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

Optimal. Leaf size=202 \[ \frac {\sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{a^2 (1-i n) \sqrt {a^2 c x^2+c}}-\frac {i 2^{\frac {3}{2}-\frac {i n}{2}} n \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} \, _2F_1\left (\frac {1}{2} (i n-1),\frac {1}{2} (i n+1);\frac {1}{2} (i n+3);\frac {1}{2} (1-i a x)\right )}{a^2 \left (n^2+1\right ) \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5085, 5082, 79, 69} \[ \frac {\sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{a^2 (1-i n) \sqrt {a^2 c x^2+c}}-\frac {i 2^{\frac {3}{2}-\frac {i n}{2}} n \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} \, _2F_1\left (\frac {1}{2} (i n-1),\frac {1}{2} (i n+1);\frac {1}{2} (i n+3);\frac {1}{2} (1-i a x)\right )}{a^2 \left (n^2+1\right ) \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

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

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Mathematica [A]  time = 0.15, size = 187, normalized size = 0.93 \[ \frac {i 2^{-\frac {1}{2}-\frac {i n}{2}} \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}} \left (2^{\frac {1}{2}+\frac {i n}{2}} (n-i) \sqrt {1+i a x}-4 n (1+i a x)^{\frac {i n}{2}} \, _2F_1\left (\frac {1}{2} (i n+1),\frac {1}{2} i (n+i);\frac {1}{2} (i n+3);\frac {1}{2} (1-i a x)\right )\right )}{a^2 \left (n^2+1\right ) \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c}}, x\right ) \]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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