3.209 \(\int e^{-3 i \tan ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=229 \[ -\frac{\left (-6 i a^2-18 a+11 i\right ) \sqrt{i a+i b x+1} (-i a-i b x+1)^{3/2}}{6 b^3}-\frac{\left (-6 i a^2-18 a+11 i\right ) \sqrt{i a+i b x+1} \sqrt{-i a-i b x+1}}{2 b^3}+\frac{\left (-6 a^2+18 i a+11\right ) \sinh ^{-1}(a+b x)}{2 b^3}-\frac{i \sqrt{i a+i b x+1} (-i a-i b x+1)^{5/2}}{3 b^3}+\frac{i (-a+i)^2 (-i a-i b x+1)^{5/2}}{b^3 \sqrt{i a+i b x+1}} \]

[Out]

(I*(I - a)^2*(1 - I*a - I*b*x)^(5/2))/(b^3*Sqrt[1 + I*a + I*b*x]) - ((11*I - 18*a - (6*I)*a^2)*Sqrt[1 - I*a -
I*b*x]*Sqrt[1 + I*a + I*b*x])/(2*b^3) - ((11*I - 18*a - (6*I)*a^2)*(1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*
x])/(6*b^3) - ((I/3)*(1 - I*a - I*b*x)^(5/2)*Sqrt[1 + I*a + I*b*x])/b^3 + ((11 + (18*I)*a - 6*a^2)*ArcSinh[a +
b*x])/(2*b^3)

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Rubi [A]  time = 0.168235, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {5095, 89, 80, 50, 53, 619, 215} \[ -\frac{\left (-6 i a^2-18 a+11 i\right ) \sqrt{i a+i b x+1} (-i a-i b x+1)^{3/2}}{6 b^3}-\frac{\left (-6 i a^2-18 a+11 i\right ) \sqrt{i a+i b x+1} \sqrt{-i a-i b x+1}}{2 b^3}+\frac{\left (-6 a^2+18 i a+11\right ) \sinh ^{-1}(a+b x)}{2 b^3}-\frac{i \sqrt{i a+i b x+1} (-i a-i b x+1)^{5/2}}{3 b^3}+\frac{i (-a+i)^2 (-i a-i b x+1)^{5/2}}{b^3 \sqrt{i a+i b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/E^((3*I)*ArcTan[a + b*x]),x]

[Out]

(I*(I - a)^2*(1 - I*a - I*b*x)^(5/2))/(b^3*Sqrt[1 + I*a + I*b*x]) - ((11*I - 18*a - (6*I)*a^2)*Sqrt[1 - I*a -
I*b*x]*Sqrt[1 + I*a + I*b*x])/(2*b^3) - ((11*I - 18*a - (6*I)*a^2)*(1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*
x])/(6*b^3) - ((I/3)*(1 - I*a - I*b*x)^(5/2)*Sqrt[1 + I*a + I*b*x])/b^3 + ((11 + (18*I)*a - 6*a^2)*ArcSinh[a +
b*x])/(2*b^3)

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]

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 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 53

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 619

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 215

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]

