3.53 \(\int e^{-3 i \tan ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=102 \[ -\frac{i (1-i a x)^3}{a^3 \sqrt{a^2 x^2+1}}-\frac{i (3-i a x)^2 \sqrt{a^2 x^2+1}}{3 a^3}-\frac{(3 a x+28 i) \sqrt{a^2 x^2+1}}{6 a^3}+\frac{11 \sinh ^{-1}(a x)}{2 a^3} \]

[Out]

((-I)*(1 - I*a*x)^3)/(a^3*Sqrt[1 + a^2*x^2]) - ((I/3)*(3 - I*a*x)^2*Sqrt[1 + a^2*x^2])/a^3 - ((28*I + 3*a*x)*S
qrt[1 + a^2*x^2])/(6*a^3) + (11*ArcSinh[a*x])/(2*a^3)

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Rubi [A]  time = 0.574355, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {5060, 1633, 1593, 12, 852, 1635, 1654, 780, 215} \[ -\frac{i (1-i a x)^3}{a^3 \sqrt{a^2 x^2+1}}-\frac{i (3-i a x)^2 \sqrt{a^2 x^2+1}}{3 a^3}-\frac{(3 a x+28 i) \sqrt{a^2 x^2+1}}{6 a^3}+\frac{11 \sinh ^{-1}(a x)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(1 - I*a*x)^3)/(a^3*Sqrt[1 + a^2*x^2]) - ((I/3)*(3 - I*a*x)^2*Sqrt[1 + a^2*x^2])/a^3 - ((28*I + 3*a*x)*S
qrt[1 + a^2*x^2])/(6*a^3) + (11*ArcSinh[a*x])/(2*a^3)

Rule 5060

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n
- 1)/2)*Sqrt[1 + a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(I*n - 1)/2]

Rule 1633

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d*e, Int[(d + e*x)^(m - 1)*
PolynomialQuotient[Pq, a*e + c*d*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + c*d*x, x], 0]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1654

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 215

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

Rubi steps

\begin{align*} \int e^{-3 i \tan ^{-1}(a x)} x^2 \, dx &=\int \frac{x^2 (1-i a x)^2}{(1+i a x) \sqrt{1+a^2 x^2}} \, dx\\ &=(i a) \int \frac{\sqrt{1+a^2 x^2} \left (-\frac{i x^2}{a}-x^3\right )}{(1+i a x)^2} \, dx\\ &=(i a) \int \frac{\left (-\frac{i}{a}-x\right ) x^2 \sqrt{1+a^2 x^2}}{(1+i a x)^2} \, dx\\ &=a^2 \int \frac{x^2 \left (1+a^2 x^2\right )^{3/2}}{a^2 (1+i a x)^3} \, dx\\ &=\int \frac{x^2 \left (1+a^2 x^2\right )^{3/2}}{(1+i a x)^3} \, dx\\ &=\int \frac{x^2 (1-i a x)^3}{\left (1+a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac{i (1-i a x)^3}{a^3 \sqrt{1+a^2 x^2}}-\int \frac{\left (-\frac{3}{a^2}+\frac{i x}{a}\right ) (1-i a x)^2}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{i (1-i a x)^3}{a^3 \sqrt{1+a^2 x^2}}-\frac{i (3-i a x)^2 \sqrt{1+a^2 x^2}}{3 a^3}+\frac{1}{3} \int \frac{\left (-\frac{3}{a^2}+\frac{i x}{a}\right ) (-5+3 i a x)}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{i (1-i a x)^3}{a^3 \sqrt{1+a^2 x^2}}-\frac{i (3-i a x)^2 \sqrt{1+a^2 x^2}}{3 a^3}-\frac{(28 i+3 a x) \sqrt{1+a^2 x^2}}{6 a^3}+\frac{11 \int \frac{1}{\sqrt{1+a^2 x^2}} \, dx}{2 a^2}\\ &=-\frac{i (1-i a x)^3}{a^3 \sqrt{1+a^2 x^2}}-\frac{i (3-i a x)^2 \sqrt{1+a^2 x^2}}{3 a^3}-\frac{(28 i+3 a x) \sqrt{1+a^2 x^2}}{6 a^3}+\frac{11 \sinh ^{-1}(a x)}{2 a^3}\\ \end{align*}

