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]

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

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Rubi [A]  time = 0.22, antiderivative size = 143, normalized size of antiderivative = 1.00, 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 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]))

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 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])

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*}

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Mathematica [A]  time = 0.06, 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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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ \frac {{\left (i \, a^{5} c^{2} x^{3} + a^{4} c^{2} x^{2} + i \, a^{3} c^{2} x + a^{2} c^{2}\right )} \sqrt {\frac {1}{a^{6} c^{3}}} \log \left (\frac {i \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a^{3} c x \sqrt {\frac {1}{a^{6} c^{3}}} + i \, a^{2} x^{3} + i \, x}{a^{3} x^{3} + i \, a^{2} x^{2} + a x + i}\right ) + {\left (-i \, a^{5} c^{2} x^{3} - a^{4} c^{2} x^{2} - i \, a^{3} c^{2} x - a^{2} c^{2}\right )} \sqrt {\frac {1}{a^{6} c^{3}}} \log \left (\frac {-i \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a^{3} c x \sqrt {\frac {1}{a^{6} c^{3}}} + i \, a^{2} x^{3} + i \, x}{a^{3} x^{3} + i \, a^{2} x^{2} + a x + i}\right ) + {\left (-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}\right )} \sqrt {\frac {1}{a^{6} c^{3}}} \log \left (\frac {i \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a^{3} c x \sqrt {\frac {1}{a^{6} c^{3}}} - i \, a^{2} x^{3} - i \, x}{a^{3} x^{3} - i \, a^{2} x^{2} + a x - i}\right ) + {\left (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}\right )} \sqrt {\frac {1}{a^{6} c^{3}}} \log \left (\frac {-i \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a^{3} c x \sqrt {\frac {1}{a^{6} c^{3}}} - i \, a^{2} x^{3} - i \, x}{a^{3} x^{3} - i \, a^{2} x^{2} + a x - i}\right ) + {\left (-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}\right )} \sqrt {\frac {1}{a^{6} c^{3}}} \log \left (\frac {\sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a^{3} c x \sqrt {\frac {1}{a^{6} c^{3}}} + a^{2} x^{3} + x}{a^{2} x^{2} + 1}\right ) + {\left (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}\right )} \sqrt {\frac {1}{a^{6} c^{3}}} \log \left (-\frac {\sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} a^{3} c x \sqrt {\frac {1}{a^{6} c^{3}}} - a^{2} x^{3} - x}{a^{2} x^{2} + 1}\right ) + 4 i \, \sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} x + 2 \, {\left (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}\right )} {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} {\left (-2 i \, a x + 1\right )}}{2 \, {\left (a^{6} c^{2} x^{4} + 2 \, a^{4} c^{2} x^{2} + a^{2} c^{2}\right )}}, x\right )}{2 \, {\left (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}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (i \, a x + 1\right )}}\,{d x} \]

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)

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maple [A]  time = 0.18, size = 86, normalized size = 0.60 \[ \frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (i \ln \left (a x +i\right ) x a +3 i \ln \left (-a x +i\right ) x a +\ln \left (a x +i\right )+3 \ln \left (-a x +i\right )+2\right )}{4 \sqrt {a^{2} x^{2}+1}\, c^{2} a^{3} \left (-a x +i\right )} \]

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

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maxima [A]  time = 0.34, size = 55, normalized size = 0.38 \[ -\frac {\sqrt {c}}{2 \, a^{4} c^{2} x - 2 i \, a^{3} c^{2}} - \frac {3 i \, \log \left (a x - i\right )}{4 \, a^{3} c^{\frac {3}{2}}} - \frac {i \, \log \left (i \, a x - 1\right )}{4 \, a^{3} c^{\frac {3}{2}}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\sqrt {a^2\,x^2+1}}{{\left (c\,a^2\,x^2+c\right )}^{3/2}\,\left (1+a\,x\,1{}\mathrm {i}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i \int \frac {x^{2} \sqrt {a^{2} x^{2} + 1}}{a^{3} c x^{3} \sqrt {a^{2} c x^{2} + c} - i a^{2} c x^{2} \sqrt {a^{2} c x^{2} + c} + a c x \sqrt {a^{2} c x^{2} + c} - i c \sqrt {a^{2} c x^{2} + c}}\, dx \]

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]

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

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