Rubi steps

\begin{align*} \int e^{-3 i \tan ^{-1}(a+b x)} x^2 \, dx &=\int \frac{x^2 (1-i a-i b x)^{3/2}}{(1+i a+i b x)^{3/2}} \, dx\\ &=\frac{i (i-a)^2 (1-i a-i b x)^{5/2}}{b^3 \sqrt{1+i a+i b x}}+\frac{i \int \frac{(1-i a-i b x)^{3/2} \left (-(i-a) (3+2 i a) b-b^2 x\right )}{\sqrt{1+i a+i b x}} \, dx}{b^3}\\ &=\frac{i (i-a)^2 (1-i a-i b x)^{5/2}}{b^3 \sqrt{1+i a+i b x}}-\frac{i (1-i a-i b x)^{5/2} \sqrt{1+i a+i b x}}{3 b^3}+\frac{\left (11+18 i a-6 a^2\right ) \int \frac{(1-i a-i b x)^{3/2}}{\sqrt{1+i a+i b x}} \, dx}{3 b^2}\\ &=\frac{i (i-a)^2 (1-i a-i b x)^{5/2}}{b^3 \sqrt{1+i a+i b x}}+\frac{\left (18 a-i \left (11-6 a^2\right )\right ) (1-i a-i b x)^{3/2} \sqrt{1+i a+i b x}}{6 b^3}-\frac{i (1-i a-i b x)^{5/2} \sqrt{1+i a+i b x}}{3 b^3}+\frac{\left (11+18 i a-6 a^2\right ) \int \frac{\sqrt{1-i a-i b x}}{\sqrt{1+i a+i b x}} \, dx}{2 b^2}\\ &=\frac{i (i-a)^2 (1-i a-i b x)^{5/2}}{b^3 \sqrt{1+i a+i b x}}+\frac{\left (18 a-i \left (11-6 a^2\right )\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{2 b^3}+\frac{\left (18 a-i \left (11-6 a^2\right )\right ) (1-i a-i b x)^{3/2} \sqrt{1+i a+i b x}}{6 b^3}-\frac{i (1-i a-i b x)^{5/2} \sqrt{1+i a+i b x}}{3 b^3}+\frac{\left (11+18 i a-6 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{i (i-a)^2 (1-i a-i b x)^{5/2}}{b^3 \sqrt{1+i a+i b x}}+\frac{\left (18 a-i \left (11-6 a^2\right )\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{2 b^3}+\frac{\left (18 a-i \left (11-6 a^2\right )\right ) (1-i a-i b x)^{3/2} \sqrt{1+i a+i b x}}{6 b^3}-\frac{i (1-i a-i b x)^{5/2} \sqrt{1+i a+i b x}}{3 b^3}+\frac{\left (11+18 i a-6 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{i (i-a)^2 (1-i a-i b x)^{5/2}}{b^3 \sqrt{1+i a+i b x}}+\frac{\left (18 a-i \left (11-6 a^2\right )\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{2 b^3}+\frac{\left (18 a-i \left (11-6 a^2\right )\right ) (1-i a-i b x)^{3/2} \sqrt{1+i a+i b x}}{6 b^3}-\frac{i (1-i a-i b x)^{5/2} \sqrt{1+i a+i b x}}{3 b^3}+\frac{\left (11+18 i a-6 a^2\right ) \operatorname{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{i (i-a)^2 (1-i a-i b x)^{5/2}}{b^3 \sqrt{1+i a+i b x}}+\frac{\left (18 a-i \left (11-6 a^2\right )\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{2 b^3}+\frac{\left (18 a-i \left (11-6 a^2\right )\right ) (1-i a-i b x)^{3/2} \sqrt{1+i a+i b x}}{6 b^3}-\frac{i (1-i a-i b x)^{5/2} \sqrt{1+i a+i b x}}{3 b^3}+\frac{\left (11+18 i a-6 a^2\right ) \sinh ^{-1}(a+b x)}{2 b^3}\\ \end{align*}

Mathematica [A]  time = 0.295245, size = 198, normalized size = 0.86 \[ \frac{a^3 (51+2 i b x)+a^2 (69 b x-50 i)+2 i a^4+a \left (2 i b^3 x^3+9 b^2 x^2-106 i b x+51\right )+i \left (2 b^4 x^4+9 i b^3 x^3-26 b^2 x^2+33 i b x-52\right )}{6 b^3 \sqrt{a^2+2 a b x+b^2 x^2+1}}+\frac{\sqrt [4]{-1} \left (-6 a^2+18 i a+11\right ) \sqrt{-i b} \sinh ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{b} \sqrt{-i (a+b x+i)}}{\sqrt{-i b}}\right )}{b^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2/E^((3*I)*ArcTan[a + b*x]),x]

[Out]

