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

Optimal. Leaf size=201 \[ -\frac{\sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{2 \left (a^2+1\right ) x^2}+\frac{(1+2 i a) b^2 \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{3/2} (a+i)^{5/2}}-\frac{(1+2 i a) b \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{2 (-a+i) (a+i)^2 x} \]

[Out]

-((1 + (2*I)*a)*b*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(2*(I - a)*(I + a)^2*x) - (Sqrt[1 - I*a - I*b*x
]*(1 + I*a + I*b*x)^(3/2))/(2*(1 + a^2)*x^2) + ((1 + (2*I)*a)*b^2*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/
(Sqrt[I - a]*Sqrt[1 - I*a - I*b*x])])/((I - a)^(3/2)*(I + a)^(5/2))

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Rubi [A]  time = 0.137747, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {5095, 96, 94, 93, 208} \[ -\frac{\sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{2 \left (a^2+1\right ) x^2}+\frac{(1+2 i a) b^2 \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{3/2} (a+i)^{5/2}}-\frac{(1+2 i a) b \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{2 (-a+i) (a+i)^2 x} \]

Antiderivative was successfully verified.

[In]

Int[E^(I*ArcTan[a + b*x])/x^3,x]

[Out]

-((1 + (2*I)*a)*b*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(2*(I - a)*(I + a)^2*x) - (Sqrt[1 - I*a - I*b*x
]*(1 + I*a + I*b*x)^(3/2))/(2*(1 + a^2)*x^2) + ((1 + (2*I)*a)*b^2*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/
(Sqrt[I - a]*Sqrt[1 - I*a - I*b*x])])/((I - a)^(3/2)*(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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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^{i \tan ^{-1}(a+b x)}}{x^3} \, dx &=\int \frac{\sqrt{1+i a+i b x}}{x^3 \sqrt{1-i a-i b x}} \, dx\\ &=-\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{2 \left (1+a^2\right ) x^2}+\frac{((i-2 a) b) \int \frac{\sqrt{1+i a+i b x}}{x^2 \sqrt{1-i a-i b x}} \, dx}{2 \left (1+a^2\right )}\\ &=-\frac{(i-2 a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{2 (1-i a) \left (1+a^2\right ) x}-\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{2 \left (1+a^2\right ) x^2}-\frac{\left ((i-2 a) b^2\right ) \int \frac{1}{x \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{2 (i+a) \left (1+a^2\right )}\\ &=-\frac{(i-2 a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{2 (1-i a) \left (1+a^2\right ) x}-\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{2 \left (1+a^2\right ) x^2}-\frac{\left ((i-2 a) b^2\right ) \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) \left (1+a^2\right )}\\ &=-\frac{(i-2 a) b \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{2 (1-i a) \left (1+a^2\right ) x}-\frac{\sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{2 \left (1+a^2\right ) x^2}+\frac{(1+2 i a) b^2 \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)^{3/2} (i+a)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.118302, size = 149, normalized size = 0.74 \[ \frac{\frac{2 (2 a-i) b^2 \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right )}{\sqrt{-1+i a} \sqrt{1+i a}}-\frac{i \left (a^2-a b x+2 i b x+1\right ) \sqrt{a^2+2 a b x+b^2 x^2+1}}{x^2}}{2 (a-i) (a+i)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.113, size = 405, normalized size = 2. \begin{align*}{\frac{-{\frac{i}{2}}a}{ \left ({a}^{2}+1 \right ){x}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{1}{ \left ( 2\,{a}^{2}+2 \right ){x}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{\frac{3\,i}{2}}{a}^{2}b}{ \left ({a}^{2}+1 \right ) ^{2}x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{3\,ab}{2\, \left ({a}^{2}+1 \right ) ^{2}x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{{\frac{3\,i}{2}}{a}^{3}{b}^{2}\ln \left ({\frac{1}{x} \left ( 2\,{a}^{2}+2+2\,xab+2\,\sqrt{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{5}{2}}}}-{\frac{3\,{a}^{2}{b}^{2}}{2}\ln \left ({\frac{1}{x} \left ( 2\,{a}^{2}+2+2\,xab+2\,\sqrt{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{5}{2}}}}+{{\frac{3\,i}{2}}{b}^{2}a\ln \left ({\frac{1}{x} \left ( 2\,{a}^{2}+2+2\,xab+2\,\sqrt{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{{b}^{2}}{2}\ln \left ({\frac{1}{x} \left ( 2\,{a}^{2}+2+2\,xab+2\,\sqrt{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{\frac{ib}{ \left ({a}^{2}+1 \right ) x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.83042, size = 1103, normalized size = 5.49 \begin{align*} \frac{{\left (i \, a + 2\right )} b^{2} x^{2} + \sqrt{\frac{{\left (4 \, a^{2} - 4 i \, a - 1\right )} b^{4}}{a^{8} + 2 i \, a^{7} + 2 \, a^{6} + 6 i \, a^{5} + 6 i \, a^{3} - 2 \, a^{2} + 2 i \, a - 1}}{\left (a^{3} + i \, a^{2} + a + i\right )} x^{2} \log \left (-\frac{{\left (2 \, a - i\right )} b^{3} x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (2 \, a - i\right )} b^{2} +{\left (a^{5} + i \, a^{4} + 2 \, a^{3} + 2 i \, a^{2} + a + i\right )} \sqrt{\frac{{\left (4 \, a^{2} - 4 i \, a - 1\right )} b^{4}}{a^{8} + 2 i \, a^{7} + 2 \, a^{6} + 6 i \, a^{5} + 6 i \, a^{3} - 2 \, a^{2} + 2 i \, a - 1}}}{{\left (2 \, a - i\right )} b^{2}}\right ) - \sqrt{\frac{{\left (4 \, a^{2} - 4 i \, a - 1\right )} b^{4}}{a^{8} + 2 i \, a^{7} + 2 \, a^{6} + 6 i \, a^{5} + 6 i \, a^{3} - 2 \, a^{2} + 2 i \, a - 1}}{\left (a^{3} + i \, a^{2} + a + i\right )} x^{2} \log \left (-\frac{{\left (2 \, a - i\right )} b^{3} x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (2 \, a - i\right )} b^{2} -{\left (a^{5} + i \, a^{4} + 2 \, a^{3} + 2 i \, a^{2} + a + i\right )} \sqrt{\frac{{\left (4 \, a^{2} - 4 i \, a - 1\right )} b^{4}}{a^{8} + 2 i \, a^{7} + 2 \, a^{6} + 6 i \, a^{5} + 6 i \, a^{3} - 2 \, a^{2} + 2 i \, a - 1}}}{{\left (2 \, a - i\right )} b^{2}}\right ) + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left ({\left (i \, a + 2\right )} b x - i \, a^{2} - i\right )}}{{\left (2 \, a^{3} + 2 i \, a^{2} + 2 \, a + 2 i\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((I*a + 2)*b^2*x^2 + sqrt((4*a^2 - 4*I*a - 1)*b^4/(a^8 + 2*I*a^7 + 2*a^6 + 6*I*a^5 + 6*I*a^3 - 2*a^2 + 2*I*a -
1))*(a^3 + I*a^2 + a + I)*x^2*log(-((2*a - I)*b^3*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a - I)*b^2 + (a^5
+ I*a^4 + 2*a^3 + 2*I*a^2 + a + I)*sqrt((4*a^2 - 4*I*a - 1)*b^4/(a^8 + 2*I*a^7 + 2*a^6 + 6*I*a^5 + 6*I*a^3 - 2
*a^2 + 2*I*a - 1)))/((2*a - I)*b^2)) - sqrt((4*a^2 - 4*I*a - 1)*b^4/(a^8 + 2*I*a^7 + 2*a^6 + 6*I*a^5 + 6*I*a^3
- 2*a^2 + 2*I*a - 1))*(a^3 + I*a^2 + a + I)*x^2*log(-((2*a - I)*b^3*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*
a - I)*b^2 - (a^5 + I*a^4 + 2*a^3 + 2*I*a^2 + a + I)*sqrt((4*a^2 - 4*I*a - 1)*b^4/(a^8 + 2*I*a^7 + 2*a^6 + 6*I
*a^5 + 6*I*a^3 - 2*a^2 + 2*I*a - 1)))/((2*a - I)*b^2)) + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*((I*a + 2)*b*x - I*
a^2 - I))/((2*a^3 + 2*I*a^2 + 2*a + 2*I)*x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i a + i b x + 1}{x^{3} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError