\(\int \frac {e^{-i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx\) [315]

Optimal result

Integrand size = 25, antiderivative size = 43 \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {i \sqrt {1+a^2 x^2} \log (i-a x)}{a \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5184, 5181, 31} \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {i \sqrt {a^2 x^2+1} \log (-a x+i)}{a \sqrt {a^2 c x^2+c}} \]

[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 31

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

Rule 5181

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 5184

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])

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+a^2 x^2} \int \frac {e^{-i \arctan (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] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {i \sqrt {1+a^2 x^2} \log (i-a x)}{a \sqrt {c+a^2 c x^2}} \]

[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] (verified)

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {i \sqrt {a^{2} x^{2}+1}\, \ln \left (-a x +i\right )}{\sqrt {c \left (a^{2} x^{2}+1\right )}\, a}\)	\(39\)
default	\(-\frac {i \sqrt {c \left (a^{2} x^{2}+1\right )}\, \ln \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}\, c a}\)	\(42\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (35) = 70\).

Time = 0.28 (sec) , antiderivative size = 253, normalized size of antiderivative = 5.88 \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\frac {1}{2} i \, \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (-i \, a^{6} x^{2} - 2 \, a^{5} x + 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (i \, a^{9} c x^{4} + 2 \, a^{8} c x^{3} + i \, a^{7} c x^{2} + 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, {\left (a^{3} x^{3} - i \, a^{2} x^{2} + a x - i\right )}}\right ) - \frac {1}{2} i \, \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (-i \, a^{6} x^{2} - 2 \, a^{5} x + 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (-i \, a^{9} c x^{4} - 2 \, a^{8} c x^{3} - i \, a^{7} c x^{2} - 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, {\left (a^{3} x^{3} - i \, a^{2} x^{2} + a x - i\right )}}\right ) \]

[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(1/8*((-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)))/(a^3*x^3 - I*a^2*x^2 + a*x - I)) - 1/2*I*sqrt
(1/(a^2*c))*log(1/8*((-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)))/(a^3*x^3 - I*a^2*x^2 + a*x - I))

Sympy [F]

\[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=- i \int \frac {\sqrt {a^{2} x^{2} + 1}}{a x \sqrt {a^{2} c x^{2} + c} - i \sqrt {a^{2} c x^{2} + c}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.35 \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=-\frac {i \, \log \left (i \, a x + 1\right )}{a \sqrt {c}} \]

[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))

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\text {Exception raised: TypeError} \]

[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]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {c+a^2 c x^2}} \, dx=\int \frac {\sqrt {a^2\,x^2+1}}{\sqrt {c\,a^2\,x^2+c}\,\left (1+a\,x\,1{}\mathrm {i}\right )} \,d x \]

[In]

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

[Out]

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