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

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Rubi [A]  time = 0.06, antiderivative size = 53, normalized size of antiderivative = 1.00, 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 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]

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 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])

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*}

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Mathematica [A]  time = 0.03, 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)

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fricas [A]  time = 0.44, size = 44, normalized size = 0.83 \[ \frac {{\left (a x - i\right )} {\left (a^{2} c x^{2} + c\right )}^{p}}{{\left (2 \, a p + a\right )} \left (-\frac {a x + i}{a x - i}\right )^{p}} \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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]

sage0*x

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maple [A]  time = 0.09, size = 41, normalized size = 0.77 \[ -\frac {\left (-a x +i\right ) {\mathrm e}^{2 i p \arctan \left (a x \right )} \left (a^{2} c \,x^{2}+c \right )^{p}}{a \left (1+2 p \right )} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a^{2} c x^{2} + c\right )}^{p} e^{\left (2 i \, p \arctan \left (a x\right )\right )}\,{d x} \]

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)

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mupad [B]  time = 0.59, size = 54, normalized size = 1.02 \[ \left (\frac {x\,{\mathrm {e}}^{p\,\mathrm {atan}\left (a\,x\right )\,2{}\mathrm {i}}}{2\,p+1}-\frac {{\mathrm {e}}^{p\,\mathrm {atan}\left (a\,x\right )\,2{}\mathrm {i}}\,1{}\mathrm {i}}{a\,\left (2\,p+1\right )}\right )\,{\left (c\,a^2\,x^2+c\right )}^p \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(p*atan(a*x)*2i)*(c + a^2*c*x^2)^p,x)

[Out]

((x*exp(p*atan(a*x)*2i))/(2*p + 1) - (exp(p*atan(a*x)*2i)*1i)/(a*(2*p + 1)))*(c + a^2*c*x^2)^p

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \frac {x}{\sqrt {c}} & \text {for}\: a = 0 \wedge p = - \frac {1}{2} \\c^{p} x & \text {for}\: a = 0 \\\int \frac {e^{- i \operatorname {atan}{\left (a x \right )}}}{\sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx & \text {for}\: p = - \frac {1}{2} \\\frac {a x \left (a^{2} c x^{2} + c\right )^{p} e^{2 i p \operatorname {atan}{\left (a x \right )}}}{2 a p + a} - \frac {i \left (a^{2} c x^{2} + c\right )^{p} e^{2 i p \operatorname {atan}{\left (a x \right )}}}{2 a p + a} & \text {otherwise} \end {cases} \]

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]

Piecewise((x/sqrt(c), Eq(a, 0) & Eq(p, -1/2)), (c**p*x, Eq(a, 0)), (Integral(exp(-I*atan(a*x))/sqrt(c*(a**2*x*
*2 + 1)), x), Eq(p, -1/2)), (a*x*(a**2*c*x**2 + c)**p*exp(2*I*p*atan(a*x))/(2*a*p + a) - I*(a**2*c*x**2 + c)**
p*exp(2*I*p*atan(a*x))/(2*a*p + a), True))

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