3.3.13 \(\int \frac {e^{-3 i \text {ArcTan}(a+b x)}}{x^2} \, dx\) [213]

Optimal. Leaf size=178 \[ \frac {6 i b \sqrt {1-i a-i b x}}{(i-a)^2 \sqrt {1+i a+i b x}}-\frac {(1-i a-i b x)^{3/2}}{(1+i a) x \sqrt {1+i a+i b x}}-\frac {6 i \sqrt {i+a} b \tanh ^{-1}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{(i-a)^{5/2}} \]

[Out]

-6*I*b*arctanh((I+a)^(1/2)*(1+I*a+I*b*x)^(1/2)/(I-a)^(1/2)/(1-I*a-I*b*x)^(1/2))*(I+a)^(1/2)/(I-a)^(5/2)-(1-I*a
-I*b*x)^(3/2)/(1+I*a)/x/(1+I*a+I*b*x)^(1/2)+6*I*b*(1-I*a-I*b*x)^(1/2)/(I-a)^2/(1+I*a+I*b*x)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5203, 96, 95, 214} \begin {gather*} -\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}} \end {gather*}

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 95

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 96

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[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 214

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

Rule 5203

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]

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) \text {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*}

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Mathematica [A]
time = 0.13, size = 145, normalized size = 0.81 \begin {gather*} \frac {\frac {\sqrt {-i (i+a+b x)} \left (1+a^2+5 i b x+a b x\right )}{x \sqrt {1+i a+i b x}}-\frac {6 i \sqrt {-1+i a} b \tanh ^{-1}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1-i a}}}{(-i+a)^2} \end {gather*}

Antiderivative was successfully verified.

[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[-1 + I*a]*b*
ArcTanh[(Sqrt[-1 - I*a]*Sqrt[(-I)*(I + a + b*x)])/(Sqrt[-1 + I*a]*Sqrt[1 + I*a + I*b*x])])/Sqrt[-1 - I*a])/(-I
+ a)^2

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1542 vs. \(2 (138 ) = 276\).
time = 0.19, size = 1543, normalized size = 8.67

method	result	size
risch	\(-\frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, \left (i+a \right )}{\left (a -i\right )^{2} x}+\frac {3 b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right ) a^{2}}{\left (a^{2}-2 i a -1\right ) \left (i-a \right ) \sqrt {a^{2}+1}}+\frac {3 b \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (a^{2}-2 i a -1\right ) \left (i-a \right ) \sqrt {a^{2}+1}}-\frac {4 \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\, a}{\left (a^{2}-2 i a -1\right ) \left (i-a \right ) \left (x -\frac {i}{b}+\frac {a}{b}\right )}+\frac {4 i \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{\left (a^{2}-2 i a -1\right ) \left (i-a \right ) \left (x -\frac {i}{b}+\frac {a}{b}\right )}\)	\(340\)
default	\(\text {Expression too large to display}\)	\(1543\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (116) = 232\).
time = 4.20, size = 389, normalized size = 2.19 \begin {gather*} -\frac {{\left (i \, a - 5\right )} b^{2} x^{2} + {\left (i \, a^{2} - 4 \, a + 5 i\right )} b x - 3 \, {\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 (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} \log \left (-\frac {b^{2} x + {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b}{b}\right ) + 3 \, {\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 (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} \log \left (-\frac {b^{2} x - {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b}{b}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (i \, a - 5\right )} b x + i \, a^{2} + i\right )}}{{\left (a^{2} - 2 i \, a - 1\right )} b x^{2} + {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x} \end {gather*}

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]

-((I*a - 5)*b^2*x^2 + (I*a^2 - 4*a + 5*I)*b*x - 3*((a^2 - 2*I*a - 1)*b*x^2 + (a^3 - 3*I*a^2 - 3*a + I)*x)*sqrt
((a + I)*b^2/(a^5 - 5*I*a^4 - 10*a^3 + 10*I*a^2 + 5*a - I))*log(-(b^2*x + (a^3 - 3*I*a^2 - 3*a + I)*sqrt((a +
I)*b^2/(a^5 - 5*I*a^4 - 10*a^3 + 10*I*a^2 + 5*a - I)) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b)/b) + 3*((a^2 - 2*
I*a - 1)*b*x^2 + (a^3 - 3*I*a^2 - 3*a + I)*x)*sqrt((a + I)*b^2/(a^5 - 5*I*a^4 - 10*a^3 + 10*I*a^2 + 5*a - I))*
log(-(b^2*x - (a^3 - 3*I*a^2 - 3*a + I)*sqrt((a + I)*b^2/(a^5 - 5*I*a^4 - 10*a^3 + 10*I*a^2 + 5*a - I)) - 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)*((I*a - 5)*b*x + I*a^2 + I))/((a^2 - 2
*I*a - 1)*b*x^2 + (a^3 - 3*I*a^2 - 3*a + I)*x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i \left (\int \frac {\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} x^{2} + 3 a^{2} b x^{3} - 3 i a^{2} x^{2} + 3 a b^{2} x^{4} - 6 i a b x^{3} - 3 a x^{2} + b^{3} x^{5} - 3 i b^{2} x^{4} - 3 b x^{3} + i x^{2}}\, dx + \int \frac {a^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} x^{2} + 3 a^{2} b x^{3} - 3 i a^{2} x^{2} + 3 a b^{2} x^{4} - 6 i a b x^{3} - 3 a x^{2} + b^{3} x^{5} - 3 i b^{2} x^{4} - 3 b x^{3} + i x^{2}}\, dx + \int \frac {b^{2} x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} x^{2} + 3 a^{2} b x^{3} - 3 i a^{2} x^{2} + 3 a b^{2} x^{4} - 6 i a b x^{3} - 3 a x^{2} + b^{3} x^{5} - 3 i b^{2} x^{4} - 3 b x^{3} + i x^{2}}\, dx + \int \frac {2 a b x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} x^{2} + 3 a^{2} b x^{3} - 3 i a^{2} x^{2} + 3 a b^{2} x^{4} - 6 i a b x^{3} - 3 a x^{2} + b^{3} x^{5} - 3 i b^{2} x^{4} - 3 b x^{3} + i x^{2}}\, dx\right ) \end {gather*}

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]

I*(Integral(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3*x**2 + 3*a**2*b*x**3 - 3*I*a**2*x**2 + 3*a*b**2*x**4 -
6*I*a*b*x**3 - 3*a*x**2 + b**3*x**5 - 3*I*b**2*x**4 - 3*b*x**3 + I*x**2), x) + Integral(a**2*sqrt(a**2 + 2*a*b
*x + b**2*x**2 + 1)/(a**3*x**2 + 3*a**2*b*x**3 - 3*I*a**2*x**2 + 3*a*b**2*x**4 - 6*I*a*b*x**3 - 3*a*x**2 + b**
3*x**5 - 3*I*b**2*x**4 - 3*b*x**3 + I*x**2), x) + Integral(b**2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**
3*x**2 + 3*a**2*b*x**3 - 3*I*a**2*x**2 + 3*a*b**2*x**4 - 6*I*a*b*x**3 - 3*a*x**2 + b**3*x**5 - 3*I*b**2*x**4 -
3*b*x**3 + I*x**2), x) + Integral(2*a*b*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3*x**2 + 3*a**2*b*x**3 - 3
*I*a**2*x**2 + 3*a*b**2*x**4 - 6*I*a*b*x**3 - 3*a*x**2 + b**3*x**5 - 3*I*b**2*x**4 - 3*b*x**3 + I*x**2), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left ({\left (a+b\,x\right )}^2+1\right )}^{3/2}}{x^2\,{\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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