3.1.21 \(\int e^{3 i \text {ArcTan}(a x)} x \, dx\) [21]

Optimal. Leaf size=92 \[ -\frac {9 \sqrt {1+a^2 x^2}}{2 a^2}-\frac {3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1-i a x)}-\frac {\left (1+a^2 x^2\right )^{5/2}}{a^2 (1-i a x)^3}+\frac {9 i \sinh ^{-1}(a x)}{2 a^2} \]

[Out]

-3/2*(a^2*x^2+1)^(3/2)/a^2/(1-I*a*x)-(a^2*x^2+1)^(5/2)/a^2/(1-I*a*x)^3+9/2*I*arcsinh(a*x)/a^2-9/2*(a^2*x^2+1)^
(1/2)/a^2

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Rubi [A]
time = 0.23, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5168, 1647, 1607, 12, 807, 679, 221} \begin {gather*} -\frac {\left (a^2 x^2+1\right )^{5/2}}{a^2 (1-i a x)^3}-\frac {3 \left (a^2 x^2+1\right )^{3/2}}{2 a^2 (1-i a x)}-\frac {9 \sqrt {a^2 x^2+1}}{2 a^2}+\frac {9 i \sinh ^{-1}(a x)}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*Sqrt[1 + a^2*x^2])/(2*a^2) - (3*(1 + a^2*x^2)^(3/2))/(2*a^2*(1 - I*a*x)) - (1 + a^2*x^2)^(5/2)/(a^2*(1 - I
*a*x)^3) + (((9*I)/2)*ArcSinh[a*x])/a^2

Rule 12

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

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 679

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

Rule 807

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

Rule 1607

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 1647

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 5168

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]

Rubi steps

\begin {align*} \int e^{3 i \tan ^{-1}(a x)} x \, dx &=\int \frac {x (1+i a x)^2}{(1-i a x) \sqrt {1+a^2 x^2}} \, dx\\ &=-\left ((i a) \int \frac {\left (\frac {i x}{a}-x^2\right ) \sqrt {1+a^2 x^2}}{(1-i a x)^2} \, dx\right )\\ &=-\left ((i a) \int \frac {\left (\frac {i}{a}-x\right ) x \sqrt {1+a^2 x^2}}{(1-i a x)^2} \, dx\right )\\ &=a^2 \int \frac {x \left (1+a^2 x^2\right )^{3/2}}{a^2 (1-i a x)^3} \, dx\\ &=\int \frac {x \left (1+a^2 x^2\right )^{3/2}}{(1-i a x)^3} \, dx\\ &=-\frac {\left (1+a^2 x^2\right )^{5/2}}{a^2 (1-i a x)^3}+\frac {(3 i) \int \frac {\left (1+a^2 x^2\right )^{3/2}}{(1-i a x)^2} \, dx}{a}\\ &=-\frac {3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1-i a x)}-\frac {\left (1+a^2 x^2\right )^{5/2}}{a^2 (1-i a x)^3}+\frac {(9 i) \int \frac {\sqrt {1+a^2 x^2}}{1-i a x} \, dx}{2 a}\\ &=-\frac {9 \sqrt {1+a^2 x^2}}{2 a^2}-\frac {3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1-i a x)}-\frac {\left (1+a^2 x^2\right )^{5/2}}{a^2 (1-i a x)^3}+\frac {(9 i) \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{2 a}\\ &=-\frac {9 \sqrt {1+a^2 x^2}}{2 a^2}-\frac {3 \left (1+a^2 x^2\right )^{3/2}}{2 a^2 (1-i a x)}-\frac {\left (1+a^2 x^2\right )^{5/2}}{a^2 (1-i a x)^3}+\frac {9 i \sinh ^{-1}(a x)}{2 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 54, normalized size = 0.59 \begin {gather*} -\frac {i \left (\frac {\sqrt {1+a^2 x^2} \left (14-5 i a x+a^2 x^2\right )}{i+a x}-9 \sinh ^{-1}(a x)\right )}{2 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*I)*((Sqrt[1 + a^2*x^2]*(14 - (5*I)*a*x + a^2*x^2))/(I + a*x) - 9*ArcSinh[a*x]))/a^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. 192 vs. \(2 (77 ) = 154\).
time = 0.10, size = 193, normalized size = 2.10

