3.85 \(\int \frac {e^{\frac {5}{2} i \tan ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=163 \[ -\frac {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}+\frac {25}{4} a^2 \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {25}{4} a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}} \]

[Out]

-25/2*a^2*(1+I*a*x)^(1/4)/(1-I*a*x)^(1/4)-5/4*I*a*(1+I*a*x)^(5/4)/x/(1-I*a*x)^(1/4)-1/2*(1+I*a*x)^(9/4)/x^2/(1
-I*a*x)^(1/4)+25/4*a^2*arctan((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))+25/4*a^2*arctanh((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4
))

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Rubi [A]  time = 0.05, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5062, 96, 94, 93, 212, 206, 203} \[ -\frac {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}+\frac {25}{4} a^2 \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {25}{4} a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}} \]

Antiderivative was successfully verified.

[In]

Int[E^(((5*I)/2)*ArcTan[a*x])/x^3,x]

[Out]

(-25*a^2*(1 + I*a*x)^(1/4))/(2*(1 - I*a*x)^(1/4)) - (((5*I)/4)*a*(1 + I*a*x)^(5/4))/(x*(1 - I*a*x)^(1/4)) - (1
+ I*a*x)^(9/4)/(2*x^2*(1 - I*a*x)^(1/4)) + (25*a^2*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/4 + (25*a^2*A
rcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[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 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 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 {5}{2} i \tan ^{-1}(a x)}}{x^3} \, dx &=\int \frac {(1+i a x)^{5/4}}{x^3 (1-i a x)^{5/4}} \, dx\\ &=-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}+\frac {1}{4} (5 i a) \int \frac {(1+i a x)^{5/4}}{x^2 (1-i a x)^{5/4}} \, dx\\ &=-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}-\frac {1}{8} \left (25 a^2\right ) \int \frac {\sqrt [4]{1+i a x}}{x (1-i a x)^{5/4}} \, dx\\ &=-\frac {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}-\frac {1}{8} \left (25 a^2\right ) \int \frac {1}{x \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \, dx\\ &=-\frac {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}-\frac {1}{2} \left (25 a^2\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 {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}+\frac {1}{4} \left (25 a^2\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}{4} \left (25 a^2\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 {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}+\frac {25}{4} a^2 \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {25}{4} a^2 \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 = 0.03, size = 99, normalized size = 0.61 \[ \frac {-129 i a^3 x^3+50 a^2 x^2 (1-i a x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {a x+i}{i-a x}\right )-102 a^2 x^2-33 i a x-6}{12 x^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(((5*I)/2)*ArcTan[a*x])/x^3,x]

[Out]

(-6 - (33*I)*a*x - 102*a^2*x^2 - (129*I)*a^3*x^3 + 50*a^2*x^2*(1 - I*a*x)*Hypergeometric2F1[3/4, 1, 7/4, (I +
a*x)/(I - a*x)])/(12*x^2*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))

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fricas [A]  time = 1.01, size = 176, normalized size = 1.08 \[ \frac {25 \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 25 i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 25 i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - 25 \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - {\left (86 \, a^{2} x^{2} + 18 i \, a x + 4\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{8 \, x^{2}} \]

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^3,x, algorithm="fricas")

[Out]

1/8*(25*a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) + 25*I*a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x
+ I)) + I) - 25*I*a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) - 25*a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 +
1)/(a*x + I)) - 1) - (86*a^2*x^2 + 18*I*a*x + 4)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/x^2

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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))^(5/2)/x^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )^{\frac {5}{2}}}{x^{3}}\, dx \]

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^3,x)

[Out]

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^3,x, algorithm="maxima")

[Out]

integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(5/2)/x^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2)/x^3,x)

[Out]

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2)/x^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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**3,x)

[Out]

Timed out

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