3.3.20 \(\int \frac {e^{\frac {1}{2} i \text {ArcTan}(a+b x)}}{x^2} \, dx\) [220]

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

[Out]

-(I+a+b*x)*(1+I*(b*x+a))^(1/4)/(I+a)/x/(1-I*(b*x+a))^(1/4)+I*b*arctan((I+a)^(1/4)*(1+I*(b*x+a))^(1/4)/(I-a)^(1
/4)/(1-I*(b*x+a))^(1/4))/(I-a)^(3/4)/(I+a)^(5/4)+I*b*arctanh((I+a)^(1/4)*(1+I*(b*x+a))^(1/4)/(I-a)^(1/4)/(1-I*
(b*x+a))^(1/4))/(I-a)^(3/4)/(I+a)^(5/4)

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Rubi [A]
time = 0.11, antiderivative size = 205, normalized size of antiderivative = 1.00, 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 = {5202, 269, 294, 218, 214, 211} \begin {gather*} \frac {i b \text {ArcTan}\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 {\sqrt [4]{1+i (a+b x)} (a+b x+i)}{(a+i) x \sqrt [4]{1-i (a+b x)}}+\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}} \end {gather*}

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 211

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

Rule 214

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 269

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 294

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 5202

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]

Rubi steps

\begin {align*} \int \frac {e^{\frac {1}{2} i \tan ^{-1}(a+b x)}}{x^2} \, dx &=(8 i b) \text {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) \text {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) \text {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 \text {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 \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 110, normalized size = 0.54 \begin {gather*} \frac {(-i (i+a+b x))^{3/4} \left (3 (i+a) (-i+a+b x)+2 i b x \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {1+a^2-i b x+a b x}{1+a^2+i b x+a b x}\right )\right )}{3 (i+a)^2 x (1+i a+i b x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}}{x^{2}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 598 vs. \(2 (141) = 282\).
time = 3.26, size = 598, normalized size = 2.92 \begin {gather*} \frac {\left (-\frac {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}\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}} + \left (-\frac {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}\right )^{\frac {1}{4}} {\left (a^{2} + 1\right )}}{b}\right ) + \left (-\frac {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}\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}} - \left (-\frac {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}\right )^{\frac {1}{4}} {\left (a^{2} + 1\right )}}{b}\right ) - \left (-\frac {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}\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}}{a^{8} + 2 i \, a^{7} + 2 \, a^{6} + 6 i \, a^{5} + 6 i \, a^{3} - 2 \, a^{2} + 2 i \, a - 1}\right )^{\frac {1}{4}} {\left (i \, a^{2} + i\right )}}{b}\right ) + \left (-\frac {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}\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}}{a^{8} + 2 i \, a^{7} + 2 \, a^{6} + 6 i \, a^{5} + 6 i \, a^{3} - 2 \, a^{2} + 2 i \, a - 1}\right )^{\frac {1}{4}} {\left (-i \, a^{2} - i\right )}}{b}\right ) - 2 \, {\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}}}{2 \, {\left (a + i\right )} x} \end {gather*}

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]

1/2*((-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))^(1/4)*(-I*a + 1)*x*log((b*sqrt(I*s
qrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) + (-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))^(1/4)*(a^2 + 1))/b) + (-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))^(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)) - (-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))^(1/4)*(a^2 + 1))/b) - (-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))^(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/(a
^8 + 2*I*a^7 + 2*a^6 + 6*I*a^5 + 6*I*a^3 - 2*a^2 + 2*I*a - 1))^(1/4)*(I*a^2 + I))/b) + (-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))^(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/(a^8 + 2*I*a^7 + 2*a^6 + 6*I*a^5 + 6*I*a^3 - 2*a^2 + 2*I*a - 1))^(1/4)*(-I*a^2 - I))/
b) - 2*(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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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