Mathematica [A]  time = 0.0490856, size = 63, normalized size = 0.62 \[ \frac{33 \sinh ^{-1}(a x)+\frac{\sqrt{a^2 x^2+1} \left (2 i a^3 x^3-7 a^2 x^2-19 i a x-52\right )}{a x-i}}{6 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[1 + a^2*x^2]*(-52 - (19*I)*a*x - 7*a^2*x^2 + (2*I)*a^3*x^3))/(-I + a*x) + 33*ArcSinh[a*x])/(6*a^3)

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Maple [B]  time = 0.079, size = 224, normalized size = 2.2 \begin{align*}{\frac{4\,i}{{a}^{5}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{\frac{i}{a}} \right ) ^{-2}}-{\frac{{\frac{11\,i}{3}}}{{a}^{3}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{11\,x}{2\,{a}^{2}}\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }}+{\frac{11}{2\,{a}^{2}}\ln \left ({ \left ( ia+{a}^{2} \left ( x-{\frac{i}{a}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{{a}^{6}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{\frac{i}{a}} \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*I/a^5/(x-I/a)^2*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(5/2)-11/3*I/a^3*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(3/2)+11/2/a^2*
(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(1/2)*x+11/2/a^2*ln((I*a+a^2*(x-I/a))/(a^2)^(1/2)+(a^2*(x-I/a)^2+2*I*a*(x-I/a))^
(1/2))/(a^2)^(1/2)+1/a^6/(x-I/a)^3*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(5/2)

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Maxima [B]  time = 1.5412, size = 244, normalized size = 2.39 \begin{align*} -\frac{i \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{5} x^{2} - 2 i \, a^{4} x - a^{3}} - \frac{i \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{i \, a^{4} x + a^{3}} - \frac{6 i \, \sqrt{a^{2} x^{2} + 1}}{i \, a^{4} x + a^{3}} - \frac{\sqrt{-a^{2} x^{2} + 4 i \, a x + 3} x}{2 \, a^{2}} + \frac{i \,{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{3 \, a^{3}} + \frac{\arcsin \left (i \, a x + 2\right )}{2 \, a^{3}} + \frac{6 \, \operatorname{arsinh}\left (a x\right )}{a^{3}} - \frac{3 i \, \sqrt{a^{2} x^{2} + 1}}{a^{3}} + \frac{i \, \sqrt{-a^{2} x^{2} + 4 i \, a x + 3}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*(a^2*x^2 + 1)^(3/2)/(a^5*x^2 - 2*I*a^4*x - a^3) - I*(a^2*x^2 + 1)^(3/2)/(I*a^4*x + a^3) - 6*I*sqrt(a^2*x^2
+ 1)/(I*a^4*x + a^3) - 1/2*sqrt(-a^2*x^2 + 4*I*a*x + 3)*x/a^2 + 1/3*I*(a^2*x^2 + 1)^(3/2)/a^3 + 1/2*arcsin(I*a
*x + 2)/a^3 + 6*arcsinh(a*x)/a^3 - 3*I*sqrt(a^2*x^2 + 1)/a^3 + I*sqrt(-a^2*x^2 + 4*I*a*x + 3)/a^3

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Fricas [A]  time = 1.66689, size = 200, normalized size = 1.96 \begin{align*} -\frac{24 \, a x +{\left (33 \, a x - 33 i\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1}\right ) -{\left (2 i \, a^{3} x^{3} - 7 \, a^{2} x^{2} - 19 i \, a x - 52\right )} \sqrt{a^{2} x^{2} + 1} - 24 i}{6 \,{\left (a^{4} x - i \, a^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(24*a*x + (33*a*x - 33*I)*log(-a*x + sqrt(a^2*x^2 + 1)) - (2*I*a^3*x^3 - 7*a^2*x^2 - 19*I*a*x - 52)*sqrt(
a^2*x^2 + 1) - 24*I)/(a^4*x - I*a^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \left (a^{2} x^{2} + 1\right )^{\frac{3}{2}}}{\left (i a x + 1\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef