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

Optimal. Leaf size=201 \[ -\frac{(-i a-i b x+1)^{3/2} \sqrt{i a+i b x+1} \left (-18 a^2-2 (-6 a+i) b x+10 i a+7\right )}{24 b^4}-\frac{\left (-8 i a^3-12 a^2+12 i a+3\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{8 b^4}-\frac{\left (8 a^3-12 i a^2-12 a+3 i\right ) \sinh ^{-1}(a+b x)}{8 b^4}+\frac{x^2 (-i a-i b x+1)^{3/2} \sqrt{i a+i b x+1}}{4 b^2} \]

[Out]

-((3 + (12*I)*a - 12*a^2 - (8*I)*a^3)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(8*b^4) + (x^2*(1 - I*a - I
*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x])/(4*b^2) - ((1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x]*(7 + (10*I)*a - 18
*a^2 - 2*(I - 6*a)*b*x))/(24*b^4) - ((3*I - 12*a - (12*I)*a^2 + 8*a^3)*ArcSinh[a + b*x])/(8*b^4)

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Rubi [A]  time = 0.194344, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {5095, 100, 147, 50, 53, 619, 215} \[ -\frac{(-i a-i b x+1)^{3/2} \sqrt{i a+i b x+1} \left (-18 a^2-2 (-6 a+i) b x+10 i a+7\right )}{24 b^4}-\frac{\left (-8 i a^3-12 a^2+12 i a+3\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{8 b^4}-\frac{\left (8 a^3-12 i a^2-12 a+3 i\right ) \sinh ^{-1}(a+b x)}{8 b^4}+\frac{x^2 (-i a-i b x+1)^{3/2} \sqrt{i a+i b x+1}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/E^(I*ArcTan[a + b*x]),x]

[Out]

-((3 + (12*I)*a - 12*a^2 - (8*I)*a^3)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(8*b^4) + (x^2*(1 - I*a - I
*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x])/(4*b^2) - ((1 - I*a - I*b*x)^(3/2)*Sqrt[1 + I*a + I*b*x]*(7 + (10*I)*a - 18
*a^2 - 2*(I - 6*a)*b*x))/(24*b^4) - ((3*I - 12*a - (12*I)*a^2 + 8*a^3)*ArcSinh[a + b*x])/(8*b^4)

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 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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^{-i \tan ^{-1}(a+b x)} x^3 \, dx &=\int \frac{x^3 \sqrt{1-i a-i b x}}{\sqrt{1+i a+i b x}} \, dx\\ &=\frac{x^2 (1-i a-i b x)^{3/2} \sqrt{1+i a+i b x}}{4 b^2}+\frac{\int \frac{x \sqrt{1-i a-i b x} \left (-2 \left (1+a^2\right )+(i-6 a) b x\right )}{\sqrt{1+i a+i b x}} \, dx}{4 b^2}\\ &=\frac{x^2 (1-i a-i b x)^{3/2} \sqrt{1+i a+i b x}}{4 b^2}-\frac{(1-i a-i b x)^{3/2} \sqrt{1+i a+i b x} \left (7+10 i a-18 a^2-2 (i-6 a) b x\right )}{24 b^4}-\frac{\left (3 i-12 a-12 i a^2+8 a^3\right ) \int \frac{\sqrt{1-i a-i b x}}{\sqrt{1+i a+i b x}} \, dx}{8 b^3}\\ &=-\frac{\left (3+12 i a-12 a^2-8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}+\frac{x^2 (1-i a-i b x)^{3/2} \sqrt{1+i a+i b x}}{4 b^2}-\frac{(1-i a-i b x)^{3/2} \sqrt{1+i a+i b x} \left (7+10 i a-18 a^2-2 (i-6 a) b x\right )}{24 b^4}-\frac{\left (3 i-12 a-12 i a^2+8 a^3\right ) \int \frac{1}{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{8 b^3}\\ &=-\frac{\left (3+12 i a-12 a^2-8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}+\frac{x^2 (1-i a-i b x)^{3/2} \sqrt{1+i a+i b x}}{4 b^2}-\frac{(1-i a-i b x)^{3/2} \sqrt{1+i a+i b x} \left (7+10 i a-18 a^2-2 (i-6 a) b x\right )}{24 b^4}-\frac{\left (3 i-12 a-12 i a^2+8 a^3\right ) \int \frac{1}{\sqrt{(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{8 b^3}\\ &=-\frac{\left (3+12 i a-12 a^2-8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}+\frac{x^2 (1-i a-i b x)^{3/2} \sqrt{1+i a+i b x}}{4 b^2}-\frac{(1-i a-i b x)^{3/2} \sqrt{1+i a+i b x} \left (7+10 i a-18 a^2-2 (i-6 a) b x\right )}{24 b^4}-\frac{\left (3 i-12 a-12 i a^2+8 a^3\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 )}{16 b^5}\\ &=-\frac{\left (3+12 i a-12 a^2-8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}+\frac{x^2 (1-i a-i b x)^{3/2} \sqrt{1+i a+i b x}}{4 b^2}-\frac{(1-i a-i b x)^{3/2} \sqrt{1+i a+i b x} \left (7+10 i a-18 a^2-2 (i-6 a) b x\right )}{24 b^4}-\frac{\left (3 i-12 a-12 i a^2+8 a^3\right ) \sinh ^{-1}(a+b x)}{8 b^4}\\ \end{align*}

