3.77 \(\int \frac{e^{\frac{3}{2} i \tan ^{-1}(a x)}}{x^4} \, dx\)

Optimal. Leaf size=170 \[ \frac{23 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{24 x}-\frac{17}{8} i a^3 \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac{17}{8} i a^3 \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac{7 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{12 x^2}-\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3} \]

[Out]

-((1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/(3*x^3) - (((7*I)/12)*a*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/x^2 + (23*
a^2*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/(24*x) - ((17*I)/8)*a^3*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)] +
((17*I)/8)*a^3*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]

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Rubi [A]  time = 0.0632094, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 9, 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, 298, 203, 206} \[ \frac{23 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{24 x}-\frac{17}{8} i a^3 \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac{17}{8} i a^3 \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac{7 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{12 x^2}-\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(((3*I)/2)*ArcTan[a*x])/x^4,x]

[Out]

-((1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/(3*x^3) - (((7*I)/12)*a*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/x^2 + (23*
a^2*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/(24*x) - ((17*I)/8)*a^3*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)] +
((17*I)/8)*a^3*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]

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 298

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

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])

Rubi steps

\begin{align*} \int \frac{e^{\frac{3}{2} i \tan ^{-1}(a x)}}{x^4} \, dx &=\int \frac{(1+i a x)^{3/4}}{x^4 (1-i a x)^{3/4}} \, dx\\ &=-\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}+\frac{1}{3} \int \frac{\frac{7 i a}{2}-2 a^2 x}{x^3 (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx\\ &=-\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}-\frac{7 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{12 x^2}-\frac{1}{6} \int \frac{\frac{23 a^2}{4}+\frac{7}{2} i a^3 x}{x^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx\\ &=-\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}-\frac{7 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{12 x^2}+\frac{23 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{24 x}+\frac{1}{6} \int -\frac{51 i a^3}{8 x (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx\\ &=-\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}-\frac{7 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{12 x^2}+\frac{23 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{24 x}-\frac{1}{16} \left (17 i a^3\right ) \int \frac{1}{x (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx\\ &=-\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}-\frac{7 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{12 x^2}+\frac{23 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{24 x}-\frac{1}{4} \left (17 i a^3\right ) \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 )\\ &=-\frac{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}-\frac{7 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{12 x^2}+\frac{23 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{24 x}+\frac{1}{8} \left (17 i a^3\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}{8} \left (17 i a^3\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{\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}-\frac{7 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{12 x^2}+\frac{23 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{24 x}-\frac{17}{8} i a^3 \tan ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac{17}{8} i a^3 \tanh ^{-1}\left (\frac{\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0226513, size = 93, normalized size = 0.55 \[ \frac{\sqrt [4]{1-i a x} \left (102 i a^3 x^3 \text{Hypergeometric2F1}\left (\frac{1}{4},1,\frac{5}{4},\frac{a x+i}{-a x+i}\right )+23 i a^3 x^3+37 a^2 x^2-22 i a x-8\right )}{24 x^3 \sqrt [4]{1+i a x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(((3*I)/2)*ArcTan[a*x])/x^4,x]

[Out]

((1 - I*a*x)^(1/4)*(-8 - (22*I)*a*x + 37*a^2*x^2 + (23*I)*a^3*x^3 + (102*I)*a^3*x^3*Hypergeometric2F1[1/4, 1,
5/4, (I + a*x)/(I - a*x)]))/(24*x^3*(1 + I*a*x)^(1/4))

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Maple [F]  time = 0.139, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \right ) ^{{\frac{3}{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))^(3/2)/x^4,x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)/x^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}\right )^{\frac{3}{2}}}{x^{4}}\,{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))^(3/2)/x^4,x, algorithm="maxima")

[Out]

integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(3/2)/x^4, x)

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Fricas [A]  time = 1.73878, size = 447, normalized size = 2.63 \begin{align*} \frac{51 i \, a^{3} x^{3} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 51 \, a^{3} x^{3} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 51 \, a^{3} x^{3} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} - i\right ) - 51 i \, a^{3} x^{3} \log \left (\sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}} - 1\right ) +{\left (46 \, a^{2} x^{2} - 28 i \, a x - 16\right )} \sqrt{a^{2} x^{2} + 1} \sqrt{\frac{i \, \sqrt{a^{2} x^{2} + 1}}{a x + i}}}{48 \, x^{3}} \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))^(3/2)/x^4,x, algorithm="fricas")

[Out]

1/48*(51*I*a^3*x^3*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) + 51*a^3*x^3*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x
+ I)) + I) - 51*a^3*x^3*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) - 51*I*a^3*x^3*log(sqrt(I*sqrt(a^2*x^2 +
1)/(a*x + I)) - 1) + (46*a^2*x^2 - 28*I*a*x - 16)*sqrt(a^2*x^2 + 1)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/x^3

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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))**(3/2)/x**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{i \, a x + 1}{\sqrt{a^{2} x^{2} + 1}}\right )^{\frac{3}{2}}}{x^{4}}\,{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))^(3/2)/x^4,x, algorithm="giac")

[Out]

integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(3/2)/x^4, x)