method	result	size
risch	\(-\frac {i \left (a x -6 i\right ) \sqrt {a^{2} x^{2}+1}}{2 a^{2}}+\frac {9 i \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 a \sqrt {a^{2}}}-\frac {4 i \sqrt {\left (x +\frac {i}{a}\right )^{2} a^{2}-2 i a \left (x +\frac {i}{a}\right )}}{a^{3} \left (x +\frac {i}{a}\right )}\)	\(104\)
meijerg	\(\frac {\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {a^{2} x^{2}+1}}}{a^{2} \sqrt {\pi }}+\frac {3 i \left (-\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\sqrt {\pi }\, \left (a^{2}\right )^{\frac {3}{2}} \arcsinh \left (a x \right )}{a^{3}}\right )}{a \sqrt {\pi }\, \sqrt {a^{2}}}-\frac {3 \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right )}{4 \sqrt {a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {\pi }}-\frac {i \left (\frac {\sqrt {\pi }\, x \left (a^{2}\right )^{\frac {5}{2}} \left (5 a^{2} x^{2}+15\right )}{10 a^{4} \sqrt {a^{2} x^{2}+1}}-\frac {3 \sqrt {\pi }\, \left (a^{2}\right )^{\frac {5}{2}} \arcsinh \left (a x \right )}{2 a^{5}}\right )}{a \sqrt {\pi }\, \sqrt {a^{2}}}\)	\(192\)
default	\(-i a^{3} \left (\frac {x^{3}}{2 a^{2} \sqrt {a^{2} x^{2}+1}}-\frac {3 \left (-\frac {x}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{a^{2} \sqrt {a^{2}}}\right )}{2 a^{2}}\right )-3 a^{2} \left (\frac {x^{2}}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {a^{2} x^{2}+1}}\right )+3 i a \left (-\frac {x}{a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{a^{2} \sqrt {a^{2}}}\right )-\frac {1}{a^{2} \sqrt {a^{2} x^{2}+1}}\)	\(193\)

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,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 76, normalized size = 0.83 \begin {gather*} -\frac {i \, a x^{3}}{2 \, \sqrt {a^{2} x^{2} + 1}} - \frac {3 \, x^{2}}{\sqrt {a^{2} x^{2} + 1}} - \frac {9 i \, x}{2 \, \sqrt {a^{2} x^{2} + 1} a} + \frac {9 i \, \operatorname {arsinh}\left (a x\right )}{2 \, a^{2}} - \frac {7}{\sqrt {a^{2} x^{2} + 1} a^{2}} \end {gather*}

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,x, algorithm="maxima")

[Out]

-1/2*I*a*x^3/sqrt(a^2*x^2 + 1) - 3*x^2/sqrt(a^2*x^2 + 1) - 9/2*I*x/(sqrt(a^2*x^2 + 1)*a) + 9/2*I*arcsinh(a*x)/
a^2 - 7/(sqrt(a^2*x^2 + 1)*a^2)

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Fricas [A]
time = 2.34, size = 72, normalized size = 0.78 \begin {gather*} \frac {-8 i \, a x - 9 \, {\left (i \, a x - 1\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + \sqrt {a^{2} x^{2} + 1} {\left (-i \, a^{2} x^{2} - 5 \, a x - 14 i\right )} + 8}{2 \, {\left (a^{3} x + i \, a^{2}\right )}} \end {gather*}

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,x, algorithm="fricas")

[Out]

1/2*(-8*I*a*x - 9*(I*a*x - 1)*log(-a*x + sqrt(a^2*x^2 + 1)) + sqrt(a^2*x^2 + 1)*(-I*a^2*x^2 - 5*a*x - 14*I) +
8)/(a^3*x + I*a^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i \left (\int \frac {i x}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x^{2}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{4}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{3}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \end {gather*}

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,x)

[Out]

-I*(Integral(I*x/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(-3*a*x**2/(a**2*x**2*sqr
t(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x) + Integral(a**3*x**4/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x*
*2 + 1)), x) + Integral(-3*I*a**2*x**3/(a**2*x**2*sqrt(a**2*x**2 + 1) + sqrt(a**2*x**2 + 1)), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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,x, algorithm="giac")

[Out]

undef

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Mupad [B]
time = 0.43, size = 104, normalized size = 1.13 \begin {gather*} -\frac {\sqrt {a^2\,x^2+1}\,\left (\frac {3\,\sqrt {a^2}}{a^2}+\frac {x\,\sqrt {a^2}\,1{}\mathrm {i}}{2\,a}\right )}{\sqrt {a^2}}+\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,9{}\mathrm {i}}{2\,a\,\sqrt {a^2}}-\frac {\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{a\,\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(asinh(x*(a^2)^(1/2))*9i)/(2*a*(a^2)^(1/2)) - ((a^2*x^2 + 1)^(1/2)*((3*(a^2)^(1/2))/a^2 + (x*(a^2)^(1/2)*1i)/(
2*a)))/(a^2)^(1/2) - ((a^2*x^2 + 1)^(1/2)*4i)/(a*(((a^2)^(1/2)*1i)/a + x*(a^2)^(1/2))*(a^2)^(1/2))

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