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

Optimal. Leaf size=72 \[ -\frac {2 (1-i a)^3 \log (a+b x+i)}{b^4}+\frac {2 i (a+i)^2 x}{b^3}+\frac {(1-i a) x^2}{b^2}+\frac {2 i x^3}{3 b}-\frac {x^4}{4} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5095, 77} \[ \frac {(1-i a) x^2}{b^2}+\frac {2 i (a+i)^2 x}{b^3}-\frac {2 (1-i a)^3 \log (a+b x+i)}{b^4}+\frac {2 i x^3}{3 b}-\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[E^((2*I)*ArcTan[a + b*x])*x^3,x]

[Out]

((2*I)*(I + a)^2*x)/b^3 + ((1 - I*a)*x^2)/b^2 + (((2*I)/3)*x^3)/b - x^4/4 - (2*(1 - I*a)^3*Log[I + a + b*x])/b
^4

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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]

Rubi steps

\begin {align*} \int e^{2 i \tan ^{-1}(a+b x)} x^3 \, dx &=\int \frac {x^3 (1+i a+i b x)}{1-i a-i b x} \, dx\\ &=\int \left (\frac {2 i (i+a)^2}{b^3}+\frac {2 (1-i a) x}{b^2}+\frac {2 i x^2}{b}-x^3+\frac {2 (-1+i a)^3}{b^3 (i+a+b x)}\right ) \, dx\\ &=\frac {2 i (i+a)^2 x}{b^3}+\frac {(1-i a) x^2}{b^2}+\frac {2 i x^3}{3 b}-\frac {x^4}{4}-\frac {2 (1-i a)^3 \log (i+a+b x)}{b^4}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 72, normalized size = 1.00 \[ -\frac {2 (1-i a)^3 \log (a+b x+i)}{b^4}+\frac {2 i (a+i)^2 x}{b^3}+\frac {(1-i a) x^2}{b^2}+\frac {2 i x^3}{3 b}-\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*(I + a)^2*x)/b^3 + ((1 - I*a)*x^2)/b^2 + (((2*I)/3)*x^3)/b - x^4/4 - (2*(1 - I*a)^3*Log[I + a + b*x])/b
^4

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fricas [A]  time = 0.47, size = 77, normalized size = 1.07 \[ -\frac {3 \, b^{4} x^{4} - 8 i \, b^{3} x^{3} + 12 \, {\left (i \, a - 1\right )} b^{2} x^{2} - {\left (24 i \, a^{2} - 48 \, a - 24 i\right )} b x - {\left (-24 i \, a^{3} + 72 \, a^{2} + 72 i \, a - 24\right )} \log \left (\frac {b x + a + i}{b}\right )}{12 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*b^4*x^4 - 8*I*b^3*x^3 + 12*(I*a - 1)*b^2*x^2 - (24*I*a^2 - 48*a - 24*I)*b*x - (-24*I*a^3 + 72*a^2 + 7
2*I*a - 24)*log((b*x + a + I)/b))/b^4

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giac [A]  time = 0.13, size = 88, normalized size = 1.22 \[ -\frac {2 \, {\left (a^{3} i - 3 \, a^{2} - 3 \, a i + 1\right )} \log \left (b x + a + i\right )}{b^{4}} - \frac {3 \, b^{4} x^{4} - 8 \, b^{3} i x^{3} + 12 \, a b^{2} i x^{2} - 24 \, a^{2} b i x - 12 \, b^{2} x^{2} + 48 \, a b x + 24 \, b i x}{12 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(a^3*i - 3*a^2 - 3*a*i + 1)*log(b*x + a + i)/b^4 - 1/12*(3*b^4*x^4 - 8*b^3*i*x^3 + 12*a*b^2*i*x^2 - 24*a^2*
b*i*x - 12*b^2*x^2 + 48*a*b*x + 24*b*i*x)/b^4

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maple [B]  time = 0.04, size = 255, normalized size = 3.54 \[ -\frac {x^{4}}{4}+\frac {2 i x^{3}}{3 b}-\frac {i x^{2} a}{b^{2}}+\frac {2 i a^{2} x}{b^{3}}+\frac {x^{2}}{b^{2}}-\frac {2 i x}{b^{3}}-\frac {4 a x}{b^{3}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}}{b^{4}}+\frac {3 i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{4}}+\frac {3 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{4}}-\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{4}}-\frac {6 i \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{2}}{b^{4}}-\frac {2 \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{3}}{b^{4}}+\frac {2 i \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b^{4}}+\frac {6 \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.43, size = 118, normalized size = 1.64 \[ -\frac {3 \, b^{3} x^{4} - 8 i \, b^{2} x^{3} + 12 \, {\left (i \, a - 1\right )} b x^{2} - {\left (24 i \, a^{2} - 48 \, a - 24 i\right )} x}{12 \, b^{3}} - \frac {{\left (2 \, a^{3} + 6 i \, a^{2} - 6 \, a - 2 i\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{4}} + \frac {{\left (-i \, a^{3} + 3 \, a^{2} + 3 i \, a - 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*b^3*x^4 - 8*I*b^2*x^3 + 12*(I*a - 1)*b*x^2 - (24*I*a^2 - 48*a - 24*I)*x)/b^3 - (2*a^3 + 6*I*a^2 - 6*a
- 2*I)*arctan((b^2*x + a*b)/b)/b^4 + (-I*a^3 + 3*a^2 + 3*I*a - 1)*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/b^4

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mupad [B]  time = 0.53, size = 153, normalized size = 2.12 \[ -x^3\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3\,b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3\,b}\right )-\frac {x^4}{4}+\ln \left (x+\frac {a+1{}\mathrm {i}}{b}\right )\,\left (\frac {6\,a^2-2}{b^4}+\frac {\left (6\,a-2\,a^3\right )\,1{}\mathrm {i}}{b^4}\right )-\frac {x^2\,\left (-1+a\,1{}\mathrm {i}\right )\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )\,1{}\mathrm {i}}{2\,b}+\frac {x\,{\left (-1+a\,1{}\mathrm {i}\right )}^2\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.38, size = 78, normalized size = 1.08 \[ - \frac {x^{4}}{4} - x^{2} \left (\frac {i a}{b^{2}} - \frac {1}{b^{2}}\right ) - x \left (- \frac {2 i a^{2}}{b^{3}} + \frac {4 a}{b^{3}} + \frac {2 i}{b^{3}}\right ) + \frac {2 i x^{3}}{3 b} - \frac {2 i \left (a + i\right )^{3} \log {\left (i a + i b x - 1 \right )}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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