3.213 \(\int \frac{e^{-3 i \tan ^{-1}(a+b x)}}{x^2} \, dx\)
Optimal. Leaf size=178 \[ -\frac{(-i a-i b x+1)^{3/2}}{(1+i a) x \sqrt{i a+i b x+1}}+\frac{6 i b \sqrt{-i a-i b x+1}}{(-a+i)^2 \sqrt{i a+i b x+1}}-\frac{6 i \sqrt{a+i} b \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{5/2}} \]
[Out]
((6*I)*b*Sqrt[1 - I*a - I*b*x])/((I - a)^2*Sqrt[1 + I*a + I*b*x]) - (1 - I*a - I*b*x)^(3/2)/((1 + I*a)*x*Sqrt[
1 + I*a + I*b*x]) - ((6*I)*Sqrt[I + a]*b*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/(Sqrt[I - a]*Sqrt[1 - I*a
- I*b*x])])/(I - a)^(5/2)
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Rubi [A] time = 0.0854949, antiderivative size = 178, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used =
{5095, 94, 93, 208} \[ -\frac{(-i a-i b x+1)^{3/2}}{(1+i a) x \sqrt{i a+i b x+1}}+\frac{6 i b \sqrt{-i a-i b x+1}}{(-a+i)^2 \sqrt{i a+i b x+1}}-\frac{6 i \sqrt{a+i} b \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[1/(E^((3*I)*ArcTan[a + b*x])*x^2),x]
[Out]
((6*I)*b*Sqrt[1 - I*a - I*b*x])/((I - a)^2*Sqrt[1 + I*a + I*b*x]) - (1 - I*a - I*b*x)^(3/2)/((1 + I*a)*x*Sqrt[
1 + I*a + I*b*x]) - ((6*I)*Sqrt[I + a]*b*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/(Sqrt[I - a]*Sqrt[1 - I*a
- I*b*x])])/(I - a)^(5/2)
Rule 5095
Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1 -
I*a*c - I*b*c*x)^((I*n)/2))/(1 + I*a*c + I*b*c*x)^((I*n)/2), x] /; FreeQ[{a, b, c, d, e, m, n}, 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 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 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]
Rubi steps
\begin{align*} \int \frac{e^{-3 i \tan ^{-1}(a+b x)}}{x^2} \, dx &=\int \frac{(1-i a-i b x)^{3/2}}{x^2 (1+i a+i b x)^{3/2}} \, dx\\ &=-\frac{(1-i a-i b x)^{3/2}}{(1+i a) x \sqrt{1+i a+i b x}}+\frac{(3 b) \int \frac{\sqrt{1-i a-i b x}}{x (1+i a+i b x)^{3/2}} \, dx}{i-a}\\ &=\frac{6 i b \sqrt{1-i a-i b x}}{(i-a)^2 \sqrt{1+i a+i b x}}-\frac{(1-i a-i b x)^{3/2}}{(1+i a) x \sqrt{1+i a+i b x}}+\frac{(3 (i+a) b) \int \frac{1}{x \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{(i-a)^2}\\ &=\frac{6 i b \sqrt{1-i a-i b x}}{(i-a)^2 \sqrt{1+i a+i b x}}-\frac{(1-i a-i b x)^{3/2}}{(1+i a) x \sqrt{1+i a+i b x}}+\frac{(6 (i+a) b) \operatorname{Subst}\left (\int \frac{1}{-1-i a-(-1+i a) x^2} \, dx,x,\frac{\sqrt{1+i a+i b x}}{\sqrt{1-i a-i b x}}\right )}{(i-a)^2}\\ &=\frac{6 i b \sqrt{1-i a-i b x}}{(i-a)^2 \sqrt{1+i a+i b x}}-\frac{(1-i a-i b x)^{3/2}}{(1+i a) x \sqrt{1+i a+i b x}}-\frac{6 i \sqrt{i+a} b \tanh ^{-1}\left (\frac{\sqrt{i+a} \sqrt{1+i a+i b x}}{\sqrt{i-a} \sqrt{1-i a-i b x}}\right )}{(i-a)^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.171053, size = 135, normalized size = 0.76 \[ \frac{\frac{\sqrt{-i (a+b x+i)} \left (a^2+a b x+5 i b x+1\right )}{x \sqrt{i a+i b x+1}}-6 i \sqrt{\frac{a+i}{a-i}} b \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right )}{(a-i)^2} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(E^((3*I)*ArcTan[a + b*x])*x^2),x]
[Out]
