\(\int e^{i \arctan (a+b x)} x^3 \, dx\) [163]

Optimal result

Integrand size = 16, antiderivative size = 201 \[ \int e^{i \arctan (a+b x)} x^3 \, dx=-\frac {\left (3-12 i a-12 a^2+8 i a^3\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{8 b^4}+\frac {x^2 \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2} \left (7-10 i a-18 a^2+2 (i+6 a) b x\right )}{24 b^4}+\frac {\left (3 i+12 a-12 i a^2-8 a^3\right ) \text {arcsinh}(a+b x)}{8 b^4} \]

[Out]

1/8*(3*I+12*a-12*I*a^2-8*a^3)*arcsinh(b*x+a)/b^4+1/4*x^2*(1+I*a+I*b*x)^(3/2)*(1-I*a-I*b*x)^(1/2)/b^2-1/24*(1+I
*a+I*b*x)^(3/2)*(7-10*I*a-18*a^2+2*(I+6*a)*b*x)*(1-I*a-I*b*x)^(1/2)/b^4-1/8*(3-12*I*a-12*a^2+8*I*a^3)*(1-I*a-I
*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/b^4

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5203, 102, 152, 52, 55, 633, 221} \[ \int e^{i \arctan (a+b x)} x^3 \, dx=-\frac {\sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2} \left (-18 a^2+2 (6 a+i) b x-10 i a+7\right )}{24 b^4}+\frac {\left (-8 a^3-12 i a^2+12 a+3 i\right ) \text {arcsinh}(a+b x)}{8 b^4}-\frac {\left (8 i a^3-12 a^2-12 i a+3\right ) \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}{8 b^4}+\frac {x^2 \sqrt {-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b^2} \]

[In]

Int[E^(I*ArcTan[a + b*x])*x^3,x]

[Out]

-1/8*((3 - (12*I)*a - 12*a^2 + (8*I)*a^3)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/b^4 + (x^2*Sqrt[1 - I*a
- I*b*x]*(1 + I*a + I*b*x)^(3/2))/(4*b^2) - (Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2)*(7 - (10*I)*a - 18
*a^2 + 2*(I + 6*a)*b*x))/(24*b^4) + ((3*I + 12*a - (12*I)*a^2 - 8*a^3)*ArcSinh[a + b*x])/(8*b^4)

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 55

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[1/Sqrt[a*c - b*(a - c)*x - b^2*x^2]
, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]

Rule 102

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

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)
^(m + 1)*((c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d
*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1
)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

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^3 \sqrt {1+i a+i b x}}{\sqrt {1-i a-i b x}} \, dx \\ & = \frac {x^2 \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}+\frac {\int \frac {x \sqrt {1+i a+i b x} \left (-2 \left (1+a^2\right )-(i+6 a) b x\right )}{\sqrt {1-i a-i b x}} \, dx}{4 b^2} \\ & = \frac {x^2 \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2} \left (7-10 i a-18 a^2+2 (i+6 a) b x\right )}{24 b^4}+\frac {\left (3 i+12 a-12 i a^2-8 a^3\right ) \int \frac {\sqrt {1+i a+i b x}}{\sqrt {1-i a-i b x}} \, dx}{8 b^3} \\ & = -\frac {\left (3-12 i a-12 a^2+8 i a^3\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{8 b^4}+\frac {x^2 \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2} \left (7-10 i a-18 a^2+2 (i+6 a) b x\right )}{24 b^4}+\frac {\left (3 i+12 a-12 i a^2-8 a^3\right ) \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{8 b^3} \\ & = -\frac {\left (3-12 i a-12 a^2+8 i a^3\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{8 b^4}+\frac {x^2 \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2} \left (7-10 i a-18 a^2+2 (i+6 a) b x\right )}{24 b^4}+\frac {\left (3 i+12 a-12 i a^2-8 a^3\right ) \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{8 b^3} \\ & = -\frac {\left (3-12 i a-12 a^2+8 i a^3\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{8 b^4}+\frac {x^2 \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2} \left (7-10 i a-18 a^2+2 (i+6 a) b x\right )}{24 b^4}+\frac {\left (3 i+12 a-12 i a^2-8 a^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{4 b^2}}} \, dx,x,2 a b+2 b^2 x\right )}{16 b^5} \\ & = -\frac {\left (3-12 i a-12 a^2+8 i a^3\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{8 b^4}+\frac {x^2 \sqrt {1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac {\sqrt {1-i a-i b x} (1+i a+i b x)^{3/2} \left (7-10 i a-18 a^2+2 (i+6 a) b x\right )}{24 b^4}+\frac {\left (3 i+12 a-12 i a^2-8 a^3\right ) \text {arcsinh}(a+b x)}{8 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.88 \[ \int e^{i \arctan (a+b x)} x^3 \, dx=\frac {\sqrt {b} \sqrt {1+a^2+2 a b x+b^2 x^2} \left (-16-6 i a^3-9 i b x+8 b^2 x^2+6 i b^3 x^3+a^2 (44+6 i b x)+a \left (39 i-20 b x-6 i b^2 x^2\right )\right )-6 \sqrt [4]{-1} \left (-3 i-12 a+12 i a^2+8 a^3\right ) \sqrt {-i b} \text {arcsinh}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{24 b^{9/2}} \]

