3.226 \(\int e^{-\frac {1}{2} i \tan ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=494 \[ \frac {\left (-8 i a^2-4 a+3 i\right ) (i a+i b x+1)^{3/4} \sqrt [4]{-i a-i b x+1}}{8 b^3}+\frac {\left (-8 i a^2-4 a+3 i\right ) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt {2} b^3}-\frac {\left (-8 i a^2-4 a+3 i\right ) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt {2} b^3}+\frac {\left (-8 i a^2-4 a+3 i\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt {2} b^3}-\frac {\left (-8 i a^2-4 a+3 i\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt {2} b^3}+\frac {(-8 a+i) (i a+i b x+1)^{3/4} (-i a-i b x+1)^{5/4}}{12 b^3}+\frac {x (i a+i b x+1)^{3/4} (-i a-i b x+1)^{5/4}}{3 b^2} \]

[Out]

1/8*(3*I-4*a-8*I*a^2)*(1-I*a-I*b*x)^(1/4)*(1+I*a+I*b*x)^(3/4)/b^3+1/12*(I-8*a)*(1-I*a-I*b*x)^(5/4)*(1+I*a+I*b*
x)^(3/4)/b^3+1/3*x*(1-I*a-I*b*x)^(5/4)*(1+I*a+I*b*x)^(3/4)/b^2+1/16*(3*I-4*a-8*I*a^2)*arctan(1-(1-I*a-I*b*x)^(
1/4)*2^(1/2)/(1+I*a+I*b*x)^(1/4))/b^3*2^(1/2)-1/16*(3*I-4*a-8*I*a^2)*arctan(1+(1-I*a-I*b*x)^(1/4)*2^(1/2)/(1+I
*a+I*b*x)^(1/4))/b^3*2^(1/2)+1/32*(3*I-4*a-8*I*a^2)*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^3*2^(1/2)-1/32*(3*I-4*a-8*I*a^2)*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^3*2^(1/2)

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Rubi [A]  time = 0.40, antiderivative size = 494, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5095, 90, 80, 50, 63, 240, 211, 1165, 628, 1162, 617, 204} \[ \frac {\left (-8 i a^2-4 a+3 i\right ) (i a+i b x+1)^{3/4} \sqrt [4]{-i a-i b x+1}}{8 b^3}+\frac {\left (-8 i a^2-4 a+3 i\right ) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt {2} b^3}-\frac {\left (-8 i a^2-4 a+3 i\right ) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt {2} b^3}+\frac {\left (-8 i a^2-4 a+3 i\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt {2} b^3}-\frac {\left (-8 i a^2-4 a+3 i\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt {2} b^3}+\frac {x (i a+i b x+1)^{3/4} (-i a-i b x+1)^{5/4}}{3 b^2}+\frac {(-8 a+i) (i a+i b x+1)^{3/4} (-i a-i b x+1)^{5/4}}{12 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/E^((I/2)*ArcTan[a + b*x]),x]

[Out]

((3*I - 4*a - (8*I)*a^2)*(1 - I*a - I*b*x)^(1/4)*(1 + I*a + I*b*x)^(3/4))/(8*b^3) + ((I - 8*a)*(1 - I*a - I*b*
x)^(5/4)*(1 + I*a + I*b*x)^(3/4))/(12*b^3) + (x*(1 - I*a - I*b*x)^(5/4)*(1 + I*a + I*b*x)^(3/4))/(3*b^2) + ((3
*I - 4*a - (8*I)*a^2)*ArcTan[1 - (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)])/(8*Sqrt[2]*b^3) -
((3*I - 4*a - (8*I)*a^2)*ArcTan[1 + (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)])/(8*Sqrt[2]*b^
3) + ((3*I - 4*a - (8*I)*a^2)*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)])/(16*Sqrt[2]*b^3) - ((3*I - 4*a - (8*I)*a^2)*Log[1 + Sqrt[1 - I*a - I*b*x]/Sq
rt[1 + I*a + I*b*x] + (Sqrt[2]*(1 - I*a - I*b*x)^(1/4))/(1 + I*a + I*b*x)^(1/4)])/(16*Sqrt[2]*b^3)

Rule 50

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 63

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 80

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 90

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 204

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

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), 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
]]))

