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

Optimal. Leaf size=164 \[ \frac {i 2^{1-\frac {i n}{2}} (1-i a x)^{\frac {i n}{2}} \, _2F_1\left (\frac {i n}{2},\frac {i n}{2};\frac {i n}{2}+1;\frac {1}{2} (1-i a x)\right )}{a^3 c}-\frac {(1+i n) (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^3 c n}+\frac {x (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^2 c} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 164, 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 = {5082, 90, 79, 69} \[ \frac {i 2^{1-\frac {i n}{2}} (1-i a x)^{\frac {i n}{2}} \, _2F_1\left (\frac {i n}{2},\frac {i n}{2};\frac {i n}{2}+1;\frac {1}{2} (1-i a x)\right )}{a^3 c}+\frac {x (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^2 c}-\frac {(1+i n) (1-i a x)^{\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}}}{a^3 c n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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])

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 121, normalized size = 0.74 \[ \frac {(1-i a x)^{\frac {i n}{2}} (2+2 i a x)^{-\frac {i n}{2}} \left (2 i n (1+i a x)^{\frac {i n}{2}} \, _2F_1\left (\frac {i n}{2},\frac {i n}{2};\frac {i n}{2}+1;\frac {1}{2} (1-i a x)\right )+2^{\frac {i n}{2}} (-1+n (a x-i))\right )}{a^3 c n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2} e^{\left (n \arctan \left (a x\right )\right )}}{a^{2} c x^{2} + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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