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

Optimal. Leaf size=139 \[ \frac{4 i a^4 \sqrt{a^2 x^2+1}}{-a x+i}-\frac{6 i a^3 \sqrt{a^2 x^2+1}}{x}+\frac{19 a^2 \sqrt{a^2 x^2+1}}{8 x^2}+\frac{i a \sqrt{a^2 x^2+1}}{x^3}-\frac{\sqrt{a^2 x^2+1}}{4 x^4}-\frac{51}{8} a^4 \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

[Out]

-Sqrt[1 + a^2*x^2]/(4*x^4) + (I*a*Sqrt[1 + a^2*x^2])/x^3 + (19*a^2*Sqrt[1 + a^2*x^2])/(8*x^2) - ((6*I)*a^3*Sqr
t[1 + a^2*x^2])/x + ((4*I)*a^4*Sqrt[1 + a^2*x^2])/(I - a*x) - (51*a^4*ArcTanh[Sqrt[1 + a^2*x^2]])/8

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Rubi [A]  time = 0.662225, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {5060, 6742, 266, 51, 63, 208, 271, 264, 651} \[ \frac{4 i a^4 \sqrt{a^2 x^2+1}}{-a x+i}-\frac{6 i a^3 \sqrt{a^2 x^2+1}}{x}+\frac{19 a^2 \sqrt{a^2 x^2+1}}{8 x^2}+\frac{i a \sqrt{a^2 x^2+1}}{x^3}-\frac{\sqrt{a^2 x^2+1}}{4 x^4}-\frac{51}{8} a^4 \tanh ^{-1}\left (\sqrt{a^2 x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((3*I)*ArcTan[a*x])*x^5),x]

[Out]

-Sqrt[1 + a^2*x^2]/(4*x^4) + (I*a*Sqrt[1 + a^2*x^2])/x^3 + (19*a^2*Sqrt[1 + a^2*x^2])/(8*x^2) - ((6*I)*a^3*Sqr
t[1 + a^2*x^2])/x + ((4*I)*a^4*Sqrt[1 + a^2*x^2])/(I - a*x) - (51*a^4*ArcTanh[Sqrt[1 + a^2*x^2]])/8

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 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 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 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^5} \, dx &=\int \frac{(1-i a x)^2}{x^5 (1+i a x) \sqrt{1+a^2 x^2}} \, dx\\ &=\int \left (\frac{1}{x^5 \sqrt{1+a^2 x^2}}-\frac{3 i a}{x^4 \sqrt{1+a^2 x^2}}-\frac{4 a^2}{x^3 \sqrt{1+a^2 x^2}}+\frac{4 i a^3}{x^2 \sqrt{1+a^2 x^2}}+\frac{4 a^4}{x \sqrt{1+a^2 x^2}}-\frac{4 a^5}{(-i+a x) \sqrt{1+a^2 x^2}}\right ) \, dx\\ &=-\left ((3 i a) \int \frac{1}{x^4 \sqrt{1+a^2 x^2}} \, dx\right )-\left (4 a^2\right ) \int \frac{1}{x^3 \sqrt{1+a^2 x^2}} \, dx+\left (4 i a^3\right ) \int \frac{1}{x^2 \sqrt{1+a^2 x^2}} \, dx+\left (4 a^4\right ) \int \frac{1}{x \sqrt{1+a^2 x^2}} \, dx-\left (4 a^5\right ) \int \frac{1}{(-i+a x) \sqrt{1+a^2 x^2}} \, dx+\int \frac{1}{x^5 \sqrt{1+a^2 x^2}} \, dx\\ &=\frac{i a \sqrt{1+a^2 x^2}}{x^3}-\frac{4 i a^3 \sqrt{1+a^2 x^2}}{x}+\frac{4 i a^4 \sqrt{1+a^2 x^2}}{i-a x}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{1+a^2 x}} \, dx,x,x^2\right )-\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1+a^2 x}} \, dx,x,x^2\right )+\left (2 i a^3\right ) \int \frac{1}{x^2 \sqrt{1+a^2 x^2}} \, dx+\left (2 a^4\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}}{4 x^4}+\frac{i a \sqrt{1+a^2 x^2}}{x^3}+\frac{2 a^2 \sqrt{1+a^2 x^2}}{x^2}-\frac{6 i a^3 \sqrt{1+a^2 x^2}}{x}+\frac{4 i a^4 \sqrt{1+a^2 x^2}}{i-a x}-\frac{1}{8} \left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1+a^2 x}} \, dx,x,x^2\right )+\left (4 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{a^2}+\frac{x^2}{a^2}} \, dx,x,\sqrt{1+a^2 x^2}\right )+a^4 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{1+a^2 x^2}}{4 x^4}+\frac{i a \sqrt{1+a^2 x^2}}{x^3}+\frac{19 a^2 \sqrt{1+a^2 x^2}}{8 x^2}-\frac{6 i a^3 \sqrt{1+a^2 x^2}}{x}+\frac{4 i a^4 \sqrt{1+a^2 x^2}}{i-a x}-4 a^4 \tanh ^{-1}\left (\sqrt{1+a^2 x^2}\right )+\left (2 a^2\right ) \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}{16} \left (3 a^4\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}}{4 x^4}+\frac{i a \sqrt{1+a^2 x^2}}{x^3}+\frac{19 a^2 \sqrt{1+a^2 x^2}}{8 x^2}-\frac{6 i a^3 \sqrt{1+a^2 x^2}}{x}+\frac{4 i a^4 \sqrt{1+a^2 x^2}}{i-a x}-6 a^4 \tanh ^{-1}\left (\sqrt{1+a^2 x^2}\right )+\frac{1}{8} \left (3 a^2\right ) \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}}{4 x^4}+\frac{i a \sqrt{1+a^2 x^2}}{x^3}+\frac{19 a^2 \sqrt{1+a^2 x^2}}{8 x^2}-\frac{6 i a^3 \sqrt{1+a^2 x^2}}{x}+\frac{4 i a^4 \sqrt{1+a^2 x^2}}{i-a x}-\frac{51}{8} a^4 \tanh ^{-1}\left (\sqrt{1+a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0772052, size = 95, normalized size = 0.68 \[ \frac{1}{8} \left (\frac{\sqrt{a^2 x^2+1} \left (-80 i a^4 x^4-29 a^3 x^3-11 i a^2 x^2+6 a x+2 i\right )}{x^4 (a x-i)}-51 a^4 \log \left (\sqrt{a^2 x^2+1}+1\right )+51 a^4 \log (x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^((3*I)*ArcTan[a*x])*x^5),x]

