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

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

[Out]

11/2*arcsinh(a*x)/a^3-I*(1-I*a*x)^3/a^3/(a^2*x^2+1)^(1/2)-1/3*I*(3-I*a*x)^2*(a^2*x^2+1)^(1/2)/a^3-1/6*(28*I+3*
a*x)*(a^2*x^2+1)^(1/2)/a^3

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Rubi [A]  time = 0.57, antiderivative size = 102, normalized size of antiderivative = 1.00, 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 12

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

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]

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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.41, size = 81, normalized size = 0.79 \[ -\frac {24 \, a x + {\left (33 \, a x - 33 i\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) - {\left (2 i \, a^{3} x^{3} - 7 \, a^{2} x^{2} - 19 i \, a x - 52\right )} \sqrt {a^{2} x^{2} + 1} - 24 i}{6 \, a^{4} x - 6 i \, a^{3}} \]

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]

-(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)/(6*a^4*x - 6*I*a^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

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

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maple [B]  time = 0.17, size = 224, normalized size = 2.20 \[ \frac {4 i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a^{5} \left (x -\frac {i}{a}\right )^{2}}-\frac {11 i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3 a^{3}}+\frac {11 \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\, x}{2 a^{2}}+\frac {11 \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{2 a^{2} \sqrt {a^{2}}}+\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a^{6} \left (x -\frac {i}{a}\right )^{3}} \]

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

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maxima [B]  time = 0.44, size = 181, normalized size = 1.77 \[ -\frac {i \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{5} x^{2} - 2 i \, a^{4} x - a^{3}} - \frac {i \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{i \, a^{4} x + a^{3}} - \frac {6 i \, \sqrt {a^{2} x^{2} + 1}}{i \, a^{4} x + a^{3}} - \frac {\sqrt {-a^{2} x^{2} + 4 i \, a x + 3} x}{2 \, a^{2}} + \frac {i \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{3 \, a^{3}} + \frac {\arcsin \left (i \, a x + 2\right )}{2 \, a^{3}} + \frac {6 \, \operatorname {arsinh}\left (a x\right )}{a^{3}} - \frac {3 i \, \sqrt {a^{2} x^{2} + 1}}{a^{3}} + \frac {i \, \sqrt {-a^{2} x^{2} + 4 i \, a x + 3}}{a^{3}} \]

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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mupad [B]  time = 0.07, size = 115, normalized size = 1.13 \[ \frac {11\,\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{2\,a^2\,\sqrt {a^2}}-\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {3\,x\,\sqrt {a^2}}{2\,a^2}+\frac {a\,14{}\mathrm {i}}{3\,{\left (a^2\right )}^{3/2}}-\frac {a^3\,x^2\,1{}\mathrm {i}}{3\,{\left (a^2\right )}^{3/2}}\right )}{\sqrt {a^2}}+\frac {4\,\sqrt {a^2\,x^2+1}}{a^2\,\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11*asinh(x*(a^2)^(1/2)))/(2*a^2*(a^2)^(1/2)) - ((a^2*x^2 + 1)^(1/2)*((a*14i)/(3*(a^2)^(3/2)) - (a^3*x^2*1i)/(
3*(a^2)^(3/2)) + (3*x*(a^2)^(1/2))/(2*a^2)))/(a^2)^(1/2) + (4*(a^2*x^2 + 1)^(1/2))/(a^2*(((a^2)^(1/2)*1i)/a -
x*(a^2)^(1/2))*(a^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i \left (\int \frac {x^{2} \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{3} - 3 i a^{2} x^{2} - 3 a x + i}\, dx + \int \frac {a^{2} x^{4} \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{3} - 3 i a^{2} x^{2} - 3 a x + i}\, dx\right ) \]

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]

I*(Integral(x**2*sqrt(a**2*x**2 + 1)/(a**3*x**3 - 3*I*a**2*x**2 - 3*a*x + I), x) + Integral(a**2*x**4*sqrt(a**
2*x**2 + 1)/(a**3*x**3 - 3*I*a**2*x**2 - 3*a*x + I), x))

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