3.2.91 \(\int e^{-i \text {ArcTan}(a+b x)} x^2 \, dx\) [191]

Optimal. Leaf size=171 \[ \frac {\left (i-2 a-2 i a^2\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \sinh ^{-1}(a+b x)}{2 b^3} \]

[Out]

-1/2*(1+2*I*a-2*a^2)*arcsinh(b*x+a)/b^3+1/6*(I-4*a)*(1-I*a-I*b*x)^(3/2)*(1+I*a+I*b*x)^(1/2)/b^3+1/3*x*(1-I*a-I
*b*x)^(3/2)*(1+I*a+I*b*x)^(1/2)/b^2+1/2*(I-2*a-2*I*a^2)*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/b^3

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5203, 92, 81, 52, 55, 633, 221} \begin {gather*} \frac {\left (-2 i a^2-2 a+i\right ) \sqrt {i a+i b x+1} \sqrt {-i a-i b x+1}}{2 b^3}-\frac {\left (-2 a^2+2 i a+1\right ) \sinh ^{-1}(a+b x)}{2 b^3}+\frac {(-4 a+i) \sqrt {i a+i b x+1} (-i a-i b x+1)^{3/2}}{6 b^3}+\frac {x \sqrt {i a+i b x+1} (-i a-i b x+1)^{3/2}}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/E^(I*ArcTan[a + b*x]),x]

[Out]

((I - 2*a - (2*I)*a^2)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(2*b^3) + ((I - 4*a)*(1 - I*a - I*b*x)^(3/
2)*Sqrt[1 + I*a + I*b*x])/(6*b^3) + (x*(1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x])/(3*b^2) - ((1 + (2*I)*a
- 2*a^2)*ArcSinh[a + b*x])/(2*b^3)

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 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 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 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*} \int e^{-i \tan ^{-1}(a+b x)} x^2 \, dx &=\int \frac {x^2 \sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}} \, dx\\ &=\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}+\frac {\int \frac {\sqrt {1-i a-i b x} \left (-1-a^2+(i-4 a) b x\right )}{\sqrt {1+i a+i b x}} \, dx}{3 b^2}\\ &=\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \int \frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}} \, dx}{2 b^2}\\ &=-\frac {\left (2 a-i \left (1-2 a^2\right )\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{2 b^2}\\ &=-\frac {\left (2 a-i \left (1-2 a^2\right )\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{2 b^2}\\ &=-\frac {\left (2 a-i \left (1-2 a^2\right )\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\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 )}{4 b^4}\\ &=-\frac {\left (2 a-i \left (1-2 a^2\right )\right ) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^3}+\frac {(i-4 a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{6 b^3}+\frac {x (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{3 b^2}-\frac {\left (1+2 i a-2 a^2\right ) \sinh ^{-1}(a+b x)}{2 b^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.21, size = 162, normalized size = 0.95 \begin {gather*} \frac {i \sqrt {1+i a+i b x} \left (4+7 a^2+2 i a^3-7 i b x-5 b^2 x^2+2 i b^3 x^3+a (5 i+8 b x)\right )}{6 b^3 \sqrt {-i (i+a+b x)}}+\frac {\sqrt [4]{-1} \left (-1-2 i a+2 a^2\right ) \sqrt {-i b} \sinh ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {b} \sqrt {-i (i+a+b x)}}{\sqrt {-i b}}\right )}{b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/6)*Sqrt[1 + I*a + I*b*x]*(4 + 7*a^2 + (2*I)*a^3 - (7*I)*b*x - 5*b^2*x^2 + (2*I)*b^3*x^3 + a*(5*I + 8*b*x))
)/(b^3*Sqrt[(-I)*(I + a + b*x)]) + ((-1)^(1/4)*(-1 - (2*I)*a + 2*a^2)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b
]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/b^(7/2)

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 484 vs. \(2 (135 ) = 270\).
time = 0.10, size = 485, normalized size = 2.84

method	result	size
risch	\(-\frac {i \left (2 b^{2} x^{2}-2 a b x +3 i b x +2 a^{2}-9 i a -4\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{6 b^{3}}-\frac {i \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right ) a}{b^{2} \sqrt {b^{2}}}+\frac {\ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right ) a^{2}}{b^{2} \sqrt {b^{2}}}-\frac {\ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}\)	\(198\)
default	\(-\frac {i \left (b \left (\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{3 b^{2}}-\frac {a \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{b}\right )+i \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )-a \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )\right )}{b^{2}}+\frac {\left (-i a^{2}-2 a +i\right ) \left (\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}+\frac {i b \ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{\sqrt {b^{2}}}\right )}{b^{3}}\)	\(485\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.47, size = 161, normalized size = 0.94 \begin {gather*} \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{b^{2}} + \frac {a^{2} \operatorname {arsinh}\left (b x + a\right )}{b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{2 \, b^{2}} - \frac {i \, a \operatorname {arsinh}\left (b x + a\right )}{b^{3}} - \frac {i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{3 \, b^{3}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{2 \, b^{3}} - \frac {\operatorname {arsinh}\left (b x + a\right )}{2 \, b^{3}} + \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*x/b^2 + a^2*arcsinh(b*x + a)/b^3 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)
*x/b^2 - I*a*arcsinh(b*x + a)/b^3 - 1/3*I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/b^3 - 3/2*sqrt(b^2*x^2 + 2*a*b*x
+ a^2 + 1)*a/b^3 - 1/2*arcsinh(b*x + a)/b^3 + I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/b^3

________________________________________________________________________________________

Fricas [A]
time = 4.60, size = 106, normalized size = 0.62 \begin {gather*} \frac {-7 i \, a^{3} - 21 \, a^{2} - 12 \, {\left (2 \, a^{2} - 2 i \, a - 1\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 i \, b^{2} x^{2} + {\left (-2 i \, a - 3\right )} b x + 2 i \, a^{2} + 9 \, a - 4 i\right )} + 9 i \, a}{24 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(-7*I*a^3 - 21*a^2 - 12*(2*a^2 - 2*I*a - 1)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - 4*sqrt(b^
2*x^2 + 2*a*b*x + a^2 + 1)*(2*I*b^2*x^2 + (-2*I*a - 3)*b*x + 2*I*a^2 + 9*a - 4*I) + 9*I*a)/b^3

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i \int \frac {x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a + b x - i}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*Integral(x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a + b*x - I), x)

________________________________________________________________________________________

Giac [A]
time = 0.47, size = 113, normalized size = 0.66 \begin {gather*} -\frac {1}{6} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (x {\left (\frac {2 i \, x}{b} - \frac {2 i \, a b^{3} + 3 \, b^{3}}{b^{5}}\right )} - \frac {-2 i \, a^{2} b^{2} - 9 \, a b^{2} + 4 i \, b^{2}}{b^{5}}\right )} - \frac {{\left (2 \, a^{2} - 2 i \, a - 1\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, b^{2} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt((b*x + a)^2 + 1)*(x*(2*I*x/b - (2*I*a*b^3 + 3*b^3)/b^5) - (-2*I*a^2*b^2 - 9*a*b^2 + 4*I*b^2)/b^5) -
1/2*(2*a^2 - 2*I*a - 1)*log(-a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/(b^2*abs(b))

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\sqrt {{\left (a+b\,x\right )}^2+1}}{1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________