\(\int \frac {e^{-i \arctan (a x)} x^2}{(c+a^2 c x^2)^{3/2}} \, dx\) [383]

Optimal result

Integrand size = 28, antiderivative size = 143 \[ \int \frac {e^{-i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1+a^2 x^2}}{2 a^3 c (i-a x) \sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \log (i-a x)}{4 a^3 c \sqrt {c+a^2 c x^2}}-\frac {i \sqrt {1+a^2 x^2} \log (i+a x)}{4 a^3 c \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {5193, 5190, 90} \[ \int \frac {e^{-i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {a^2 x^2+1}}{2 a^3 c (-a x+i) \sqrt {a^2 c x^2+c}}-\frac {3 i \sqrt {a^2 x^2+1} \log (-a x+i)}{4 a^3 c \sqrt {a^2 c x^2+c}}-\frac {i \sqrt {a^2 x^2+1} \log (a x+i)}{4 a^3 c \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

Sqrt[1 + a^2*x^2]/(2*a^3*c*(I - a*x)*Sqrt[c + a^2*c*x^2]) - (((3*I)/4)*Sqrt[1 + a^2*x^2]*Log[I - a*x])/(a^3*c*
Sqrt[c + a^2*c*x^2]) - ((I/4)*Sqrt[1 + a^2*x^2]*Log[I + a*x])/(a^3*c*Sqrt[c + a^2*c*x^2])

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 5190

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 - I
*a*x)^(p + I*(n/2))*(1 + I*a*x)^(p - I*(n/2)), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[d, a^2*c] && (Int
egerQ[p] || GtQ[c, 0])

Rule 5193

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[c^IntPart[p]*((c + d
*x^2)^FracPart[p]/(1 + a^2*x^2)^FracPart[p]), Int[x^m*(1 + a^2*x^2)^p*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c
, d, m, 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)} x^2}{\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 {x^2}{(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 a^2 (-i+a x)^2}-\frac {3 i}{4 a^2 (-i+a x)}-\frac {i}{4 a^2 (i+a x)}\right ) \, dx}{c \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2}}{2 a^3 c (i-a x) \sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \log (i-a x)}{4 a^3 c \sqrt {c+a^2 c x^2}}-\frac {i \sqrt {1+a^2 x^2} \log (i+a x)}{4 a^3 c \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.52 \[ \int \frac {e^{-i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1+a^2 x^2} \left (\frac {2}{i-a x}-3 i \log (i-a x)-i \log (i+a x)\right )}{4 a^3 c \sqrt {c+a^2 c x^2}} \]

[In]

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

[Out]

(Sqrt[1 + a^2*x^2]*(2/(I - a*x) - (3*I)*Log[I - a*x] - I*Log[I + a*x]))/(4*a^3*c*Sqrt[c + a^2*c*x^2])

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (3 i \ln \left (-a x +i\right ) a x +i \ln \left (a x +i\right ) a x +3 \ln \left (-a x +i\right )+\ln \left (a x +i\right )+2\right )}{4 \sqrt {a^{2} x^{2}+1}\, c^{2} a^{3} \left (-a x +i\right )}\)	\(86\)
risch	\(-\frac {\sqrt {a^{2} x^{2}+1}}{2 c \sqrt {c \left (a^{2} x^{2}+1\right )}\, a^{3} \left (a x -i\right )}-\frac {i \sqrt {a^{2} x^{2}+1}\, \ln \left (i a x -1\right )}{4 c \sqrt {c \left (a^{2} x^{2}+1\right )}\, a^{3}}-\frac {3 i \sqrt {a^{2} x^{2}+1}\, \ln \left (-i a x -1\right )}{4 c \sqrt {c \left (a^{2} x^{2}+1\right )}\, a^{3}}\)	\(124\)

[In]

int(x^2/(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)*(3*I*ln(I-a*x)*a*x+I*ln(I+a*x)*a*x+3*ln(I-a*x)+ln(I+a*x)+2)/c^2/a^
3/(I-a*x)

Fricas [F]

\[ \int \frac {e^{-i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (i \, a x + 1\right )}} \,d x } \]

