3.329 \(\int \frac{e^{4 i \tan ^{-1}(a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\)
Optimal. Leaf size=69 \[ \frac{i c (1+i a x)^5}{15 a \left (a^2 c x^2+c\right )^{5/2}}-\frac{i c (1+i a x)^4}{3 a \left (a^2 c x^2+c\right )^{5/2}} \]
[Out]
((-I/3)*c*(1 + I*a*x)^4)/(a*(c + a^2*c*x^2)^(5/2)) + ((I/15)*c*(1 + I*a*x)^5)/(a*(c + a^2*c*x^2)^(5/2))
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Rubi [A] time = 0.0652056, antiderivative size = 69, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used =
{5075, 659, 651} \[ \frac{i c (1+i a x)^5}{15 a \left (a^2 c x^2+c\right )^{5/2}}-\frac{i c (1+i a x)^4}{3 a \left (a^2 c x^2+c\right )^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[E^((4*I)*ArcTan[a*x])/(c + a^2*c*x^2)^(3/2),x]
[Out]
((-I/3)*c*(1 + I*a*x)^4)/(a*(c + a^2*c*x^2)^(5/2)) + ((I/15)*c*(1 + I*a*x)^5)/(a*(c + a^2*c*x^2)^(5/2))
Rule 5075
Int[E^(ArcTan[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c^((I*n)/2), Int[(c + d*x^2)^(
p + (I*n)/2)/(1 + I*a*x)^(I*n), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[d, a^2*c] && !(IntegerQ[p] || GtQ[c,
0]) && ILtQ[(I*n)/2, 0]
Rule 659
Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]
Rule 651
Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]
Rubi steps
\begin{align*} \int \frac{e^{4 i \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx &=c^2 \int \frac{(1+i a x)^4}{\left (c+a^2 c x^2\right )^{7/2}} \, dx\\ &=-\frac{i c (1+i a x)^4}{3 a \left (c+a^2 c x^2\right )^{5/2}}-\frac{1}{3} c^2 \int \frac{(1+i a x)^5}{\left (c+a^2 c x^2\right )^{7/2}} \, dx\\ &=-\frac{i c (1+i a x)^4}{3 a \left (c+a^2 c x^2\right )^{5/2}}+\frac{i c (1+i a x)^5}{15 a \left (c+a^2 c x^2\right )^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.0336332, size = 77, normalized size = 1.12 \[ \frac{(1+i a x)^{3/2} (a x+4 i) \sqrt{a^2 x^2+1}}{15 a c \sqrt{1-i a x} (a x+i)^2 \sqrt{a^2 c x^2+c}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[E^((4*I)*ArcTan[a*x])/(c + a^2*c*x^2)^(3/2),x]
[Out]
((1 + I*a*x)^(3/2)*(4*I + a*x)*Sqrt[1 + a^2*x^2])/(15*a*c*Sqrt[1 - I*a*x]*(I + a*x)^2*Sqrt[c + a^2*c*x^2])
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Maple [B] time = 0.082, size = 940, normalized size = 13.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x)
[Out]
x/c/(a^2*c*x^2+c)^(1/2)-2/a*(I*(-a^2)^(1/2)-a)/(-a^2)^(1/2)*(1/3/c/(-a^2)^(1/2)/(x+(-a^2)^(1/2)/a^2)/((x+(-a^2
)^(1/2)/a^2)^2*a^2*c-2*c*(-a^2)^(1/2)*(x+(-a^2)^(1/2)/a^2))^(1/2)+1/3/c^2/(-a^2)^(1/2)*(2*(x+(-a^2)^(1/2)/a^2)
*a^2*c-2*c*(-a^2)^(1/2))/((x+(-a^2)^(1/2)/a^2)^2*a^2*c-2*c*(-a^2)^(1/2)*(x+(-a^2)^(1/2)/a^2))^(1/2))+2/a^3*(I*
(-a^2)^(1/2)-a)*(1/5/c/(-a^2)^(1/2)/(x+(-a^2)^(1/2)/a^2)^2/((x+(-a^2)^(1/2)/a^2)^2*a^2*c-2*c*(-a^2)^(1/2)*(x+(
-a^2)^(1/2)/a^2))^(1/2)+3/5*a^2/(-a^2)^(1/2)*(1/3/c/(-a^2)^(1/2)/(x+(-a^2)^(1/2)/a^2)/((x+(-a^2)^(1/2)/a^2)^2*
a^2*c-2*c*(-a^2)^(1/2)*(x+(-a^2)^(1/2)/a^2))^(1/2)+1/3/c^2/(-a^2)^(1/2)*(2*(x+(-a^2)^(1/2)/a^2)*a^2*c-2*c*(-a^
