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

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

[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])

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Rubi [A]  time = 0.0602542, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {5075, 653, 217, 206} \[ -\frac{\tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{a \sqrt{c}}-\frac{2 i (1+i a x)}{a \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[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 5075

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/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]) && ILtQ[(I*n)/2, 0]

Rule 653

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 217

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 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{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}}-\operatorname{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*}

Mathematica [A]  time = 0.0272302, size = 91, normalized size = 1.44 \[ -\frac{2 i \sqrt{a^2 x^2+1} \left (\sqrt{1+i a x}+\sqrt{1-i a x} \sin ^{-1}\left (\frac{\sqrt{1-i a x}}{\sqrt{2}}\right )\right )}{a \sqrt{1-i a x} \sqrt{a^2 c x^2+c}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.172, size = 204, normalized size = 3.2 \begin{align*} -{\ln \left ({{a}^{2}cx{\frac{1}{\sqrt{{a}^{2}c}}}}+\sqrt{{a}^{2}c{x}^{2}+c} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}-{\frac{1}{{a}^{3}c} \left ( i\sqrt{-{a}^{2}}-a \right ) \sqrt{ \left ( x+{\frac{1}{{a}^{2}}\sqrt{-{a}^{2}}} \right ) ^{2}{a}^{2}c-2\,c\sqrt{-{a}^{2}} \left ( x+{\frac{\sqrt{-{a}^{2}}}{{a}^{2}}} \right ) } \left ( x+{\frac{1}{{a}^{2}}\sqrt{-{a}^{2}}} \right ) ^{-1}}+{\frac{1}{{a}^{3}c} \left ( i\sqrt{-{a}^{2}}+a \right ) \sqrt{ \left ( x-{\frac{1}{{a}^{2}}\sqrt{-{a}^{2}}} \right ) ^{2}{a}^{2}c+2\,c\sqrt{-{a}^{2}} \left ( x-{\frac{\sqrt{-{a}^{2}}}{{a}^{2}}} \right ) } \left ( x-{\frac{1}{{a}^{2}}\sqrt{-{a}^{2}}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(x*a^2*c/(a^2*c)^(1/2)+(a^2*c*x^2+c)^(1/2))/(a^2*c)^(1/2)-1/a^3*(I*(-a^2)^(1/2)-a)/c/(x+(-a^2)^(1/2)/a^2)*(
(x+(-a^2)^(1/2)/a^2)^2*a^2*c-2*c*(-a^2)^(1/2)*(x+(-a^2)^(1/2)/a^2))^(1/2)+1/a^3*(I*(-a^2)^(1/2)+a)/c/(x-(-a^2)
^(1/2)/a^2)*((x-(-a^2)^(1/2)/a^2)^2*a^2*c+2*c*(-a^2)^(1/2)*(x-(-a^2)^(1/2)/a^2))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.00894, size = 332, normalized size = 5.27 \begin{align*} -\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 \, a^{2} c x + 2 i \, a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((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
))/(2*a^2*c*x + 2*I*a*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14073, size = 97, normalized size = 1.54 \begin{align*} \frac{\log \left ({\left | -\sqrt{a^{2} c} x + \sqrt{a^{2} c x^{2} + c} \right |}\right )}{a \sqrt{c}} - \frac{4}{{\left ({\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} + c}\right )} i - \sqrt{c}\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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/(((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))*i - sqr
t(c))*a)