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

Optimal. Leaf size=49 \[ \frac{c x^{m+1} F_1\left (m+1;-\frac{i n}{2}-1,\frac{i n}{2}-1;m+2;i a x,-i a x\right )}{m+1} \]

[Out]

(c*x^(1 + m)*AppellF1[1 + m, -1 - (I/2)*n, -1 + (I/2)*n, 2 + m, I*a*x, (-I)*a*x])/(1 + m)

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Rubi [A]  time = 0.0681657, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {5082, 133} \[ \frac{c x^{m+1} F_1\left (m+1;-\frac{i n}{2}-1,\frac{i n}{2}-1;m+2;i a x,-i a x\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x^(1 + m)*AppellF1[1 + m, -1 - (I/2)*n, -1 + (I/2)*n, 2 + m, I*a*x, (-I)*a*x])/(1 + m)

Rule 5082

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 - I
*a*x)^(p + (I*n)/2)*(1 + I*a*x)^(p - (I*n)/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[d, a^2*c] && (Int
egerQ[p] || GtQ[c, 0])

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int e^{n \tan ^{-1}(a x)} x^m \left (c+a^2 c x^2\right ) \, dx &=c \int x^m (1-i a x)^{1+\frac{i n}{2}} (1+i a x)^{1-\frac{i n}{2}} \, dx\\ &=\frac{c x^{1+m} F_1\left (1+m;-1-\frac{i n}{2},-1+\frac{i n}{2};2+m;i a x,-i a x\right )}{1+m}\\ \end{align*}

Mathematica [F]  time = 0.459059, size = 0, normalized size = 0. \[ \int e^{n \tan ^{-1}(a x)} x^m \left (c+a^2 c x^2\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.112, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n\arctan \left ( ax \right ) }}{x}^{m} \left ({a}^{2}c{x}^{2}+c \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*c*x^2 + c)*x^m*e^(n*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(exp(n*atan(a*x))*x**m*(a**2*c*x**2+c),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 )} x^{m} e^{\left (n \arctan \left (a x\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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