3.236 \(\int e^{n \tan ^{-1}(a+b x)} x^m \, dx\)

Optimal. Leaf size=140 \[ \frac{x^{m+1} (-i a-i b x+1)^{\frac{i n}{2}} (i a+i b x+1)^{-\frac{i n}{2}} \left (1-\frac{b x}{-a+i}\right )^{\frac{i n}{2}} \left (1+\frac{b x}{a+i}\right )^{-\frac{i n}{2}} F_1\left (m+1;-\frac{i n}{2},\frac{i n}{2};m+2;-\frac{b x}{a+i},\frac{b x}{i-a}\right )}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.0744353, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5095, 135, 133} \[ \frac{x^{m+1} (-i a-i b x+1)^{\frac{i n}{2}} (i a+i b x+1)^{-\frac{i n}{2}} \left (1-\frac{b x}{-a+i}\right )^{\frac{i n}{2}} \left (1+\frac{b x}{a+i}\right )^{-\frac{i n}{2}} F_1\left (m+1;-\frac{i n}{2},\frac{i n}{2};m+2;-\frac{b x}{a+i},\frac{b x}{i-a}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTan[a + b*x])*x^m,x]

[Out]

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

Rule 5095

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1 -
I*a*c - I*b*c*x)^((I*n)/2))/(1 + I*a*c + I*b*c*x)^((I*n)/2), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !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+b x)} x^m \, dx &=\int x^m (1-i a-i b x)^{\frac{i n}{2}} (1+i a+i b x)^{-\frac{i n}{2}} \, dx\\ &=\left ((1-i a-i b x)^{\frac{i n}{2}} \left (1-\frac{i b x}{1-i a}\right )^{-\frac{i n}{2}}\right ) \int x^m (1+i a+i b x)^{-\frac{i n}{2}} \left (1-\frac{i b x}{1-i a}\right )^{\frac{i n}{2}} \, dx\\ &=\left ((1-i a-i b x)^{\frac{i n}{2}} (1+i a+i b x)^{-\frac{i n}{2}} \left (1-\frac{i b x}{1-i a}\right )^{-\frac{i n}{2}} \left (1+\frac{i b x}{1+i a}\right )^{\frac{i n}{2}}\right ) \int x^m \left (1-\frac{i b x}{1-i a}\right )^{\frac{i n}{2}} \left (1+\frac{i b x}{1+i a}\right )^{-\frac{i n}{2}} \, dx\\ &=\frac{x^{1+m} (1-i a-i b x)^{\frac{i n}{2}} (1+i a+i b x)^{-\frac{i n}{2}} \left (1-\frac{b x}{i-a}\right )^{\frac{i n}{2}} \left (1+\frac{b x}{i+a}\right )^{-\frac{i n}{2}} F_1\left (1+m;-\frac{i n}{2},\frac{i n}{2};2+m;-\frac{b x}{i+a},\frac{b x}{i-a}\right )}{1+m}\\ \end{align*}

Mathematica [F]  time = 0.807305, size = 0, normalized size = 0. \[ \int e^{n \tan ^{-1}(a+b x)} x^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.131, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n\arctan \left ( bx+a \right ) }}{x}^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^m*e^(n*arctan(b*x + a)), 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(b*x+a))*x**m,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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