[Out]

((Sqrt[1 + a^2*x^2]*(2*I + 6*a*x - (11*I)*a^2*x^2 - 29*a^3*x^3 - (80*I)*a^4*x^4))/(x^4*(-I + a*x)) + 51*a^4*Lo
g[x] - 51*a^4*Log[1 + Sqrt[1 + a^2*x^2]])/8

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Maple [B]  time = 0.089, size = 416, normalized size = 3. \begin{align*} -{\frac{1}{4\,{x}^{4}} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+3\,{{a}^{2} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{5/2} \left ( x-{\frac{i}{a}} \right ) ^{-2}}+{\frac{23\,{a}^{2}}{8\,{x}^{2}} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{12\,i{a}^{5}\ln \left ({ \left ( ia+{a}^{2} \left ( x-{\frac{i}{a}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{12\,i{a}^{5}\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+12\,i{a}^{5}x\sqrt{{a}^{2}{x}^{2}+1}+{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 ) ^{-3}}+{\frac{ia}{{x}^{3}} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+8\,i{a}^{5}x \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}-{\frac{51\,{a}^{4}}{8}{\it Artanh} \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}} \right ) }+{\frac{51\,{a}^{4}}{8}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{8\,i{a}^{3}}{x} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+{\frac{17\,{a}^{4}}{8} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-8\,{a}^{4} \left ({a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) \right ) ^{3/2}-12\,i{a}^{5}\sqrt{{a}^{2} \left ( x-{\frac{i}{a}} \right ) ^{2}+2\,ia \left ( x-{\frac{i}{a}} \right ) }x \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^5,x)

[Out]

-1/4/x^4*(a^2*x^2+1)^(5/2)+3*a^2/(x-I/a)^2*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(5/2)+23/8*a^2/x^2*(a^2*x^2+1)^(5/2)-
12*I*a^5*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)+12*I*a^5*ln(a^2*x/(
a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)+12*I*a^5*x*(a^2*x^2+1)^(1/2)+I*a/(x-I/a)^3*(a^2*(x-I/a)^2+2*I*a*(x-I
/a))^(5/2)+I*a/x^3*(a^2*x^2+1)^(5/2)+8*I*a^5*x*(a^2*x^2+1)^(3/2)-51/8*a^4*arctanh(1/(a^2*x^2+1)^(1/2))+51/8*a^
4*(a^2*x^2+1)^(1/2)-8*I*a^3/x*(a^2*x^2+1)^(5/2)+17/8*a^4*(a^2*x^2+1)^(3/2)-8*a^4*(a^2*(x-I/a)^2+2*I*a*(x-I/a))
^(3/2)-12*I*a^5*(a^2*(x-I/a)^2+2*I*a*(x-I/a))^(1/2)*x

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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^{5}}\,{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^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.77248, size = 346, normalized size = 2.49 \begin{align*} \frac{-80 i \, a^{5} x^{5} - 80 \, a^{4} x^{4} -{\left (51 \, a^{5} x^{5} - 51 i \, a^{4} x^{4}\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} + 1\right ) +{\left (51 \, a^{5} x^{5} - 51 i \, a^{4} x^{4}\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1} - 1\right ) +{\left (-80 i \, a^{4} x^{4} - 29 \, a^{3} x^{3} - 11 i \, a^{2} x^{2} + 6 \, a x + 2 i\right )} \sqrt{a^{2} x^{2} + 1}}{8 \,{\left (a x^{5} - i \, x^{4}\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^5,x, algorithm="fricas")

[Out]

1/8*(-80*I*a^5*x^5 - 80*a^4*x^4 - (51*a^5*x^5 - 51*I*a^4*x^4)*log(-a*x + sqrt(a^2*x^2 + 1) + 1) + (51*a^5*x^5
- 51*I*a^4*x^4)*log(-a*x + sqrt(a^2*x^2 + 1) - 1) + (-80*I*a^4*x^4 - 29*a^3*x^3 - 11*I*a^2*x^2 + 6*a*x + 2*I)*
sqrt(a^2*x^2 + 1))/(a*x^5 - I*x^4)

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

[Out]

Timed out

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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^5,x, algorithm="giac")

[Out]

undef