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

Optimal. Leaf size=51 \[ -\frac{\left (\frac{1}{5}-\frac{2 i}{5}\right ) 2^{1+i} c (1-i a x)^{2-i} \, _2F_1\left (-1-i,2-i;3-i;\frac{1}{2} (1-i a x)\right )}{a} \]

[Out]

((-1/5 + (2*I)/5)*2^(1 + I)*c*(1 - I*a*x)^(2 - I)*Hypergeometric2F1[-1 - I, 2 - I, 3 - I, (1 - I*a*x)/2])/a

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Rubi [A]  time = 0.0258661, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {5073, 69} \[ -\frac{\left (\frac{1}{5}-\frac{2 i}{5}\right ) 2^{1+i} c (1-i a x)^{2-i} \, _2F_1\left (-1-i,2-i;3-i;\frac{1}{2} (1-i a x)\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/5 + (2*I)/5)*2^(1 + I)*c*(1 - I*a*x)^(2 - I)*Hypergeometric2F1[-1 - I, 2 - I, 3 - I, (1 - I*a*x)/2])/a

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 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int e^{-2 \tan ^{-1}(a x)} \left (c+a^2 c x^2\right ) \, dx &=c \int (1-i a x)^{1-i} (1+i a x)^{1+i} \, dx\\ &=-\frac{\left (\frac{1}{5}-\frac{2 i}{5}\right ) 2^{1+i} c (1-i a x)^{2-i} \, _2F_1\left (-1-i,2-i;3-i;\frac{1}{2} (1-i a x)\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.0118533, size = 51, normalized size = 1. \[ -\frac{\left (\frac{1}{5}-\frac{2 i}{5}\right ) 2^{1+i} c (1-i a x)^{2-i} \, _2F_1\left (-1-i,2-i;3-i;\frac{1}{2} (1-i a x)\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/5 + (2*I)/5)*2^(1 + I)*c*(1 - I*a*x)^(2 - I)*Hypergeometric2F1[-1 - I, 2 - I, 3 - I, (1 - I*a*x)/2])/a

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{\frac{{a}^{2}c{x}^{2}+c}{{{\rm e}^{2\,\arctan \left ( ax \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{2} c x^{2} + c\right )} e^{\left (-2 \, \arctan \left (a x\right )\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*c*x^2 + c)*e^(-2*arctan(a*x)), x)

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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)/exp(2*atan(a*x)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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