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

Optimal. Leaf size=291 \[ \frac {x \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{2 a^2 \sqrt {a^2 c x^2+c}}-\frac {i 2^{\frac {1}{2}-\frac {i n}{2}} \left (1-n^2\right ) \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} \, _2F_1\left (\frac {1}{2} (i n-1),\frac {1}{2} (i n+1);\frac {1}{2} (i n+3);\frac {1}{2} (1-i a x)\right )}{a^3 \left (n^2+1\right ) \sqrt {a^2 c x^2+c}}-\frac {(1+i n) \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{2 a^3 (n+i) \sqrt {a^2 c x^2+c}} \]

[Out]

-1/2*(1+I*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)/a^3/(I+n)/(a^2*c*x^2+c)^(1/2)+1
/2*x*(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/(a^2*c*x^2+c)^(1/2)-I*2^(1/2-1/2*I*
n)*(-n^2+1)*(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^3/(n^2+1)/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.34, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {5085, 5082, 90, 79, 69} \[ -\frac {i 2^{\frac {1}{2}-\frac {i n}{2}} \left (1-n^2\right ) \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} \, _2F_1\left (\frac {1}{2} (i n-1),\frac {1}{2} (i n+1);\frac {1}{2} (i n+3);\frac {1}{2} (1-i a x)\right )}{a^3 \left (n^2+1\right ) \sqrt {a^2 c x^2+c}}+\frac {x \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{2 a^2 \sqrt {a^2 c x^2+c}}-\frac {(1+i n) \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2} (1+i n)} (1+i a x)^{\frac {1}{2} (1-i n)}}{2 a^3 (n+i) \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.18, size = 206, normalized size = 0.71 \[ \frac {2^{-1-\frac {i n}{2}} \sqrt {a^2 x^2+1} (1-i a x)^{\frac {1}{2}+\frac {i n}{2}} (1+i a x)^{-\frac {i n}{2}} \left (2 i \sqrt {2} \left (n^2-1\right ) (1+i a x)^{\frac {i n}{2}} \, _2F_1\left (\frac {1}{2} (i n+1),\frac {1}{2} i (n+i);\frac {1}{2} (i n+3);\frac {1}{2} (1-i a x)\right )+2^{\frac {i n}{2}} (n-i) \sqrt {1+i a x} (n (a x-i)+i a x-1)\right )}{a^3 \left (n^2+1\right ) \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2} e^{\left (n \arctan \left (a x\right )\right )}}{\sqrt {a^{2} c x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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