\(\int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx\) [359]

Optimal result

Integrand size = 26, antiderivative size = 281 \[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=-\frac {(1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{2 x^2 \sqrt {c+a^2 c x^2}}-\frac {a n (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)} \sqrt {1+a^2 x^2}}{2 x \sqrt {c+a^2 c x^2}}+\frac {a^2 \left (1-n^2\right ) (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (-1-i n)} \sqrt {1+a^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (1+i n),\frac {1}{2} (3+i n),\frac {1-i a x}{1+i a x}\right )}{(1+i n) \sqrt {c+a^2 c x^2}} \]

[Out]

-1/2*(1-I*a*x)^(1/2+1/2*I*n)*(1+I*a*x)^(1/2-1/2*I*n)*(a^2*x^2+1)^(1/2)/x^2/(a^2*c*x^2+c)^(1/2)-1/2*a*n*(1-I*a*
x)^(1/2+1/2*I*n)*(1+I*a*x)^(1/2-1/2*I*n)*(a^2*x^2+1)^(1/2)/x/(a^2*c*x^2+c)^(1/2)+a^2*(-n^2+1)*(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)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5193, 5190, 105, 156, 12, 133} \[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\frac {a^2 \left (1-n^2\right ) \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (-1-i n)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (i n+1),\frac {1}{2} (i n+3),\frac {1-i a x}{i a x+1}\right )}{(1+i n) \sqrt {a^2 c x^2+c}}-\frac {a n \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{2 x \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{2 x^2 \sqrt {a^2 c x^2+c}} \]

[In]

Int[E^(n*ArcTan[a*x])/(x^3*Sqrt[c + a^2*c*x^2]),x]

[Out]

-1/2*((1 - I*a*x)^((1 + I*n)/2)*(1 + I*a*x)^((1 - I*n)/2)*Sqrt[1 + a^2*x^2])/(x^2*Sqrt[c + a^2*c*x^2]) - (a*n*
(1 - I*a*x)^((1 + I*n)/2)*(1 + I*a*x)^((1 - I*n)/2)*Sqrt[1 + a^2*x^2])/(2*x*Sqrt[c + a^2*c*x^2]) + (a^2*(1 - n
^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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 105

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[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*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

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 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

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])

Rule 5193

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*} \text {integral}& = \frac {\sqrt {1+a^2 x^2} \int \frac {e^{n \arctan (a x)}}{x^3 \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^3} \, dx}{\sqrt {c+a^2 c x^2}} \\ & = -\frac {(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}}{2 x^2 \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}} \left (-a n+a^2 x\right )}{x^2} \, dx}{2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {(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}}{2 x^2 \sqrt {c+a^2 c x^2}}-\frac {a n (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}}{2 x \sqrt {c+a^2 c x^2}}-\frac {\sqrt {1+a^2 x^2} \int \frac {a^2 \left (1-n^2\right ) (1-i a x)^{-\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {1}{2}-\frac {i n}{2}}}{x} \, dx}{2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {(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}}{2 x^2 \sqrt {c+a^2 c x^2}}-\frac {a n (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}}{2 x \sqrt {c+a^2 c x^2}}-\frac {\left (a^2 \left (1-n^2\right ) \sqrt {1+a^2 x^2}\right ) \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}{2 \sqrt {c+a^2 c x^2}} \\ & = -\frac {(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}}{2 x^2 \sqrt {c+a^2 c x^2}}-\frac {a n (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}}{2 x \sqrt {c+a^2 c x^2}}+\frac {a^2 \left (1-n^2\right ) (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} \operatorname {Hypergeometric2F1}\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*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.57 \[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\frac {i (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {1}{2}-\frac {i n}{2}} \sqrt {1+a^2 x^2} \left (-((-i+n) (-i+a x) (1+a n x))+2 a^2 \left (-1+n^2\right ) x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+\frac {i n}{2},\frac {3}{2}+\frac {i n}{2},\frac {i+a x}{i-a x}\right )\right )}{2 (-i+n) x^2 \sqrt {c+a^2 c x^2}} \]

[In]

Integrate[E^(n*ArcTan[a*x])/(x^3*Sqrt[c + a^2*c*x^2]),x]

[Out]

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

Maple [F]

\[\int \frac {{\mathrm e}^{n \arctan \left (a x \right )}}{x^{3} \sqrt {a^{2} c \,x^{2}+c}}d x\]

[In]

int(exp(n*arctan(a*x))/x^3/(a^2*c*x^2+c)^(1/2),x)

[Out]

int(exp(n*arctan(a*x))/x^3/(a^2*c*x^2+c)^(1/2),x)

Fricas [F]

\[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c} x^{3}} \,d x } \]

[In]

integrate(exp(n*arctan(a*x))/x^3/(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^5 + c*x^3), x)

Sympy [F]

\[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int \frac {e^{n \operatorname {atan}{\left (a x \right )}}}{x^{3} \sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \]

[In]

integrate(exp(n*atan(a*x))/x**3/(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(exp(n*atan(a*x))/(x**3*sqrt(c*(a**2*x**2 + 1))), x)

Maxima [F]

\[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c} x^{3}} \,d x } \]

[In]

integrate(exp(n*arctan(a*x))/x^3/(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^3), x)

Giac [F]

\[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int { \frac {e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c} x^{3}} \,d x } \]

[In]

integrate(exp(n*arctan(a*x))/x^3/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{n \arctan (a x)}}{x^3 \sqrt {c+a^2 c x^2}} \, dx=\int \frac {{\mathrm {e}}^{n\,\mathrm {atan}\left (a\,x\right )}}{x^3\,\sqrt {c\,a^2\,x^2+c}} \,d x \]

[In]

int(exp(n*atan(a*x))/(x^3*(c + a^2*c*x^2)^(1/2)),x)

[Out]

int(exp(n*atan(a*x))/(x^3*(c + a^2*c*x^2)^(1/2)), x)