3.333 \(\int \frac {e^{-i \tan ^{-1}(a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\)

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

[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 = 0.08, 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 = {5076, 5073, 44, 203} \[ \frac {\sqrt {a^2 x^2+1} \tan ^{-1}(a x)}{2 a c \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 x^2+1}}{2 a c (-a x+i) \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 + a^2*x^2]/(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 44

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] && L
tQ[m + n + 2, 0])

Rule 203

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

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)}}{\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 = 0.04, size = 60, normalized size = 0.67 \[ \frac {\sqrt {a^2 x^2+1} \left (\frac {\tan ^{-1}(a x)}{2 a}-\frac {1}{2 a (-a x+i)}\right )}{c \sqrt {a^2 c x^2+c}} \]

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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fricas [B]  time = 0.48, size = 317, normalized size = 3.56 \[ \frac {{\left (-i \, a^{3} c^{2} x^{3} - a^{2} c^{2} x^{2} - i \, a c^{2} x - c^{2}\right )} \sqrt {\frac {1}{a^{2} c^{3}}} \log \left (\frac {8 \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a^{6} x + {\left (4 i \, a^{10} c^{2} x^{4} - 4 i \, a^{6} c^{2}\right )} \sqrt {\frac {1}{a^{2} c^{3}}}}{2 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )}}\right ) + {\left (i \, a^{3} c^{2} x^{3} + a^{2} c^{2} x^{2} + i \, a c^{2} x + c^{2}\right )} \sqrt {\frac {1}{a^{2} c^{3}}} \log \left (\frac {8 \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a^{6} x + {\left (-4 i \, a^{10} c^{2} x^{4} + 4 i \, a^{6} c^{2}\right )} \sqrt {\frac {1}{a^{2} c^{3}}}}{2 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )}}\right ) - 4 i \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} x}{2 \, {\left (4 \, a^{3} c^{2} x^{3} - 4 i \, a^{2} c^{2} x^{2} + 4 \, a c^{2} x - 4 i \, c^{2}\right )}} \]

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]

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

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

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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.20, size = 86, normalized size = 0.97 \[ -\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (i \ln \left (a x +i\right ) x a -i \ln \left (-a x +i\right ) x a +\ln \left (a x +i\right )-\ln \left (-a x +i\right )+2\right )}{4 \sqrt {a^{2} x^{2}+1}\, c^{2} a \left (-a x +i\right )} \]

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)

[Out]

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

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maxima [A]  time = 0.33, size = 52, normalized size = 0.58 \[ \frac {\sqrt {c}}{2 \, a^{2} c^{2} x - 2 i \, a c^{2}} - \frac {i \, \log \left (a x - i\right )}{4 \, a c^{\frac {3}{2}}} + \frac {i \, \log \left (i \, a x - 1\right )}{4 \, a c^{\frac {3}{2}}} \]

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]

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

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

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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