\(\int \frac {e^{i \arctan (a x)} x^2}{(c+a^2 c x^2)^{3/2}} \, dx\) [382]
Optimal result
Integrand size = 28, antiderivative size = 142 \[
\int \frac {e^{i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {1+a^2 x^2}}{2 a^3 c (i+a x) \sqrt {c+a^2 c x^2}}+\frac {i \sqrt {1+a^2 x^2} \log (i-a x)}{4 a^3 c \sqrt {c+a^2 c x^2}}+\frac {3 i \sqrt {1+a^2 x^2} \log (i+a x)}{4 a^3 c \sqrt {c+a^2 c x^2}}
\]
[Out]
-1/2*(a^2*x^2+1)^(1/2)/a^3/c/(I+a*x)/(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)+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)
Rubi [A] (verified)
Time = 0.17 (sec) , antiderivative size = 142, 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}
\[
\int \frac {e^{i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {a^2 x^2+1}}{2 a^3 c (a x+i) \sqrt {a^2 c x^2+c}}+\frac {i \sqrt {a^2 x^2+1} \log (-a x+i)}{4 a^3 c \sqrt {a^2 c x^2+c}}+\frac {3 i \sqrt {a^2 x^2+1} \log (a x+i)}{4 a^3 c \sqrt {a^2 c x^2+c}}
\]
[In]
Int[(E^(I*ArcTan[a*x])*x^2)/(c + a^2*c*x^2)^(3/2),x]
[Out]
-1/2*Sqrt[1 + a^2*x^2]/(a^3*c*(I + a*x)*Sqrt[c + a^2*c*x^2]) + ((I/4)*Sqrt[1 + a^2*x^2]*Log[I - a*x])/(a^3*c*S
qrt[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])
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)^2 (1+i a x)} \, dx}{c \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2} \int \left (\frac {i}{4 a^2 (-i+a x)}+\frac {1}{2 a^2 (i+a x)^2}+\frac {3 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 {i \sqrt {1+a^2 x^2} \log (i-a x)}{4 a^3 c \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}} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.52
\[
\int \frac {e^{i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1+a^2 x^2} \left (-\frac {2}{i+a x}+i \log (i-a x)+3 i \log (i+a x)\right )}{4 a^3 c \sqrt {c+a^2 c x^2}}
\]
[In]
Integrate[(E^(I*ArcTan[a*x])*x^2)/(c + a^2*c*x^2)^(3/2),x]
[Out]
(Sqrt[1 + a^2*x^2]*(-2/(I + a*x) + I*Log[I - a*x] + (3*I)*Log[I + a*x]))/(4*a^3*c*Sqrt[c + a^2*c*x^2])
Maple [A] (verified)
Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.61
| | |
| method | result | size |
| | |
| default |
\(\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (i \ln \left (-a x +i\right ) a x +3 i \ln \left (a x +i\right ) a x -\ln \left (-a x +i\right )-3 \ln \left (a x +i\right )-2\right )}{4 \sqrt {a^{2} x^{2}+1}\, c^{2} a^{3} \left (a x +i\right )}\) |
\(87\) |
| risch |
\(-\frac {\sqrt {a^{2} x^{2}+1}}{2 c \sqrt {c \left (a^{2} x^{2}+1\right )}\, a^{3} \left (a x +i\right )}+\frac {3 i \sqrt {a^{2} x^{2}+1}\, \ln \left (i a x -1\right )}{4 c \sqrt {c \left (a^{2} x^{2}+1\right )}\, a^{3}}+\frac {i \sqrt {a^{2} x^{2}+1}\, \ln \left (-i a x -1\right )}{4 c \sqrt {c \left (a^{2} x^{2}+1\right )}\, a^{3}}\) |
\(124\) |
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[In]
int((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^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+3*I*ln(I+a*x)*a*x-ln(I-a*x)-3*ln(I+a*x)-2)/c^2/a^
3/(I+a*x)
Fricas [F]
\[
\int \frac {e^{i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (i \, a x + 1\right )} x^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {a^{2} x^{2} + 1}} \,d x }
\]
[In]
integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^2/(a^2*c*x^2+c)^(3/2),x, algorithm="fricas")
[Out]
-1/8*(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)*sqr
t(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)*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*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)*sq
rt(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]
\[
\int \frac {e^{i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=i \left (\int \left (- \frac {i x^{2}}{a^{2} c x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + c \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\right )\, dx + \int \frac {a x^{3}}{a^{2} c x^{2} \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c} + c \sqrt {a^{2} x^{2} + 1} \sqrt {a^{2} c x^{2} + c}}\, dx\right )
\]
[In]
integrate((1+I*a*x)/(a**2*x**2+1)**(1/2)*x**2/(a**2*c*x**2+c)**(3/2),x)
[Out]
I*(Integral(-I*x**2/(a**2*c*x**2*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + c*sqrt(a**2*x**2 + 1)*sqrt(a**2*c
*x**2 + c)), x) + Integral(a*x**3/(a**2*c*x**2*sqrt(a**2*x**2 + 1)*sqrt(a**2*c*x**2 + c) + c*sqrt(a**2*x**2 +
1)*sqrt(a**2*c*x**2 + c)), x))
Maxima [F(-2)]
Exception generated. \[
\int \frac {e^{i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: RuntimeError}
\]
[In]
integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^2/(a^2*c*x^2+c)^(3/2),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.
Giac [F]
\[
\int \frac {e^{i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {{\left (i \, a x + 1\right )} x^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \sqrt {a^{2} x^{2} + 1}} \,d x }
\]
[In]
integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)*x^2/(a^2*c*x^2+c)^(3/2),x, algorithm="giac")
[Out]
integrate((I*a*x + 1)*x^2/((a^2*c*x^2 + c)^(3/2)*sqrt(a^2*x^2 + 1)), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {e^{i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^2\,\left (1+a\,x\,1{}\mathrm {i}\right )}{{\left (c\,a^2\,x^2+c\right )}^{3/2}\,\sqrt {a^2\,x^2+1}} \,d x
\]
[In]
int((x^2*(a*x*1i + 1))/((c + a^2*c*x^2)^(3/2)*(a^2*x^2 + 1)^(1/2)),x)
[Out]
int((x^2*(a*x*1i + 1))/((c + a^2*c*x^2)^(3/2)*(a^2*x^2 + 1)^(1/2)), x)