3.4.33 \(\int \frac {e^{-i \text {ArcTan}(a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\) [333]

Optimal. Leaf size=89 \[ -\frac {\sqrt {1+a^2 x^2}}{2 a c (i-a x) \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \text {ArcTan}(a x)}{2 a c \sqrt {c+a^2 c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {5184, 5181, 46, 209} \begin {gather*} \frac {\sqrt {a^2 x^2+1} \text {ArcTan}(a x)}{2 a c \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 x^2+1}}{2 a c (-a x+i) \sqrt {a^2 c x^2+c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(I*ArcTan[a*x])*(c + a^2*c*x^2)^(3/2)),x]

[Out]

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

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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*} \int \frac {e^{-i \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx &=\frac {\sqrt {1+a^2 x^2} \int \frac {e^{-i \tan ^{-1}(a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx}{c \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \int \frac {1}{(1-i a x) (1+i a x)^2} \, dx}{c \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \int \left (-\frac {1}{2 (-i+a x)^2}+\frac {1}{2 \left (1+a^2 x^2\right )}\right ) \, dx}{c \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {1+a^2 x^2}}{2 a c (i-a x) \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \int \frac {1}{1+a^2 x^2} \, dx}{2 c \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {1+a^2 x^2}}{2 a c (i-a x) \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \tan ^{-1}(a x)}{2 a c \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 60, normalized size = 0.67 \begin {gather*} \frac {\sqrt {1+a^2 x^2} \left (-\frac {1}{2 a (i-a x)}+\frac {\text {ArcTan}(a x)}{2 a}\right )}{c \sqrt {c+a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(I*ArcTan[a*x])*(c + a^2*c*x^2)^(3/2)),x]

[Out]

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

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Maple [A]
time = 0.08, size = 86, normalized size = 0.97

method	result	size
default	\(\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (i \ln \left (-a x +i\right ) a x -i \ln \left (a x +i\right ) a x +\ln \left (-a x +i\right )-\ln \left (a x +i\right )-2\right )}{4 \sqrt {a^{2} x^{2}+1}\, c^{2} \left (-a x +i\right ) a}\)	\(86\)
risch	\(\frac {\sqrt {a^{2} x^{2}+1}}{2 c \sqrt {c \left (a^{2} x^{2}+1\right )}\, a \left (a x -i\right )}+\frac {i \sqrt {a^{2} x^{2}+1}\, \ln \left (i a x -1\right )}{4 c \sqrt {c \left (a^{2} x^{2}+1\right )}\, a}-\frac {i \sqrt {a^{2} x^{2}+1}\, \ln \left (-i a x -1\right )}{4 c \sqrt {c \left (a^{2} x^{2}+1\right )}\, a}\)	\(124\)

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)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/4/(a^2*x^2+1)^(1/2)*(c*(a^2*x^2+1))^(1/2)*(I*ln(I-a*x)*a*x-I*ln(I+a*x)*a*x+ln(I-a*x)-ln(I+a*x)-2)/c^2/(I-a*x
)/a

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Maxima [A]
time = 0.27, size = 52, normalized size = 0.58 \begin {gather*} \frac {\sqrt {c}}{2 \, {\left (a^{2} c^{2} x - i \, a c^{2}\right )}} - \frac {i \, \log \left (a x - i\right )}{4 \, a c^{\frac {3}{2}}} + \frac {i \, \log \left (i \, a x - 1\right )}{4 \, a c^{\frac {3}{2}}} \end {gather*}

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)^(3/2),x, algorithm="maxima")

[Out]

1/2*sqrt(c)/(a^2*c^2*x - I*a*c^2) - 1/4*I*log(a*x - I)/(a*c^(3/2)) + 1/4*I*log(I*a*x - 1)/(a*c^(3/2))

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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. 317 vs. \(2 (74) = 148\).
time = 3.31, size = 317, normalized size = 3.56 \begin {gather*} \frac {{\left (i \, a^{3} c^{2} x^{3} + a^{2} c^{2} x^{2} + i \, a c^{2} x + c^{2}\right )} \sqrt {\frac {1}{a^{2} c^{3}}} \log \left (\frac {2 \, {\left (2 \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a^{6} x - {\left (i \, a^{10} c^{2} x^{4} - i \, a^{6} c^{2}\right )} \sqrt {\frac {1}{a^{2} c^{3}}}\right )}}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) + {\left (-i \, a^{3} c^{2} x^{3} - a^{2} c^{2} x^{2} - i \, a c^{2} x - c^{2}\right )} \sqrt {\frac {1}{a^{2} c^{3}}} \log \left (\frac {2 \, {\left (2 \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a^{6} x - {\left (-i \, a^{10} c^{2} x^{4} + i \, a^{6} c^{2}\right )} \sqrt {\frac {1}{a^{2} c^{3}}}\right )}}{a^{4} x^{4} + 2 \, a^{2} x^{2} + 1}\right ) - 4 i \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} x}{8 \, {\left (a^{3} c^{2} x^{3} - i \, a^{2} c^{2} x^{2} + a c^{2} x - i \, c^{2}\right )}} \end {gather*}

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)^(3/2),x, algorithm="fricas")

[Out]

1/8*((I*a^3*c^2*x^3 + a^2*c^2*x^2 + I*a*c^2*x + c^2)*sqrt(1/(a^2*c^3))*log(2*(2*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x
^2 + 1)*a^6*x - (I*a^10*c^2*x^4 - I*a^6*c^2)*sqrt(1/(a^2*c^3)))/(a^4*x^4 + 2*a^2*x^2 + 1)) + (-I*a^3*c^2*x^3 -
a^2*c^2*x^2 - I*a*c^2*x - c^2)*sqrt(1/(a^2*c^3))*log(2*(2*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*a^6*x - (-I*a
^10*c^2*x^4 + I*a^6*c^2)*sqrt(1/(a^2*c^3)))/(a^4*x^4 + 2*a^2*x^2 + 1)) - 4*I*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2
+ 1)*x)/(a^3*c^2*x^3 - I*a^2*c^2*x^2 + a*c^2*x - I*c^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i \int \frac {\sqrt {a^{2} x^{2} + 1}}{a^{3} c x^{3} \sqrt {a^{2} c x^{2} + c} - i a^{2} c x^{2} \sqrt {a^{2} c x^{2} + c} + a c x \sqrt {a^{2} c x^{2} + c} - i c \sqrt {a^{2} c x^{2} + c}}\, dx \end {gather*}

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)**(3/2),x)

[Out]

-I*Integral(sqrt(a**2*x**2 + 1)/(a**3*c*x**3*sqrt(a**2*c*x**2 + c) - I*a**2*c*x**2*sqrt(a**2*c*x**2 + c) + a*c
*x*sqrt(a**2*c*x**2 + c) - I*c*sqrt(a**2*c*x**2 + c)), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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)^(3/2),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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a^2\,x^2+1}}{{\left (c\,a^2\,x^2+c\right )}^{3/2}\,\left (1+a\,x\,1{}\mathrm {i}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*x^2 + 1)^(1/2)/((c + a^2*c*x^2)^(3/2)*(a*x*1i + 1)),x)

[Out]

int((a^2*x^2 + 1)^(1/2)/((c + a^2*c*x^2)^(3/2)*(a*x*1i + 1)), x)

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