3.381 \(\int \frac {e^{3 i \tan ^{-1}(a x)} x^2}{(c+a^2 c x^2)^{11/2}} \, dx\)

Optimal. Leaf size=65 \[ -\frac {(3 a x+i) \sqrt {a^2 x^2+1}}{24 a^3 c^5 (1-i a x)^6 (1+i a x)^3 \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {5085, 5082, 81} \[ -\frac {(3 a x+i) \sqrt {a^2 x^2+1}}{24 a^3 c^5 (1-i a x)^6 (1+i a x)^3 \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[(E^((3*I)*ArcTan[a*x])*x^2)/(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)^6*(1 + I*a*x)^3*Sqrt[c + a^2*c*x^2])

Rule 81

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 5082

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 5085

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

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Mathematica [A]  time = 0.11, size = 65, normalized size = 1.00 \[ \frac {i (3 a x+i) \sqrt {a^2 x^2+1}}{24 a^3 c^5 (a x-i)^3 (a x+i)^6 \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.49, size = 193, normalized size = 2.97 \[ \frac {{\left (i \, a^{6} x^{9} - 3 \, a^{5} x^{8} - 8 \, a^{3} x^{6} - 6 i \, a^{2} x^{5} - 6 \, a x^{4} - 8 i \, x^{3}\right )} \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1}}{24 \, a^{11} c^{6} x^{11} + 72 i \, a^{10} c^{6} x^{10} + 24 \, a^{9} c^{6} x^{9} + 264 i \, a^{8} c^{6} x^{8} - 144 \, a^{7} c^{6} x^{7} + 336 i \, a^{6} c^{6} x^{6} - 336 \, a^{5} c^{6} x^{5} + 144 i \, a^{4} c^{6} x^{4} - 264 \, a^{3} c^{6} x^{3} - 24 i \, a^{2} c^{6} x^{2} - 72 \, a c^{6} x - 24 i \, c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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)/(2
4*a^11*c^6*x^11 + 72*I*a^10*c^6*x^10 + 24*a^9*c^6*x^9 + 264*I*a^8*c^6*x^8 - 144*a^7*c^6*x^7 + 336*I*a^6*c^6*x^
6 - 336*a^5*c^6*x^5 + 144*I*a^4*c^6*x^4 - 264*a^3*c^6*x^3 - 24*I*a^2*c^6*x^2 - 72*a*c^6*x - 24*I*c^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a x + 1\right )}^{3} x^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {11}{2}} {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.15, size = 58, normalized size = 0.89 \[ \frac {\left (-a x +i\right ) \left (a x +i\right ) \left (3 a x +i\right ) \left (i a x +1\right )^{3}}{24 a^{3} \left (a^{2} x^{2}+1\right )^{\frac {3}{2}} \left (a^{2} c \,x^{2}+c \right )^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(-a*x+I)*(I+a*x)*(I+3*a*x)*(1+I*a*x)^3/a^3/(a^2*x^2+1)^(3/2)/(a^2*c*x^2+c)^(11/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [B]  time = 1.60, size = 48, normalized size = 0.74 \[ \frac {\sqrt {c\,\left (a^2\,x^2+1\right )}\,{\left (a\,x-\mathrm {i}\right )}^3\,\left (3\,a\,x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{24\,a^3\,c^6\,{\left (a^2\,x^2+1\right )}^{13/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c*(a^2*x^2 + 1))^(1/2)*(a*x - 1i)^3*(3*a*x + 1i)*1i)/(24*a^3*c^6*(a^2*x^2 + 1)^(13/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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