3.3.10 \(\int e^{-3 i \text {ArcTan}(a+b x)} x \, dx\) [210]

Optimal. Leaf size=163 \[ -\frac {(1+i a) (1-i a-i b x)^{5/2}}{b^2 \sqrt {1+i a+i b x}}-\frac {3 (3+2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}-\frac {(3+2 i a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}-\frac {3 (3 i-2 a) \sinh ^{-1}(a+b x)}{2 b^2} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5203, 79, 52, 55, 633, 221} \begin {gather*} -\frac {(1+i a) (-i a-i b x+1)^{5/2}}{b^2 \sqrt {i a+i b x+1}}-\frac {(3+2 i a) \sqrt {i a+i b x+1} (-i a-i b x+1)^{3/2}}{2 b^2}-\frac {3 (3+2 i a) \sqrt {i a+i b x+1} \sqrt {-i a-i b x+1}}{2 b^2}-\frac {3 (-2 a+3 i) \sinh ^{-1}(a+b x)}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/E^((3*I)*ArcTan[a + b*x]),x]

[Out]

-(((1 + I*a)*(1 - I*a - I*b*x)^(5/2))/(b^2*Sqrt[1 + I*a + I*b*x])) - (3*(3 + (2*I)*a)*Sqrt[1 - I*a - I*b*x]*Sq
rt[1 + I*a + I*b*x])/(2*b^2) - ((3 + (2*I)*a)*(1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x])/(2*b^2) - (3*(3*I
- 2*a)*ArcSinh[a + b*x])/(2*b^2)

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 79

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

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 \, dx &=\int \frac {x (1-i a-i b x)^{3/2}}{(1+i a+i b x)^{3/2}} \, dx\\ &=-\frac {(1+i a) (1-i a-i b x)^{5/2}}{b^2 \sqrt {1+i a+i b x}}-\frac {(3 i-2 a) \int \frac {(1-i a-i b x)^{3/2}}{\sqrt {1+i a+i b x}} \, dx}{b}\\ &=-\frac {(1+i a) (1-i a-i b x)^{5/2}}{b^2 \sqrt {1+i a+i b x}}-\frac {(3+2 i a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}-\frac {(3 (3 i-2 a)) \int \frac {\sqrt {1-i a-i b x}}{\sqrt {1+i a+i b x}} \, dx}{2 b}\\ &=-\frac {(1+i a) (1-i a-i b x)^{5/2}}{b^2 \sqrt {1+i a+i b x}}-\frac {3 (3+2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}-\frac {(3+2 i a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}-\frac {(3 (3 i-2 a)) \int \frac {1}{\sqrt {1-i a-i b x} \sqrt {1+i a+i b x}} \, dx}{2 b}\\ &=-\frac {(1+i a) (1-i a-i b x)^{5/2}}{b^2 \sqrt {1+i a+i b x}}-\frac {3 (3+2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}-\frac {(3+2 i a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}-\frac {(3 (3 i-2 a)) \int \frac {1}{\sqrt {(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{2 b}\\ &=-\frac {(1+i a) (1-i a-i b x)^{5/2}}{b^2 \sqrt {1+i a+i b x}}-\frac {3 (3+2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}-\frac {(3+2 i a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}-\frac {(3 (3 i-2 a)) \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^3}\\ &=-\frac {(1+i a) (1-i a-i b x)^{5/2}}{b^2 \sqrt {1+i a+i b x}}-\frac {3 (3+2 i a) \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{2 b^2}-\frac {(3+2 i a) (1-i a-i b x)^{3/2} \sqrt {1+i a+i b x}}{2 b^2}-\frac {3 (3 i-2 a) \sinh ^{-1}(a+b x)}{2 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 157, normalized size = 0.96 \begin {gather*} \frac {i \left (14 i-a^3+9 b x+6 i b^2 x^2+b^3 x^3+a^2 (14 i-b x)+a \left (-1+20 i b x+b^2 x^2\right )\right )}{2 b^2 \sqrt {1+a^2+2 a b x+b^2 x^2}}+\frac {3 \sqrt [4]{-1} (-3 i+2 a) \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^{5/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((I/2)*(14*I - a^3 + 9*b*x + (6*I)*b^2*x^2 + b^3*x^3 + a^2*(14*I - b*x) + a*(-1 + (20*I)*b*x + b^2*x^2)))/(b^2
*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]) + (3*(-1)^(1/4)*(-3*I + 2*a)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqr
t[(-I)*(I + a + b*x)])/Sqrt[(-I)*b]])/b^(5/2)

