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

Optimal. Leaf size=205 \[ -\frac{\sqrt [4]{1+i (a+b x)} (a+b x+i)}{(a+i) x \sqrt [4]{1-i (a+b x)}}+\frac{i b \tan ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{(-a+i)^{3/4} (a+i)^{5/4}}+\frac{i b \tanh ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{(-a+i)^{3/4} (a+i)^{5/4}} \]

[Out]

-(((I + a + b*x)*(1 + I*(a + b*x))^(1/4))/((I + a)*x*(1 - I*(a + b*x))^(1/4))) + (I*b*ArcTan[((I + a)^(1/4)*(1
+ I*(a + b*x))^(1/4))/((I - a)^(1/4)*(1 - I*(a + b*x))^(1/4))])/((I - a)^(3/4)*(I + a)^(5/4)) + (I*b*ArcTanh[
((I + a)^(1/4)*(1 + I*(a + b*x))^(1/4))/((I - a)^(1/4)*(1 - I*(a + b*x))^(1/4))])/((I - a)^(3/4)*(I + a)^(5/4)
)

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Rubi [A]  time = 0.102695, antiderivative size = 205, 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 = {5094, 263, 288, 212, 208, 205} \[ -\frac{\sqrt [4]{1+i (a+b x)} (a+b x+i)}{(a+i) x \sqrt [4]{1-i (a+b x)}}+\frac{i b \tan ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{(-a+i)^{3/4} (a+i)^{5/4}}+\frac{i b \tanh ^{-1}\left (\frac{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{(-a+i)^{3/4} (a+i)^{5/4}} \]

Antiderivative was successfully verified.

[In]

Int[E^((I/2)*ArcTan[a + b*x])/x^2,x]

[Out]

-(((I + a + b*x)*(1 + I*(a + b*x))^(1/4))/((I + a)*x*(1 - I*(a + b*x))^(1/4))) + (I*b*ArcTan[((I + a)^(1/4)*(1
+ I*(a + b*x))^(1/4))/((I - a)^(1/4)*(1 - I*(a + b*x))^(1/4))])/((I - a)^(3/4)*(I + a)^(5/4)) + (I*b*ArcTanh[
((I + a)^(1/4)*(1 + I*(a + b*x))^(1/4))/((I - a)^(1/4)*(1 - I*(a + b*x))^(1/4))])/((I - a)^(3/4)*(I + a)^(5/4)
)

Rule 5094

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_))*(x_)^(m_), x_Symbol] :> Dist[4/(I^m*n*b^(m + 1)*c^(m + 1)), Sub
st[Int[(x^(2/(I*n))*(1 - I*a*c - (1 + I*a*c)*x^(2/(I*n)))^m)/(1 + x^(2/(I*n)))^(m + 2), x], x, (1 - I*c*(a + b
*x))^((I*n)/2)/(1 + I*c*(a + b*x))^((I*n)/2)], x] /; FreeQ[{a, b, c}, x] && ILtQ[m, 0] && LtQ[-1, I*n, 1]

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, 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{1}{2} i \tan ^{-1}(a+b x)}}{x^2} \, dx &=(8 i b) \operatorname{Subst}\left (\int \frac{1}{\left (1-i a-\frac{1+i a}{x^4}\right )^2 x^4} \, dx,x,\frac{\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )\\ &=(8 i b) \operatorname{Subst}\left (\int \frac{x^4}{\left (-1-i a+(1-i a) x^4\right )^2} \, dx,x,\frac{\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )\\ &=-\frac{(i+a+b x) \sqrt [4]{1+i (a+b x)}}{(i+a) x \sqrt [4]{1-i (a+b x)}}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{-1-i a+(1-i a) x^4} \, dx,x,\frac{\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{i+a}\\ &=-\frac{(i+a+b x) \sqrt [4]{1+i (a+b x)}}{(i+a) x \sqrt [4]{1-i (a+b x)}}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{i-a}-\sqrt{i+a} x^2} \, dx,x,\frac{\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt{i-a} (1-i a)}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{i-a}+\sqrt{i+a} x^2} \, dx,x,\frac{\sqrt [4]{1+i (a+b x)}}{\sqrt [4]{1-i (a+b x)}}\right )}{\sqrt{i-a} (1-i a)}\\ &=-\frac{(i+a+b x) \sqrt [4]{1+i (a+b x)}}{(i+a) x \sqrt [4]{1-i (a+b x)}}+\frac{i b \tan ^{-1}\left (\frac{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{(i-a)^{3/4} (i+a)^{5/4}}+\frac{i b \tanh ^{-1}\left (\frac{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{(i-a)^{3/4} (i+a)^{5/4}}\\ \end{align*}

Mathematica [C]  time = 0.0268317, size = 110, normalized size = 0.54 \[ \frac{(-i (a+b x+i))^{3/4} \left (2 i b x \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};\frac{a^2+b x a-i b x+1}{a^2+b x a+i b x+1}\right )+3 (a+i) (a+b x-i)\right )}{3 (a+i)^2 x (i a+i b x+1)^{3/4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.227, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}}\sqrt{{(1+i \left ( bx+a \right ) ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}}}}\, dx \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))^(1/2)/x^2,x)

