\(\int \frac {e^{-3 i \arctan (a x)} x^2}{(c+a^2 c x^2)^{11/2}} \, dx\) [384]

Optimal result

Integrand size = 28, antiderivative size = 65 \[ \int \frac {e^{-3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\frac {(i-3 a x) \sqrt {1+a^2 x^2}}{24 a^3 c^5 (1-i a x)^3 (1+i a x)^6 \sqrt {c+a^2 c x^2}} \]

[Out]

1/24*(I-3*a*x)*(a^2*x^2+1)^(1/2)/a^3/c^5/(1-I*a*x)^3/(1+I*a*x)^6/(a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {5193, 5190, 82} \[ \int \frac {e^{-3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\frac {(-3 a x+i) \sqrt {a^2 x^2+1}}{24 a^3 c^5 (1-i a x)^3 (1+i a x)^6 \sqrt {a^2 c x^2+c}} \]

[In]

Int[x^2/(E^((3*I)*ArcTan[a*x])*(c + a^2*c*x^2)^(11/2)),x]

[Out]

((I - 3*a*x)*Sqrt[1 + a^2*x^2])/(24*a^3*c^5*(1 - I*a*x)^3*(1 + I*a*x)^6*Sqrt[c + a^2*c*x^2])

Rule 82

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x
)^(n + 1)*(e + f*x)^(p + 1)*((2*a*d*f*(n + p + 3) - b*(d*e*(n + 2) + c*f*(p + 2)) + b*d*f*(n + p + 2)*x)/(d^2*
f^2*(n + p + 2)*(n + p + 3))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && NeQ[n + p + 3,
0] && EqQ[d*f*(n + p + 2)*(a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1)))) - b*(d*e*(n + 1)
+ c*f*(p + 1))*(a*d*f*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2))), 0]

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^{-3 i \arctan (a x)} x^2}{\left (1+a^2 x^2\right )^{11/2}} \, dx}{c^5 \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2} \int \frac {x^2}{(1-i a x)^4 (1+i a x)^7} \, dx}{c^5 \sqrt {c+a^2 c x^2}} \\ & = \frac {(i-3 a x) \sqrt {1+a^2 x^2}}{24 a^3 c^5 (1-i a x)^3 (1+i a x)^6 \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=-\frac {i (-i+3 a x) \sqrt {1+a^2 x^2}}{24 a^3 c^5 (-i+a x)^6 (i+a x)^3 \sqrt {c+a^2 c x^2}} \]

[In]

Integrate[x^2/(E^((3*I)*ArcTan[a*x])*(c + a^2*c*x^2)^(11/2)),x]

[Out]

((-1/24*I)*(-I + 3*a*x)*Sqrt[1 + a^2*x^2])/(a^3*c^5*(-I + a*x)^6*(I + a*x)^3*Sqrt[c + a^2*c*x^2])

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77

method	result	size
risch	\(\frac {-\frac {i x}{8 a^{2}}-\frac {1}{24 a^{3}}}{c^{5} \left (a^{2} x^{2}+1\right )^{\frac {5}{2}} \sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (a x -i\right )^{3}}\)	\(50\)
default	\(-\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (3 i a x +1\right )}{24 \sqrt {a^{2} x^{2}+1}\, c^{6} a^{3} \left (-a x +i\right )^{6} \left (a x +i\right )^{3}}\)	\(57\)
gosper	\(-\frac {\left (-a x +i\right ) \left (a x +i\right ) \left (-3 a x +i\right ) \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{24 a^{3} \left (i a x +1\right )^{3} \left (a^{2} c \,x^{2}+c \right )^{\frac {11}{2}}}\)	\(58\)

[In]

int(x^2/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/(a^2*c*x^2+c)^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/c^5/(a^2*x^2+1)^(5/2)/(c*(a^2*x^2+1))^(1/2)*(-1/8*I/a^2*x-1/24/a^3)/(a*x-I)^3

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (53) = 106\).

Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.95 \[ \int \frac {e^{-3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\frac {{\left (-i \, a^{6} x^{9} - 3 \, a^{5} x^{8} - 8 \, a^{3} x^{6} + 6 i \, a^{2} x^{5} - 6 \, a x^{4} + 8 i \, x^{3}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1}}{24 \, {\left (a^{11} c^{6} x^{11} - 3 i \, a^{10} c^{6} x^{10} + a^{9} c^{6} x^{9} - 11 i \, a^{8} c^{6} x^{8} - 6 \, a^{7} c^{6} x^{7} - 14 i \, a^{6} c^{6} x^{6} - 14 \, a^{5} c^{6} x^{5} - 6 i \, a^{4} c^{6} x^{4} - 11 \, a^{3} c^{6} x^{3} + i \, a^{2} c^{6} x^{2} - 3 \, a c^{6} x + i \, c^{6}\right )}} \]

[In]

integrate(x^2/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/(a^2*c*x^2+c)^(11/2),x, algorithm="fricas")

[Out]

1/24*(-I*a^6*x^9 - 3*a^5*x^8 - 8*a^3*x^6 + 6*I*a^2*x^5 - 6*a*x^4 + 8*I*x^3)*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 +
1)/(a^11*c^6*x^11 - 3*I*a^10*c^6*x^10 + a^9*c^6*x^9 - 11*I*a^8*c^6*x^8 - 6*a^7*c^6*x^7 - 14*I*a^6*c^6*x^6 - 1
4*a^5*c^6*x^5 - 6*I*a^4*c^6*x^4 - 11*a^3*c^6*x^3 + I*a^2*c^6*x^2 - 3*a*c^6*x + I*c^6)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate(x**2/(1+I*a*x)**3*(a**2*x**2+1)**(3/2)/(a**2*c*x**2+c)**(11/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.43 \[ \int \frac {e^{-3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\frac {3 \, a x - i}{24 i \, a^{12} c^{\frac {11}{2}} x^{9} + 72 \, a^{11} c^{\frac {11}{2}} x^{8} + 192 \, a^{9} c^{\frac {11}{2}} x^{6} - 144 i \, a^{8} c^{\frac {11}{2}} x^{5} + 144 \, a^{7} c^{\frac {11}{2}} x^{4} - 192 i \, a^{6} c^{\frac {11}{2}} x^{3} - 72 i \, a^{4} c^{\frac {11}{2}} x - 24 \, a^{3} c^{\frac {11}{2}}} \]

[In]

integrate(x^2/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/(a^2*c*x^2+c)^(11/2),x, algorithm="maxima")

[Out]

(3*a*x - I)/(24*I*a^12*c^(11/2)*x^9 + 72*a^11*c^(11/2)*x^8 + 192*a^9*c^(11/2)*x^6 - 144*I*a^8*c^(11/2)*x^5 + 1
44*a^7*c^(11/2)*x^4 - 192*I*a^6*c^(11/2)*x^3 - 72*I*a^4*c^(11/2)*x - 24*a^3*c^(11/2))

Giac [F]

\[ \int \frac {e^{-3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\int { \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {11}{2}} {\left (i \, a x + 1\right )}^{3}} \,d x } \]

[In]

integrate(x^2/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/(a^2*c*x^2+c)^(11/2),x, algorithm="giac")

[Out]

integrate((a^2*x^2 + 1)^(3/2)*x^2/((a^2*c*x^2 + c)^(11/2)*(I*a*x + 1)^3), x)

Mupad [B] (verification not implemented)

Time = 1.80 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-3 i \arctan (a x)} x^2}{\left (c+a^2 c x^2\right )^{11/2}} \, dx=\frac {\sqrt {c\,\left (a^2\,x^2+1\right )}\,\sqrt {a^2\,x^2+1}\,\left (1+a\,x\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{24\,a^3\,c^6\,{\left (a\,x+1{}\mathrm {i}\right )}^4\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^7} \]

[In]

int((x^2*(a^2*x^2 + 1)^(3/2))/((c + a^2*c*x^2)^(11/2)*(a*x*1i + 1)^3),x)

[Out]

((c*(a^2*x^2 + 1))^(1/2)*(a^2*x^2 + 1)^(1/2)*(a*x*3i + 1)*1i)/(24*a^3*c^6*(a*x + 1i)^4*(a*x*1i + 1)^7)