((Sqrt[(-I)*(I + a + b*x)]*(1 + a^2 + (5*I)*b*x + a*b*x))/(x*Sqrt[1 + I*a + I*b*x]) - (6*I)*Sqrt[(I + a)/(-I +
a)]*b*ArcTan[Sqrt[(-I)*(I + a + b*x)]/(Sqrt[(I + a)/(-I + a)]*Sqrt[1 + I*a + I*b*x])])/(-I + a)^2
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Maple [B] time = 0.132, size = 1917, normalized size = 10.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2)/x^2,x)
[Out]
2/(I-a)^3*b*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(3/2)-2*I/(I-a)^2/b/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*(x-
(I-a)/b)*b)^(5/2)-3*b^2/(I-a)^2*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/
2))/(b^2)^(1/2)-2*I/(I-a)^3*a*b/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-3*I/(I-a)^3*a^4*b^2/(a^2+1)*ln((b^2*x+a*
b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)+3*I/(I-a)^3*a*b/(a^2+1)^(1/2)*ln((2*a^2+2+2*x*a*b+2*
(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/x)-I/(I-a)^3*b^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)*x-3/2*I/(I
-a)^3*b^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-3/2*I/(I-a)^3*b^2/(a^2+1)*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^
2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)-3*I/(I-a)^4*b^2*a^3*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/
2))/(b^2)^(1/2)+3*I/(I-a)^4*b*(a^2+1)^(1/2)*ln((2*a^2+2+2*x*a*b+2*(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))
/x)*a^2-3/2*I/(I-a)^4*b^2*a*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-9/2*I/(I-a)^4*b^2*a*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^
2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)-9/2*I/(I-a)^3*a^3*b/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-9/2*I/(I-a)^
3*a*b/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3*I/(I-a)^3*a^3*b/(a^2+1)^(1/2)*ln((2*a^2+2+2*x*a*b+2*(a^2+1)^(1/2
)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/x)-3/2*I/(I-a)^3*a^2*b^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-9/2*I/(I-a)^
3*a^2*b^2/(a^2+1)*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)-1/(I-a)^2/b^2/(x-I/b+a
/b)^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(5/2)-3/(I-a)^2*b^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*x-
3/(I-a)^2*b*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*a+2*I/(I-a)^2*b*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^
(3/2)-2/(I-a)^3/b/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(5/2)+I/(I-a)^4*b*((x-(I-a)/b)^2*b^2+2*I
*(x-(I-a)/b)*b)^(3/2)-3/2/(I-a)^4*b^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*x-3/2/(I-a)^4*b*((x-(I-a)/b)
^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*a-3/2/(I-a)^4*b^2*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((x-(I-a)/b)^2*b^2+2*I*
(x-(I-a)/b)*b)^(1/2))/(b^2)^(1/2)-I/(I-a)^4*b*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-3*I/(I-a)^4*b*(b^2*x^2+2*a*b*x+a^2
+1)^(1/2)+3*I/(I-a)^3*b^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*x+3*I/(I-a)^3*b*((x-(I-a)/b)^2*b^2+2*I*(
x-(I-a)/b)*b)^(1/2)*a+3*I/(I-a)^3*b^2*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*