[Out]

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

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

[Out]

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

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Fricas [B]  time = 2.37556, size = 1575, normalized size = 7.68 \begin{align*} \frac{\left (-\frac{b^{4}}{16 \, a^{8} + 32 i \, a^{7} + 32 \, a^{6} + 96 i \, a^{5} + 96 i \, a^{3} - 32 \, a^{2} + 32 i \, a - 16}\right )^{\frac{1}{4}}{\left (-i \, a + 1\right )} x \log \left (\frac{b \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + 2 \, \left (-\frac{b^{4}}{16 \, a^{8} + 32 i \, a^{7} + 32 \, a^{6} + 96 i \, a^{5} + 96 i \, a^{3} - 32 \, a^{2} + 32 i \, a - 16}\right )^{\frac{1}{4}}{\left (a^{2} + 1\right )}}{b}\right ) + \left (-\frac{b^{4}}{16 \, a^{8} + 32 i \, a^{7} + 32 \, a^{6} + 96 i \, a^{5} + 96 i \, a^{3} - 32 \, a^{2} + 32 i \, a - 16}\right )^{\frac{1}{4}}{\left (i \, a - 1\right )} x \log \left (\frac{b \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - 2 \, \left (-\frac{b^{4}}{16 \, a^{8} + 32 i \, a^{7} + 32 \, a^{6} + 96 i \, a^{5} + 96 i \, a^{3} - 32 \, a^{2} + 32 i \, a - 16}\right )^{\frac{1}{4}}{\left (a^{2} + 1\right )}}{b}\right ) + \left (-\frac{b^{4}}{16 \, a^{8} + 32 i \, a^{7} + 32 \, a^{6} + 96 i \, a^{5} + 96 i \, a^{3} - 32 \, a^{2} + 32 i \, a - 16}\right )^{\frac{1}{4}}{\left (a + i\right )} x \log \left (\frac{b \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + \left (-\frac{b^{4}}{16 \, a^{8} + 32 i \, a^{7} + 32 \, a^{6} + 96 i \, a^{5} + 96 i \, a^{3} - 32 \, a^{2} + 32 i \, a - 16}\right )^{\frac{1}{4}}{\left (2 i \, a^{2} + 2 i\right )}}{b}\right ) - \left (-\frac{b^{4}}{16 \, a^{8} + 32 i \, a^{7} + 32 \, a^{6} + 96 i \, a^{5} + 96 i \, a^{3} - 32 \, a^{2} + 32 i \, a - 16}\right )^{\frac{1}{4}}{\left (a + i\right )} x \log \left (\frac{b \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + \left (-\frac{b^{4}}{16 \, a^{8} + 32 i \, a^{7} + 32 \, a^{6} + 96 i \, a^{5} + 96 i \, a^{3} - 32 \, a^{2} + 32 i \, a - 16}\right )^{\frac{1}{4}}{\left (-2 i \, a^{2} - 2 i\right )}}{b}\right ) -{\left (b x + a + i\right )} \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{{\left (a + i\right )} x} \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))^(1/2)/x^2,x, algorithm="fricas")

[Out]

((-b^4/(16*a^8 + 32*I*a^7 + 32*a^6 + 96*I*a^5 + 96*I*a^3 - 32*a^2 + 32*I*a - 16))^(1/4)*(-I*a + 1)*x*log((b*sq
rt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + 2*(-b^4/(16*a^8 + 32*I*a^7 + 32*a^6 + 96*I*a^5 + 96*I*
a^3 - 32*a^2 + 32*I*a - 16))^(1/4)*(a^2 + 1))/b) + (-b^4/(16*a^8 + 32*I*a^7 + 32*a^6 + 96*I*a^5 + 96*I*a^3 - 3
2*a^2 + 32*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 + 32*I*a^7 + 32*a^6 + 96*I*a^5 + 96*I*a^3 - 32*a^2 + 32*I*a - 16))^(1/4)*(a^2 + 1))/b) + (-b^4/(16*
a^8 + 32*I*a^7 + 32*a^6 + 96*I*a^5 + 96*I*a^3 - 32*a^2 + 32*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 + 32*I*a^7 + 32*a^6 + 96*I*a^5 + 96*I*a^3 - 32*a^2 + 3
2*I*a - 16))^(1/4)*(2*I*a^2 + 2*I))/b) - (-b^4/(16*a^8 + 32*I*a^7 + 32*a^6 + 96*I*a^5 + 96*I*a^3 - 32*a^2 + 32
*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 + 3
2*I*a^7 + 32*a^6 + 96*I*a^5 + 96*I*a^3 - 32*a^2 + 32*I*a - 16))^(1/4)*(-2*I*a^2 - 2*I))/b) - (b*x + a + I)*sqr
t(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+I*(b*x+a))/(1+(b*x+a)**2)**(1/2))**(1/2)/x**2,x)

[Out]

Timed out

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

[Out]

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