3.357 \(\int \frac {e^{n \tan ^{-1}(a x)}}{x \sqrt {c+a^2 c x^2}} \, dx\)

Optimal. Leaf size=121 \[ -\frac {2 \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (-1-i n)} \, _2F_1\left (1,\frac {1}{2} (i n+1);\frac {1}{2} (i n+3);\frac {1-i a x}{i a x+1}\right )}{(1+i n) \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {5085, 5082, 131} \[ -\frac {2 \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (-1-i n)} \, _2F_1\left (1,\frac {1}{2} (i n+1);\frac {1}{2} (i n+3);\frac {1-i a x}{i a x+1}\right )}{(1+i n) \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 5085

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 120, normalized size = 0.99 \[ \frac {2 \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {1}{2}-\frac {i n}{2}} \, _2F_1\left (1,\frac {i n}{2}+\frac {1}{2};\frac {i n}{2}+\frac {3}{2};\frac {a x+i}{i-a x}\right )}{(-1-i n) \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} e^{\left (n \arctan \left (a x\right )\right )}}{a^{2} c x^{3} + c x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(a*x))/x/(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(a*x))/x/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

sage0*x

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctan \left (a x \right )}}{x \sqrt {a^{2} c \,x^{2}+c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctan(a*x))/x/(a^2*c*x^2+c)^(1/2),x)

[Out]

int(exp(n*arctan(a*x))/x/(a^2*c*x^2+c)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(a*x))/x/(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{x\,\sqrt {c\,a^2\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atan(a*x))/(x*(c + a^2*c*x^2)^(1/2)),x)

[Out]

int(exp(n*atan(a*x))/(x*(c + a^2*c*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{n \operatorname {atan}{\left (a x \right )}}}{x \sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atan(a*x))/x/(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(exp(n*atan(a*x))/(x*sqrt(c*(a**2*x**2 + 1))), x)

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