Mathematica [A]  time = 0.472465, size = 202, normalized size = 1. \[ \frac{\sqrt{i a+i b x+1} \left (5 a^2 (1-6 i b x)+6 a^4-38 i a^3+i a \left (18 b^2 x^2+50 i b x-23\right )-6 b^4 x^4-14 i b^3 x^3+17 b^2 x^2+25 i b x-16\right )}{24 b^4 \sqrt{-i (a+b x+i)}}+\frac{(-1)^{3/4} \left (8 i a^3+12 a^2-12 i a-3\right ) \sqrt{-i b} \sinh ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{b} \sqrt{-i (a+b x+i)}}{\sqrt{-i b}}\right )}{4 b^{9/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 + I*a + I*b*x]*(-16 - (38*I)*a^3 + 6*a^4 + (25*I)*b*x + 17*b^2*x^2 - (14*I)*b^3*x^3 - 6*b^4*x^4 + 5*a^
2*(1 - (6*I)*b*x) + I*a*(-23 + (50*I)*b*x + 18*b^2*x^2)))/(24*b^4*Sqrt[(-I)*(I + a + b*x)]) + ((-1)^(3/4)*(-3
- (12*I)*a + 12*a^2 + (8*I)*a^3)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)
*b]])/(4*b^(9/2))

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Maple [B]  time = 0.122, size = 894, 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^3/(1+I*(b*x+a))*(1+(b*x+a)^2)^(1/2),x)

[Out]

-3/2*I/b^3*a^2*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)+3*I/b^3*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^2-3/2*I/b^4*a^3*(b^2*x^2+2*a*b*
x+a^2+1)^(1/2)+I/b^4*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*a^3-3/2*I/b^3*a^2*x*(b^2*x^2+2*a*b*x+a^2+1)^(
1/2)+5/8*I/b^3*x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3/2*a/b^3*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1
/2))/(b^2)^(1/2)+3/b^4*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*a^2-3/2/b^4*a^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/
2)+1/3/b^4*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-I/b^3*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)-1/b^3*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^3+3/b^3*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+5/8*I/b^3*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)+5/8*I/b^4*(b^
2*x^2+2*a*b*x+a^2+1)^(1/2)*a+3/4*I/b^4*a*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-1/4*I/b^3*x*(b^2*x^2+2*a*b*x+a^2+1)^(3/
2)-3/2/b^3*a*x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3*I/b^4*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*a-1/b^4*((x-(
I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)

