Integrand size = 14, antiderivative size = 338 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} \, dx=\frac {i (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{b}-\frac {i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b}+\frac {i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{\sqrt {2} b}+\frac {i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b}-\frac {i \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 \sqrt {2} b} \]
I*(1-I*a-I*b*x)^(3/4)*(1+I*a+I*b*x)^(1/4)/b-1/2*I*arctan(1-(1-I*a-I*b*x)^( 1/4)*2^(1/2)/(1+I*a+I*b*x)^(1/4))/b*2^(1/2)+1/2*I*arctan(1+(1-I*a-I*b*x)^( 1/4)*2^(1/2)/(1+I*a+I*b*x)^(1/4))/b*2^(1/2)+1/4*I*ln(1-(1-I*a-I*b*x)^(1/4) *2^(1/2)/(1+I*a+I*b*x)^(1/4)+(1-I*a-I*b*x)^(1/2)/(1+I*a+I*b*x)^(1/2))/b*2^ (1/2)-1/4*I*ln(1+(1-I*a-I*b*x)^(1/4)*2^(1/2)/(1+I*a+I*b*x)^(1/4)+(1-I*a-I* b*x)^(1/2)/(1+I*a+I*b*x)^(1/2))/b*2^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.13 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} \, dx=-\frac {8 i e^{\frac {5}{2} i \arctan (a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {9}{4},-e^{2 i \arctan (a+b x)}\right )}{5 b} \]
(((-8*I)/5)*E^(((5*I)/2)*ArcTan[a + b*x])*Hypergeometric2F1[5/4, 2, 9/4, - E^((2*I)*ArcTan[a + b*x])])/b
Time = 0.40 (sec) , antiderivative size = 299, normalized size of antiderivative = 0.88, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {5616, 60, 73, 854, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int e^{\frac {1}{2} i \arctan (a+b x)} \, dx\) |
| \(\Big \downarrow \) 5616 |
| \(\displaystyle \int \frac {\sqrt [4]{i a+i b x+1}}{\sqrt [4]{-i a-i b x+1}}dx\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{\sqrt [4]{-i a-i b x+1} (i a+i b x+1)^{3/4}}dx+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 i \int \frac {\sqrt {-i a-i b x+1}}{(i a+i b x+1)^{3/4}}d\sqrt [4]{-i a-i b x+1}}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\) |
| \(\Big \downarrow \) 854 |
| \(\displaystyle \frac {2 i \int \frac {\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle \frac {2 i \left (\frac {1}{2} \int \frac {\sqrt {-i a-i b x+1}+1}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}-\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {2 i \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+\frac {1}{2} \int \frac {1}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )-\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {2 i \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\sqrt {-i a-i b x+1}-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\sqrt {-i a-i b x+1}-1}d\left (\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {2 i \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\sqrt {-i a-i b x+1}}{-i a-i b x+2}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {2 i \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 i \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 i \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}{\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1}d\frac {\sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {2 i \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {-i a-i b x+1}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {-i a-i b x+1}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{2 \sqrt {2}}\right )\right )}{b}+\frac {i (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{b}\) |
(I*(1 - I*a - I*b*x)^(3/4)*(1 + I*a + I*b*x)^(1/4))/b + ((2*I)*((-(ArcTan[ 1 - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sqrt[2]) + ArcTan[1 + (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/Sqrt [2])/2 + (Log[1 + Sqrt[1 - I*a - I*b*x] - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4) )/(1 + I*a + I*b*x)^(1/4)]/(2*Sqrt[2]) - Log[1 + Sqrt[1 - I*a - I*b*x] + ( Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)]/(2*Sqrt[2]))/2)) /b
3.3.18.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b]]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 1)/n) Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n )^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 2^(-1)] && IntegersQ[m, p + (m + 1)/n]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.)), x_Symbol] :> Int[(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, n}, x]
\[\int \sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}d x\]
Time = 0.27 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.75 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} \, dx=\frac {b \sqrt {\frac {i}{b^{2}}} \log \left (i \, b \sqrt {\frac {i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - b \sqrt {\frac {i}{b^{2}}} \log \left (-i \, b \sqrt {\frac {i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + b \sqrt {-\frac {i}{b^{2}}} \log \left (i \, b \sqrt {-\frac {i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) - b \sqrt {-\frac {i}{b^{2}}} \log \left (-i \, b \sqrt {-\frac {i}{b^{2}}} + \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}\right ) + 2 \, {\left (b x + a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{2 \, b} \]
1/2*(b*sqrt(I/b^2)*log(I*b*sqrt(I/b^2) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a ^2 + 1)/(b*x + a + I))) - b*sqrt(I/b^2)*log(-I*b*sqrt(I/b^2) + sqrt(I*sqrt (b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) + b*sqrt(-I/b^2)*log(I*b*sqr t(-I/b^2) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) - b*s qrt(-I/b^2)*log(-I*b*sqrt(-I/b^2) + sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I))) + 2*(b*x + a + I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/b
\[ \int e^{\frac {1}{2} i \arctan (a+b x)} \, dx=\int \sqrt {\frac {i \left (a + b x\right ) + 1}{\sqrt {\left (a + b x\right )^{2} + 1}}}\, dx \]
\[ \int e^{\frac {1}{2} i \arctan (a+b x)} \, dx=\int { \sqrt {\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}} \,d x } \]
Exception generated. \[ \int e^{\frac {1}{2} i \arctan (a+b x)} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int e^{\frac {1}{2} i \arctan (a+b x)} \, dx=\int \sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}} \,d x \]