[In]

Integrate[E^(I*ArcTan[a + b*x])*x^3,x]

[Out]

(Sqrt[b]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(-16 - (6*I)*a^3 - (9*I)*b*x + 8*b^2*x^2 + (6*I)*b^3*x^3 + a^2*(44
+ (6*I)*b*x) + a*(39*I - 20*b*x - (6*I)*b^2*x^2)) - 6*(-1)^(1/4)*(-3*I - 12*a + (12*I)*a^2 + 8*a^3)*Sqrt[(-I)*
b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/(24*b^(9/2))

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {i \left (-6 b^{3} x^{3}+6 a \,b^{2} x^{2}+8 i b^{2} x^{2}-6 a^{2} b x -20 i a b x +6 a^{3}+44 i a^{2}+9 b x -39 a -16 i\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{24 b^{4}}-\frac {\left (8 a^{3}+12 i a^{2}-12 a -3 i\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{3} \sqrt {b^{2}}}\)	\(150\)
default	\(i b \left (\frac {x^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}-\frac {7 a \left (\frac {x^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{3 b^{2}}-\frac {5 a \left (\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\right )}{3 b}-\frac {2 \left (a^{2}+1\right ) \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{3 b^{2}}\right )}{4 b}-\frac {3 \left (a^{2}+1\right ) \left (\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\right )}{4 b^{2}}\right )+\left (i a +1\right ) \left (\frac {x^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{3 b^{2}}-\frac {5 a \left (\frac {x \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {3 a \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{2 b}-\frac {\left (a^{2}+1\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\right )}{3 b}-\frac {2 \left (a^{2}+1\right ) \left (\frac {\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b^{2}}-\frac {a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}\right )}{3 b^{2}}\right )\)	\(749\)

[In]

int((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2)*x^3,x,method=_RETURNVERBOSE)

[Out]

-1/24*I*(-6*b^3*x^3+8*I*b^2*x^2+6*a*b^2*x^2-20*I*a*b*x-6*a^2*b*x+44*I*a^2+6*a^3+9*b*x-16*I-39*a)*(b^2*x^2+2*a*
b*x+a^2+1)^(1/2)/b^4-1/8*(12*I*a^2+8*a^3-3*I-12*a)/b^3*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2
))/(b^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.69 \[ \int e^{i \arctan (a+b x)} x^3 \, dx=\frac {-45 i \, a^{4} + 224 \, a^{3} + 192 i \, a^{2} + 24 \, {\left (8 \, a^{3} + 12 i \, a^{2} - 12 \, a - 3 i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 8 \, {\left (-6 i \, b^{3} x^{3} + 2 \, {\left (3 i \, a - 4\right )} b^{2} x^{2} + 6 i \, a^{3} + {\left (-6 i \, a^{2} + 20 \, a + 9 i\right )} b x - 44 \, a^{2} - 39 i \, a + 16\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 72 \, a}{192 \, b^{4}} \]

