3.3.9 \(\int e^{-3 i \text {ArcTan}(a+b x)} x^2 \, dx\) [209]

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

[Out]

1/2*(11+18*I*a-6*a^2)*arcsinh(b*x+a)/b^3+I*(I-a)^2*(1-I*a-I*b*x)^(5/2)/b^3/(1+I*a+I*b*x)^(1/2)-1/6*(11*I-18*a-
6*I*a^2)*(1-I*a-I*b*x)^(3/2)*(1+I*a+I*b*x)^(1/2)/b^3-1/3*I*(1-I*a-I*b*x)^(5/2)*(1+I*a+I*b*x)^(1/2)/b^3-1/2*(11
*I-18*a-6*I*a^2)*(1-I*a-I*b*x)^(1/2)*(1+I*a+I*b*x)^(1/2)/b^3

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Rubi [A]
time = 0.12, antiderivative size = 229, normalized size of antiderivative = 1.00, 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 = {5203, 91, 81, 52, 55, 633, 221} \begin {gather*} -\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}} \end {gather*}

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 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 91

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 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^{-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 ) \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 {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*}

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Mathematica [A]
time = 0.24, size = 198, normalized size = 0.86 \begin {gather*} \frac {2 i a^4+a^3 (51+2 i b x)+a^2 (-50 i+69 b x)+a \left (51-106 i b x+9 b^2 x^2+2 i b^3 x^3\right )+i \left (-52+33 i b x-26 b^2 x^2+9 i b^3 x^3+2 b^4 x^4\right )}{6 b^3 \sqrt {1+a^2+2 a b x+b^2 x^2}}+\frac {\sqrt [4]{-1} \left (11+18 i a-6 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*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 798 vs. \(2 (181 ) = 362\).
time = 0.17, size = 799, normalized size = 3.49

method	result	size
risch	\(\frac {i \left (2 b^{2} x^{2}-2 a b x +9 i b x +2 a^{2}-27 i a -28\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{6 b^{3}}+\frac {11 \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{2} \sqrt {b^{2}}}+\frac {9 i \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right ) a}{b^{2} \sqrt {b^{2}}}-\frac {3 \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right ) a^{2}}{b^{2} \sqrt {b^{2}}}+\frac {4 \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\, a^{2}}{b^{4} \left (x -\frac {i}{b}+\frac {a}{b}\right )}-\frac {4 \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{b^{4} \left (x -\frac {i}{b}+\frac {a}{b}\right )}-\frac {8 i \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\, a}{b^{4} \left (x -\frac {i}{b}+\frac {a}{b}\right )}\)	\(381\)
default	\(\frac {2 i \left (i-a \right ) \left (-\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{2}}+3 i b \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\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 )}{2 \sqrt {b^{2}}}\right )\right )\right )}{b^{4}}+\frac {i \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\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 )}{2 \sqrt {b^{2}}}\right )\right )}{b^{3}}+\frac {i \left (i-a \right )^{2} \left (\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{3}}-2 i b \left (-\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{2}}+3 i b \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\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 )}{2 \sqrt {b^{2}}}\right )\right )\right )\right )}{b^{5}}\)	\(799\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 624 vs. \(2 (155) = 310\).
time = 0.48, size = 624, normalized size = 2.72 \begin {gather*} \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 {gather*}

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 = 2.99, size = 174, normalized size = 0.76 \begin {gather*} \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} + 12 \, {\left (6 \, a^{3} + {\left (6 \, a^{2} - 18 i \, a - 11\right )} b x - 24 i \, a^{2} - 29 \, a + 11 i\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 4 \, {\left (-2 i \, b^{3} x^{3} + 7 \, b^{2} x^{2} - 2 i \, a^{3} - {\left (16 \, a - 19 i\right )} b x - 53 \, a^{2} + 103 i \, a + 52\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 345 \, a + 96 i}{24 \, {\left (b^{4} x + {\left (a - i\right )} b^{3}\right )}} \end {gather*}

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]

1/24*(7*I*a^4 + 166*a^3 + (7*I*a^3 + 159*a^2 - 249*I*a - 96)*b*x - 408*I*a^2 + 12*(6*a^3 + (6*a^2 - 18*I*a - 1
1)*b*x - 24*I*a^2 - 29*a + 11*I)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - 4*(-2*I*b^3*x^3 + 7*b^2*x
^2 - 2*I*a^3 - (16*a - 19*I)*b*x - 53*a^2 + 103*I*a + 52)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 345*a + 96*I)/(b
^4*x + (a - I)*b^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i \left (\int \frac {x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx + \int \frac {a^{2} x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx + \int \frac {b^{2} x^{4} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx + \int \frac {2 a b x^{3} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{a^{3} + 3 a^{2} b x - 3 i a^{2} + 3 a b^{2} x^{2} - 6 i a b x - 3 a + b^{3} x^{3} - 3 i b^{2} x^{2} - 3 b x + i}\, dx\right ) \end {gather*}

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]

I*(Integral(x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 + 3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*
x - 3*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x) + Integral(a**2*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)
/(a**3 + 3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x)
+ Integral(b**2*x**4*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)/(a**3 + 3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*
a*b*x - 3*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I), x) + Integral(2*a*b*x**3*sqrt(a**2 + 2*a*b*x + b**2*x**2
+ 1)/(a**3 + 3*a**2*b*x - 3*I*a**2 + 3*a*b**2*x**2 - 6*I*a*b*x - 3*a + b**3*x**3 - 3*I*b**2*x**2 - 3*b*x + I)
, x))

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Giac [A]
time = 0.46, size = 243, normalized size = 1.06 \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^{6} + 9 \, b^{6}}{b^{8}}\right )} + \frac {-2 i \, a^{2} b^{5} - 27 \, a b^{5} + 28 i \, b^{5}}{b^{8}}\right )} + \frac {{\left (6 \, a^{2} - 18 i \, a - 11\right )} \log \left (3 i \, {\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} {\left | b \right |} - 3 \, {\left (-i \, x {\left | b \right |} + i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} {\left | b \right |} + 2 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b + 2 \, a^{2} b + 4 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} - i \, a b + {\left (-i \, x {\left | b \right |} + i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{6 \, 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))^3*(1+(b*x+a)^2)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2\,{\left ({\left (a+b\,x\right )}^2+1\right )}^{3/2}}{{\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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