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

Optimal. Leaf size=211 \[ -\frac{(-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{(1+i a) x}-\frac{3 i b \tan ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{(-a+i)^{7/4} \sqrt [4]{a+i}}-\frac{3 i b \tanh ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{(-a+i)^{7/4} \sqrt [4]{a+i}} \]

[Out]

-(((1 - I*a - I*b*x)^(3/4)*(1 + I*a + I*b*x)^(1/4))/((1 + I*a)*x)) - ((3*I)*b*ArcTan[((I + a)^(1/4)*(1 + I*a +
I*b*x)^(1/4))/((I - a)^(1/4)*(1 - I*a - I*b*x)^(1/4))])/((I - a)^(7/4)*(I + a)^(1/4)) - ((3*I)*b*ArcTanh[((I
+ a)^(1/4)*(1 + I*a + I*b*x)^(1/4))/((I - a)^(1/4)*(1 - I*a - I*b*x)^(1/4))])/((I - a)^(7/4)*(I + a)^(1/4))

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Rubi [A]  time = 0.118121, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5095, 94, 93, 212, 208, 205} \[ -\frac{(-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{(1+i a) x}-\frac{3 i b \tan ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{(-a+i)^{7/4} \sqrt [4]{a+i}}-\frac{3 i b \tanh ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{(-a+i)^{7/4} \sqrt [4]{a+i}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(((3*I)/2)*ArcTan[a + b*x])*x^2),x]

[Out]

-(((1 - I*a - I*b*x)^(3/4)*(1 + I*a + I*b*x)^(1/4))/((1 + I*a)*x)) - ((3*I)*b*ArcTan[((I + a)^(1/4)*(1 + I*a +
I*b*x)^(1/4))/((I - a)^(1/4)*(1 - I*a - I*b*x)^(1/4))])/((I - a)^(7/4)*(I + a)^(1/4)) - ((3*I)*b*ArcTanh[((I
+ a)^(1/4)*(1 + I*a + I*b*x)^(1/4))/((I - a)^(1/4)*(1 - I*a - I*b*x)^(1/4))])/((I - a)^(7/4)*(I + a)^(1/4))

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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

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]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{e^{-\frac{3}{2} i \tan ^{-1}(a+b x)}}{x^2} \, dx &=\int \frac{(1-i a-i b x)^{3/4}}{x^2 (1+i a+i b x)^{3/4}} \, dx\\ &=-\frac{(1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{(1+i a) x}+\frac{(3 b) \int \frac{1}{x \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}} \, dx}{2 (i-a)}\\ &=-\frac{(1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{(1+i a) x}+\frac{(6 b) \operatorname{Subst}\left (\int \frac{1}{-1-i a-(-1+i a) x^4} \, dx,x,\frac{\sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}}\right )}{i-a}\\ &=-\frac{(1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{(1+i a) x}-\frac{(3 i b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{i-a}-\sqrt{i+a} x^2} \, dx,x,\frac{\sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}}\right )}{(i-a)^{3/2}}-\frac{(3 i b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{i-a}+\sqrt{i+a} x^2} \, dx,x,\frac{\sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}}\right )}{(i-a)^{3/2}}\\ &=-\frac{(1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{(1+i a) x}-\frac{3 i b \tan ^{-1}\left (\frac{\sqrt [4]{i+a} \sqrt [4]{1+i a+i b x}}{\sqrt [4]{i-a} \sqrt [4]{1-i a-i b x}}\right )}{(i-a)^{7/4} \sqrt [4]{i+a}}-\frac{3 i b \tanh ^{-1}\left (\frac{\sqrt [4]{i+a} \sqrt [4]{1+i a+i b x}}{\sqrt [4]{i-a} \sqrt [4]{1-i a-i b x}}\right )}{(i-a)^{7/4} \sqrt [4]{i+a}}\\ \end{align*}

Mathematica [C]  time = 0.0258798, size = 107, normalized size = 0.51 \[ -\frac{(-i (a+b x+i))^{3/4} \left (-2 i b x \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};\frac{a^2+b x a-i b x+1}{a^2+b x a+i b x+1}\right )+a^2+a b x+i b x+1\right )}{\left (a^2+1\right ) x (i a+i b x+1)^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(((3*I)/2)*ArcTan[a + b*x])*x^2),x]

[Out]

-((((-I)*(I + a + b*x))^(3/4)*(1 + a^2 + I*b*x + a*b*x - (2*I)*b*x*Hypergeometric2F1[3/4, 1, 7/4, (1 + a^2 - I
*b*x + a*b*x)/(1 + a^2 + I*b*x + a*b*x)]))/((1 + a^2)*x*(1 + I*a + I*b*x)^(3/4)))

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Maple [F]  time = 0.232, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \left ({(1+i \left ( bx+a \right ) ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}} \right ) ^{-{\frac{3}{2}}}}\, dx \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))^(3/2)/x^2,x)

[Out]

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

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

[Out]

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

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Fricas [B]  time = 2.62222, size = 1667, normalized size = 7.9 \begin{align*} \text{result too large to display} \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))^(3/2)/x^2,x, algorithm="fricas")

[Out]

(3*(-b^4/(16*a^8 - 96*I*a^7 - 224*a^6 + 224*I*a^5 + 224*I*a^3 + 224*a^2 - 96*I*a - 16))^(1/4)*(-I*a - 1)*x*log
((b*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + 2*(-b^4/(16*a^8 - 96*I*a^7 - 224*a^6 + 224*I*a^5
+ 224*I*a^3 + 224*a^2 - 96*I*a - 16))^(1/4)*(a^2 - 2*I*a - 1))/b) + 3*(-b^4/(16*a^8 - 96*I*a^7 - 224*a^6 + 22
4*I*a^5 + 224*I*a^3 + 224*a^2 - 96*I*a - 16))^(1/4)*(I*a + 1)*x*log((b*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1
)/(b*x + a + I)) - 2*(-b^4/(16*a^8 - 96*I*a^7 - 224*a^6 + 224*I*a^5 + 224*I*a^3 + 224*a^2 - 96*I*a - 16))^(1/4
)*(a^2 - 2*I*a - 1))/b) + 3*(-b^4/(16*a^8 - 96*I*a^7 - 224*a^6 + 224*I*a^5 + 224*I*a^3 + 224*a^2 - 96*I*a - 16
))^(1/4)*(a - I)*x*log((b*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (-b^4/(16*a^8 - 96*I*a^7 -
224*a^6 + 224*I*a^5 + 224*I*a^3 + 224*a^2 - 96*I*a - 16))^(1/4)*(2*I*a^2 + 4*a - 2*I))/b) - 3*(-b^4/(16*a^8 -
96*I*a^7 - 224*a^6 + 224*I*a^5 + 224*I*a^3 + 224*a^2 - 96*I*a - 16))^(1/4)*(a - I)*x*log((b*sqrt(I*sqrt(b^2*x
^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (-b^4/(16*a^8 - 96*I*a^7 - 224*a^6 + 224*I*a^5 + 224*I*a^3 + 224*a^2
- 96*I*a - 16))^(1/4)*(-2*I*a^2 - 4*a + 2*I))/b) + (b*x + a + I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x
+ a + I)))/((a - 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))/(1+(b*x+a)**2)**(1/2))**(3/2)/x**2,x)

[Out]

Timed out

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

[Out]

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