3.3.25 \(\int \frac {e^{\frac {3}{2} i \arctan (a+b x)}}{x^2} \, dx\) [225]

3.3.25.1 Optimal result
3.3.25.2 Mathematica [C] (verified)
3.3.25.3 Rubi [A] (verified)
3.3.25.4 Maple [F]
3.3.25.5 Fricas [B] (verification not implemented)
3.3.25.6 Sympy [F(-1)]
3.3.25.7 Maxima [F]
3.3.25.8 Giac [F(-2)]
3.3.25.9 Mupad [F(-1)]

3.3.25.1 Optimal result

Integrand size = 18, antiderivative size = 211 \[ \int \frac {e^{\frac {3}{2} i \arctan (a+b x)}}{x^2} \, dx=-\frac {\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{(1-i a) x}-\frac {3 i b \arctan \left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i a+i b x}}{\sqrt [4]{i-a} \sqrt [4]{1-i a-i b x}}\right )}{\sqrt [4]{i-a} (i+a)^{7/4}}+\frac {3 i b \text {arctanh}\left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i a+i b x}}{\sqrt [4]{i-a} \sqrt [4]{1-i a-i b x}}\right )}{\sqrt [4]{i-a} (i+a)^{7/4}} \]

output
-(1-I*a-I*b*x)^(1/4)*(1+I*a+I*b*x)^(3/4)/(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)^(1/4)/(I+a)^( 
7/4)+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)^(1/4)/(I+a)^(7/4)
 
3.3.25.2 Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.50 \[ \int \frac {e^{\frac {3}{2} i \arctan (a+b x)}}{x^2} \, dx=\frac {\sqrt [4]{-i (i+a+b x)} \left (1+a^2+i b x+a b x+6 i b x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {1+a^2-i b x+a b x}{1+a^2+i b x+a b x}\right )\right )}{(i+a)^2 x \sqrt [4]{1+i a+i b x}} \]

input
Integrate[E^(((3*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 + (6*I)*b*x*Hypergeom 
etric2F1[1/4, 1, 5/4, (1 + a^2 - I*b*x + a*b*x)/(1 + a^2 + I*b*x + a*b*x)] 
))/((I + a)^2*x*(1 + I*a + I*b*x)^(1/4))
 
3.3.25.3 Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5618, 105, 104, 25, 827, 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.

\(\displaystyle \int \frac {e^{\frac {3}{2} i \arctan (a+b x)}}{x^2} \, dx\)

\(\Big \downarrow \) 5618

\(\displaystyle \int \frac {(i a+i b x+1)^{3/4}}{x^2 (-i a-i b x+1)^{3/4}}dx\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {3 b \int \frac {1}{x (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}dx}{2 (a+i)}-\frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{(1-i a) x}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {6 b \int -\frac {\sqrt {i a+i b x+1}}{\sqrt {-i a-i b x+1} \left (i a-\frac {(1-i a) (i a+i b x+1)}{-i a-i b x+1}+1\right )}d\frac {\sqrt [4]{i a+i b x+1}}{\sqrt [4]{-i a-i b x+1}}}{a+i}-\frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{(1-i a) x}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {6 b \int \frac {\sqrt {i a+i b x+1}}{\sqrt {-i a-i b x+1} \left (i a-\frac {(1-i a) (i a+i b x+1)}{-i a-i b x+1}+1\right )}d\frac {\sqrt [4]{i a+i b x+1}}{\sqrt [4]{-i a-i b x+1}}}{a+i}-\frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{(1-i a) x}\)

\(\Big \downarrow \) 827

\(\displaystyle -\frac {6 b \left (\frac {i \int \frac {1}{\sqrt {i-a}+\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-i a-i b x+1}}}d\frac {\sqrt [4]{i a+i b x+1}}{\sqrt [4]{-i a-i b x+1}}}{2 \sqrt {a+i}}-\frac {i \int \frac {1}{\sqrt {i-a}-\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-i a-i b x+1}}}d\frac {\sqrt [4]{i a+i b x+1}}{\sqrt [4]{-i a-i b x+1}}}{2 \sqrt {a+i}}\right )}{a+i}-\frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{(1-i a) x}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {6 b \left (\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{2 \sqrt [4]{-a+i} (a+i)^{3/4}}-\frac {i \int \frac {1}{\sqrt {i-a}-\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-i a-i b x+1}}}d\frac {\sqrt [4]{i a+i b x+1}}{\sqrt [4]{-i a-i b x+1}}}{2 \sqrt {a+i}}\right )}{a+i}-\frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{(1-i a) x}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {6 b \left (\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{2 \sqrt [4]{-a+i} (a+i)^{3/4}}-\frac {i \text {arctanh}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{i a+i b x+1}}{\sqrt [4]{-a+i} \sqrt [4]{-i a-i b x+1}}\right )}{2 \sqrt [4]{-a+i} (a+i)^{3/4}}\right )}{a+i}-\frac {\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}{(1-i a) x}\)

input
Int[E^(((3*I)/2)*ArcTan[a + b*x])/x^2,x]
 
output
-(((1 - I*a - I*b*x)^(1/4)*(1 + I*a + I*b*x)^(3/4))/((1 - I*a)*x)) - (6*b* 
(((I/2)*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)^(1/4)*(I + a)^(3/4)) - ((I/2)*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)^(1/4)*(I + a)^(3/4))))/(I + a)
 

