3.4.13 \(\int \frac {e^{2 i \text {ArcTan}(a x)}}{\sqrt {c+a^2 c x^2}} \, dx\) [313]

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 = 0.04, 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 = {5183, 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[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 5183

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]

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 = 0.03, size = 91, normalized size = 1.44 \begin {gather*} -\frac {2 i \sqrt {1+a^2 x^2} \left (\sqrt {1+i a x}+\sqrt {1-i a x} \text {ArcSin}\left (\frac {\sqrt {1-i a x}}{\sqrt {2}}\right )\right )}{a \sqrt {1-i a x} \sqrt {c+a^2 c x^2}} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (53 ) = 106\).
time = 0.08, size = 204, normalized size = 3.24

method	result	size
default	\(-\frac {\ln \left (\frac {a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}+c}\right )}{\sqrt {a^{2} c}}-\frac {\left (i \sqrt {-a^{2}}-a \right ) \sqrt {\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )^{2} a^{2} c -2 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}{a^{3} c \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}+\frac {\left (i \sqrt {-a^{2}}+a \right ) \sqrt {\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )^{2} a^{2} c +2 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}{a^{3} c \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}\)	\(204\)

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,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)-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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] 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 = 2.74, 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+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} + \sqrt {a^{2} c x^{2} + c}}\, dx - \int \left (- \frac {2 i a x}{a^{2} x^{2} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} c x^{2} + c}}\right )\, dx - \int \left (- \frac {1}{a^{2} x^{2} \sqrt {a^{2} c x^{2} + c} + \sqrt {a^{2} c x^{2} + c}}\right )\, dx \end {gather*}

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

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Giac [A]
time = 0.46, 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+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) - sqr
t(c))*a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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