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

Rule 617

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 628

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 1162

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 1165

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 5095

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^{-\frac {1}{2} i \tan ^{-1}(a+b x)} x^2 \, dx &=\int \frac {x^2 \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}} \, dx\\ &=\frac {x (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{3 b^2}+\frac {\int \frac {\sqrt [4]{1-i a-i b x} \left (-1-a^2+\frac {1}{2} (i-8 a) b x\right )}{\sqrt [4]{1+i a+i b x}} \, dx}{3 b^2}\\ &=\frac {(i-8 a) (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{12 b^3}+\frac {x (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{3 b^2}-\frac {\left (3+4 i a-8 a^2\right ) \int \frac {\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}} \, dx}{8 b^2}\\ &=-\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{8 b^3}+\frac {(i-8 a) (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{12 b^3}+\frac {x (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{3 b^2}-\frac {\left (3+4 i a-8 a^2\right ) \int \frac {1}{(1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}} \, dx}{16 b^2}\\ &=-\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{8 b^3}+\frac {(i-8 a) (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{12 b^3}+\frac {x (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{3 b^2}-\frac {\left (i \left (3+4 i a-8 a^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-i a-i b x}\right )}{4 b^3}\\ &=-\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{8 b^3}+\frac {(i-8 a) (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{12 b^3}+\frac {x (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{3 b^2}-\frac {\left (i \left (3+4 i a-8 a^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{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^3}\\ &=-\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{8 b^3}+\frac {(i-8 a) (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{12 b^3}+\frac {x (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{3 b^2}-\frac {\left (i \left (3+4 i a-8 a^2\right )\right ) \operatorname {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 )}{8 b^3}-\frac {\left (i \left (3+4 i a-8 a^2\right )\right ) \operatorname {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 )}{8 b^3}\\ &=-\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{8 b^3}+\frac {(i-8 a) (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{12 b^3}+\frac {x (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{3 b^2}-\frac {\left (i \left (3+4 i a-8 a^2\right )\right ) \operatorname {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 )}{16 b^3}-\frac {\left (i \left (3+4 i a-8 a^2\right )\right ) \operatorname {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 )}{16 b^3}+\frac {\left (i \left (3+4 i a-8 a^2\right )\right ) \operatorname {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 )}{16 \sqrt {2} b^3}+\frac {\left (i \left (3+4 i a-8 a^2\right )\right ) \operatorname {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 )}{16 \sqrt {2} b^3}\\ &=-\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{8 b^3}+\frac {(i-8 a) (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{12 b^3}+\frac {x (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{3 b^2}-\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \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 )}{16 \sqrt {2} b^3}+\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \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 )}{16 \sqrt {2} b^3}-\frac {\left (i \left (3+4 i a-8 a^2\right )\right ) \operatorname {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 )}{8 \sqrt {2} b^3}+\frac {\left (i \left (3+4 i a-8 a^2\right )\right ) \operatorname {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 )}{8 \sqrt {2} b^3}\\ &=-\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{8 b^3}+\frac {(i-8 a) (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{12 b^3}+\frac {x (1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{3 b^2}-\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^3}+\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt {2} b^3}-\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \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 )}{16 \sqrt {2} b^3}+\frac {\left (4 a-i \left (3-8 a^2\right )\right ) \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 )}{16 \sqrt {2} b^3}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 99, normalized size = 0.20 \[ \frac {(-i (a+b x+i))^{5/4} \left (3\ 2^{3/4} \left (8 i a^2+4 a-3 i\right ) \, _2F_1\left (\frac {1}{4},\frac {5}{4};\frac {9}{4};-\frac {1}{2} i (a+b x+i)\right )+5 (i a+i b x+1)^{3/4} (-8 a+4 b x+i)\right )}{60 b^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((-I)*(I + a + b*x))^(5/4)*(5*(1 + I*a + I*b*x)^(3/4)*(I - 8*a + 4*b*x) + 3*2^(3/4)*(-3*I + 4*a + (8*I)*a^2)*
Hypergeometric2F1[1/4, 5/4, 9/4, (-1/2*I)*(I + a + b*x)]))/(60*b^3)

