3.328 \(\int \frac{e^{5 i \tan ^{-1}(a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=95 \[ -\frac{i \sqrt{a^2 x^2+1}}{2 a c (a x+i)^2 \sqrt{a^2 c x^2+c}}-\frac{2 \sqrt{a^2 x^2+1}}{3 a c (a x+i)^3 \sqrt{a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.0799853, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {5076, 5073, 43} \[ -\frac{i \sqrt{a^2 x^2+1}}{2 a c (a x+i)^2 \sqrt{a^2 c x^2+c}}-\frac{2 \sqrt{a^2 x^2+1}}{3 a c (a x+i)^3 \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Int[E^((5*I)*ArcTan[a*x])/(c + a^2*c*x^2)^(3/2),x]

[Out]

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

Rule 5076

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

Rule 5073

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{e^{5 i \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx &=\frac{\sqrt{1+a^2 x^2} \int \frac{e^{5 i \tan ^{-1}(a x)}}{\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{1+i a x}{(1-i a x)^4} \, dx}{c \sqrt{c+a^2 c x^2}}\\ &=\frac{\sqrt{1+a^2 x^2} \int \left (\frac{2}{(i+a x)^4}+\frac{i}{(i+a x)^3}\right ) \, dx}{c \sqrt{c+a^2 c x^2}}\\ &=-\frac{2 \sqrt{1+a^2 x^2}}{3 a c (i+a x)^3 \sqrt{c+a^2 c x^2}}-\frac{i \sqrt{1+a^2 x^2}}{2 a c (i+a x)^2 \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0290565, size = 56, normalized size = 0.59 \[ -\frac{i (3 a x-i) \sqrt{a^2 x^2+1}}{6 a c (a x+i)^3 \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((5*I)*ArcTan[a*x])/(c + a^2*c*x^2)^(3/2),x]

[Out]

((-I/6)*(-I + 3*a*x)*Sqrt[1 + a^2*x^2])/(a*c*(I + a*x)^3*Sqrt[c + a^2*c*x^2])

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Maple [A]  time = 0.15, size = 48, normalized size = 0.5 \begin{align*} -{\frac{3\,iax+1}{6\,a{c}^{2} \left ( ax+i \right ) ^{3}}\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((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(3/2),x)

[Out]

-1/6/(a^2*x^2+1)^(1/2)*(c*(a^2*x^2+1))^(1/2)*(3*I*a*x+1)/c^2/a/(a*x+I)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a x + 1\right )}^{5}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}}{\left (a^{2} x^{2} + 1\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((I*a*x + 1)^5/((a^2*c*x^2 + c)^(3/2)*(a^2*x^2 + 1)^(5/2)), x)

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Fricas [A]  time = 2.04973, size = 220, normalized size = 2.32 \begin{align*} \frac{\sqrt{a^{2} c x^{2} + c}{\left (i \, a^{2} x^{3} - 3 \, a x^{2} - 6 i \, x\right )} \sqrt{a^{2} x^{2} + 1}}{6 \, a^{5} c^{2} x^{5} + 18 i \, a^{4} c^{2} x^{4} - 12 \, a^{3} c^{2} x^{3} + 12 i \, a^{2} c^{2} x^{2} - 18 \, a c^{2} x - 6 i \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

sqrt(a^2*c*x^2 + c)*(I*a^2*x^3 - 3*a*x^2 - 6*I*x)*sqrt(a^2*x^2 + 1)/(6*a^5*c^2*x^5 + 18*I*a^4*c^2*x^4 - 12*a^3
*c^2*x^3 + 12*I*a^2*c^2*x^2 - 18*a*c^2*x - 6*I*c^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**5/(a**2*x**2+1)**(5/2)/(a**2*c*x**2+c)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a x + 1\right )}^{5}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}}{\left (a^{2} x^{2} + 1\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^5/(a^2*x^2+1)^(5/2)/(a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*x + 1)^5/((a^2*c*x^2 + c)^(3/2)*(a^2*x^2 + 1)^(5/2)), x)