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

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)

________________________________________________________________________________________

Rubi [A]  time = 0.0610917, 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[E^((2*I)*p*ArcTan[a*x])*(c + a^2*c*x^2)^p,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 = 0.0248986, 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[E^((2*I)*p*ArcTan[a*x])*(c + a^2*c*x^2)^p,x]

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a^2*c*x^2 + c)^p*e^(2*I*p*arctan(a*x)), x)

________________________________________________________________________________________

Fricas [A]  time = 2.21179, size = 92, normalized size = 1.74 \begin{align*} \frac{{\left (a x - i\right )}{\left (a^{2} c x^{2} + c\right )}^{p}}{{\left (2 \, a p + a\right )} \left (-\frac{a x + i}{a x - i}\right )^{p}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.14425, size = 78, normalized size = 1.47 \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(exp(2*I*p*arctan(a*x))*(a^2*c*x^2+c)^p,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)