[In]

integrate(x^2/(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^5*c^2*x^3 + a^4*c^2*x^2 + I*a^3*c^2*x + a^2*c^2)*sqrt(1/(a^6*c^3))*log((I*sqrt(a^2*c*x^2 + c)*sqrt(a
^2*x^2 + 1)*a^3*c*x*sqrt(1/(a^6*c^3)) + I*a^2*x^3 + I*x)/(a^3*x^3 + I*a^2*x^2 + a*x + I)) + (-I*a^5*c^2*x^3 -
a^4*c^2*x^2 - I*a^3*c^2*x - a^2*c^2)*sqrt(1/(a^6*c^3))*log((-I*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*a^3*c*x*s
qrt(1/(a^6*c^3)) + I*a^2*x^3 + I*x)/(a^3*x^3 + I*a^2*x^2 + a*x + I)) - 3*(I*a^5*c^2*x^3 + a^4*c^2*x^2 + I*a^3*
c^2*x + a^2*c^2)*sqrt(1/(a^6*c^3))*log((I*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*a^3*c*x*sqrt(1/(a^6*c^3)) - I*
a^2*x^3 - I*x)/(a^3*x^3 - I*a^2*x^2 + a*x - I)) - 3*(-I*a^5*c^2*x^3 - a^4*c^2*x^2 - I*a^3*c^2*x - a^2*c^2)*sqr
t(1/(a^6*c^3))*log((-I*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*a^3*c*x*sqrt(1/(a^6*c^3)) - I*a^2*x^3 - I*x)/(a^3
*x^3 - I*a^2*x^2 + a*x - I)) - 4*(I*a^5*c^2*x^3 + a^4*c^2*x^2 + I*a^3*c^2*x + a^2*c^2)*sqrt(1/(a^6*c^3))*log((
sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*a^3*c*x*sqrt(1/(a^6*c^3)) + a^2*x^3 + x)/(a^2*x^2 + 1)) - 4*(-I*a^5*c^2*
x^3 - a^4*c^2*x^2 - I*a^3*c^2*x - a^2*c^2)*sqrt(1/(a^6*c^3))*log(-(sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*a^3*c
*x*sqrt(1/(a^6*c^3)) - a^2*x^3 - x)/(a^2*x^2 + 1)) + 4*I*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*x + 8*(a^5*c^2*
x^3 - I*a^4*c^2*x^2 + a^3*c^2*x - I*a^2*c^2)*integral(1/2*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*(-2*I*a*x + 1)
/(a^6*c^2*x^4 + 2*a^4*c^2*x^2 + a^2*c^2), x))/(a^5*c^2*x^3 - I*a^4*c^2*x^2 + a^3*c^2*x - I*a^2*c^2)

Sympy [F]

\[ \int \frac {e^{-i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=- i \int \frac {x^{2} \sqrt {a^{2} x^{2} + 1}}{a^{3} c x^{3} \sqrt {a^{2} c x^{2} + c} - i a^{2} c x^{2} \sqrt {a^{2} c x^{2} + c} + a c x \sqrt {a^{2} c x^{2} + c} - i c \sqrt {a^{2} c x^{2} + c}}\, dx \]

[In]

integrate(x**2/(1+I*a*x)*(a**2*x**2+1)**(1/2)/(a**2*c*x**2+c)**(3/2),x)

[Out]

-I*Integral(x**2*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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.38 \[ \int \frac {e^{-i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {c}}{2 \, {\left (a^{4} c^{2} x - i \, a^{3} c^{2}\right )}} - \frac {3 i \, \log \left (a x - i\right )}{4 \, a^{3} c^{\frac {3}{2}}} - \frac {i \, \log \left (i \, a x - 1\right )}{4 \, a^{3} c^{\frac {3}{2}}} \]

[In]

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

Giac [F]

\[ \int \frac {e^{-i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (i \, a x + 1\right )}} \,d x } \]

[In]

integrate(x^2/(1+I*a*x)*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(a^2*x^2 + 1)*x^2/((a^2*c*x^2 + c)^(3/2)*(I*a*x + 1)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^2\,\sqrt {a^2\,x^2+1}}{{\left (c\,a^2\,x^2+c\right )}^{3/2}\,\left (1+a\,x\,1{}\mathrm {i}\right )} \,d x \]

[In]

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

[Out]

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