\(\int e^{\frac {1}{2} i \arctan (a+b x)} x \, dx\) [217]

Optimal result

Integrand size = 16, antiderivative size = 410 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x \, dx=\frac {(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac {(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}-\frac {(1-4 i a) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt {2} b^2}+\frac {(1-4 i a) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt {2} b^2}+\frac {(1-4 i a) \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^2}-\frac {(1-4 i a) \log \left (1+\frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}}+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^2} \]

[Out]

1/4*(1-4*I*a)*(1-I*a-I*b*x)^(3/4)*(1+I*a+I*b*x)^(1/4)/b^2+1/2*(1-I*a-I*b*x)^(3/4)*(1+I*a+I*b*x)^(5/4)/b^2-1/8*
(1-4*I*a)*arctan(1-(1-I*a-I*b*x)^(1/4)*2^(1/2)/(1+I*a+I*b*x)^(1/4))/b^2*2^(1/2)+1/8*(1-4*I*a)*arctan(1+(1-I*a-
I*b*x)^(1/4)*2^(1/2)/(1+I*a+I*b*x)^(1/4))/b^2*2^(1/2)+1/16*(1-4*I*a)*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*2^(1/2)-1/16*(1-4*I*a)*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*2^(1/2)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {5203, 81, 52, 65, 338, 303, 1176, 631, 210, 1179, 642} \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x \, dx=-\frac {(1-4 i a) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{4 \sqrt {2} b^2}+\frac {(1-4 i a) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{4 \sqrt {2} b^2}+\frac {(-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b^2}+\frac {(1-4 i a) (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{4 b^2}+\frac {(1-4 i a) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{8 \sqrt {2} b^2}-\frac {(1-4 i a) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{8 \sqrt {2} b^2} \]

[In]

Int[E^((I/2)*ArcTan[a + b*x])*x,x]

[Out]

((1 - (4*I)*a)*(1 - I*a - I*b*x)^(3/4)*(1 + I*a + I*b*x)^(1/4))/(4*b^2) + ((1 - I*a - I*b*x)^(3/4)*(1 + I*a +
I*b*x)^(5/4))/(2*b^2) - ((1 - (4*I)*a)*ArcTan[1 - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)])/
(4*Sqrt[2]*b^2) + ((1 - (4*I)*a)*ArcTan[1 + (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)])/(4*Sqr
t[2]*b^2) + ((1 - (4*I)*a)*Log[1 + Sqrt[1 - I*a - I*b*x]/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)])/(8*Sqrt[2]*b^2) - ((1 - (4*I)*a)*Log[1 + Sqrt[1 - I*a - I*b*x]/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)])/(8*Sqrt[2]*b^2)

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 210

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])

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 5203

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}} \, dx \\ & = \frac {(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}-\frac {(i+4 a) \int \frac {\sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}} \, dx}{4 b} \\ & = \frac {(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac {(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}-\frac {(i+4 a) \int \frac {1}{\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}} \, dx}{8 b} \\ & = \frac {(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac {(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}+\frac {(1-4 i a) \text {Subst}\left (\int \frac {x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-i a-i b x}\right )}{2 b^2} \\ & = \frac {(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac {(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}+\frac {(1-4 i a) \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 b^2} \\ & = \frac {(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac {(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}-\frac {(1-4 i a) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 b^2}+\frac {(1-4 i a) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 b^2} \\ & = \frac {(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac {(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}+\frac {(1-4 i a) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 b^2}+\frac {(1-4 i a) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 b^2}+\frac {(1-4 i a) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^2}+\frac {(1-4 i a) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^2} \\ & = \frac {(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac {(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}+\frac {(1-4 i a) \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 )}{8 \sqrt {2} b^2}-\frac {(1-4 i a) \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 )}{8 \sqrt {2} b^2}+\frac {(1-4 i a) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt {2} b^2}-\frac {(1-4 i a) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt {2} b^2} \\ & = \frac {(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac {(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}-\frac {(1-4 i a) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt {2} b^2}+\frac {(1-4 i a) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt {2} b^2}+\frac {(1-4 i a) \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 )}{8 \sqrt {2} b^2}-\frac {(1-4 i a) \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 )}{8 \sqrt {2} b^2} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.20 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x \, dx=\frac {(-i (i+a+b x))^{3/4} \left (3 (1+i a+i b x)^{5/4}+2 \sqrt [4]{2} (1-4 i a) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{4},\frac {7}{4},-\frac {1}{2} i (i+a+b x)\right )\right )}{6 b^2} \]