2)^(1/2))/((x+(-a^2)^(1/2)/a^2)^2*a^2*c-2*c*(-a^2)^(1/2)*(x+(-a^2)^(1/2)/a^2))^(1/2)))-2/a*(I*(-a^2)^(1/2)+a)/
(-a^2)^(1/2)*(-1/3/c/(-a^2)^(1/2)/(x-(-a^2)^(1/2)/a^2)/((x-(-a^2)^(1/2)/a^2)^2*a^2*c+2*c*(-a^2)^(1/2)*(x-(-a^2
)^(1/2)/a^2))^(1/2)-1/3/c^2/(-a^2)^(1/2)*(2*(x-(-a^2)^(1/2)/a^2)*a^2*c+2*c*(-a^2)^(1/2))/((x-(-a^2)^(1/2)/a^2)
^2*a^2*c+2*c*(-a^2)^(1/2)*(x-(-a^2)^(1/2)/a^2))^(1/2))-2/a^3*(I*(-a^2)^(1/2)+a)*(-1/5/c/(-a^2)^(1/2)/(x-(-a^2)
^(1/2)/a^2)^2/((x-(-a^2)^(1/2)/a^2)^2*a^2*c+2*c*(-a^2)^(1/2)*(x-(-a^2)^(1/2)/a^2))^(1/2)-3/5*a^2/(-a^2)^(1/2)*
(-1/3/c/(-a^2)^(1/2)/(x-(-a^2)^(1/2)/a^2)/((x-(-a^2)^(1/2)/a^2)^2*a^2*c+2*c*(-a^2)^(1/2)*(x-(-a^2)^(1/2)/a^2))
^(1/2)-1/3/c^2/(-a^2)^(1/2)*(2*(x-(-a^2)^(1/2)/a^2)*a^2*c+2*c*(-a^2)^(1/2))/((x-(-a^2)^(1/2)/a^2)^2*a^2*c+2*c*
(-a^2)^(1/2)*(x-(-a^2)^(1/2)/a^2))^(1/2)))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a x + 1\right )}^{4}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}}{\left (a^{2} x^{2} + 1\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x, algorithm="maxima")
[Out]
integrate((I*a*x + 1)^4/((a^2*c*x^2 + c)^(3/2)*(a^2*x^2 + 1)^2), x)
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Fricas [A] time = 2.83624, size = 149, normalized size = 2.16 \begin{align*} -\frac{\sqrt{a^{2} c x^{2} + c}{\left (a^{2} x^{2} + 3 i \, a x + 4\right )}}{15 \, a^{4} c^{2} x^{3} + 45 i \, a^{3} c^{2} x^{2} - 45 \, a^{2} c^{2} x - 15 i \, a c^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x, algorithm="fricas")
[Out]
-sqrt(a^2*c*x^2 + c)*(a^2*x^2 + 3*I*a*x + 4)/(15*a^4*c^2*x^3 + 45*I*a^3*c^2*x^2 - 45*a^2*c^2*x - 15*I*a*c^2)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (i a x + 1\right )^{4}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac{3}{2}} \left (a^{2} x^{2} + 1\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+I*a*x)**4/(a**2*x**2+1)**2/(a**2*c*x**2+c)**(3/2),x)
[Out]
Integral((I*a*x + 1)**4/((c*(a**2*x**2 + 1))**(3/2)*(a**2*x**2 + 1)**2), x)
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Giac [B] time = 1.13021, size = 182, normalized size = 2.64 \begin{align*} -\frac{2 \,{\left (5 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} + c}\right )}^{2} c i - 15 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} + c}\right )}^{3} \sqrt{c} + c^{2} i + 5 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} + c}\right )} c^{\frac{3}{2}}\right )}}{15 \,{\left (\sqrt{c} i + \sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} + c}\right )}^{5} a c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x, algorithm="giac")
[Out]
-2/15*(5*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))^2*c*i - 15*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))^3*sqrt(c) + c^
2*i + 5*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))*c^(3/2))/((sqrt(c)*i + sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))^5*a*
c)