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

Optimal. Leaf size=118 \[ -\frac{4 a^3 \sqrt{a^2 x^2+1}}{-a x+i}+\frac{14 a^2 \sqrt{a^2 x^2+1}}{3 x}+\frac{3 i a \sqrt{a^2 x^2+1}}{2 x^2}-\frac{\sqrt{a^2 x^2+1}}{3 x^3}-\frac{11}{2} i a^3 \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

[Out]

-Sqrt[1 + a^2*x^2]/(3*x^3) + (((3*I)/2)*a*Sqrt[1 + a^2*x^2])/x^2 + (14*a^2*Sqrt[1 + a^2*x^2])/(3*x) - (4*a^3*S
qrt[1 + a^2*x^2])/(I - a*x) - ((11*I)/2)*a^3*ArcTanh[Sqrt[1 + a^2*x^2]]

________________________________________________________________________________________

Rubi [A]  time = 0.603894, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {5060, 6742, 271, 264, 266, 51, 63, 208, 651} \[ -\frac{4 a^3 \sqrt{a^2 x^2+1}}{-a x+i}+\frac{14 a^2 \sqrt{a^2 x^2+1}}{3 x}+\frac{3 i a \sqrt{a^2 x^2+1}}{2 x^2}-\frac{\sqrt{a^2 x^2+1}}{3 x^3}-\frac{11}{2} i a^3 \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((3*I)*ArcTan[a*x])*x^4),x]

[Out]

-Sqrt[1 + a^2*x^2]/(3*x^3) + (((3*I)/2)*a*Sqrt[1 + a^2*x^2])/x^2 + (14*a^2*Sqrt[1 + a^2*x^2])/(3*x) - (4*a^3*S
qrt[1 + a^2*x^2])/(I - a*x) - ((11*I)/2)*a^3*ArcTanh[Sqrt[1 + a^2*x^2]]

Rule 5060

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n
- 1)/2)*Sqrt[1 + a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(I*n - 1)/2]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rubi steps

\begin{align*} \int \frac{e^{-3 i \tan ^{-1}(a x)}}{x^4} \, dx &=\int \frac{(1-i a x)^2}{x^4 (1+i a x) \sqrt{1+a^2 x^2}} \, dx\\ &=\int \left (\frac{1}{x^4 \sqrt{1+a^2 x^2}}-\frac{3 i a}{x^3 \sqrt{1+a^2 x^2}}-\frac{4 a^2}{x^2 \sqrt{1+a^2 x^2}}+\frac{4 i a^3}{x \sqrt{1+a^2 x^2}}-\frac{4 i a^4}{(-i+a x) \sqrt{1+a^2 x^2}}\right ) \, dx\\ &=-\left ((3 i a) \int \frac{1}{x^3 \sqrt{1+a^2 x^2}} \, dx\right )-\left (4 a^2\right ) \int \frac{1}{x^2 \sqrt{1+a^2 x^2}} \, dx+\left (4 i a^3\right ) \int \frac{1}{x \sqrt{1+a^2 x^2}} \, dx-\left (4 i a^4\right ) \int \frac{1}{(-i+a x) \sqrt{1+a^2 x^2}} \, dx+\int \frac{1}{x^4 \sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 x^3}+\frac{4 a^2 \sqrt{1+a^2 x^2}}{x}-\frac{4 a^3 \sqrt{1+a^2 x^2}}{i-a x}-\frac{1}{2} (3 i a) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1+a^2 x}} \, dx,x,x^2\right )-\frac{1}{3} \left (2 a^2\right ) \int \frac{1}{x^2 \sqrt{1+a^2 x^2}} \, dx+\left (2 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 x^3}+\frac{3 i a \sqrt{1+a^2 x^2}}{2 x^2}+\frac{14 a^2 \sqrt{1+a^2 x^2}}{3 x}-\frac{4 a^3 \sqrt{1+a^2 x^2}}{i-a x}+(4 i a) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{a^2}+\frac{x^2}{a^2}} \, dx,x,\sqrt{1+a^2 x^2}\right )+\frac{1}{4} \left (3 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 x^3}+\frac{3 i a \sqrt{1+a^2 x^2}}{2 x^2}+\frac{14 a^2 \sqrt{1+a^2 x^2}}{3 x}-\frac{4 a^3 \sqrt{1+a^2 x^2}}{i-a x}-4 i a^3 \tanh ^{-1}\left (\sqrt{1+a^2 x^2}\right )+\frac{1}{2} (3 i a) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{a^2}+\frac{x^2}{a^2}} \, dx,x,\sqrt{1+a^2 x^2}\right )\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 x^3}+\frac{3 i a \sqrt{1+a^2 x^2}}{2 x^2}+\frac{14 a^2 \sqrt{1+a^2 x^2}}{3 x}-\frac{4 a^3 \sqrt{1+a^2 x^2}}{i-a x}-\frac{11}{2} i a^3 \tanh ^{-1}\left (\sqrt{1+a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0697248, size = 89, normalized size = 0.75 \[ \frac{1}{6} \left (\frac{\sqrt{a^2 x^2+1} \left (52 a^3 x^3-19 i a^2 x^2+7 a x+2 i\right )}{x^3 (a x-i)}-33 i a^3 \log \left (\sqrt{a^2 x^2+1}+1\right )+33 i a^3 \log (x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^((3*I)*ArcTan[a*x])*x^4),x]

