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

Optimal. Leaf size=102 \[ \frac{i 2^{p+\left (1-\frac{i}{2}\right )} (1-i a x)^{p+\left (1+\frac{i}{2}\right )} \left (a^2 x^2+1\right )^{-p} \left (a^2 c x^2+c\right )^p \, _2F_1\left (\frac{i}{2}-p,p+\left (1+\frac{i}{2}\right );p+\left (2+\frac{i}{2}\right );\frac{1}{2} (1-i a x)\right )}{a (2 p+(2+i))} \]

[Out]

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

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Rubi [A]  time = 0.0751491, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {5076, 5073, 69} \[ \frac{i 2^{p+\left (1-\frac{i}{2}\right )} (1-i a x)^{p+\left (1+\frac{i}{2}\right )} \left (a^2 x^2+1\right )^{-p} \left (a^2 c x^2+c\right )^p \, _2F_1\left (\frac{i}{2}-p,p+\left (1+\frac{i}{2}\right );p+\left (2+\frac{i}{2}\right );\frac{1}{2} (1-i a x)\right )}{a (2 p+(2+i))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*2^((1 - I/2) + p)*(1 - I*a*x)^((1 + I/2) + p)*(c + a^2*c*x^2)^p*Hypergeometric2F1[I/2 - p, (1 + I/2) + p, (
2 + I/2) + p, (1 - I*a*x)/2])/(a*((2 + I) + 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 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^{\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^{\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)^{\frac{i}{2}+p} (1+i a x)^{-\frac{i}{2}+p} \, dx\\ &=\frac{i 2^{\left (1-\frac{i}{2}\right )+p} (1-i a x)^{\left (1+\frac{i}{2}\right )+p} \left (1+a^2 x^2\right )^{-p} \left (c+a^2 c x^2\right )^p \, _2F_1\left (\frac{i}{2}-p,\left (1+\frac{i}{2}\right )+p;\left (2+\frac{i}{2}\right )+p;\frac{1}{2} (1-i a x)\right )}{a ((2+i)+2 p)}\\ \end{align*}

Mathematica [A]  time = 0.0281641, size = 102, normalized size = 1. \[ \frac{i 2^{p-\frac{i}{2}} (1-i a x)^{p+\left (1+\frac{i}{2}\right )} \left (a^2 x^2+1\right )^{-p} \left (a^2 c x^2+c\right )^p \, _2F_1\left (\frac{i}{2}-p,p+\left (1+\frac{i}{2}\right );p+\left (2+\frac{i}{2}\right );\frac{1}{2} (1-i a x)\right )}{a \left (p+\left (1+\frac{i}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.322, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{\arctan \left ( ax \right ) }} \left ({a}^{2}c{x}^{2}+c \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(arctan(a*x))*(a^2*c*x^2+c)^p,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 )}^{p} e^{\left (\arctan \left (a x\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a^2*c*x^2 + c)^p*e^(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 )}^{p} e^{\left (\arctan \left (a x\right )\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \left (a^{2} x^{2} + 1\right )\right )^{p} e^{\operatorname{atan}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*(a**2*x**2 + 1))**p*exp(atan(a*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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