[In]

integrate((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2)*x^3,x, algorithm="fricas")

[Out]

1/192*(-45*I*a^4 + 224*a^3 + 192*I*a^2 + 24*(8*a^3 + 12*I*a^2 - 12*a - 3*I)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*
b*x + a^2 + 1)) - 8*(-6*I*b^3*x^3 + 2*(3*I*a - 4)*b^2*x^2 + 6*I*a^3 + (-6*I*a^2 + 20*a + 9*I)*b*x - 44*a^2 - 3
9*I*a + 16)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 72*a)/b^4

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 857 vs. \(2 (173) = 346\).

Time = 1.60 (sec) , antiderivative size = 857, normalized size of antiderivative = 4.26 \[ \int e^{i \arctan (a+b x)} x^3 \, dx=\begin {cases} \frac {\left (- \frac {a \left (- \frac {3 a \left (- \frac {5 a \left (- \frac {3 i a}{4} + 1\right )}{3 b} - \frac {i \left (3 a^{2} + 3\right )}{4 b}\right )}{2 b} - \frac {\left (2 a^{2} + 2\right ) \left (- \frac {3 i a}{4} + 1\right )}{3 b^{2}}\right )}{b} - \frac {\left (a^{2} + 1\right ) \left (- \frac {5 a \left (- \frac {3 i a}{4} + 1\right )}{3 b} - \frac {i \left (3 a^{2} + 3\right )}{4 b}\right )}{2 b^{2}}\right ) \log {\left (2 a b + 2 b^{2} x + 2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \sqrt {b^{2}} \right )}}{\sqrt {b^{2}}} + \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \left (\frac {i x^{3}}{4 b} + \frac {x^{2} \left (- \frac {3 i a}{4} + 1\right )}{3 b^{2}} + \frac {x \left (- \frac {5 a \left (- \frac {3 i a}{4} + 1\right )}{3 b} - \frac {i \left (3 a^{2} + 3\right )}{4 b}\right )}{2 b^{2}} + \frac {- \frac {3 a \left (- \frac {5 a \left (- \frac {3 i a}{4} + 1\right )}{3 b} - \frac {i \left (3 a^{2} + 3\right )}{4 b}\right )}{2 b} - \frac {\left (2 a^{2} + 2\right ) \left (- \frac {3 i a}{4} + 1\right )}{3 b^{2}}}{b^{2}}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {i \left (- a^{6} \sqrt {a^{2} + 2 a b x + 1} - 3 a^{4} \sqrt {a^{2} + 2 a b x + 1} - 3 a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (- 3 a^{2} - 3\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}} \cdot \left (3 a^{4} + 6 a^{2} + 3\right )}{3} - \sqrt {a^{2} + 2 a b x + 1}\right )}{4 a^{2} b^{3}} + \frac {- a^{6} \sqrt {a^{2} + 2 a b x + 1} - 3 a^{4} \sqrt {a^{2} + 2 a b x + 1} - 3 a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (- 3 a^{2} - 3\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}} \cdot \left (3 a^{4} + 6 a^{2} + 3\right )}{3} - \sqrt {a^{2} + 2 a b x + 1}}{4 a^{3} b^{3}} + \frac {i \left (a^{8} \sqrt {a^{2} + 2 a b x + 1} + 4 a^{6} \sqrt {a^{2} + 2 a b x + 1} + 6 a^{4} \sqrt {a^{2} + 2 a b x + 1} + 4 a^{2} \sqrt {a^{2} + 2 a b x + 1} + \frac {\left (- 4 a^{2} - 4\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {9}{2}}}{9} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}} \cdot \left (6 a^{4} + 12 a^{2} + 6\right )}{5} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}} \left (- 4 a^{6} - 12 a^{4} - 12 a^{2} - 4\right )}{3} + \sqrt {a^{2} + 2 a b x + 1}\right )}{8 a^{4} b^{3}}}{2 a b} & \text {for}\: a b \neq 0 \\\frac {\frac {i a x^{4}}{4} + \frac {i b x^{5}}{5} + \frac {x^{4}}{4}}{\sqrt {a^{2} + 1}} & \text {otherwise} \end {cases} \]

