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

Optimal. Leaf size=137 \[ -\frac{i x^3 \sqrt{a^2 x^2+1}}{4 a}-\frac{x^2 \sqrt{a^2 x^2+1}}{a^2}-\frac{9 i (-3 a x+2 i) \sqrt{a^2 x^2+1}}{8 a^4}+\frac{27 \sqrt{a^2 x^2+1}}{4 a^4}+\frac{(1+i a x)^3}{a^4 \sqrt{a^2 x^2+1}}-\frac{51 i \sinh ^{-1}(a x)}{8 a^4} \]

[Out]

(1 + I*a*x)^3/(a^4*Sqrt[1 + a^2*x^2]) + (27*Sqrt[1 + a^2*x^2])/(4*a^4) - (x^2*Sqrt[1 + a^2*x^2])/a^2 - ((I/4)*
x^3*Sqrt[1 + a^2*x^2])/a - (((9*I)/8)*(2*I - 3*a*x)*Sqrt[1 + a^2*x^2])/a^4 - (((51*I)/8)*ArcSinh[a*x])/a^4

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Rubi [A]  time = 0.621367, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.786, Rules used = {5060, 1633, 1593, 12, 852, 1635, 1815, 27, 743, 641, 215} \[ -\frac{i x^3 \sqrt{a^2 x^2+1}}{4 a}-\frac{x^2 \sqrt{a^2 x^2+1}}{a^2}-\frac{9 i (-3 a x+2 i) \sqrt{a^2 x^2+1}}{8 a^4}+\frac{27 \sqrt{a^2 x^2+1}}{4 a^4}+\frac{(1+i a x)^3}{a^4 \sqrt{a^2 x^2+1}}-\frac{51 i \sinh ^{-1}(a x)}{8 a^4} \]

Antiderivative was successfully verified.

[In]

Int[E^((3*I)*ArcTan[a*x])*x^3,x]

[Out]

(1 + I*a*x)^3/(a^4*Sqrt[1 + a^2*x^2]) + (27*Sqrt[1 + a^2*x^2])/(4*a^4) - (x^2*Sqrt[1 + a^2*x^2])/a^2 - ((I/4)*
x^3*Sqrt[1 + a^2*x^2])/a - (((9*I)/8)*(2*I - 3*a*x)*Sqrt[1 + a^2*x^2])/a^4 - (((51*I)/8)*ArcSinh[a*x])/a^4

Rule 5060

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n
- 1)/2)*Sqrt[1 + a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(I*n - 1)/2]

Rule 1633

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d*e, Int[(d + e*x)^(m - 1)*
PolynomialQuotient[Pq, a*e + c*d*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + c*d*x, x], 0]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 852

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

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

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 x)} x^3 \, dx &=\int \frac{x^3 (1+i a x)^2}{(1-i a x) \sqrt{1+a^2 x^2}} \, dx\\ &=-\left ((i a) \int \frac{\sqrt{1+a^2 x^2} \left (\frac{i x^3}{a}-x^4\right )}{(1-i a x)^2} \, dx\right )\\ &=-\left ((i a) \int \frac{\left (\frac{i}{a}-x\right ) x^3 \sqrt{1+a^2 x^2}}{(1-i a x)^2} \, dx\right )\\ &=a^2 \int \frac{x^3 \left (1+a^2 x^2\right )^{3/2}}{a^2 (1-i a x)^3} \, dx\\ &=\int \frac{x^3 \left (1+a^2 x^2\right )^{3/2}}{(1-i a x)^3} \, dx\\ &=\int \frac{x^3 (1+i a x)^3}{\left (1+a^2 x^2\right )^{3/2}} \, dx\\ &=\frac{(1+i a x)^3}{a^4 \sqrt{1+a^2 x^2}}-\int \frac{(1+i a x)^2 \left (\frac{3 i}{a^3}-\frac{x}{a^2}-\frac{i x^2}{a}\right )}{\sqrt{1+a^2 x^2}} \, dx\\ &=\frac{(1+i a x)^3}{a^4 \sqrt{1+a^2 x^2}}-\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{\int \frac{\frac{12 i}{a}-28 x-27 i a x^2+12 a^2 x^3}{\sqrt{1+a^2 x^2}} \, dx}{4 a^2}\\ &=\frac{(1+i a x)^3}{a^4 \sqrt{1+a^2 x^2}}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}-\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{\int \frac{36 i a-108 a^2 x-81 i a^3 x^2}{\sqrt{1+a^2 x^2}} \, dx}{12 a^4}\\ &=\frac{(1+i a x)^3}{a^4 \sqrt{1+a^2 x^2}}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}-\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{\int -\frac{9 i a (-2 i+3 a x)^2}{\sqrt{1+a^2 x^2}} \, dx}{12 a^4}\\ &=\frac{(1+i a x)^3}{a^4 \sqrt{1+a^2 x^2}}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}-\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}+\frac{(3 i) \int \frac{(-2 i+3 a x)^2}{\sqrt{1+a^2 x^2}} \, dx}{4 a^3}\\ &=\frac{(1+i a x)^3}{a^4 \sqrt{1+a^2 x^2}}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}-\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{9 i (2 i-3 a x) \sqrt{1+a^2 x^2}}{8 a^4}+\frac{(3 i) \int \frac{-17 a^2-18 i a^3 x}{\sqrt{1+a^2 x^2}} \, dx}{8 a^5}\\ &=\frac{(1+i a x)^3}{a^4 \sqrt{1+a^2 x^2}}+\frac{27 \sqrt{1+a^2 x^2}}{4 a^4}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}-\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{9 i (2 i-3 a x) \sqrt{1+a^2 x^2}}{8 a^4}-\frac{(51 i) \int \frac{1}{\sqrt{1+a^2 x^2}} \, dx}{8 a^3}\\ &=\frac{(1+i a x)^3}{a^4 \sqrt{1+a^2 x^2}}+\frac{27 \sqrt{1+a^2 x^2}}{4 a^4}-\frac{x^2 \sqrt{1+a^2 x^2}}{a^2}-\frac{i x^3 \sqrt{1+a^2 x^2}}{4 a}-\frac{9 i (2 i-3 a x) \sqrt{1+a^2 x^2}}{8 a^4}-\frac{51 i \sinh ^{-1}(a x)}{8 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0600751, size = 80, normalized size = 0.58 \[ \sqrt{a^2 x^2+1} \left (-\frac{x^2}{a^2}+\frac{19 i x}{8 a^3}+\frac{4 i}{a^4 (a x+i)}+\frac{6}{a^4}-\frac{i x^3}{4 a}\right )-\frac{51 i \sinh ^{-1}(a x)}{8 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 + a^2*x^2]*(6/a^4 + (((19*I)/8)*x)/a^3 - x^2/a^2 - ((I/4)*x^3)/a + (4*I)/(a^4*(I + a*x))) - (((51*I)/8)
*ArcSinh[a*x])/a^4

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Maple [A]  time = 0.079, size = 143, normalized size = 1. \begin{align*}{-{\frac{i}{4}}a{x}^{5}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{{\frac{17\,i}{8}}{x}^{3}}{a}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{{\frac{51\,i}{8}}x}{{a}^{3}}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-{\frac{{\frac{51\,i}{8}}}{{a}^{3}}\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{{x}^{4}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+5\,{\frac{{x}^{2}}{{a}^{2}\sqrt{{a}^{2}{x}^{2}+1}}}+10\,{\frac{1}{{a}^{4}\sqrt{{a}^{2}{x}^{2}+1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*a*x^5/(a^2*x^2+1)^(1/2)+17/8*I/a*x^3/(a^2*x^2+1)^(1/2)+51/8*I/a^3*x/(a^2*x^2+1)^(1/2)-51/8*I/a^3*ln(a^2
*x/(a^2)^(1/2)+(a^2*x^2+1)^(1/2))/(a^2)^(1/2)-x^4/(a^2*x^2+1)^(1/2)+5*x^2/a^2/(a^2*x^2+1)^(1/2)+10/a^4/(a^2*x^
2+1)^(1/2)

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Maxima [A]  time = 1.01132, size = 170, normalized size = 1.24 \begin{align*} -\frac{i \, a x^{5}}{4 \, \sqrt{a^{2} x^{2} + 1}} - \frac{x^{4}}{\sqrt{a^{2} x^{2} + 1}} + \frac{17 i \, x^{3}}{8 \, \sqrt{a^{2} x^{2} + 1} a} + \frac{5 \, x^{2}}{\sqrt{a^{2} x^{2} + 1} a^{2}} + \frac{51 i \, x}{8 \, \sqrt{a^{2} x^{2} + 1} a^{3}} - \frac{51 i \, \operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{8 \, \sqrt{a^{2}} a^{3}} + \frac{10}{\sqrt{a^{2} x^{2} + 1} a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*a*x^5/sqrt(a^2*x^2 + 1) - x^4/sqrt(a^2*x^2 + 1) + 17/8*I*x^3/(sqrt(a^2*x^2 + 1)*a) + 5*x^2/(sqrt(a^2*x^
2 + 1)*a^2) + 51/8*I*x/(sqrt(a^2*x^2 + 1)*a^3) - 51/8*I*arcsinh(a^2*x/sqrt(a^2))/(sqrt(a^2)*a^3) + 10/(sqrt(a^
2*x^2 + 1)*a^4)

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Fricas [A]  time = 1.72582, size = 220, normalized size = 1.61 \begin{align*} \frac{32 i \, a x - 51 \,{\left (-i \, a x + 1\right )} \log \left (-a x + \sqrt{a^{2} x^{2} + 1}\right ) +{\left (-2 i \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 11 i \, a^{2} x^{2} + 29 \, a x + 80 i\right )} \sqrt{a^{2} x^{2} + 1} - 32}{8 \,{\left (a^{5} x + i \, a^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(32*I*a*x - 51*(-I*a*x + 1)*log(-a*x + sqrt(a^2*x^2 + 1)) + (-2*I*a^4*x^4 - 6*a^3*x^3 + 11*I*a^2*x^2 + 29*
a*x + 80*I)*sqrt(a^2*x^2 + 1) - 32)/(a^5*x + I*a^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef