3.113 \(\int \frac {e^{-\frac {5}{2} i \tan ^{-1}(a x)}}{x^4} \, dx\)

Optimal. Leaf size=203 \[ \frac {287 i a^3 \sqrt [4]{1-i a x}}{24 \sqrt [4]{1+i a x}}+\frac {55}{8} i a^3 \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {55}{8} i a^3 \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {61 a^2 \sqrt [4]{1-i a x}}{24 x \sqrt [4]{1+i a x}}-\frac {\sqrt [4]{1-i a x}}{3 x^3 \sqrt [4]{1+i a x}}+\frac {13 i a \sqrt [4]{1-i a x}}{12 x^2 \sqrt [4]{1+i a x}} \]

[Out]

287/24*I*a^3*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)-1/3*(1-I*a*x)^(1/4)/x^3/(1+I*a*x)^(1/4)+13/12*I*a*(1-I*a*x)^(1/4)
/x^2/(1+I*a*x)^(1/4)+61/24*a^2*(1-I*a*x)^(1/4)/x/(1+I*a*x)^(1/4)+55/8*I*a^3*arctan((1+I*a*x)^(1/4)/(1-I*a*x)^(
1/4))-55/8*I*a^3*arctanh((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))

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Rubi [A]  time = 0.08, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {5062, 98, 151, 155, 12, 93, 298, 203, 206} \[ \frac {287 i a^3 \sqrt [4]{1-i a x}}{24 \sqrt [4]{1+i a x}}+\frac {61 a^2 \sqrt [4]{1-i a x}}{24 x \sqrt [4]{1+i a x}}+\frac {55}{8} i a^3 \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {55}{8} i a^3 \tanh ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {13 i a \sqrt [4]{1-i a x}}{12 x^2 \sqrt [4]{1+i a x}}-\frac {\sqrt [4]{1-i a x}}{3 x^3 \sqrt [4]{1+i a x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(((5*I)/2)*ArcTan[a*x])*x^4),x]

[Out]

(((287*I)/24)*a^3*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4) - (1 - I*a*x)^(1/4)/(3*x^3*(1 + I*a*x)^(1/4)) + (((13*I
)/12)*a*(1 - I*a*x)^(1/4))/(x^2*(1 + I*a*x)^(1/4)) + (61*a^2*(1 - I*a*x)^(1/4))/(24*x*(1 + I*a*x)^(1/4)) + ((5
5*I)/8)*a^3*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)] - ((55*I)/8)*a^3*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)
^(1/4)]

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 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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 155

