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

Optimal. Leaf size=92 \[ -\frac{\left (a^2 x^2+1\right )^{5/2}}{a^2 (1+i a x)^3}-\frac{3 \left (a^2 x^2+1\right )^{3/2}}{2 a^2 (1+i a x)}-\frac{9 \sqrt{a^2 x^2+1}}{2 a^2}-\frac{9 i \sinh ^{-1}(a x)}{2 a^2} \]

[Out]

(-9*Sqrt[1 + a^2*x^2])/(2*a^2) - (3*(1 + a^2*x^2)^(3/2))/(2*a^2*(1 + I*a*x)) - (1 + a^2*x^2)^(5/2)/(a^2*(1 + I
*a*x)^3) - (((9*I)/2)*ArcSinh[a*x])/a^2

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Rubi [A]  time = 0.323644, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {5060, 1633, 1593, 12, 793, 665, 215} \[ -\frac{\left (a^2 x^2+1\right )^{5/2}}{a^2 (1+i a x)^3}-\frac{3 \left (a^2 x^2+1\right )^{3/2}}{2 a^2 (1+i a x)}-\frac{9 \sqrt{a^2 x^2+1}}{2 a^2}-\frac{9 i \sinh ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*Sqrt[1 + a^2*x^2])/(2*a^2) - (3*(1 + a^2*x^2)^(3/2))/(2*a^2*(1 + I*a*x)) - (1 + a^2*x^2)^(5/2)/(a^2*(1 + I
*a*x)^3) - (((9*I)/2)*ArcSinh[a*x])/a^2

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 793

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

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] - Dist[(2*c*d*p)/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

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 \, dx &=\int \frac{x (1-i a x)^2}{(1+i a x) \sqrt{1+a^2 x^2}} \, dx\\ &=(i a) \int \frac{\left (-\frac{i x}{a}-x^2\right ) \sqrt{1+a^2 x^2}}{(1+i a x)^2} \, dx\\ &=(i a) \int \frac{\left (-\frac{i}{a}-x\right ) x \sqrt{1+a^2 x^2}}{(1+i a x)^2} \, dx\\ &=a^2 \int \frac{x \left (1+a^2 x^2\right )^{3/2}}{a^2 (1+i a x)^3} \, dx\\ &=\int \frac{x \left (1+a^2 x^2\right )^{3/2}}{(1+i a x)^3} \, dx\\ &=-\frac{\left (1+a^2 x^2\right )^{5/2}}{a^2 (1+i a x)^3}-\frac{(3 i) \int \frac{\left (1+a^2 x^2\right )^{3/2}}{(1+i a x)^2} \, dx}{a}\\ &=-\frac{3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1+i a x)}-\frac{\left (1+a^2 x^2\right )^{5/2}}{a^2 (1+i a x)^3}-\frac{(9 i) \int \frac{\sqrt{1+a^2 x^2}}{1+i a x} \, dx}{2 a}\\ &=-\frac{9 \sqrt{1+a^2 x^2}}{2 a^2}-\frac{3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1+i a x)}-\frac{\left (1+a^2 x^2\right )^{5/2}}{a^2 (1+i a x)^3}-\frac{(9 i) \int \frac{1}{\sqrt{1+a^2 x^2}} \, dx}{2 a}\\ &=-\frac{9 \sqrt{1+a^2 x^2}}{2 a^2}-\frac{3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1+i a x)}-\frac{\left (1+a^2 x^2\right )^{5/2}}{a^2 (1+i a x)^3}-\frac{9 i \sinh ^{-1}(a x)}{2 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0396151, size = 60, normalized size = 0.65 \[ \sqrt{a^2 x^2+1} \left (\frac{4 i}{a^2 (a x-i)}-\frac{3}{a^2}+\frac{i x}{2 a}\right )-\frac{9 i \sinh ^{-1}(a x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 + a^2*x^2]*(-3/a^2 + ((I/2)*x)/a + (4*I)/(a^2*(-I + a*x))) - (((9*I)/2)*ArcSinh[a*x])/a^2

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Maple [B]  time = 0.078, size = 226, normalized size = 2.5 \begin{align*} 3\,{\frac{1}{{a}^{4}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{5/2} \left ( x-{\frac{i}{a}} \right ) ^{-2}}-3\,{\frac{1}{{a}^{2}} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{3/2}}-{\frac{{\frac{9\,i}{2}}x}{a}\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }}-{\frac{{\frac{9\,i}{2}}}{a}\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{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 ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/a^4/(x-I/a)^2*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(5/2)-3/a^2*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(3/2)-9/2*I/a*(a^2*(x-
I/a)^2+2*I*a*(x-I/a))^(1/2)*x-9/2*I/a*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)-I/a^5/(x-I/a)^3*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(5/2)

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Maxima [A]  time = 1.55776, size = 151, normalized size = 1.64 \begin{align*} -\frac{{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{4} x^{2} - 2 i \, a^{3} x - a^{2}} - \frac{{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{2 i \, a^{3} x + 2 \, a^{2}} - \frac{6 \, \sqrt{a^{2} x^{2} + 1}}{i \, a^{3} x + a^{2}} - \frac{9 i \, \operatorname{arsinh}\left (a x\right )}{2 \, a^{2}} - \frac{3 \, \sqrt{a^{2} x^{2} + 1}}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*x^2 + 1)^(3/2)/(a^4*x^2 - 2*I*a^3*x - a^2) - (a^2*x^2 + 1)^(3/2)/(2*I*a^3*x + 2*a^2) - 6*sqrt(a^2*x^2 +
1)/(I*a^3*x + a^2) - 9/2*I*arcsinh(a*x)/a^2 - 3/2*sqrt(a^2*x^2 + 1)/a^2

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Fricas [A]  time = 1.65666, size = 174, normalized size = 1.89 \begin{align*} \frac{8 i \, a x - 9 \,{\left (-i \, a x - 1\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1}\right ) + \sqrt{a^{2} x^{2} + 1}{\left (i \, a^{2} x^{2} - 5 \, a x + 14 i\right )} + 8}{2 \,{\left (a^{3} x - i \, a^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(8*I*a*x - 9*(-I*a*x - 1)*log(-a*x + sqrt(a^2*x^2 + 1)) + sqrt(a^2*x^2 + 1)*(I*a^2*x^2 - 5*a*x + 14*I) + 8
)/(a^3*x - I*a^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \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/(1+I*a*x)**3*(a**2*x**2+1)**(3/2),x)

[Out]

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

[Out]

undef