((2*I)*a^4 + a^3*(51 + (2*I)*b*x) + a^2*(-50*I + 69*b*x) + a*(51 - (106*I)*b*x + 9*b^2*x^2 + (2*I)*b^3*x^3) +
I*(-52 + (33*I)*b*x - 26*b^2*x^2 + (9*I)*b^3*x^3 + 2*b^4*x^4))/(6*b^3*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]) + ((-
1)^(1/4)*(11 + (18*I)*a - 6*a^2)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)
*b]])/b^(7/2)

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Maple [B]  time = 0.094, size = 1026, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*I/b^5/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(5/2)+9*I/b^2*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)
+((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2))/(b^2)^(1/2)*a+1/b^6/(x-I/b+a/b)^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a
)/b)*b)^(5/2)-1/b^6/(x-I/b+a/b)^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(5/2)*a^2+2*I/b^6/(x-I/b+a/b)^3*((x-(I
-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(5/2)*a-2*I/b^5/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(5/2)*a^2+
9*I/b^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*x*a-11/3*I/b^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(3/2)
-3/b^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*a^3-3/b^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*x*a^2
-3/b^2*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2))/(b^2)^(1/2)*a^2+9*I/b
^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*a^2+2*I/b^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(3/2)*a^2+6/b
^3*a*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(3/2)+11/2/b^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*x+11/2/b
^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*a+11/2/b^2*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((x-(I-a)/b)^2*
b^2+2*I*(x-(I-a)/b)*b)^(1/2))/(b^2)^(1/2)-6/b^5*a/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(5/2)

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Maxima [B]  time = 1.56663, size = 842, normalized size = 3.68 \begin{align*} \frac{i \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}} a^{2}}{b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3} - 2 i \, b^{4} x - 2 i \, a b^{3} - b^{3}} + \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}} a}{b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3} - 2 i \, b^{4} x - 2 i \, a b^{3} - b^{3}} + \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}} a}{2 i \, b^{4} x + 2 i \, a b^{3} + 2 \, b^{3}} + \frac{6 i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{i \, b^{4} x + i \, a b^{3} + b^{3}} - \frac{i \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}}}{b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3} - 2 i \, b^{4} x - 2 i \, a b^{3} - b^{3}} - \frac{2 i \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}}}{2 i \, b^{4} x + 2 i \, a b^{3} + 2 \, b^{3}} + \frac{12 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{i \, b^{4} x + i \, a b^{3} + b^{3}} - \frac{3 \, a^{2} \operatorname{arsinh}\left (b x + a\right )}{b^{3}} - \frac{6 i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{i \, b^{4} x + i \, a b^{3} + b^{3}} - \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 4 i \, b x + 4 i \, a + 3} x}{2 \, b^{2}} + \frac{9 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}{b^{3}} - \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 4 i \, b x + 4 i \, a + 3} a}{2 \, b^{3}} + \frac{\arcsin \left (i \, b x + i \, a + 2\right )}{2 \, b^{3}} + \frac{6 \, \operatorname{arsinh}\left (b x + a\right )}{b^{3}} - \frac{3 i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b^{3}} + \frac{i \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 4 i \, b x + 4 i \, a + 3}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a^2/(b^5*x^2 + 2*a*b^4*x + a^2*b^3 - 2*I*b^4*x - 2*I*a*b^3 - b^3) + 2*(b
^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a/(b^5*x^2 + 2*a*b^4*x + a^2*b^3 - 2*I*b^4*x - 2*I*a*b^3 - b^3) + 2*(b^2*x^2
+ 2*a*b*x + a^2 + 1)^(3/2)*a/(2*I*b^4*x + 2*I*a*b^3 + 2*b^3) + 6*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2/(I*b
^4*x + I*a*b^3 + b^3) - I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/(b^5*x^2 + 2*a*b^4*x + a^2*b^3 - 2*I*b^4*x - 2*I
*a*b^3 - b^3) - 2*I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/(2*I*b^4*x + 2*I*a*b^3 + 2*b^3) + 12*sqrt(b^2*x^2 + 2*
a*b*x + a^2 + 1)*a/(I*b^4*x + I*a*b^3 + b^3) - 3*a^2*arcsinh(b*x + a)/b^3 - 6*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 +
1)/(I*b^4*x + I*a*b^3 + b^3) - 1/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)*x/b^2 + 9*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*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a/b^3 - 1/
2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)*a/b^3 + 1/2*arcsin(I*b*x + I*a + 2)/b^3 + 6*arcsinh(b*x
+ a)/b^3 - 3*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/b^3 + I*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 4*I*b*x + 4*I*a + 3)
/b^3