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] && ILtQ[m + n + p + 2, 0] && NeQ[m, -1] && (Sum
SimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 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 {5}{2} i \tan ^{-1}(a x)}}{x^4} \, dx &=\int \frac {(1-i a x)^{5/4}}{x^4 (1+i a x)^{5/4}} \, dx\\ &=-\frac {\sqrt [4]{1-i a x}}{3 x^3 \sqrt [4]{1+i a x}}-\frac {1}{3} \int \frac {\frac {13 i a}{2}+6 a^2 x}{x^3 (1-i a x)^{3/4} (1+i a x)^{5/4}} \, dx\\ &=-\frac {\sqrt [4]{1-i a x}}{3 x^3 \sqrt [4]{1+i a x}}+\frac {13 i a \sqrt [4]{1-i a x}}{12 x^2 \sqrt [4]{1+i a x}}+\frac {1}{6} \int \frac {-\frac {61 a^2}{4}+13 i a^3 x}{x^2 (1-i a x)^{3/4} (1+i a x)^{5/4}} \, dx\\ &=-\frac {\sqrt [4]{1-i a x}}{3 x^3 \sqrt [4]{1+i a x}}+\frac {13 i a \sqrt [4]{1-i a x}}{12 x^2 \sqrt [4]{1+i a x}}+\frac {61 a^2 \sqrt [4]{1-i a x}}{24 x \sqrt [4]{1+i a x}}-\frac {1}{6} \int \frac {-\frac {165 i a^3}{8}-\frac {61 a^4 x}{4}}{x (1-i a x)^{3/4} (1+i a x)^{5/4}} \, dx\\ &=\frac {287 i a^3 \sqrt [4]{1-i a x}}{24 \sqrt [4]{1+i a x}}-\frac {\sqrt [4]{1-i a x}}{3 x^3 \sqrt [4]{1+i a x}}+\frac {13 i a \sqrt [4]{1-i a x}}{12 x^2 \sqrt [4]{1+i a x}}+\frac {61 a^2 \sqrt [4]{1-i a x}}{24 x \sqrt [4]{1+i a x}}+\frac {i \int \frac {165 a^4}{16 x (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx}{3 a}\\ &=\frac {287 i a^3 \sqrt [4]{1-i a x}}{24 \sqrt [4]{1+i a x}}-\frac {\sqrt [4]{1-i a x}}{3 x^3 \sqrt [4]{1+i a x}}+\frac {13 i a \sqrt [4]{1-i a x}}{12 x^2 \sqrt [4]{1+i a x}}+\frac {61 a^2 \sqrt [4]{1-i a x}}{24 x \sqrt [4]{1+i a x}}+\frac {1}{16} \left (55 i a^3\right ) \int \frac {1}{x (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx\\ &=\frac {287 i a^3 \sqrt [4]{1-i a x}}{24 \sqrt [4]{1+i a x}}-\frac {\sqrt [4]{1-i a x}}{3 x^3 \sqrt [4]{1+i a x}}+\frac {13 i a \sqrt [4]{1-i a x}}{12 x^2 \sqrt [4]{1+i a x}}+\frac {61 a^2 \sqrt [4]{1-i a x}}{24 x \sqrt [4]{1+i a x}}+\frac {1}{4} \left (55 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 {287 i a^3 \sqrt [4]{1-i a x}}{24 \sqrt [4]{1+i a x}}-\frac {\sqrt [4]{1-i a x}}{3 x^3 \sqrt [4]{1+i a x}}+\frac {13 i a \sqrt [4]{1-i a x}}{12 x^2 \sqrt [4]{1+i a x}}+\frac {61 a^2 \sqrt [4]{1-i a x}}{24 x \sqrt [4]{1+i a x}}-\frac {1}{8} \left (55 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 (55 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 {287 i a^3 \sqrt [4]{1-i a x}}{24 \sqrt [4]{1+i a x}}-\frac {\sqrt [4]{1-i a x}}{3 x^3 \sqrt [4]{1+i a x}}+\frac {13 i a \sqrt [4]{1-i a x}}{12 x^2 \sqrt [4]{1+i a x}}+\frac {61 a^2 \sqrt [4]{1-i a x}}{24 x \sqrt [4]{1+i a x}}+\frac {55}{8} i a^3 \tan ^{-1}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {55}{8} i a^3 \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 = 93, normalized size = 0.46 \[ \frac {\sqrt [4]{1-i a x} \left (-330 i a^3 x^3 \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {a x+i}{i-a x}\right )+287 i a^3 x^3+61 a^2 x^2+26 i a x-8\right )}{24 x^3 \sqrt [4]{1+i a x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(((5*I)/2)*ArcTan[a*x])*x^4),x]

[Out]

((1 - I*a*x)^(1/4)*(-8 + (26*I)*a*x + 61*a^2*x^2 + (287*I)*a^3*x^3 - (330*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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fricas [A]  time = 0.75, size = 246, normalized size = 1.21 \[ \frac {{\left (574 \, a^{3} x^{3} - 122 i \, a^{2} x^{2} + 52 \, a x + 16 i\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 165 \, {\left (i \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - {\left (165 \, a^{4} x^{4} - 165 i \, a^{3} x^{3}\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + {\left (165 \, a^{4} x^{4} - 165 i \, a^{3} x^{3}\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - 165 \, {\left (-i \, a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right )}{48 \, a x^{4} - 48 i \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^4,x, algorithm="fricas")

[Out]

((574*a^3*x^3 - 122*I*a^2*x^2 + 52*a*x + 16*I)*sqrt(a^2*x^2 + 1)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - 165*(I*
a^4*x^4 + a^3*x^3)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) - (165*a^4*x^4 - 165*I*a^3*x^3)*log(sqrt(I*sqr
t(a^2*x^2 + 1)/(a*x + I)) + I) + (165*a^4*x^4 - 165*I*a^3*x^3)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) -
165*(-I*a^4*x^4 - a^3*x^3)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - 1))/(48*a*x^4 - 48*I*x^3)

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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/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^4,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 ValueWarning, need to choos
e 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 pol
ynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0]index.cc index_m operator + E
rror: Bad Argument ValueDone

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )^{\frac {5}{2}} x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^4,x)

[Out]

int(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^4,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{4} \left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^4,x, algorithm="maxima")

[Out]

integrate(1/(x^4*((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(5/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^4\,{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2)),x)

[Out]

int(1/(x^4*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2)), 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/((1+I*a*x)/(a**2*x**2+1)**(1/2))**(5/2)/x**4,x)

[Out]

Timed out

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