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fricas [A]  time = 0.48, size = 561, normalized size = 1.14 \[ \frac {3 \, b^{3} \sqrt {\frac {64 i \, a^{4} + 64 \, a^{3} - 64 i \, a^{2} - 24 \, a + 9 i}{b^{6}}} \log \left (\frac {b^{3} \sqrt {\frac {64 i \, a^{4} + 64 \, a^{3} - 64 i \, a^{2} - 24 \, a + 9 i}{b^{6}}} + {\left (8 \, a^{2} - 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} - 4 i \, a - 3}\right ) - 3 \, b^{3} \sqrt {\frac {64 i \, a^{4} + 64 \, a^{3} - 64 i \, a^{2} - 24 \, a + 9 i}{b^{6}}} \log \left (-\frac {b^{3} \sqrt {\frac {64 i \, a^{4} + 64 \, a^{3} - 64 i \, a^{2} - 24 \, a + 9 i}{b^{6}}} - {\left (8 \, a^{2} - 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} - 4 i \, a - 3}\right ) - 3 \, b^{3} \sqrt {\frac {-64 i \, a^{4} - 64 \, a^{3} + 64 i \, a^{2} + 24 \, a - 9 i}{b^{6}}} \log \left (\frac {b^{3} \sqrt {\frac {-64 i \, a^{4} - 64 \, a^{3} + 64 i \, a^{2} + 24 \, a - 9 i}{b^{6}}} + {\left (8 \, a^{2} - 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} - 4 i \, a - 3}\right ) + 3 \, b^{3} \sqrt {\frac {-64 i \, a^{4} - 64 \, a^{3} + 64 i \, a^{2} + 24 \, a - 9 i}{b^{6}}} \log \left (-\frac {b^{3} \sqrt {\frac {-64 i \, a^{4} - 64 \, a^{3} + 64 i \, a^{2} + 24 \, a - 9 i}{b^{6}}} - {\left (8 \, a^{2} - 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} - 4 i \, a - 3}\right ) + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-8 i \, b^{2} x^{2} - 2 \, {\left (-4 i \, a - 5\right )} b x - 8 i \, a^{2} - 26 \, a + 11 i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{48 \, b^{3}} \]

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))^(1/2),x, algorithm="fricas")

[Out]

1/48*(3*b^3*sqrt((64*I*a^4 + 64*a^3 - 64*I*a^2 - 24*a + 9*I)/b^6)*log((b^3*sqrt((64*I*a^4 + 64*a^3 - 64*I*a^2
- 24*a + 9*I)/b^6) + (8*a^2 - 4*I*a - 3)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(8*a^2 - 4*I
*a - 3)) - 3*b^3*sqrt((64*I*a^4 + 64*a^3 - 64*I*a^2 - 24*a + 9*I)/b^6)*log(-(b^3*sqrt((64*I*a^4 + 64*a^3 - 64*
I*a^2 - 24*a + 9*I)/b^6) - (8*a^2 - 4*I*a - 3)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/(8*a^2
- 4*I*a - 3)) - 3*b^3*sqrt((-64*I*a^4 - 64*a^3 + 64*I*a^2 + 24*a - 9*I)/b^6)*log((b^3*sqrt((-64*I*a^4 - 64*a^
3 + 64*I*a^2 + 24*a - 9*I)/b^6) + (8*a^2 - 4*I*a - 3)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))
/(8*a^2 - 4*I*a - 3)) + 3*b^3*sqrt((-64*I*a^4 - 64*a^3 + 64*I*a^2 + 24*a - 9*I)/b^6)*log(-(b^3*sqrt((-64*I*a^4
- 64*a^3 + 64*I*a^2 + 24*a - 9*I)/b^6) - (8*a^2 - 4*I*a - 3)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x +
a + I)))/(8*a^2 - 4*I*a - 3)) + 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(-8*I*b^2*x^2 - 2*(-4*I*a - 5)*b*x - 8*I*a
^2 - 26*a + 11*I)*sqrt(I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/(b*x + a + I)))/b^3

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming 0=[0,0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch f
or the root of a polynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0,0]index.c
c index_m operator + Error: Bad Argument ValueWarning, need to choose a branch for the root of a polynomial wi
th parameters. This might be wrong.The choice was done assuming 0=[0,0,0]index.cc index_m operator + Error: Ba
d Argument ValueWarning, need to choose a branch for the root of a polynomial with parameters. This might be w
rong.The choice was done assuming 0=[0,0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need
to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was done assum
ing 0=[0,0,0]index.cc index_m operator + Error: Bad Argument ValueWarning, need to choose a branch for the roo
t of a polynomial with parameters. This might be wrong.The choice was done assuming 0=[0,0,0]index.cc index_m
operator + Error: Bad Argument ValueEvaluation time: 1.67Done

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}}\, dx \]

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))^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}}}\,{d x} \]

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))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {\frac {i \left (a + b x - i\right )}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}}}\, dx \]

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))**(1/2),x)

[Out]

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

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