[In]

integrate((1+I*(b*x+a))/(1+(b*x+a)**2)**(1/2)*x**3,x)

[Out]

Piecewise(((-a*(-3*a*(-5*a*(-3*I*a/4 + 1)/(3*b) - I*(3*a**2 + 3)/(4*b))/(2*b) - (2*a**2 + 2)*(-3*I*a/4 + 1)/(3
*b**2))/b - (a**2 + 1)*(-5*a*(-3*I*a/4 + 1)/(3*b) - I*(3*a**2 + 3)/(4*b))/(2*b**2))*log(2*a*b + 2*b**2*x + 2*s
qrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*sqrt(b**2))/sqrt(b**2) + sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*(I*x**3/(4*b
) + x**2*(-3*I*a/4 + 1)/(3*b**2) + x*(-5*a*(-3*I*a/4 + 1)/(3*b) - I*(3*a**2 + 3)/(4*b))/(2*b**2) + (-3*a*(-5*a
*(-3*I*a/4 + 1)/(3*b) - I*(3*a**2 + 3)/(4*b))/(2*b) - (2*a**2 + 2)*(-3*I*a/4 + 1)/(3*b**2))/b**2), Ne(b**2, 0)
), ((I*(-a**6*sqrt(a**2 + 2*a*b*x + 1) - 3*a**4*sqrt(a**2 + 2*a*b*x + 1) - 3*a**2*sqrt(a**2 + 2*a*b*x + 1) + (
-3*a**2 - 3)*(a**2 + 2*a*b*x + 1)**(5/2)/5 + (a**2 + 2*a*b*x + 1)**(7/2)/7 + (a**2 + 2*a*b*x + 1)**(3/2)*(3*a*
*4 + 6*a**2 + 3)/3 - sqrt(a**2 + 2*a*b*x + 1))/(4*a**2*b**3) + (-a**6*sqrt(a**2 + 2*a*b*x + 1) - 3*a**4*sqrt(a
**2 + 2*a*b*x + 1) - 3*a**2*sqrt(a**2 + 2*a*b*x + 1) + (-3*a**2 - 3)*(a**2 + 2*a*b*x + 1)**(5/2)/5 + (a**2 + 2
*a*b*x + 1)**(7/2)/7 + (a**2 + 2*a*b*x + 1)**(3/2)*(3*a**4 + 6*a**2 + 3)/3 - sqrt(a**2 + 2*a*b*x + 1))/(4*a**3
*b**3) + I*(a**8*sqrt(a**2 + 2*a*b*x + 1) + 4*a**6*sqrt(a**2 + 2*a*b*x + 1) + 6*a**4*sqrt(a**2 + 2*a*b*x + 1)
+ 4*a**2*sqrt(a**2 + 2*a*b*x + 1) + (-4*a**2 - 4)*(a**2 + 2*a*b*x + 1)**(7/2)/7 + (a**2 + 2*a*b*x + 1)**(9/2)/
9 + (a**2 + 2*a*b*x + 1)**(5/2)*(6*a**4 + 12*a**2 + 6)/5 + (a**2 + 2*a*b*x + 1)**(3/2)*(-4*a**6 - 12*a**4 - 12
*a**2 - 4)/3 + sqrt(a**2 + 2*a*b*x + 1))/(8*a**4*b**3))/(2*a*b), Ne(a*b, 0)), ((I*a*x**4/4 + I*b*x**5/5 + x**4
/4)/sqrt(a**2 + 1), True))

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (145) = 290\).