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Maxima [B]  time = 1.56296, size = 416, normalized size = 2.07 \begin{align*} -\frac{3 i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2} x}{2 \, b^{3}} - \frac{a^{3} \operatorname{arsinh}\left (b x + a\right )}{b^{4}} - \frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{3}}{2 \, b^{4}} - \frac{i \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}} x}{4 \, b^{3}} - \frac{3 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{2 \, b^{3}} + \frac{3 i \, a^{2} \operatorname{arsinh}\left (b x + a\right )}{2 \, b^{4}} + \frac{3 i \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}} a}{4 \, b^{4}} + \frac{3 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{2 \, b^{4}} + \frac{5 i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{8 \, b^{3}} + \frac{3 \, a \operatorname{arsinh}\left (b x + a\right )}{2 \, b^{4}} + \frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}}}{3 \, b^{4}} - \frac{19 i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{8 \, b^{4}} - \frac{3 i \, \operatorname{arsinh}\left (b x + a\right )}{8 \, b^{4}} - \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2*x/b^3 - a^3*arcsinh(b*x + a)/b^4 - 1/2*I*sqrt(b^2*x^2 + 2*a*b*x +
a^2 + 1)*a^3/b^4 - 1/4*I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*x/b^3 - 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*
x/b^3 + 3/2*I*a^2*arcsinh(b*x + a)/b^4 + 3/4*I*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a/b^4 + 3/2*sqrt(b^2*x^2 +
2*a*b*x + a^2 + 1)*a^2/b^4 + 5/8*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*x/b^3 + 3/2*a*arcsinh(b*x + a)/b^4 + 1/3*
(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/b^4 - 19/8*I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a/b^4 - 3/8*I*arcsinh(b*x +
a)/b^4 - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/b^4

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Fricas [A]  time = 2.23941, size = 389, normalized size = 1.94 \begin{align*} \frac{45 i \, a^{4} + 224 \, a^{3} - 192 i \, a^{2} +{\left (192 \, a^{3} - 288 i \, a^{2} - 288 \, a + 72 i\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) +{\left (-48 i \, b^{3} x^{3} - 16 \,{\left (-3 i \, a - 4\right )} b^{2} x^{2} + 48 i \, a^{3} +{\left (-48 i \, a^{2} - 160 \, a + 72 i\right )} b x + 352 \, a^{2} - 312 i \, a - 128\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 72 \, a}{192 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(45*I*a^4 + 224*a^3 - 192*I*a^2 + (192*a^3 - 288*I*a^2 - 288*a + 72*I)*log(-b*x - a + sqrt(b^2*x^2 + 2*a
*b*x + a^2 + 1)) + (-48*I*b^3*x^3 - 16*(-3*I*a - 4)*b^2*x^2 + 48*I*a^3 + (-48*I*a^2 - 160*a + 72*I)*b*x + 352*
a^2 - 312*I*a - 128)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 72*a)/b^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{i a + i b x + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13196, size = 220, normalized size = 1.09 \begin{align*} -\frac{1}{24} \, \sqrt{{\left (b x + a\right )}^{2} + 1}{\left ({\left (2 \,{\left (\frac{3 \, i x}{b} - \frac{3 \, a b^{5} i + 4 \, b^{5}}{b^{7}}\right )} x + \frac{6 \, a^{2} b^{4} i + 20 \, a b^{4} - 9 \, b^{4} i}{b^{7}}\right )} x - \frac{6 \, a^{3} b^{3} i + 44 \, a^{2} b^{3} - 39 \, a b^{3} i - 16 \, b^{3}}{b^{7}}\right )} + \frac{{\left (8 \, a^{3} - 12 \, a^{2} i - 12 \, a + 3 \, i\right )} \log \left (-a b -{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}{\left | b \right |}\right )}{8 \, b^{3}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*sqrt((b*x + a)^2 + 1)*((2*(3*i*x/b - (3*a*b^5*i + 4*b^5)/b^7)*x + (6*a^2*b^4*i + 20*a*b^4 - 9*b^4*i)/b^7
)*x - (6*a^3*b^3*i + 44*a^2*b^3 - 39*a*b^3*i - 16*b^3)/b^7) + 1/8*(8*a^3 - 12*a^2*i - 12*a + 3*i)*log(-a*b - (
x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/(b^3*abs(b))