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Fricas [A]  time = 2.43877, size = 485, normalized size = 2.12 \begin{align*} \frac{7 i \, a^{4} + 166 \, a^{3} +{\left (7 i \, a^{3} + 159 \, a^{2} - 249 i \, a - 96\right )} b x - 408 i \, a^{2} +{\left (72 \, a^{3} + 12 \,{\left (6 \, a^{2} - 18 i \, a - 11\right )} b x - 288 i \, a^{2} - 348 \, a + 132 i\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) +{\left (8 i \, b^{3} x^{3} - 28 \, b^{2} x^{2} + 8 i \, a^{3} +{\left (64 \, a - 76 i\right )} b x + 212 \, a^{2} - 412 i \, a - 208\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 345 \, a + 96 i}{24 \, b^{4} x +{\left (24 \, a - 24 i\right )} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*I*a^4 + 166*a^3 + (7*I*a^3 + 159*a^2 - 249*I*a - 96)*b*x - 408*I*a^2 + (72*a^3 + 12*(6*a^2 - 18*I*a - 11)*b
*x - 288*I*a^2 - 348*a + 132*I)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + (8*I*b^3*x^3 - 28*b^2*x^2
+ 8*I*a^3 + (64*a - 76*I)*b*x + 212*a^2 - 412*I*a - 208)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 345*a + 96*I)/(24
*b^4*x + (24*a - 24*I)*b^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.1758, size = 386, normalized size = 1.69 \begin{align*} \frac{1}{6} \, \sqrt{{\left (b x + a\right )}^{2} + 1}{\left ({\left (\frac{2 \, i x}{b} - \frac{2 \, a b^{6} i + 9 \, b^{6}}{b^{8}}\right )} x + \frac{2 \, a^{2} b^{5} i + 27 \, a b^{5} - 28 \, b^{5} i}{b^{8}}\right )} + \frac{{\left (6 \, a^{2} i + 18 \, a - 11 \, i\right )} \log \left (3 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b i + a^{3} b i +{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}^{3} i{\left | b \right |} + 3 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )} a^{2} i{\left | b \right |} + 2 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}^{2} b + 2 \, a^{2} b - a b i + 4 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )} a{\left | b \right |} -{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )} i{\left | b \right |}\right )}{6 \, b^{2} i{\left | b \right |}} + \frac{{\left (6 \, a^{2}{\left | b \right |} - 18 \, a i{\left | b \right |} - 11 \,{\left | b \right |}\right )} \log \left (8 \, b^{3}\right )}{3 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt((b*x + a)^2 + 1)*((2*i*x/b - (2*a*b^6*i + 9*b^6)/b^8)*x + (2*a^2*b^5*i + 27*a*b^5 - 28*b^5*i)/b^8) +
1/6*(6*a^2*i + 18*a - 11*i)*log(3*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a*b*i + a^3*b*i + (x*abs(b) - sqrt((b*x
+ a)^2 + 1))^3*i*abs(b) + 3*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a^2*i*abs(b) + 2*(x*abs(b) - sqrt((b*x + a)^2
+ 1))^2*b + 2*a^2*b - a*b*i + 4*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a*abs(b) - (x*abs(b) - sqrt((b*x + a)^2 + 1
))*i*abs(b))/(b^2*i*abs(b)) + 1/3*(6*a^2*abs(b) - 18*a*i*abs(b) - 11*abs(b))*log(8*b^3)/b^4