Time = 0.21 (sec) , antiderivative size = 529, normalized size of antiderivative = 2.63 \[ \int e^{i \arctan (a+b x)} x^3 \, dx=\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x^{3}}{4 \, b} - \frac {7 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x^{2}}{12 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-i \, a - 1\right )} x^{2}}{3 \, b^{2}} + \frac {35 i \, a^{4} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{8 \, b^{4}} - \frac {5 \, a^{3} {\left (i \, a + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{4}} + \frac {35 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2} x}{24 \, b^{3}} - \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a {\left (i \, a + 1\right )} x}{6 \, b^{3}} - \frac {15 i \, {\left (a^{2} + 1\right )} a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{4 \, b^{4}} + \frac {3 \, {\left (a^{2} + 1\right )} a {\left (i \, a + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{4}} - \frac {35 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{3}}{8 \, b^{4}} + \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2} {\left (i \, a + 1\right )}}{2 \, b^{4}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (i \, a^{2} + i\right )} x}{8 \, b^{3}} + \frac {3 i \, {\left (a^{2} + 1\right )}^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{8 \, b^{4}} + \frac {55 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + 1\right )} a}{24 \, b^{4}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + 1\right )} {\left (i \, a + 1\right )}}{3 \, b^{4}} \]

[In]

integrate((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2)*x^3,x, algorithm="maxima")

[Out]

1/4*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*x^3/b - 7/12*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*x^2/b^2 - 1/3*sqrt(
b^2*x^2 + 2*a*b*x + a^2 + 1)*(-I*a - 1)*x^2/b^2 + 35/8*I*a^4*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2
+ 1)*b^2))/b^4 - 5/2*a^3*(I*a + 1)*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^4 + 35/24*I*s
qrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2*x/b^3 - 5/6*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*(I*a + 1)*x/b^3 - 15/4*I*
(a^2 + 1)*a^2*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^4 + 3/2*(a^2 + 1)*a*(I*a + 1)*arcs
inh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^4 - 35/8*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^3/b^4
+ 5/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2*(I*a + 1)/b^4 - 3/8*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(I*a^2 + I)
*x/b^3 + 3/8*I*(a^2 + 1)^2*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^4 + 55/24*I*sqrt(b^2*
x^2 + 2*a*b*x + a^2 + 1)*(a^2 + 1)*a/b^4 - 2/3*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(a^2 + 1)*(I*a + 1)/b^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.77 \[ \int e^{i \arctan (a+b x)} x^3 \, dx=-\frac {1}{24} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left ({\left (2 \, x {\left (-\frac {3 i \, x}{b} - \frac {-3 i \, a b^{5} + 4 \, b^{5}}{b^{7}}\right )} - \frac {6 i \, a^{2} b^{4} - 20 \, a b^{4} - 9 i \, b^{4}}{b^{7}}\right )} x - \frac {-6 i \, a^{3} b^{3} + 44 \, a^{2} b^{3} + 39 i \, a b^{3} - 16 \, b^{3}}{b^{7}}\right )} + \frac {{\left (8 \, a^{3} + 12 i \, a^{2} - 12 \, a - 3 i\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{8 \, b^{3} {\left | b \right |}} \]

[In]

integrate((1+I*(b*x+a))/(1+(b*x+a)^2)^(1/2)*x^3,x, algorithm="giac")

[Out]

-1/24*sqrt((b*x + a)^2 + 1)*((2*x*(-3*I*x/b - (-3*I*a*b^5 + 4*b^5)/b^7) - (6*I*a^2*b^4 - 20*a*b^4 - 9*I*b^4)/b
^7)*x - (-6*I*a^3*b^3 + 44*a^2*b^3 + 39*I*a*b^3 - 16*b^3)/b^7) + 1/8*(8*a^3 + 12*I*a^2 - 12*a - 3*I)*log(-a*b
- (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/(b^3*abs(b))

Mupad [F(-1)]

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

[In]

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

[Out]

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