3.238 \(\int e^{n \tan ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=220 \[ \frac {2^{-\frac {i n}{2}} \left (-6 a^2-6 a n-n^2+2\right ) (-i a-i b x+1)^{1+\frac {i n}{2}} \, _2F_1\left (\frac {i n}{2}+1,\frac {i n}{2};\frac {i n}{2}+2;\frac {1}{2} (-i a-i b x+1)\right )}{3 b^3 (-n+2 i)}-\frac {(4 a+n) (-i a-i b x+1)^{1+\frac {i n}{2}} (i a+i b x+1)^{1-\frac {i n}{2}}}{6 b^3}+\frac {x (-i a-i b x+1)^{1+\frac {i n}{2}} (i a+i b x+1)^{1-\frac {i n}{2}}}{3 b^2} \]

[Out]

-1/6*(4*a+n)*(1-I*a-I*b*x)^(1+1/2*I*n)*(1+I*a+I*b*x)^(1-1/2*I*n)/b^3+1/3*x*(1-I*a-I*b*x)^(1+1/2*I*n)*(1+I*a+I*
b*x)^(1-1/2*I*n)/b^2+1/3*(-6*a^2-6*a*n-n^2+2)*(1-I*a-I*b*x)^(1+1/2*I*n)*hypergeom([1/2*I*n, 1+1/2*I*n],[2+1/2*
I*n],1/2-1/2*I*a-1/2*I*b*x)/(2^(1/2*I*n))/b^3/(2*I-n)

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Rubi [A]  time = 0.14, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5095, 90, 80, 69} \[ \frac {2^{-\frac {i n}{2}} \left (-6 a^2-6 a n-n^2+2\right ) (-i a-i b x+1)^{1+\frac {i n}{2}} \, _2F_1\left (\frac {i n}{2}+1,\frac {i n}{2};\frac {i n}{2}+2;\frac {1}{2} (-i a-i b x+1)\right )}{3 b^3 (-n+2 i)}-\frac {(4 a+n) (-i a-i b x+1)^{1+\frac {i n}{2}} (i a+i b x+1)^{1-\frac {i n}{2}}}{6 b^3}+\frac {x (-i a-i b x+1)^{1+\frac {i n}{2}} (i a+i b x+1)^{1-\frac {i n}{2}}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTan[a + b*x])*x^2,x]

[Out]

-((4*a + n)*(1 - I*a - I*b*x)^(1 + (I/2)*n)*(1 + I*a + I*b*x)^(1 - (I/2)*n))/(6*b^3) + (x*(1 - I*a - I*b*x)^(1
+ (I/2)*n)*(1 + I*a + I*b*x)^(1 - (I/2)*n))/(3*b^2) + ((2 - 6*a^2 - 6*a*n - n^2)*(1 - I*a - I*b*x)^(1 + (I/2)
*n)*Hypergeometric2F1[1 + (I/2)*n, (I/2)*n, 2 + (I/2)*n, (1 - I*a - I*b*x)/2])/(3*2^((I/2)*n)*b^3*(2*I - n))

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 5095

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1 -
I*a*c - I*b*c*x)^((I*n)/2))/(1 + I*a*c + I*b*c*x)^((I*n)/2), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

Rubi steps

\begin {align*} \int e^{n \tan ^{-1}(a+b x)} x^2 \, dx &=\int x^2 (1-i a-i b x)^{\frac {i n}{2}} (1+i a+i b x)^{-\frac {i n}{2}} \, dx\\ &=\frac {x (1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{1-\frac {i n}{2}}}{3 b^2}+\frac {\int (1-i a-i b x)^{\frac {i n}{2}} (1+i a+i b x)^{-\frac {i n}{2}} \left (-1-a^2-b (4 a+n) x\right ) \, dx}{3 b^2}\\ &=-\frac {(4 a+n) (1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{1-\frac {i n}{2}}}{6 b^3}+\frac {x (1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{1-\frac {i n}{2}}}{3 b^2}-\frac {\left (2-6 a^2-6 a n-n^2\right ) \int (1-i a-i b x)^{\frac {i n}{2}} (1+i a+i b x)^{-\frac {i n}{2}} \, dx}{6 b^2}\\ &=-\frac {(4 a+n) (1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{1-\frac {i n}{2}}}{6 b^3}+\frac {x (1-i a-i b x)^{1+\frac {i n}{2}} (1+i a+i b x)^{1-\frac {i n}{2}}}{3 b^2}+\frac {2^{-\frac {i n}{2}} \left (2-6 a^2-6 a n-n^2\right ) (1-i a-i b x)^{1+\frac {i n}{2}} \, _2F_1\left (1+\frac {i n}{2},\frac {i n}{2};2+\frac {i n}{2};\frac {1}{2} (1-i a-i b x)\right )}{3 b^3 (2 i-n)}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 160, normalized size = 0.73 \[ \frac {(-i (a+b x+i))^{1+\frac {i n}{2}} \left (\frac {2^{1-\frac {i n}{2}} \left (6 a^2+6 a n+n^2-2\right ) \, _2F_1\left (\frac {i n}{2}+1,\frac {i n}{2};\frac {i n}{2}+2;-\frac {1}{2} i (a+b x+i)\right )}{n-2 i}-\left ((4 a+n) (i a+i b x+1)^{1-\frac {i n}{2}}\right )+2 b x (i a+i b x+1)^{1-\frac {i n}{2}}\right )}{6 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*(I + a + b*x))^(1 + (I/2)*n)*(-((4*a + n)*(1 + I*a + I*b*x)^(1 - (I/2)*n)) + 2*b*x*(1 + I*a + I*b*x)^(1
- (I/2)*n) + (2^(1 - (I/2)*n)*(-2 + 6*a^2 + 6*a*n + n^2)*Hypergeometric2F1[1 + (I/2)*n, (I/2)*n, 2 + (I/2)*n,
(-1/2*I)*(I + a + b*x)])/(-2*I + n)))/(6*b^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*e^(n*arctan(b*x + a)), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(b*x+a))*x^2,x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{n \arctan \left (b x +a \right )} x^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*exp(n*atan(a + b*x)),x)

[Out]

int(x^2*exp(n*atan(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*exp(n*atan(a + b*x)), x)

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