3.3.30 \(\int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx\) [230]

3.3.30.1 Optimal result
3.3.30.2 Mathematica [C] (verified)
3.3.30.3 Rubi [A] (verified)
3.3.30.4 Maple [F]
3.3.30.5 Fricas [B] (verification not implemented)
3.3.30.6 Sympy [F]
3.3.30.7 Maxima [F]
3.3.30.8 Giac [F(-2)]
3.3.30.9 Mupad [F(-1)]

3.3.30.1 Optimal result

Integrand size = 18, antiderivative size = 210 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=-\frac {(i-a-b x) \sqrt [4]{1-i (a+b x)}}{(i-a) x \sqrt [4]{1+i (a+b x)}}-\frac {i b \arctan \left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{(i-a)^{5/4} (i+a)^{3/4}}-\frac {i b \text {arctanh}\left (\frac {\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}\right )}{(i-a)^{5/4} (i+a)^{3/4}} \]

output
-(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)^(5/4)/( 
I+a)^(3/4)-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)^(5/4)/(I+a)^(3/4)
 
3.3.30.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.51 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=-\frac {\sqrt [4]{-i (i+a+b x)} \left (1+a^2+i b x+a b x-2 i b x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {1+a^2-i b x+a b x}{1+a^2+i b x+a b x}\right )\right )}{\left (1+a^2\right ) x \sqrt [4]{1+i a+i b x}} \]

input
Integrate[1/(E^((I/2)*ArcTan[a + b*x])*x^2),x]
 
output
-((((-I)*(I + a + b*x))^(1/4)*(1 + a^2 + I*b*x + a*b*x - (2*I)*b*x*Hyperge 
ometric2F1[1/4, 1, 5/4, (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)^(1/4)))
 
3.3.30.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5617, 817, 756, 218, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 5617

\(\displaystyle -8 i b \int \frac {1-i (a+b x)}{(i (a+b x)+1) \left (-i a-\frac {(i a+1) (1-i (a+b x))}{i (a+b x)+1}+1\right )^2}d\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i (a+b x)+1}}\)

\(\Big \downarrow \) 817

\(\displaystyle -8 i b \left (\frac {\sqrt [4]{1-i (a+b x)}}{4 (1+i a) \sqrt [4]{1+i (a+b x)} \left (-\frac {(1+i a) (1-i (a+b x))}{1+i (a+b x)}-i a+1\right )}-\frac {\int \frac {1}{-i a-\frac {(i a+1) (1-i (a+b x))}{i (a+b x)+1}+1}d\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i (a+b x)+1}}}{4 (1+i a)}\right )\)

\(\Big \downarrow \) 756

\(\displaystyle -8 i b \left (\frac {\sqrt [4]{1-i (a+b x)}}{4 (1+i a) \sqrt [4]{1+i (a+b x)} \left (-\frac {(1+i a) (1-i (a+b x))}{1+i (a+b x)}-i a+1\right )}-\frac {\frac {i \int \frac {1}{\sqrt {a+i}-\frac {\sqrt {i-a} \sqrt {1-i (a+b x)}}{\sqrt {i (a+b x)+1}}}d\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i (a+b x)+1}}}{2 \sqrt {a+i}}+\frac {i \int \frac {1}{\sqrt {a+i}+\frac {\sqrt {i-a} \sqrt {1-i (a+b x)}}{\sqrt {i (a+b x)+1}}}d\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i (a+b x)+1}}}{2 \sqrt {a+i}}}{4 (1+i a)}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle -8 i b \left (\frac {\sqrt [4]{1-i (a+b x)}}{4 (1+i a) \sqrt [4]{1+i (a+b x)} \left (-\frac {(1+i a) (1-i (a+b x))}{1+i (a+b x)}-i a+1\right )}-\frac {\frac {i \int \frac {1}{\sqrt {a+i}-\frac {\sqrt {i-a} \sqrt {1-i (a+b x)}}{\sqrt {i (a+b x)+1}}}d\frac {\sqrt [4]{1-i (a+b x)}}{\sqrt [4]{i (a+b x)+1}}}{2 \sqrt {a+i}}+\frac {i \arctan \left (\frac {\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}\right )}{2 \sqrt [4]{-a+i} (a+i)^{3/4}}}{4 (1+i a)}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle -8 i b \left (\frac {\sqrt [4]{1-i (a+b x)}}{4 (1+i a) \sqrt [4]{1+i (a+b x)} \left (-\frac {(1+i a) (1-i (a+b x))}{1+i (a+b x)}-i a+1\right )}-\frac {\frac {i \arctan \left (\frac {\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}\right )}{2 \sqrt [4]{-a+i} (a+i)^{3/4}}+\frac {i \text {arctanh}\left (\frac {\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}{\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}\right )}{2 \sqrt [4]{-a+i} (a+i)^{3/4}}}{4 (1+i a)}\right )\)

