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

Optimal. Leaf size=92 \[ -\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}+3 i a \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-3 i a \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

[Out]

-(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/x+3*I*a*arctan((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))-3*I*a*arctanh((1+I*a*x)^(1/4)
/(1-I*a*x)^(1/4))

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Rubi [A]  time = 0.03, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5062, 94, 93, 298, 203, 206} \[ -\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}+3 i a \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-3 i a \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/x) + (3*I)*a*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)] - (3*I)*a*Arc
Tanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]

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 94

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[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

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])

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 68, normalized size = 0.74 \[ -\frac {i \sqrt [4]{1-i a x} \left (6 a x \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {a x+i}{i-a x}\right )+a x-i\right )}{x \sqrt [4]{1+i a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*(1 - I*a*x)^(1/4)*(-I + a*x + 6*a*x*Hypergeometric2F1[1/4, 1, 5/4, (I + a*x)/(I - a*x)]))/(x*(1 + I*a*x)
^(1/4))

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fricas [B]  time = 0.74, size = 157, normalized size = 1.71 \[ \frac {-3 i \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - 3 \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + 3 \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) + 3 i \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, \sqrt {a^{2} x^{2} + 1} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{2 \, x} \]

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^2,x, algorithm="fricas")

[Out]

1/2*(-3*I*a*x*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) - 3*a*x*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I
) + 3*a*x*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) + 3*I*a*x*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - 1)
- 2*sqrt(a^2*x^2 + 1)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/x

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming 0=[0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch for
the root of a polynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0]index.cc in
dex_m operator + Error: Bad Argument ValueWarning, need to choose a branch for the root of a polynomial with p
arameters. This might be wrong.The choice was done assuming 0=[0,0]index.cc index_m operator + Error: Bad Argu
ment ValueWarning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.T
he choice was done assuming 0=[0,0]index.cc index_m operator + Error: Bad Argument ValueDone

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )^{\frac {3}{2}}}{x^{2}}\, dx \]

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^2,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}{x^{2}}\,{d x} \]

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^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{3/2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {3}{2}}}{x^{2}}\, dx \]

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**2,x)

[Out]

Integral((I*(a*x - I)/sqrt(a**2*x**2 + 1))**(3/2)/x**2, x)

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