3.372 \(\int e^{-2 i p \tan ^{-1}(a x)} (c+a^2 c x^2)^p \, dx\)

Optimal. Leaf size=53 \[ \frac{i (1-i a x)^{2 p+1} \left (a^2 x^2+1\right )^{-p} \left (a^2 c x^2+c\right )^p}{a (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0649463, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5076, 5073, 32} \[ \frac{i (1-i a x)^{2 p+1} \left (a^2 x^2+1\right )^{-p} \left (a^2 c x^2+c\right )^p}{a (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0283349, size = 39, normalized size = 0.74 \[ \frac{(a x+i) \left (a^2 c x^2+c\right )^p e^{-2 i p \tan ^{-1}(a x)}}{2 a p+a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I + a*x)*(c + a^2*c*x^2)^p)/(E^((2*I)*p*ArcTan[a*x])*(a + 2*a*p))

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Maple [A]  time = 0.046, size = 41, normalized size = 0.8 \begin{align*}{\frac{ \left ( ax+i \right ) \left ({a}^{2}c{x}^{2}+c \right ) ^{p}}{a \left ( 1+2\,p \right ){{\rm e}^{2\,ip\arctan \left ( ax \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^p/exp(2*I*p*arctan(a*x)),x)

[Out]

(a*x+I)/a/(1+2*p)*(a^2*c*x^2+c)^p/exp(2*I*p*arctan(a*x))

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Maxima [A]  time = 1.0347, size = 103, normalized size = 1.94 \begin{align*} \frac{{\left (a c^{p} x + i \, c^{p}\right )}{\left (a^{2} x^{2} + 1\right )}^{p} \cos \left (2 \, p \arctan \left (a x\right )\right ) -{\left (i \, a c^{p} x - c^{p}\right )}{\left (a^{2} x^{2} + 1\right )}^{p} \sin \left (2 \, p \arctan \left (a x\right )\right )}{2 \, a p + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^p/exp(2*I*p*arctan(a*x)),x, algorithm="maxima")

[Out]

((a*c^p*x + I*c^p)*(a^2*x^2 + 1)^p*cos(2*p*arctan(a*x)) - (I*a*c^p*x - c^p)*(a^2*x^2 + 1)^p*sin(2*p*arctan(a*x
)))/(2*a*p + a)

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Fricas [A]  time = 2.28301, size = 89, normalized size = 1.68 \begin{align*} \frac{{\left (a x + i\right )}{\left (a^{2} c x^{2} + c\right )}^{p} \left (-\frac{a x + i}{a x - i}\right )^{p}}{2 \, a p + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^p/exp(2*I*p*arctan(a*x)),x, algorithm="fricas")

[Out]

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

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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((a**2*c*x**2+c)**p/exp(2*I*p*atan(a*x)),x)

[Out]

Timed out

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Giac [A]  time = 1.1315, size = 74, normalized size = 1.4 \begin{align*} \frac{a x e^{\left (-\pi i p + 2 \, p \log \left (a x + i\right ) + p \log \left (c\right )\right )} + i e^{\left (-\pi i p + 2 \, p \log \left (a x + i\right ) + p \log \left (c\right )\right )}}{2 \, a p + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^p/exp(2*I*p*arctan(a*x)),x, algorithm="giac")

[Out]

(a*x*e^(-pi*i*p + 2*p*log(a*x + i) + p*log(c)) + i*e^(-pi*i*p + 2*p*log(a*x + i) + p*log(c)))/(2*a*p + a)