3.243 \(\int \frac{e^{n \tan ^{-1}(a+b x)}}{x^3} \, dx\)

Optimal. Leaf size=207 \[ -\frac{(-i a-i b x+1)^{1+\frac{i n}{2}} (i a+i b x+1)^{1-\frac{i n}{2}}}{2 \left (a^2+1\right ) x^2}-\frac{2 b^2 (2 a-n) (-i a-i b x+1)^{1+\frac{i n}{2}} (i a+i b x+1)^{-1-\frac{i n}{2}} \, _2F_1\left (2,\frac{i n}{2}+1;\frac{i n}{2}+2;\frac{(i-a) (-i a-i b x+1)}{(a+i) (i a+i b x+1)}\right )}{(-a+i) (a+i)^3 (-n+2 i)} \]

[Out]

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

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Rubi [A]  time = 0.108783, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5095, 96, 131} \[ -\frac{(-i a-i b x+1)^{1+\frac{i n}{2}} (i a+i b x+1)^{1-\frac{i n}{2}}}{2 \left (a^2+1\right ) x^2}-\frac{2 b^2 (2 a-n) (-i a-i b x+1)^{1+\frac{i n}{2}} (i a+i b x+1)^{-1-\frac{i n}{2}} \, _2F_1\left (2,\frac{i n}{2}+1;\frac{i n}{2}+2;\frac{(i-a) (-i a-i b x+1)}{(a+i) (i a+i b x+1)}\right )}{(-a+i) (a+i)^3 (-n+2 i)} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTan[a + b*x])/x^3,x]

[Out]

-((1 - I*a - I*b*x)^(1 + (I/2)*n)*(1 + I*a + I*b*x)^(1 - (I/2)*n))/(2*(1 + a^2)*x^2) - (2*b^2*(2*a - n)*(1 - I
*a - I*b*x)^(1 + (I/2)*n)*(1 + I*a + I*b*x)^(-1 - (I/2)*n)*Hypergeometric2F1[2, 1 + (I/2)*n, 2 + (I/2)*n, ((I
- a)*(1 - I*a - I*b*x))/((I + a)*(1 + I*a + I*b*x))])/((I - a)*(I + a)^3*(2*I - 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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 131

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*c -
a*d)^n*(a + b*x)^(m + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rubi steps

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

Mathematica [A]  time = 0.0644245, size = 173, normalized size = 0.84 \[ -\frac{i (i a+i b x+1)^{-\frac{i n}{2}} (-i (a+b x+i))^{1+\frac{i n}{2}} \left (4 b^2 x^2 (n-2 a) \, _2F_1\left (2,\frac{i n}{2}+1;\frac{i n}{2}+2;\frac{a^2+b x a-i b x+1}{a^2+b x a+i b x+1}\right )+(a+i)^2 (n-2 i) (a+b x-i)^2\right )}{2 (a-i) (a+i)^3 (n-2 i) x^2 (a+b x-i)} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTan[a + b*x])/x^3,x]

[Out]

((-I/2)*((-I)*(I + a + b*x))^(1 + (I/2)*n)*((I + a)^2*(-2*I + n)*(-I + a + b*x)^2 + 4*b^2*(-2*a + n)*x^2*Hyper
geometric2F1[2, 1 + (I/2)*n, 2 + (I/2)*n, (1 + a^2 - I*b*x + a*b*x)/(1 + a^2 + I*b*x + a*b*x)]))/((-I + a)*(I
+ a)^3*(-2*I + n)*x^2*(1 + I*a + I*b*x)^((I/2)*n)*(-I + a + b*x))

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Maple [F]  time = 0.205, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{n\arctan \left ( bx+a \right ) }}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctan(b*x+a))/x^3,x)

[Out]

int(exp(n*arctan(b*x+a))/x^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (n \arctan \left (b x + a\right )\right )}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(b*x+a))/x^3,x, algorithm="maxima")

[Out]

integrate(e^(n*arctan(b*x + a))/x^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{e^{\left (n \arctan \left (b x + a\right )\right )}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(b*x+a))/x^3,x, algorithm="fricas")

[Out]

integral(e^(n*arctan(b*x + a))/x^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{n \operatorname{atan}{\left (a + b x \right )}}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atan(b*x+a))/x**3,x)

[Out]

Integral(exp(n*atan(a + b*x))/x**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (n \arctan \left (b x + a\right )\right )}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(b*x+a))/x^3,x, algorithm="giac")

[Out]

integrate(e^(n*arctan(b*x + a))/x^3, x)