input
Int[1/(E^((I/2)*ArcTan[a + b*x])*x^2),x]
 
output
(-8*I)*b*((1 - I*(a + b*x))^(1/4)/(4*(1 + I*a)*(1 + I*(a + b*x))^(1/4)*(1 
- I*a - ((1 + I*a)*(1 - I*(a + b*x)))/(1 + I*(a + b*x)))) - (((I/2)*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)^(1/4)*(I + a)^(3/4)) + ((I/2)*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)^( 
1/4)*(I + a)^(3/4)))/(4*(1 + I*a)))
 

3.3.30.3.1 Defintions of rubi rules used

rule 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 221
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 756
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 817
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] - Simp[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 5617
Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_))*(x_)^(m_), x_Symbol] :> Simp 
[4/(I^m*n*b^(m + 1)*c^(m + 1))   Subst[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] && ILt 
Q[m, 0] && LtQ[-1, I*n, 1]
 
3.3.30.4 Maple [F]

\[\int \frac {1}{\sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}\, x^{2}}d x\]

input
int(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x^2,x)
 
output
int(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x^2,x)
 
3.3.30.5 Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 707 vs. \(2 (141) = 282\).

Time = 0.28 (sec) , antiderivative size = 707, normalized size of antiderivative = 3.37 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=\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^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} + {\left (a^{6} - 2 i \, a^{5} + a^{4} - 4 i \, a^{3} - a^{2} - 2 i \, a - 1\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 {3}{4}}}{b^{3}}\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^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - {\left (a^{6} - 2 i \, a^{5} + a^{4} - 4 i \, a^{3} - a^{2} - 2 i \, a - 1\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 {3}{4}}}{b^{3}}\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^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - {\left (i \, a^{6} + 2 \, a^{5} + i \, a^{4} + 4 \, a^{3} - i \, a^{2} + 2 \, a - i\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 {3}{4}}}{b^{3}}\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^{3} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}} - {\left (-i \, a^{6} - 2 \, a^{5} - i \, a^{4} - 4 \, a^{3} + i \, a^{2} - 2 \, a + i\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 {3}{4}}}{b^{3}}\right ) + 2 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \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} \]

input
integrate(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x^2,x, algorithm="fr 
icas")
 
output
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^3*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b* 
x + a + I)) + (a^6 - 2*I*a^5 + a^4 - 4*I*a^3 - a^2 - 2*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))^(3/4))/b^3) + 
 (-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^3*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a 
 + I)) - (a^6 - 2*I*a^5 + a^4 - 4*I*a^3 - a^2 - 2*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))^(3/4))/b^3) + (-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^3*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) 
- (I*a^6 + 2*a^5 + I*a^4 + 4*a^3 - I*a^2 + 2*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))^(3/4))/b^3) - (-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^3*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)) - (-I*a 
^6 - 2*a^5 - I*a^4 - 4*a^3 + I*a^2 - 2*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))^(3/4))/b^3) + 2*I*sqrt(b^2*x^2 
+ 2*a*b*x + a^2 + 1)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I 
)))/((a - I)*x)
 
3.3.30.6 Sympy [F]

\[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=\int \frac {1}{x^{2} \sqrt {\frac {i \left (a + b x - i\right )}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}}}\, dx \]

input
integrate(1/((1+I*(b*x+a))/(1+(b*x+a)**2)**(1/2))**(1/2)/x**2,x)
 
output
Integral(1/(x**2*sqrt(I*(a + b*x - I)/sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) 
)), x)
 
3.3.30.7 Maxima [F]

\[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=\int { \frac {1}{x^{2} \sqrt {\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}}} \,d x } \]

input
integrate(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x^2,x, algorithm="ma 
xima")
 
output
integrate(1/(x^2*sqrt((I*b*x + I*a + 1)/sqrt((b*x + a)^2 + 1))), x)
 
3.3.30.8 Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=\text {Exception raised: TypeError} \]

input
integrate(1/((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)/x^2,x, algorithm="gi 
ac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0,0 
,0]Warning, replacing 0 by 14, a substitution variable should perhaps be p 
urged.War
 
3.3.30.9 Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=\int \frac {1}{x^2\,\sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}}} \,d x \]

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