b)^(1/2))/(b^2)^(1/2)+I/(I-a)^3/(a^2+1)/x*(b^2*x^2+2*a*b*x+a^2+1)^(5/2)+3*I/(I-a)^4*b*(a^2+1)^(1/2)*ln((2*a^2+
2+2*x*a*b+2*(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/x)-9/2*I/(I-a)^4*b*a^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left ({\left (b x + a\right )}^{2} + 1\right )}^{\frac{3}{2}}}{{\left (i \, b x + i \, a + 1\right )}^{3} x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2)/x^2,x, algorithm="maxima")
[Out]
integrate(((b*x + a)^2 + 1)^(3/2)/((I*b*x + I*a + 1)^3*x^2), x)
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Fricas [B] time = 2.37605, size = 1029, normalized size = 5.78 \begin{align*} -\frac{2 \,{\left (i \, a - 5\right )} b^{2} x^{2} -{\left (-2 i \, a^{2} + 8 \, a - 10 i\right )} b x -{\left ({\left (a^{2} - 2 i \, a - 1\right )} b x^{2} +{\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x\right )} \sqrt{\frac{{\left (36 \, a + 36 i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} \log \left (-\frac{6 \, b^{2} x +{\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} \sqrt{\frac{{\left (36 \, a + 36 i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} - 6 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b}{6 \, b}\right ) +{\left ({\left (a^{2} - 2 i \, a - 1\right )} b x^{2} +{\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x\right )} \sqrt{\frac{{\left (36 \, a + 36 i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} \log \left (-\frac{6 \, b^{2} x -{\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} \sqrt{\frac{{\left (36 \, a + 36 i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} - 6 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b}{6 \, b}\right ) + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (2 \,{\left (i \, a - 5\right )} b x + 2 i \, a^{2} + 2 i\right )}}{2 \,{\left (a^{2} - 2 i \, a - 1\right )} b x^{2} +{\left (2 \, a^{3} - 6 i \, a^{2} - 6 \, a + 2 i\right )} x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2)/x^2,x, algorithm="fricas")
[Out]
-(2*(I*a - 5)*b^2*x^2 - (-2*I*a^2 + 8*a - 10*I)*b*x - ((a^2 - 2*I*a - 1)*b*x^2 + (a^3 - 3*I*a^2 - 3*a + I)*x)*
sqrt((36*a + 36*I)*b^2/(a^5 - 5*I*a^4 - 10*a^3 + 10*I*a^2 + 5*a - I))*log(-1/6*(6*b^2*x + (a^3 - 3*I*a^2 - 3*a
+ I)*sqrt((36*a + 36*I)*b^2/(a^5 - 5*I*a^4 - 10*a^3 + 10*I*a^2 + 5*a - I)) - 6*sqrt(b^2*x^2 + 2*a*b*x + a^2 +
1)*b)/b) + ((a^2 - 2*I*a - 1)*b*x^2 + (a^3 - 3*I*a^2 - 3*a + I)*x)*sqrt((36*a + 36*I)*b^2/(a^5 - 5*I*a^4 - 10
*a^3 + 10*I*a^2 + 5*a - I))*log(-1/6*(6*b^2*x - (a^3 - 3*I*a^2 - 3*a + I)*sqrt((36*a + 36*I)*b^2/(a^5 - 5*I*a^
4 - 10*a^3 + 10*I*a^2 + 5*a - I)) - 6*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b)/b) + sqrt(b^2*x^2 + 2*a*b*x + a^2 +
1)*(2*(I*a - 5)*b*x + 2*I*a^2 + 2*I))/(2*(a^2 - 2*I*a - 1)*b*x^2 + (2*a^3 - 6*I*a^2 - 6*a + 2*I)*x)
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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*(b*x+a))**3*(1+(b*x+a)**2)**(3/2)/x**2,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*(b*x+a))^3*(1+(b*x+a)^2)^(3/2)/x^2,x, algorithm="giac")
[Out]
undef