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

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

[Out]

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

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Rubi [A]  time = 0.117133, 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} \[ -\frac{(-i a-i b x+1)^{3/2} \sqrt{i a+i b x+1}}{2 \left (a^2+1\right ) x^2}+\frac{(1-2 i a) b^2 \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{5/2} (a+i)^{3/2}}+\frac{(1-2 i a) b \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{2 (-a+i)^2 (a+i) x} \]

Antiderivative was successfully verified.

[In]

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

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

Mathematica [A]  time = 0.111014, size = 149, normalized size = 0.74 \[ \frac{\frac{i \left (a^2-a b x-2 i b x+1\right ) \sqrt{a^2+2 a b x+b^2 x^2+1}}{x^2}+\frac{2 (2 a+i) b^2 \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right )}{\sqrt{-1+i a} \sqrt{1+i a}}}{2 (a-i)^2 (a+i)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(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)^2*(
I + a))

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Maple [B]  time = 0.124, size = 1146, normalized size = 5.7 \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))*(1+(b*x+a)^2)^(1/2)/x^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{{\left (b x + a\right )}^{2} + 1}}{{\left (i \, b x + i \, a + 1\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.36476, size = 1106, normalized size = 5.5 \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/(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{\sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{x^{3} \left (i a + i b x + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: NotImplementedError