3.383 \(\int \frac{e^{-i \tan ^{-1}(a x)} x^2}{(c+a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=143 \[ \frac{\sqrt{a^2 x^2+1}}{2 a^3 c (-a x+i) \sqrt{a^2 c x^2+c}}-\frac{3 i \sqrt{a^2 x^2+1} \log (-a x+i)}{4 a^3 c \sqrt{a^2 c x^2+c}}-\frac{i \sqrt{a^2 x^2+1} \log (a x+i)}{4 a^3 c \sqrt{a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.223591, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {5085, 5082, 88} \[ \frac{\sqrt{a^2 x^2+1}}{2 a^3 c (-a x+i) \sqrt{a^2 c x^2+c}}-\frac{3 i \sqrt{a^2 x^2+1} \log (-a x+i)}{4 a^3 c \sqrt{a^2 c x^2+c}}-\frac{i \sqrt{a^2 x^2+1} \log (a x+i)}{4 a^3 c \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(E^(I*ArcTan[a*x])*(c + a^2*c*x^2)^(3/2)),x]

[Out]

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

Rule 5085

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c + d
*x^2)^FracPart[p])/(1 + a^2*x^2)^FracPart[p], Int[x^m*(1 + a^2*x^2)^p*E^(n*ArcTan[a*x]), x], x] /; FreeQ[{a, c
, d, m, n, p}, x] && EqQ[d, a^2*c] &&  !(IntegerQ[p] || GtQ[c, 0])

Rule 5082

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 - I
*a*x)^(p + (I*n)/2)*(1 + I*a*x)^(p - (I*n)/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[d, a^2*c] && (Int
egerQ[p] || GtQ[c, 0])

Rule 88

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

Rubi steps

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

Mathematica [A]  time = 0.056661, size = 75, normalized size = 0.52 \[ \frac{\sqrt{a^2 x^2+1} \left (\frac{2}{-a x+i}-3 i \log (-a x+i)-i \log (a x+i)\right )}{4 a^3 c \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.152, size = 86, normalized size = 0.6 \begin{align*}{\frac{3\,i\ln \left ( -ax+i \right ) xa+i\ln \left ( ax+i \right ) xa+3\,\ln \left ( -ax+i \right ) +\ln \left ( ax+i \right ) +2}{4\,{c}^{2}{a}^{3} \left ( -ax+i \right ) }\sqrt{c \left ({a}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00657, size = 74, normalized size = 0.52 \begin{align*} -\frac{\sqrt{c}}{2 \, a^{4} c^{2} x - 2 i \, a^{3} c^{2}} - \frac{3 i \, \log \left (a x - i\right )}{4 \, a^{3} c^{\frac{3}{2}}} - \frac{i \, \log \left (i \, a x - 1\right )}{4 \, a^{3} c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(c)/(2*a^4*c^2*x - 2*I*a^3*c^2) - 3/4*I*log(a*x - I)/(a^3*c^(3/2)) - 1/4*I*log(I*a*x - 1)/(a^3*c^(3/2))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((I*a^5*c^2*x^3 + a^4*c^2*x^2 + I*a^3*c^2*x + a^2*c^2)*sqrt(1/(a^6*c^3))*log((I*sqrt(a^2*c*x^2 + c)*sqrt(a
^2*x^2 + 1)*a^3*c*x*sqrt(1/(a^6*c^3)) + I*a^2*x^3 + I*x)/(a^3*x^3 + I*a^2*x^2 + a*x + I)) + (-I*a^5*c^2*x^3 -
a^4*c^2*x^2 - I*a^3*c^2*x - a^2*c^2)*sqrt(1/(a^6*c^3))*log((-I*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*a^3*c*x*s
qrt(1/(a^6*c^3)) + I*a^2*x^3 + I*x)/(a^3*x^3 + I*a^2*x^2 + a*x + I)) + (-3*I*a^5*c^2*x^3 - 3*a^4*c^2*x^2 - 3*I
*a^3*c^2*x - 3*a^2*c^2)*sqrt(1/(a^6*c^3))*log((I*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*a^3*c*x*sqrt(1/(a^6*c^3
)) - I*a^2*x^3 - I*x)/(a^3*x^3 - I*a^2*x^2 + a*x - I)) + (3*I*a^5*c^2*x^3 + 3*a^4*c^2*x^2 + 3*I*a^3*c^2*x + 3*
a^2*c^2)*sqrt(1/(a^6*c^3))*log((-I*sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*a^3*c*x*sqrt(1/(a^6*c^3)) - I*a^2*x^3
- I*x)/(a^3*x^3 - I*a^2*x^2 + a*x - I)) + (-4*I*a^5*c^2*x^3 - 4*a^4*c^2*x^2 - 4*I*a^3*c^2*x - 4*a^2*c^2)*sqrt
(1/(a^6*c^3))*log((sqrt(a^2*c*x^2 + c)*sqrt(a^2*x^2 + 1)*a^3*c*x*sqrt(1/(a^6*c^3)) + a^2*x^3 + x)/(a^2*x^2 + 1
)) + (4*I*a^5*c^2*x^3 + 4*a^4*c^2*x^2 + 4*I*a^3*c^2*x + 4*a^2*c^2)*sqrt(1/(a^6*c^3))*log(-(sqrt(a^2*c*x^2 + c)
*sqrt(a^2*x^2 + 1)*a^3*c*x*sqrt(1/(a^6*c^3)) - a^2*x^3 - x)/(a^2*x^2 + 1)) + 4*I*sqrt(a^2*c*x^2 + c)*sqrt(a^2*
x^2 + 1)*x + 2*(4*a^5*c^2*x^3 - 4*I*a^4*c^2*x^2 + 4*a^3*c^2*x - 4*I*a^2*c^2)*integral(1/2*sqrt(a^2*c*x^2 + c)*
sqrt(a^2*x^2 + 1)*(-2*I*a*x + 1)/(a^6*c^2*x^4 + 2*a^4*c^2*x^2 + a^2*c^2), x))/(4*a^5*c^2*x^3 - 4*I*a^4*c^2*x^2
+ 4*a^3*c^2*x - 4*I*a^2*c^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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