3.4.44 \(\int \frac {e^{n \text {ArcTan}(a x)}}{x (c+a^2 c x^2)} \, dx\) [344]

Optimal. Leaf size=65 \[ \frac {i e^{n \text {ArcTan}(a x)}}{c n}-\frac {2 i e^{n \text {ArcTan}(a x)} \, _2F_1\left (1,-\frac {i n}{2};1-\frac {i n}{2};e^{2 i \text {ArcTan}(a x)}\right )}{c n} \]

[Out]

I*exp(n*arctan(a*x))/c/n-2*I*exp(n*arctan(a*x))*hypergeom([1, -1/2*I*n],[1-1/2*I*n],(1+I*a*x)^2/(a^2*x^2+1))/c
/n

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Rubi [A]
time = 0.07, antiderivative size = 122, normalized size of antiderivative = 1.88, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5190, 98, 133} \begin {gather*} \frac {i (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{c n}-\frac {2 i (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}} \, _2F_1\left (1,-\frac {i n}{2};1-\frac {i n}{2};\frac {i a x+1}{1-i a x}\right )}{c n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 98

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 133

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)/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2,
(-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 0]

Rule 5190

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])

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 120, normalized size = 1.85 \begin {gather*} \frac {(1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}} \left ((2+i n) (-i+a x)+2 (n-i a n x) \, _2F_1\left (1,1+\frac {i n}{2};2+\frac {i n}{2};\frac {i+a x}{i-a x}\right )\right )}{c n (-2 i+n) (-i+a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{n \arctan \left (a x \right )}}{x \left (a^{2} c \,x^{2}+c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(e^(n*arctan(a*x))/(a^2*c*x^3 + c*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {e^{n \operatorname {atan}{\left (a x \right )}}}{a^{2} x^{3} + x}\, dx}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(n*atan(a*x))/(a**2*x**3 + x), x)/c

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{x\,\left (c\,a^2\,x^2+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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