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

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

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Rubi [A]  time = 0.568457, 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[E^((3*I)*ArcTan[a*x])*x^2,x]

[Out]

(I*(1 + I*a*x)^3)/(a^3*Sqrt[1 + a^2*x^2]) + ((28*I - 3*a*x)*Sqrt[1 + a^2*x^2])/(6*a^3) + ((I/3)*(3 + I*a*x)^2*
Sqrt[1 + a^2*x^2])/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\\ &=-\left ((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\right )\\ &=-\left ((i a) \int \frac{\left (\frac{i}{a}-x\right ) x^2 \sqrt{1+a^2 x^2}}{(1-i a x)^2} \, dx\right )\\ &=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{(28 i-3 a x) \sqrt{1+a^2 x^2}}{6 a^3}+\frac{i (3+i a x)^2 \sqrt{1+a^2 x^2}}{3 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{(28 i-3 a x) \sqrt{1+a^2 x^2}}{6 a^3}+\frac{i (3+i a x)^2 \sqrt{1+a^2 x^2}}{3 a^3}+\frac{11 \sinh ^{-1}(a x)}{2 a^3}\\ \end{align*}

Mathematica [A]  time = 0.0512884, 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[E^((3*I)*ArcTan[a*x])*x^2,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 [A]  time = 0.073, size = 123, normalized size = 1.2 \begin{align*}{-{\frac{i}{3}}a{x}^{4}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{{\frac{13\,i}{3}}{x}^{2}}{a}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{{\frac{26\,i}{3}}}{{a}^{3}}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-{\frac{3\,{x}^{3}}{2}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-{\frac{11\,x}{2\,{a}^{2}}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{11}{2\,{a}^{2}}\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

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,x)

[Out]

-1/3*I*a*x^4/(a^2*x^2+1)^(1/2)+13/3*I/a*x^2/(a^2*x^2+1)^(1/2)+26/3*I/a^3/(a^2*x^2+1)^(1/2)-3/2*x^3/(a^2*x^2+1)
^(1/2)-11/2*x/a^2/(a^2*x^2+1)^(1/2)+11/2/a^2*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)

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Maxima [A]  time = 1.02009, size = 144, normalized size = 1.41 \begin{align*} -\frac{i \, a x^{4}}{3 \, \sqrt{a^{2} x^{2} + 1}} - \frac{3 \, x^{3}}{2 \, \sqrt{a^{2} x^{2} + 1}} + \frac{13 i \, x^{2}}{3 \, \sqrt{a^{2} x^{2} + 1} a} - \frac{11 \, x}{2 \, \sqrt{a^{2} x^{2} + 1} a^{2}} + \frac{11 \, \operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{2 \, \sqrt{a^{2}} a^{2}} + \frac{26 i}{3 \, \sqrt{a^{2} x^{2} + 1} a^{3}} \end{align*}

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,x, algorithm="maxima")

[Out]

-1/3*I*a*x^4/sqrt(a^2*x^2 + 1) - 3/2*x^3/sqrt(a^2*x^2 + 1) + 13/3*I*x^2/(sqrt(a^2*x^2 + 1)*a) - 11/2*x/(sqrt(a
^2*x^2 + 1)*a^2) + 11/2*arcsinh(a^2*x/sqrt(a^2))/(sqrt(a^2)*a^2) + 26/3*I/(sqrt(a^2*x^2 + 1)*a^3)

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Fricas [A]  time = 1.67389, size = 201, normalized size = 1.97 \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((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x^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 (i a x + 1\right )^{3}}{\left (a^{2} x^{2} + 1\right )^{\frac{3}{2}}}\, dx \end{align*}

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,x)

[Out]

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

[Out]

undef