3.81 \(\int e^{\frac{5}{2} i \tan ^{-1}(a x)} x \, dx\)

Optimal. Leaf size=324 \[ -\frac{2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}-\frac{5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac{25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac{25 \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt{2} a^2}+\frac{25 \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt{2} a^2}+\frac{25 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2}-\frac{25 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2} \]

[Out]

(-25*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/(4*a^2) - (5*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(5/4))/(2*a^2) - (2*(1 +
I*a*x)^(9/4))/(a^2*(1 - I*a*x)^(1/4)) + (25*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(4*Sqrt
[2]*a^2) - (25*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(4*Sqrt[2]*a^2) - (25*Log[1 + Sqrt[1
- I*a*x]/Sqrt[1 + I*a*x] - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(8*Sqrt[2]*a^2) + (25*Log[1 + Sqrt
[1 - I*a*x]/Sqrt[1 + I*a*x] + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(8*Sqrt[2]*a^2)

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Rubi [A]  time = 0.209591, antiderivative size = 324, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.786, Rules used = {5062, 78, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ -\frac{2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}-\frac{5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac{25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac{25 \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt{2} a^2}+\frac{25 \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{8 \sqrt{2} a^2}+\frac{25 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2}-\frac{25 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(((5*I)/2)*ArcTan[a*x])*x,x]

[Out]

(-25*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/(4*a^2) - (5*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(5/4))/(2*a^2) - (2*(1 +
I*a*x)^(9/4))/(a^2*(1 - I*a*x)^(1/4)) + (25*ArcTan[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(4*Sqrt
[2]*a^2) - (25*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(4*Sqrt[2]*a^2) - (25*Log[1 + Sqrt[1
- I*a*x]/Sqrt[1 + I*a*x] - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(8*Sqrt[2]*a^2) + (25*Log[1 + Sqrt
[1 - I*a*x]/Sqrt[1 + I*a*x] + (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(8*Sqrt[2]*a^2)

Rule 5062

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 - I*a*x)^((I*n)/2))/(1 + I*a*x)^((I*n)/2
), x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int e^{\frac{5}{2} i \tan ^{-1}(a x)} x \, dx &=\int \frac{x (1+i a x)^{5/4}}{(1-i a x)^{5/4}} \, dx\\ &=-\frac{2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac{(5 i) \int \frac{(1+i a x)^{5/4}}{\sqrt [4]{1-i a x}} \, dx}{a}\\ &=-\frac{5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac{2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac{(25 i) \int \frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}} \, dx}{4 a}\\ &=-\frac{25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac{5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac{2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac{(25 i) \int \frac{1}{\sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx}{8 a}\\ &=-\frac{25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac{5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac{2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}-\frac{25 \operatorname{Subst}\left (\int \frac{x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{2 a^2}\\ &=-\frac{25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac{5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac{2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}-\frac{25 \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{2 a^2}\\ &=-\frac{25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac{5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac{2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac{25 \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 a^2}-\frac{25 \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 a^2}\\ &=-\frac{25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac{5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac{2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}-\frac{25 \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 a^2}-\frac{25 \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 a^2}-\frac{25 \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^2}-\frac{25 \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^2}\\ &=-\frac{25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac{5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac{2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}-\frac{25 \log \left (1+\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^2}+\frac{25 \log \left (1+\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^2}-\frac{25 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2}+\frac{25 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2}\\ &=-\frac{25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac{5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac{2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac{25 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2}-\frac{25 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt{2} a^2}-\frac{25 \log \left (1+\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^2}+\frac{25 \log \left (1+\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{8 \sqrt{2} a^2}\\ \end{align*}

Mathematica [C]  time = 0.0343106, size = 72, normalized size = 0.22 \[ \frac{2 \left (20 i \sqrt [4]{2} (a x+i) \text{Hypergeometric2F1}\left (-\frac{5}{4},\frac{3}{4},\frac{7}{4},\frac{1}{2} (1-i a x)\right )-3 (1+i a x)^{9/4}\right )}{3 a^2 \sqrt [4]{1-i a x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(((5*I)/2)*ArcTan[a*x])*x,x]

[Out]

(2*(-3*(1 + I*a*x)^(9/4) + (20*I)*2^(1/4)*(I + a*x)*Hypergeometric2F1[-5/4, 3/4, 7/4, (1 - I*a*x)/2]))/(3*a^2*
(1 - I*a*x)^(1/4))

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Maple [F]  time = 0.151, size = 0, normalized size = 0. \begin{align*} \int \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) ^{{\frac{5}{2}}}x\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x,x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x,x, algorithm="maxima")

[Out]

integrate(x*((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(5/2), x)

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Fricas [A]  time = 1.7549, size = 637, normalized size = 1.97 \begin{align*} \frac{2 \, a^{2} \sqrt{\frac{625 i}{16 \, a^{4}}} \log \left (\frac{4}{25} \, a^{2} \sqrt{\frac{625 i}{16 \, a^{4}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, a^{2} \sqrt{\frac{625 i}{16 \, a^{4}}} \log \left (-\frac{4}{25} \, a^{2} \sqrt{\frac{625 i}{16 \, a^{4}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) + 2 \, a^{2} \sqrt{-\frac{625 i}{16 \, a^{4}}} \log \left (\frac{4}{25} \, a^{2} \sqrt{-\frac{625 i}{16 \, a^{4}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, a^{2} \sqrt{-\frac{625 i}{16 \, a^{4}}} \log \left (-\frac{4}{25} \, a^{2} \sqrt{-\frac{625 i}{16 \, a^{4}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) -{\left (2 \, a^{2} x^{2} - 9 i \, a x + 43\right )} \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}}{4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x,x, algorithm="fricas")

[Out]

1/4*(2*a^2*sqrt(625/16*I/a^4)*log(4/25*a^2*sqrt(625/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 2*a^2*s
qrt(625/16*I/a^4)*log(-4/25*a^2*sqrt(625/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + 2*a^2*sqrt(-625/16
*I/a^4)*log(4/25*a^2*sqrt(-625/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 2*a^2*sqrt(-625/16*I/a^4)*lo
g(-4/25*a^2*sqrt(-625/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - (2*a^2*x^2 - 9*I*a*x + 43)*sqrt(I*sqr
t(a^2*x^2 + 1)/(a*x + I)))/a^2

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(5/2)*x,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)*x,x, algorithm="giac")

[Out]

integrate(x*((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(5/2), x)