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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. 594 vs. \(2 (129 ) = 258\).
time = 0.16, size = 595, normalized size = 3.65

method	result	size
risch	\(-\frac {i \left (-b x +a -6 i\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{2}}-\frac {9 i \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b \sqrt {b^{2}}}+\frac {3 a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b \sqrt {b^{2}}}-\frac {4 \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\, a}{b^{3} \left (x -\frac {i}{b}+\frac {a}{b}\right )}+\frac {4 i \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{b^{3} \left (x -\frac {i}{b}+\frac {a}{b}\right )}\)	\(249\)
default	\(\frac {i \left (-\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{2}}+3 i b \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{2 \sqrt {b^{2}}}\right )\right )\right )}{b^{3}}+\frac {i \left (i-a \right ) \left (\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{3}}-2 i b \left (-\frac {i \left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {5}{2}}}{b \left (x -\frac {i-a}{b}\right )^{2}}+3 i b \left (\frac {\left (\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )\right )^{\frac {3}{2}}}{3}+i b \left (\frac {\left (2 \left (x -\frac {i-a}{b}\right ) b^{2}+2 i b \right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{4 b^{2}}+\frac {\ln \left (\frac {i b +\left (x -\frac {i-a}{b}\right ) b^{2}}{\sqrt {b^{2}}}+\sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}\right )}{2 \sqrt {b^{2}}}\right )\right )\right )\right )}{b^{4}}\)	\(595\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I/b^3*(-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*(I-a)/b^4*(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. 293 vs. \(2 (113) = 226\).
time = 0.47, size = 293, normalized size = 1.80 \begin {gather*} -\frac {i \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a}{b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2} - 2 i \, b^{3} x - 2 i \, a b^{2} - b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2} - 2 i \, b^{3} x - 2 i \, a b^{2} - b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{2 i \, b^{3} x + 2 i \, a b^{2} + 2 \, b^{2}} - \frac {6 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{i \, b^{3} x + i \, a b^{2} + b^{2}} + \frac {3 \, a \operatorname {arsinh}\left (b x + a\right )}{b^{2}} - \frac {6 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{i \, b^{3} x + i \, a b^{2} + b^{2}} - \frac {9 i \, \operatorname {arsinh}\left (b x + a\right )}{2 \, b^{2}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 1.90, size = 136, normalized size = 0.83 \begin {gather*} \frac {-3 i \, a^{3} + {\left (-3 i \, a^{2} - 44 \, a + 32 i\right )} b x - 47 \, a^{2} - 12 \, {\left ({\left (2 \, a - 3 i\right )} b x + 2 \, a^{2} - 5 i \, a - 3\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 (-i \, b^{2} x^{2} + i \, a^{2} + 5 \, b x + 15 \, a - 14 i\right )} + 76 i \, a + 32}{8 \, {\left (b^{3} x + {\left (a - i\right )} b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(-3*I*a^3 + (-3*I*a^2 - 44*a + 32*I)*b*x - 47*a^2 - 12*((2*a - 3*I)*b*x + 2*a^2 - 5*I*a - 3)*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)*(-I*b^2*x^2 + I*a^2 + 5*b*x + 15*a
- 14*I) + 76*I*a + 32)/(b^3*x + (a - I)*b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(Integral(x*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*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) + Inte
gral(b**2*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) + Integral(2*a*b*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))

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Giac [A]
time = 0.46, size = 210, normalized size = 1.29 \begin {gather*} -\frac {1}{2} \, \sqrt {{\left (b x + a\right )}^{2} + 1} {\left (-\frac {i \, x}{b} - \frac {-i \, a b^{2} - 6 \, b^{2}}{b^{4}}\right )} - \frac {{\left (2 \, a - 3 i\right )} \log \left (3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b + a^{3} b + {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{3} {\left | b \right |} + 3 \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a^{2} {\left | b \right |} - 2 i \, {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}^{2} b - 2 i \, a^{2} b + 4 \, {\left (-i \, x {\left | b \right |} + i \, \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} a {\left | b \right |} - a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, b {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,{\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*((a + b*x)^2 + 1)^(3/2))/(a*1i + b*x*1i + 1)^3,x)

[Out]

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

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