Integrand size = 18, antiderivative size = 205 \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=-\frac {(i+a+b x) \sqrt [4]{1+i (a+b x)}}{(i+a) x \sqrt [4]{1-i (a+b x)}}+\frac {i b \arctan \left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{(i-a)^{3/4} (i+a)^{5/4}}+\frac {i b \text {arctanh}\left (\frac {\sqrt [4]{i+a} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{i-a} \sqrt [4]{1-i (a+b x)}}\right )}{(i-a)^{3/4} (i+a)^{5/4}} \]
-(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)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.54 \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=\frac {(-i (i+a+b x))^{3/4} \left (3 (i+a) (-i+a+b x)+2 i b x \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {1+a^2-i b x+a b x}{1+a^2+i b x+a b x}\right )\right )}{3 (i+a)^2 x (1+i a+i b x)^{3/4}} \]
(((-I)*(I + a + b*x))^(3/4)*(3*(I + a)*(-I + a + b*x) + (2*I)*b*x*Hypergeo metric2F1[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))
Time = 0.30 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.27, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5617, 795, 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 |
\(\displaystyle 8 i b \int \frac {1-i (a+b x)}{(i (a+b x)+1) \left (-i a-\frac {(i a+1) (1-i (a+b x))}{i (a+b x)+1}+1\right )^2}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\) |
\(\Big \downarrow \) 795 |
\(\displaystyle 8 i b \int \frac {i (a+b x)+1}{(1-i (a+b x)) \left (-i a+\frac {(1-i a) (i (a+b x)+1)}{1-i (a+b x)}-1\right )^2}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}\) |
\(\Big \downarrow \) 817 |
\(\displaystyle 8 i b \left (\frac {\int \frac {1}{-i a+\frac {(1-i a) (i (a+b x)+1)}{1-i (a+b x)}-1}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{4 (1-i a)}+\frac {\sqrt [4]{1+i (a+b x)}}{4 (1-i a) \sqrt [4]{1-i (a+b x)} \left (-\frac {(1-i a) (1+i (a+b x))}{1-i (a+b x)}+i a+1\right )}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle 8 i b \left (\frac {-\frac {i \int \frac {1}{\sqrt {i-a}-\frac {\sqrt {a+i} \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{2 \sqrt {-a+i}}-\frac {i \int \frac {1}{\sqrt {i-a}+\frac {\sqrt {a+i} \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{2 \sqrt {-a+i}}}{4 (1-i a)}+\frac {\sqrt [4]{1+i (a+b x)}}{4 (1-i a) \sqrt [4]{1-i (a+b x)} \left (-\frac {(1-i a) (1+i (a+b x))}{1-i (a+b x)}+i a+1\right )}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle 8 i b \left (\frac {-\frac {i \int \frac {1}{\sqrt {i-a}-\frac {\sqrt {a+i} \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}}d\frac {\sqrt [4]{i (a+b x)+1}}{\sqrt [4]{1-i (a+b x)}}}{2 \sqrt {-a+i}}-\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}}{4 (1-i a)}+\frac {\sqrt [4]{1+i (a+b x)}}{4 (1-i a) \sqrt [4]{1-i (a+b x)} \left (-\frac {(1-i a) (1+i (a+b x))}{1-i (a+b x)}+i a+1\right )}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle 8 i b \left (\frac {-\frac {i \arctan \left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}-\frac {i \text {arctanh}\left (\frac {\sqrt [4]{a+i} \sqrt [4]{1+i (a+b x)}}{\sqrt [4]{-a+i} \sqrt [4]{1-i (a+b x)}}\right )}{2 (-a+i)^{3/4} \sqrt [4]{a+i}}}{4 (1-i a)}+\frac {\sqrt [4]{1+i (a+b x)}}{4 (1-i a) \sqrt [4]{1-i (a+b x)} \left (-\frac {(1-i a) (1+i (a+b x))}{1-i (a+b x)}+i a+1\right )}\right )\) |
(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)))) + (((-1/2*I)*ArcT an[((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)^(1/4)) - ((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) ^(3/4)*(I + a)^(1/4)))/(4*(1 - I*a)))
3.3.20.3.1 Defintions of rubi rules used
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]
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]
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]
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]
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]
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]
\[\int \frac {\sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}}{x^{2}}d x\]
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 = 0.27 (sec) , antiderivative size = 598, normalized size of antiderivative = 2.92 \[ \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 \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} \]
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*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)*(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)
Timed out. \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=\text {Timed out} \]
\[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=\int { \frac {\sqrt {\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}}}{x^{2}} \,d x } \]
Exception generated. \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=\text {Exception raised: TypeError} \]
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 -27, a substitution variable should perhaps be purged.Wa
Timed out. \[ \int \frac {e^{\frac {1}{2} i \arctan (a+b x)}}{x^2} \, dx=\int \frac {\sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}}}{x^2} \,d x \]