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

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 = 0.325647, 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[E^((3*I)*ArcTan[a*x])*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\\ &=-\left ((i a) \int \frac{\left (\frac{i x}{a}-x^2\right ) \sqrt{1+a^2 x^2}}{(1-i a x)^2} \, dx\right )\\ &=-\left ((i a) \int \frac{\left (\frac{i}{a}-x\right ) x \sqrt{1+a^2 x^2}}{(1-i a x)^2} \, dx\right )\\ &=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 = 0.0594771, size = 54, normalized size = 0.59 \[ -\frac{i \left (-9 \sinh ^{-1}(a x)+\frac{\sqrt{a^2 x^2+1} \left (a^2 x^2-5 i a x+14\right )}{a x+i}\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*((Sqrt[1 + a^2*x^2]*(14 - (5*I)*a*x + a^2*x^2))/(I + a*x) - 9*ArcSinh[a*x]))/a^2

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Maple [A]  time = 0.071, size = 104, normalized size = 1.1 \begin{align*}{-{\frac{i}{2}}a{x}^{3}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-{\frac{{\frac{9\,i}{2}}x}{a}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{{\frac{9\,i}{2}}}{a}\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-3\,{\frac{{x}^{2}}{\sqrt{{a}^{2}{x}^{2}+1}}}-7\,{\frac{1}{{a}^{2}\sqrt{{a}^{2}{x}^{2}+1}}} \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,x)

[Out]

-1/2*I*a*x^3/(a^2*x^2+1)^(1/2)-9/2*I/a*x/(a^2*x^2+1)^(1/2)+9/2*I/a*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^
2)^(1/2)-3*x^2/(a^2*x^2+1)^(1/2)-7/a^2/(a^2*x^2+1)^(1/2)

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Maxima [A]  time = 1.03676, size = 119, normalized size = 1.29 \begin{align*} -\frac{i \, a x^{3}}{2 \, \sqrt{a^{2} x^{2} + 1}} - \frac{3 \, x^{2}}{\sqrt{a^{2} x^{2} + 1}} - \frac{9 i \, x}{2 \, \sqrt{a^{2} x^{2} + 1} a} + \frac{9 i \, \operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{2 \, \sqrt{a^{2}} a} - \frac{7}{\sqrt{a^{2} x^{2} + 1} a^{2}} \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,x, algorithm="maxima")

[Out]

-1/2*I*a*x^3/sqrt(a^2*x^2 + 1) - 3*x^2/sqrt(a^2*x^2 + 1) - 9/2*I*x/(sqrt(a^2*x^2 + 1)*a) + 9/2*I*arcsinh(a^2*x
/sqrt(a^2))/(sqrt(a^2)*a) - 7/(sqrt(a^2*x^2 + 1)*a^2)

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Fricas [A]  time = 1.71467, size = 176, normalized size = 1.91 \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((1+I*a*x)^3/(a^2*x^2+1)^(3/2)*x,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 (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,x)

[Out]

Integral(x*(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,x, algorithm="giac")

[Out]

undef