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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.119072, antiderivative size = 163, normalized size of antiderivative = 1., 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 = {5095, 78, 50, 53, 619, 215} \[ -\frac{(1+i a) (-i a-i b x+1)^{5/2}}{b^2 \sqrt{i a+i b x+1}}-\frac{(3+2 i a) \sqrt{i a+i b x+1} (-i a-i b x+1)^{3/2}}{2 b^2}-\frac{3 (3+2 i a) \sqrt{i a+i b x+1} \sqrt{-i a-i b x+1}}{2 b^2}-\frac{3 (-2 a+3 i) \sinh ^{-1}(a+b x)}{2 b^2} \]

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 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 78

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] || LtQ
[p, n]))))

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 \, 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)) \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^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*}

Mathematica [A]  time = 0.299525, size = 157, normalized size = 0.96 \[ \frac{i \left (a^2 (-b x+14 i)-a^3+a \left (b^2 x^2+20 i b x-1\right )+b^3 x^3+6 i b^2 x^2+9 b x+14 i\right )}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2+1}}+\frac{3 \sqrt [4]{-1} (2 a-3 i) \sqrt{-i b} \sinh ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{b} \sqrt{-i (a+b x+i)}}{\sqrt{-i b}}\right )}{b^{5/2}} \]

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)

________________________________________________________________________________________

Maple [B]  time = 0.086, size = 676, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.55039, size = 396, normalized size = 2.43 \begin{align*} -\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{align*}

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

________________________________________________________________________________________

Fricas [A]  time = 2.34392, size = 367, normalized size = 2.25 \begin{align*} \frac{-3 i \, a^{3} +{\left (-3 i \, a^{2} - 44 \, a + 32 i\right )} b x - 47 \, a^{2} -{\left ({\left (24 \, a - 36 i\right )} b x + 24 \, a^{2} - 60 i \, a - 36\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (4 i \, b^{2} x^{2} - 4 i \, a^{2} - 20 \, b x - 60 \, a + 56 i\right )} + 76 i \, a + 32}{8 \, b^{3} x +{\left (8 \, a - 8 i\right )} b^{2}} \end{align*}

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]

(-3*I*a^3 + (-3*I*a^2 - 44*a + 32*I)*b*x - 47*a^2 - ((24*a - 36*I)*b*x + 24*a^2 - 60*I*a - 36)*log(-b*x - a +
sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(4*I*b^2*x^2 - 4*I*a^2 - 20*b*x - 60*a
+ 56*I) + 76*I*a + 32)/(8*b^3*x + (8*a - 8*I)*b^2)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.15398, size = 325, normalized size = 1.99 \begin{align*} -\frac{1}{2} \, \sqrt{{\left (b x + a\right )}^{2} + 1}{\left (\frac{x}{b i} - \frac{a b^{2} - 6 \, b^{2} i}{b^{4} i}\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 - 2 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}^{2} b i - 2 \, a^{2} b i +{\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 |} - 4 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )} a i{\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 |}} - \frac{{\left (2 \, a{\left | b \right |} - 3 \, i{\left | b \right |}\right )} \log \left (24 \, b^{2}\right )}{b^{3}} \end{align*}

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)*(x/(b*i) - (a*b^2 - 6*b^2*i)/(b^4*i)) - 1/2*(2*a - 3*i)*log(3*(x*abs(b) - sqrt((b*x
+ a)^2 + 1))^2*a*b + a^3*b - 2*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*b*i - 2*a^2*b*i + (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) - 4*(x*abs(b) - sqrt((b*x + a)^2 + 1))*
a*i*abs(b) - a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/(b*abs(b)) - (2*a*abs(b) - 3*i*abs(b))*log(24*b^
2)/b^3