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

Optimal. Leaf size=299 \[ -\frac{4 i (1+i a x)^{5/4}}{a \sqrt [4]{1-i a x}}-\frac{5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}-\frac{5 i \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt{2} a}+\frac{5 i \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt{2} a}+\frac{5 i \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt{2} a}-\frac{5 i \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt{2} a} \]

[Out]

((-5*I)*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/a - ((4*I)*(1 + I*a*x)^(5/4))/(a*(1 - I*a*x)^(1/4)) + ((5*I)*ArcT
an[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a) - ((5*I)*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1
/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a) - (((5*I)/2)*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)])/(Sqrt[2]*a) + (((5*I)/2)*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)])/(Sqrt[2]*a)

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Rubi [A]  time = 0.178278, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.917, Rules used = {5061, 47, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ -\frac{4 i (1+i a x)^{5/4}}{a \sqrt [4]{1-i a x}}-\frac{5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}-\frac{5 i \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt{2} a}+\frac{5 i \log \left (\frac{\sqrt{1-i a x}}{\sqrt{1+i a x}}+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}+1\right )}{2 \sqrt{2} a}+\frac{5 i \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt{2} a}-\frac{5 i \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt{2} a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-5*I)*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/a - ((4*I)*(1 + I*a*x)^(5/4))/(a*(1 - I*a*x)^(1/4)) + ((5*I)*ArcT
an[1 - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a) - ((5*I)*ArcTan[1 + (Sqrt[2]*(1 - I*a*x)^(1
/4))/(1 + I*a*x)^(1/4)])/(Sqrt[2]*a) - (((5*I)/2)*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)])/(Sqrt[2]*a) + (((5*I)/2)*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)])/(Sqrt[2]*a)

Rule 5061

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

Rule 47

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

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)} \, dx &=\int \frac{(1+i a x)^{5/4}}{(1-i a x)^{5/4}} \, dx\\ &=-\frac{4 i (1+i a x)^{5/4}}{a \sqrt [4]{1-i a x}}-5 \int \frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}} \, dx\\ &=-\frac{5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}-\frac{4 i (1+i a x)^{5/4}}{a \sqrt [4]{1-i a x}}-\frac{5}{2} \int \frac{1}{\sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx\\ &=-\frac{5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}-\frac{4 i (1+i a x)^{5/4}}{a \sqrt [4]{1-i a x}}-\frac{(10 i) \operatorname{Subst}\left (\int \frac{x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-i a x}\right )}{a}\\ &=-\frac{5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}-\frac{4 i (1+i a x)^{5/4}}{a \sqrt [4]{1-i a x}}-\frac{(10 i) \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 )}{a}\\ &=-\frac{5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}-\frac{4 i (1+i a x)^{5/4}}{a \sqrt [4]{1-i a x}}+\frac{(5 i) \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 )}{a}-\frac{(5 i) \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 )}{a}\\ &=-\frac{5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}-\frac{4 i (1+i a x)^{5/4}}{a \sqrt [4]{1-i a x}}-\frac{(5 i) \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 )}{2 a}-\frac{(5 i) \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 )}{2 a}-\frac{(5 i) \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 )}{2 \sqrt{2} a}-\frac{(5 i) \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 )}{2 \sqrt{2} a}\\ &=-\frac{5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}-\frac{4 i (1+i a x)^{5/4}}{a \sqrt [4]{1-i a x}}-\frac{5 i \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 )}{2 \sqrt{2} a}+\frac{5 i \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 )}{2 \sqrt{2} a}-\frac{(5 i) \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 )}{\sqrt{2} a}+\frac{(5 i) \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 )}{\sqrt{2} a}\\ &=-\frac{5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}-\frac{4 i (1+i a x)^{5/4}}{a \sqrt [4]{1-i a x}}+\frac{5 i \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt{2} a}-\frac{5 i \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt{2} a}-\frac{5 i \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 )}{2 \sqrt{2} a}+\frac{5 i \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 )}{2 \sqrt{2} a}\\ \end{align*}

Mathematica [C]  time = 0.0379242, size = 41, normalized size = 0.14 \[ -\frac{8 i e^{\frac{9}{2} i \tan ^{-1}(a x)} \text{Hypergeometric2F1}\left (2,\frac{9}{4},\frac{13}{4},-e^{2 i \tan ^{-1}(a x)}\right )}{9 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((-8*I)/9)*E^(((9*I)/2)*ArcTan[a*x])*Hypergeometric2F1[2, 9/4, 13/4, -E^((2*I)*ArcTan[a*x])])/a

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Maple [F]  time = 0.13, size = 0, normalized size = 0. \begin{align*} \int \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) ^{{\frac{5}{2}}}\, 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)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.64298, size = 549, normalized size = 1.84 \begin{align*} -\frac{a \sqrt{\frac{25 i}{a^{2}}} \log \left (\frac{1}{5} i \, a \sqrt{\frac{25 i}{a^{2}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - a \sqrt{\frac{25 i}{a^{2}}} \log \left (-\frac{1}{5} i \, a \sqrt{\frac{25 i}{a^{2}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) + a \sqrt{-\frac{25 i}{a^{2}}} \log \left (\frac{1}{5} i \, a \sqrt{-\frac{25 i}{a^{2}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) - a \sqrt{-\frac{25 i}{a^{2}}} \log \left (-\frac{1}{5} i \, a \sqrt{-\frac{25 i}{a^{2}}} + \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}\right ) +{\left (2 \, a x + 18 i\right )} \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}}{2 \, a} \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, algorithm="fricas")

[Out]

-1/2*(a*sqrt(25*I/a^2)*log(1/5*I*a*sqrt(25*I/a^2) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - a*sqrt(25*I/a^2)*lo
g(-1/5*I*a*sqrt(25*I/a^2) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + a*sqrt(-25*I/a^2)*log(1/5*I*a*sqrt(-25*I/a^
2) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - a*sqrt(-25*I/a^2)*log(-1/5*I*a*sqrt(-25*I/a^2) + sqrt(I*sqrt(a^2*x
^2 + 1)/(a*x + I))) + (2*a*x + 18*I)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/a

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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)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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, algorithm="giac")

[Out]

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