3.1.20 \(\int e^{3 i \text {ArcTan}(a x)} x^2 \, dx\) [20]

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

[Out]

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

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Rubi [A]
time = 0.40, 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 = {5168, 1647, 1607, 12, 866, 1649, 1668, 794, 221} \begin {gather*} \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} \end {gather*}

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 12

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

Rule 221

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 794

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 866

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 1607

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 1647

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 1649

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 1668

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 5168

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\\ &=-\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*}

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Mathematica [A]
time = 0.04, size = 63, normalized size = 0.62 \begin {gather*} \frac {\frac {\sqrt {1+a^2 x^2} \left (-52+19 i a x-7 a^2 x^2-2 i a^3 x^3\right )}{i+a x}+33 \sinh ^{-1}(a x)}{6 a^3} \end {gather*}

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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (85 ) = 170\).
time = 0.12, size = 236, normalized size = 2.31

method	result	size
risch	\(-\frac {i \left (2 a^{2} x^{2}-9 i a x -28\right ) \sqrt {a^{2} x^{2}+1}}{6 a^{3}}+\frac {11 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 a^{2} \sqrt {a^{2}}}-\frac {4 \sqrt {\left (x +\frac {i}{a}\right )^{2} a^{2}-2 i a \left (x +\frac {i}{a}\right )}}{a^{4} \left (x +\frac {i}{a}\right )}\)	\(111\)
meijerg	\(\frac {-\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\sqrt {\pi }\, \left (a^{2}\right )^{\frac {3}{2}} \arcsinh \left (a x \right )}{a^{3}}}{a^{2} \sqrt {\pi }\, \sqrt {a^{2}}}+\frac {3 i \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right )}{4 \sqrt {a^{2} x^{2}+1}}\right )}{a^{3} \sqrt {\pi }}-\frac {3 \left (\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{10 a^{4} \sqrt {a^{2} x^{2}+1}}-\frac {3 \sqrt {\pi }\, \left (a^{2}\right )^{\frac {5}{2}} \arcsinh \left (a x \right )}{2 a^{5}}\right )}{a^{2} \sqrt {\pi }\, \sqrt {a^{2}}}-\frac {i \left (\frac {8 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-2 a^{4} x^{4}+8 a^{2} x^{2}+16\right )}{6 \sqrt {a^{2} x^{2}+1}}\right )}{a^{3} \sqrt {\pi }}\)	\(212\)
default	\(-i a^{3} \left (\frac {x^{4}}{3 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {4 \left (\frac {x^{2}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {a^{2} x^{2}+1}}\right )}{3 a^{2}}\right )-3 a^{2} \left (\frac {x^{3}}{2 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {3 \left (-\frac {x}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{a^{2} \sqrt {a^{2}}}\right )}{2 a^{2}}\right )+3 i a \left (\frac {x^{2}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {a^{2} x^{2}+1}}\right )-\frac {x}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{a^{2} \sqrt {a^{2}}}\)	\(236\)

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,method=_RETURNVERBOSE)

[Out]

-I*a^3*(1/3*x^4/a^2/(a^2*x^2+1)^(1/2)-4/3/a^2*(x^2/a^2/(a^2*x^2+1)^(1/2)+2/a^4/(a^2*x^2+1)^(1/2)))-3*a^2*(1/2*
x^3/a^2/(a^2*x^2+1)^(1/2)-3/2/a^2*(-x/a^2/(a^2*x^2+1)^(1/2)+1/a^2*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2
)^(1/2)))+3*I*a*(x^2/a^2/(a^2*x^2+1)^(1/2)+2/a^4/(a^2*x^2+1)^(1/2))-x/a^2/(a^2*x^2+1)^(1/2)+1/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 = 0.27, size = 95, normalized size = 0.93 \begin {gather*} -\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 (a x\right )}{2 \, a^{3}} + \frac {26 i}{3 \, \sqrt {a^{2} x^{2} + 1} a^{3}} \end {gather*}

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*x)/a^3 + 26/3*I/(sqrt(a^2*x^2 + 1)*a^3)

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Fricas [A]
time = 2.64, size = 80, normalized size = 0.78 \begin {gather*} -\frac {24 \, a x + 33 \, {\left (a x + 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 {gather*}

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 + 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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i \left (\int \frac {i x^{2}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x^{3}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{5}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{4}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \end {gather*}

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]

-I*(Integral(I*x**2/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(-3*a*x**3/(a**2*x**2*
sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(a**3*x**5/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2
*x**2 + 1)), x) + Integral(-3*I*a**2*x**4/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [B]
time = 0.06, size = 114, normalized size = 1.12 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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