[In]

Integrate[E^((I/2)*ArcTan[a + b*x])*x,x]

[Out]

(((-I)*(I + a + b*x))^(3/4)*(3*(1 + I*a + I*b*x)^(5/4) + 2*2^(1/4)*(1 - (4*I)*a)*Hypergeometric2F1[-1/4, 3/4,
7/4, (-1/2*I)*(I + a + b*x)]))/(6*b^2)

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.01 \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x \, dx=-\frac {b^{2} \sqrt {\frac {16 i \, a^{2} - 8 \, a - i}{b^{4}}} \log \left (\frac {i \, b^{2} \sqrt {\frac {16 i \, a^{2} - 8 \, a - i}{b^{4}}} + {\left (4 \, a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + i}\right ) - b^{2} \sqrt {\frac {16 i \, a^{2} - 8 \, a - i}{b^{4}}} \log \left (\frac {-i \, b^{2} \sqrt {\frac {16 i \, a^{2} - 8 \, a - i}{b^{4}}} + {\left (4 \, a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + i}\right ) + b^{2} \sqrt {\frac {-16 i \, a^{2} + 8 \, a + i}{b^{4}}} \log \left (\frac {i \, b^{2} \sqrt {\frac {-16 i \, a^{2} + 8 \, a + i}{b^{4}}} + {\left (4 \, a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + i}\right ) - b^{2} \sqrt {\frac {-16 i \, a^{2} + 8 \, a + i}{b^{4}}} \log \left (\frac {-i \, b^{2} \sqrt {\frac {-16 i \, a^{2} + 8 \, a + i}{b^{4}}} + {\left (4 \, a + i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + i}\right ) - 2 \, {\left (2 \, b^{2} x^{2} - 2 \, a^{2} - i \, b x - 5 i \, a + 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, b^{2}} \]

[In]

integrate(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)*x,x, algorithm="fricas")

[Out]

-1/8*(b^2*sqrt((16*I*a^2 - 8*a - I)/b^4)*log((I*b^2*sqrt((16*I*a^2 - 8*a - I)/b^4) + (4*a + I)*sqrt(I*sqrt(b^2
*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(4*a + I)) - b^2*sqrt((16*I*a^2 - 8*a - I)/b^4)*log((-I*b^2*sqrt((16
*I*a^2 - 8*a - I)/b^4) + (4*a + I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(4*a + I)) + b^2*s
qrt((-16*I*a^2 + 8*a + I)/b^4)*log((I*b^2*sqrt((-16*I*a^2 + 8*a + I)/b^4) + (4*a + I)*sqrt(I*sqrt(b^2*x^2 + 2*
a*b*x + a^2 + 1)/(b*x + a + I)))/(4*a + I)) - b^2*sqrt((-16*I*a^2 + 8*a + I)/b^4)*log((-I*b^2*sqrt((-16*I*a^2
+ 8*a + I)/b^4) + (4*a + I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(4*a + I)) - 2*(2*b^2*x^2
- 2*a^2 - I*b*x - 5*I*a + 3)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/b^2

Sympy [F(-1)]

Timed out. \[ \int e^{\frac {1}{2} i \arctan (a+b x)} x \, dx=\text {Timed out} \]

[In]

integrate(((1+I*(b*x+a))/(1+(b*x+a)**2)**(1/2))**(1/2)*x,x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)*x,x, algorithm="maxima")

[Out]

integrate(x*sqrt((I*b*x + I*a + 1)/sqrt((b*x + a)^2 + 1)), x)

Giac [F(-2)]

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

[In]

integrate(((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2))^(1/2)*x,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:The choice was done assuming 0=[0,0,0]Warning, replacing 0 by -27, a substitution variable should perhaps b
e purged.Wa

Mupad [F(-1)]

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

[In]

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

[Out]

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