3.3.25.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 104
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[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] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 105
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[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[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

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 827
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 5618
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]
 
3.3.25.4 Maple [F]

\[\int \frac {{\left (\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}^{\frac {3}{2}}}{x^{2}}d x\]

input
int(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x^2,x)
 
output
int(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x^2,x)
 
3.3.25.5 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. 694 vs. \(2 (137) = 274\).

Time = 0.27 (sec) , antiderivative size = 694, normalized size of antiderivative = 3.29 \[ \int \frac {e^{\frac {3}{2} i \arctan (a+b x)}}{x^2} \, dx=\frac {3 \, \left (-\frac {b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 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} + 4 i \, a^{5} - 5 \, a^{4} - 5 \, a^{2} - 4 i \, a + 1\right )} \left (-\frac {b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 i \, a - 1}\right )^{\frac {3}{4}}}{b^{3}}\right ) + 3 \, \left (-\frac {b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 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} + 4 i \, a^{5} - 5 \, a^{4} - 5 \, a^{2} - 4 i \, a + 1\right )} \left (-\frac {b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 i \, a - 1}\right )^{\frac {3}{4}}}{b^{3}}\right ) + 3 \, \left (-\frac {b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 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} - 4 \, a^{5} - 5 i \, a^{4} - 5 i \, a^{2} + 4 \, a + i\right )} \left (-\frac {b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 i \, a - 1}\right )^{\frac {3}{4}}}{b^{3}}\right ) - 3 \, \left (-\frac {b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 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} + 4 \, a^{5} + 5 i \, a^{4} + 5 i \, a^{2} - 4 \, a - i\right )} \left (-\frac {b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 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+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x^2,x, algorithm="fric 
as")
 
output
1/2*(3*(-b^4/(a^8 + 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^2 + 6*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 + 4*I*a^5 - 5*a^4 - 5*a^2 - 4*I*a + 1)*(-b^4/(a^8 
 + 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^2 + 6*I*a - 1))^(3/4))/b^ 
3) + 3*(-b^4/(a^8 + 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^2 + 6*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 + 4*I*a^5 - 5*a^4 - 5*a^2 - 4*I*a + 1)*(-b^4/(a^8 
+ 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^2 + 6*I*a - 1))^(3/4))/b^3 
) + 3*(-b^4/(a^8 + 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^2 + 6*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 - 4*a^5 - 5*I*a^4 - 5*I*a^2 + 4*a + I)*(-b^4/(a^8 + 
 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^2 + 6*I*a - 1))^(3/4))/b^3) 
 - 3*(-b^4/(a^8 + 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^2 + 6*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 + 4*a^5 + 5*I*a^4 + 5*I*a^2 - 4*a - I)*(-b^4/(a^8 + 
 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^2 + 6*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)
 
3.3.25.6 Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {3}{2} i \arctan (a+b x)}}{x^2} \, dx=\text {Timed out} \]

input
integrate(((1+I*(b*x+a))/(1+(b*x+a)**2)**(1/2))**(3/2)/x**2,x)
 
output
Timed out
 
3.3.25.7 Maxima [F]

\[ \int \frac {e^{\frac {3}{2} i \arctan (a+b x)}}{x^2} \, dx=\int { \frac {\left (\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}\right )^{\frac {3}{2}}}{x^{2}} \,d x } \]

input
integrate(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x^2,x, algorithm="maxi 
ma")
 
output
integrate(((I*b*x + I*a + 1)/sqrt((b*x + a)^2 + 1))^(3/2)/x^2, x)
 
3.3.25.8 Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {3}{2} i \arctan (a+b x)}}{x^2} \, dx=\text {Exception raised: TypeError} \]

input
integrate(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(3/2)/x^2,x, algorithm="giac 
")
 
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
 
3.3.25.9 Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {3}{2} i \arctan (a+b x)}}{x^2} \, dx=\int \frac {{\left (\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}\right )}^{3/2}}{x^2} \,d x \]

input
int(((a*1i + b*x*1i + 1)/((a + b*x)^2 + 1)^(1/2))^(3/2)/x^2,x)
 
output
int(((a*1i + b*x*1i + 1)/((a + b*x)^2 + 1)^(1/2))^(3/2)/x^2, x)