3.186 \(\int \frac{e^{3 i \tan ^{-1}(a+b x)}}{x^2} \, dx\)

Optimal. Leaf size=176 \[ -\frac{(i a+i b x+1)^{3/2}}{(1-i a) x \sqrt{-i a-i b x+1}}-\frac{6 i b \sqrt{i a+i b x+1}}{(a+i)^2 \sqrt{-i a-i b x+1}}+\frac{6 i \sqrt{-a+i} b \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(a+i)^{5/2}} \]

[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 = 0.0833436, antiderivative size = 176, 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[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 = 0.15052, size = 143, normalized size = 0.81 \[ \frac{\frac{\sqrt{i a+i b x+1} \left (a^2+a b x-5 i b x+1\right )}{x \sqrt{-i (a+b x+i)}}+\frac{6 (a-i) b \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right )}{\sqrt{-1+i a} \sqrt{1+i a}}}{(a+i)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((3*I)*ArcTan[a + b*x])/x^2,x]

[Out]

((Sqrt[1 + I*a + I*b*x]*(1 + a^2 - (5*I)*b*x + a*b*x))/(x*Sqrt[(-I)*(I + a + b*x)]) + (6*(-I + a)*b*ArcTan[Sqr
t[(-I)*(I + a + b*x)]/(Sqrt[(I + a)/(-I + a)]*Sqrt[1 + I*a + I*b*x])])/(Sqrt[-1 + I*a]*Sqrt[1 + I*a]))/(I + a)
^2

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Maple [B]  time = 0.117, size = 1358, normalized size = 7.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x*b^2*a^2-3*I/(a^2+1)/x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a+12*I*a^4*b/(a
^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-9*I*a^2*b/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3*I*a^6*b/(a^2+1)^2/(b
^2*x^2+2*a*b*x+a^2+1)^(1/2)-3*I*a^4*b/(a^2+1)^(5/2)*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*I*a^2*b/(a^2+1)^(5/2)*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/(a
^2+1)/x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a^3-9*a^4*b^2/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+I*b^2*a/(b^2*x^2
+2*a*b*x+a^2+1)^(1/2)*x+3*a^2*b^2/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+I*b/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-
1/(a^2+1)/x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+5*I/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a^4*b+3*I*a^2*b/(a^2+1)^(3
/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)-6*I*a*b^2*(2*b^2*x+2*a*b)/(4*b^2*(a^
2+1)-4*a^2*b^2)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+5*I/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x*b^2*a^3-9*I/(a^2+1)/
(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x*b^2*a+9*I*a^3*b^2/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-3*I*a^5*b^2/(a^2+1
)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-12*I/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a^2*b-2/(a^2+1)/(b^2*x^2+2*a*b*
x+a^2+1)^(1/2)*x*b^2-8/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a*b+3/(a^2+1)/x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a^2
-9*a^5*b/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-9*a^3*b/(a^2+1)^(5/2)*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)+12/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a^3*b+I*b*a^2/(b^2*x^2+2*a*b*x+a^2+1)^
(1/2)-3*a*b/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+12*a^3*b/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+6*a*b/(a^
2+1)^(3/2)*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)-6*b^2*(2*b^2*x+2*a*b)/(4*b^2*
(a^2+1)-4*a^2*b^2)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3*a*b/(a^2+1)^(5/2)*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*I*b/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3*I*b/(a^2+1)^(3/2)*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)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.40508, size = 1030, normalized size = 5.85 \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+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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (i a + i b x + 1\right )^{3}}{x^{2} \left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

undef