(a*sqrt(c))

________________________________________________________________________________________

Fricas [B]  time = 2.25501, size = 578, normalized size = 13.44 \begin{align*} \frac{1}{2} i \, \sqrt{\frac{1}{a^{2} c}} \log \left (\frac{{\left (-i \, a^{6} x^{2} - 2 \, a^{5} x + 2 i \, a^{4}\right )} \sqrt{a^{2} c x^{2} + c} \sqrt{a^{2} x^{2} + 1} +{\left (i \, a^{9} c x^{4} + 2 \, a^{8} c x^{3} + i \, a^{7} c x^{2} + 2 \, a^{6} c x\right )} \sqrt{\frac{1}{a^{2} c}}}{8 \, a^{3} x^{3} - 8 i \, a^{2} x^{2} + 8 \, a x - 8 i}\right ) - \frac{1}{2} i \, \sqrt{\frac{1}{a^{2} c}} \log \left (\frac{{\left (-i \, a^{6} x^{2} - 2 \, a^{5} x + 2 i \, a^{4}\right )} \sqrt{a^{2} c x^{2} + c} \sqrt{a^{2} x^{2} + 1} +{\left (-i \, a^{9} c x^{4} - 2 \, a^{8} c x^{3} - i \, a^{7} c x^{2} - 2 \, a^{6} c x\right )} \sqrt{\frac{1}{a^{2} c}}}{8 \, a^{3} x^{3} - 8 i \, a^{2} x^{2} + 8 \, a x - 8 i}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*sqrt(1/(a^2*c))*log(((-I*a^6*x^2 - 2*a^5*x + 2*I*a^4)*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1) + (I*a^9*c*x
^4 + 2*a^8*c*x^3 + I*a^7*c*x^2 + 2*a^6*c*x)*sqrt(1/(a^2*c)))/(8*a^3*x^3 - 8*I*a^2*x^2 + 8*a*x - 8*I)) - 1/2*I*
sqrt(1/(a^2*c))*log(((-I*a^6*x^2 - 2*a^5*x + 2*I*a^4)*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1) + (-I*a^9*c*x^4 -
2*a^8*c*x^3 - I*a^7*c*x^2 - 2*a^6*c*x)*sqrt(1/(a^2*c)))/(8*a^3*x^3 - 8*I*a^2*x^2 + 8*a*x - 8*I))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a^{2} x^{2} + 1}}{\sqrt{c \left (a^{2} x^{2} + 1\right )} \left (i a x + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a^{2} x^{2} + 1}}{\sqrt{a^{2} c x^{2} + c}{\left (i \, a x + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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