[Out]

((Sqrt[1 + a^2*x^2]*(2*I + 7*a*x - (19*I)*a^2*x^2 + 52*a^3*x^3))/(x^3*(-I + a*x)) + (33*I)*a^3*Log[x] - (33*I)
*a^3*Log[1 + Sqrt[1 + a^2*x^2]])/6

________________________________________________________________________________________

Maple [B]  time = 0.09, size = 392, normalized size = 3.3 \begin{align*} -{\frac{16\,i}{3}}{a}^{3} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{3}{2}}}-{\frac{11\,i}{2}}{a}^{3}{\it Artanh} \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) +8\,{a}^{4}\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }x+8\,{\frac{{a}^{4}}{\sqrt{{a}^{2}}}\ln \left ({\frac{1}{\sqrt{{a}^{2}}} \left ( ia+{a}^{2} \left ( x-{\frac{i}{a}} \right ) \right ) }+\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) } \right ) }-{ \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{\frac{i}{a}} \right ) ^{-3}}+{\frac{11\,i}{2}}{a}^{3}\sqrt{{a}^{2}{x}^{2}+1}+{2\,ia \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{\frac{i}{a}} \right ) ^{-2}}+{\frac{{\frac{3\,i}{2}}a}{{x}^{2}} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+{\frac{11\,i}{6}}{a}^{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}+{\frac{16\,{a}^{2}}{3\,x} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{\frac{16\,{a}^{4}x}{3} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-8\,{a}^{4}x\sqrt{{a}^{2}{x}^{2}+1}-8\,{\frac{{a}^{4}}{\sqrt{{a}^{2}}}\ln \left ({\frac{{a}^{2}x}{\sqrt{{a}^{2}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ) }-{\frac{1}{3\,{x}^{3}} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/3*I*a^3*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(3/2)-11/2*I*a^3*arctanh(1/(a^2*x^2+1)^(1/2))+8*a^4*(a^2*(x-I/a)^2+2
*I*a*(x-I/a))^(1/2)*x+8*a^4*ln((I*a+a^2*(x-I/a))/(a^2)^(1/2)+(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(1/2))/(a^2)^(1/2)-
1/(x-I/a)^3*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(5/2)+11/2*I*a^3*(a^2*x^2+1)^(1/2)+2*I*a/(x-I/a)^2*(a^2*(x-I/a)^2+2*
I*a*(x-I/a))^(5/2)+3/2*I*a/x^2*(a^2*x^2+1)^(5/2)+11/6*I*a^3*(a^2*x^2+1)^(3/2)+16/3*a^2/x*(a^2*x^2+1)^(5/2)-16/
3*a^4*x*(a^2*x^2+1)^(3/2)-8*a^4*x*(a^2*x^2+1)^(1/2)-8*a^4*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)-
1/3/x^3*(a^2*x^2+1)^(5/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.75583, size = 316, normalized size = 2.68 \begin{align*} \frac{52 \, a^{4} x^{4} - 52 i \, a^{3} x^{3} - 33 \,{\left (i \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} + 1\right ) - 33 \,{\left (-i \, a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} - 1\right ) +{\left (52 \, a^{3} x^{3} - 19 i \, a^{2} x^{2} + 7 \, a x + 2 i\right )} \sqrt{a^{2} x^{2} + 1}}{6 \,{\left (a x^{4} - i \, x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^4,x, algorithm="fricas")

[Out]

1/6*(52*a^4*x^4 - 52*I*a^3*x^3 - 33*(I*a^4*x^4 + a^3*x^3)*log(-a*x + sqrt(a^2*x^2 + 1) + 1) - 33*(-I*a^4*x^4 -
a^3*x^3)*log(-a*x + sqrt(a^2*x^2 + 1) - 1) + (52*a^3*x^3 - 19*I*a^2*x^2 + 7*a*x + 2*I)*sqrt(a^2*x^2 + 1))/(a*
x^4 - I*x^3)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^4,x, algorithm="giac")

[Out]

undef