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

Optimal. Leaf size=202 \[ \frac{29 a^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{96 x^2}+\frac{83 i a^3 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{192 x}-\frac{11}{64} a^4 \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac{11}{64} a^4 \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac{7 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 x^3}-\frac{(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4} \]

[Out]

-((1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/(4*x^4) - (((7*I)/24)*a*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/x^3 + (29*
a^2*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/(96*x^2) + (((83*I)/192)*a^3*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/x -
(11*a^4*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/64 - (11*a^4*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)
])/64

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Rubi [A]  time = 0.0790615, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5062, 99, 151, 12, 93, 212, 206, 203} \[ \frac{29 a^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{96 x^2}+\frac{83 i a^3 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{192 x}-\frac{11}{64} a^4 \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac{11}{64} a^4 \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac{7 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 x^3}-\frac{(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/(4*x^4) - (((7*I)/24)*a*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/x^3 + (29*
a^2*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/(96*x^2) + (((83*I)/192)*a^3*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/x -
(11*a^4*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/64 - (11*a^4*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)
])/64

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 99

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

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 12

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 212

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

Rule 206

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

Rule 203

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

Rubi steps

\begin{align*} \int \frac{e^{\frac{1}{2} i \tan ^{-1}(a x)}}{x^5} \, dx &=\int \frac{\sqrt [4]{1+i a x}}{x^5 \sqrt [4]{1-i a x}} \, dx\\ &=-\frac{(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}+\frac{1}{4} \int \frac{\frac{7 i a}{2}-3 a^2 x}{x^4 \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx\\ &=-\frac{(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}-\frac{7 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 x^3}-\frac{1}{12} \int \frac{\frac{29 a^2}{4}+7 i a^3 x}{x^3 \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx\\ &=-\frac{(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}-\frac{7 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 x^3}+\frac{29 a^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{96 x^2}+\frac{1}{24} \int \frac{-\frac{83 i a^3}{8}+\frac{29 a^4 x}{4}}{x^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx\\ &=-\frac{(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}-\frac{7 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 x^3}+\frac{29 a^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{96 x^2}+\frac{83 i a^3 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{192 x}-\frac{1}{24} \int -\frac{33 a^4}{16 x \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx\\ &=-\frac{(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}-\frac{7 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 x^3}+\frac{29 a^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{96 x^2}+\frac{83 i a^3 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{192 x}+\frac{1}{128} \left (11 a^4\right ) \int \frac{1}{x \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx\\ &=-\frac{(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}-\frac{7 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 x^3}+\frac{29 a^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{96 x^2}+\frac{83 i a^3 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{192 x}+\frac{1}{32} \left (11 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^4} \, dx,x,\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\\ &=-\frac{(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}-\frac{7 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 x^3}+\frac{29 a^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{96 x^2}+\frac{83 i a^3 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{192 x}-\frac{1}{64} \left (11 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac{1}{64} \left (11 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\\ &=-\frac{(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}-\frac{7 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 x^3}+\frac{29 a^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{96 x^2}+\frac{83 i a^3 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{192 x}-\frac{11}{64} a^4 \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac{11}{64} a^4 \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0278389, size = 99, normalized size = 0.49 \[ -\frac{(1-i a x)^{3/4} \left (22 a^4 x^4 \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},\frac{a x+i}{-a x+i}\right )+83 a^4 x^4-141 i a^3 x^3-114 a^2 x^2+104 i a x+48\right )}{192 x^4 (1+i a x)^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((1 - I*a*x)^(3/4)*(48 + (104*I)*a*x - 114*a^2*x^2 - (141*I)*a^3*x^3 + 83*a^4*x^4 + 22*a^4*x^4*Hypergeometric
2F1[3/4, 1, 7/4, (I + a*x)/(I - a*x)]))/(192*x^4*(1 + I*a*x)^(3/4))

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Maple [F]  time = 0.141, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{5}}\sqrt{{(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}}}\, 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))^(1/2)/x^5,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}}}{x^{5}}\,{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))^(1/2)/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.79748, size = 463, normalized size = 2.29 \begin{align*} -\frac{33 \, a^{4} x^{4} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 33 i \, a^{4} x^{4} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 33 i \, a^{4} x^{4} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} - i\right ) - 33 \, a^{4} x^{4} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} - 1\right ) -{\left (166 \, a^{4} x^{4} + 50 i \, a^{3} x^{3} + 4 \, a^{2} x^{2} - 16 i \, a x - 96\right )} \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}}{384 \, x^{4}} \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))^(1/2)/x^5,x, algorithm="fricas")

[Out]

-1/384*(33*a^4*x^4*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) + 33*I*a^4*x^4*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a
*x + I)) + I) - 33*I*a^4*x^4*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) - 33*a^4*x^4*log(sqrt(I*sqrt(a^2*x^2
+ 1)/(a*x + I)) - 1) - (166*a^4*x^4 + 50*I*a^3*x^3 + 4*a^2*x^2 - 16*I*a*x - 96)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x
+ I)))/x^4

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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))**(1/2)/x**5,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}}}{x^{5}}\,{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))^(1/2)/x^5,x, algorithm="giac")

[Out]

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