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

Optimal. Leaf size=79 \[ \frac {\sqrt {a^2 x^2+1} x^{m+1} F_1\left (m+1;\frac {1}{2} (1-i n),\frac {1}{2} (i n+1);m+2;i a x,-i a x\right )}{(m+1) \sqrt {a^2 c x^2+c}} \]

[Out]

x^(1+m)*AppellF1(1+m,1/2+1/2*I*n,1/2-1/2*I*n,2+m,-I*a*x,I*a*x)*(a^2*x^2+1)^(1/2)/(1+m)/(a^2*c*x^2+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 79, 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, 133} \[ \frac {\sqrt {a^2 x^2+1} x^{m+1} F_1\left (m+1;\frac {1}{2} (1-i n),\frac {1}{2} (i n+1);m+2;i a x,-i a x\right )}{(m+1) \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Sqrt[1 + a^2*x^2]*AppellF1[1 + m, (1 - I*n)/2, (1 + I*n)/2, 2 + m, I*a*x, (-I)*a*x])/((1 + m)*Sqrt[
c + a^2*c*x^2])

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 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^m}{\sqrt {c+a^2 c x^2}} \, dx &=\frac {\sqrt {1+a^2 x^2} \int \frac {e^{n \tan ^{-1}(a x)} x^m}{\sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \int x^m (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 {x^{1+m} \sqrt {1+a^2 x^2} F_1\left (1+m;\frac {1}{2} (1-i n),\frac {1}{2} (1+i n);2+m;i a x,-i a x\right )}{(1+m) \sqrt {c+a^2 c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \frac {e^{n \tan ^{-1}(a x)} x^m}{\sqrt {c+a^2 c x^2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{m} e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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^m/(a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

sage0*x

________________________________________________________________________________________

maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctan \left (a x \right )} x^{m}}{\sqrt {a^{2} c \,x^{2}+c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________