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

Optimal. Leaf size=42 \[ \frac {i \sqrt {a^2 x^2+1} \log (a x+i)}{a \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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[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 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 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])

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*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 1.00 \[ \frac {i \sqrt {a^2 x^2+1} \log (a x+i)}{a \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[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])

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fricas [B]  time = 0.48, size = 255, normalized size = 6.07 \[ \frac {1}{2} i \, \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (i \, a^{6} x^{2} - 2 \, a^{5} x - 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (i \, a^{9} c x^{4} - 2 \, a^{8} c x^{3} + i \, a^{7} c x^{2} - 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, a^{3} x^{3} + 8 i \, a^{2} x^{2} + 8 \, a x + 8 i}\right ) - \frac {1}{2} i \, \sqrt {\frac {1}{a^{2} c}} \log \left (\frac {{\left (i \, a^{6} x^{2} - 2 \, a^{5} x - 2 i \, a^{4}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} + {\left (-i \, a^{9} c x^{4} + 2 \, a^{8} c x^{3} - i \, a^{7} c x^{2} + 2 \, a^{6} c x\right )} \sqrt {\frac {1}{a^{2} c}}}{8 \, a^{3} x^{3} + 8 i \, a^{2} x^{2} + 8 \, a x + 8 i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*s
qrt(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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {i \, a x + 1}{\sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 53, normalized size = 1.26 \[ \frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (i \ln \left (a^{2} x^{2}+1\right )+2 \arctan \left (a x \right )\right )}{2 \sqrt {a^{2} x^{2}+1}\, c a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {c\,a^2\,x^2+c}\,\sqrt {a^2\,x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i \left (\int \left (- \frac {i}{\sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx + \int \frac {a x}{\sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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