3.237 \(\int e^{n \tan ^{-1}(a+b x)} x^3 \, dx\)

Optimal. Leaf size=260 \[ \frac{2^{-2-\frac{i n}{2}} \left (36 a^2 n+24 a^3-12 a \left (2-n^2\right )-n \left (8-n^2\right )\right ) (-i a-i b x+1)^{1+\frac{i n}{2}} \, _2F_1\left (\frac{i n}{2}+1,\frac{i n}{2};\frac{i n}{2}+2;\frac{1}{2} (-i a-i b x+1)\right )}{3 b^4 (-n+2 i)}-\frac{(-i a-i b x+1)^{1+\frac{i n}{2}} \left (-18 a^2+2 b x (6 a+n)-10 a n-n^2+6\right ) (i a+i b x+1)^{1-\frac{i n}{2}}}{24 b^4}+\frac{x^2 (-i a-i b x+1)^{1+\frac{i n}{2}} (i a+i b x+1)^{1-\frac{i n}{2}}}{4 b^2} \]

[Out]

(x^2*(1 - I*a - I*b*x)^(1 + (I/2)*n)*(1 + I*a + I*b*x)^(1 - (I/2)*n))/(4*b^2) - ((1 - I*a - I*b*x)^(1 + (I/2)*
n)*(1 + I*a + I*b*x)^(1 - (I/2)*n)*(6 - 18*a^2 - 10*a*n - n^2 + 2*b*(6*a + n)*x))/(24*b^4) + (2^(-2 - (I/2)*n)
*(24*a^3 + 36*a^2*n - 12*a*(2 - n^2) - n*(8 - 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*b^4*(2*I - n))

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Rubi [A]  time = 0.185642, antiderivative size = 260, normalized size of antiderivative = 1., 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, 100, 147, 69} \[ \frac{2^{-2-\frac{i n}{2}} \left (36 a^2 n+24 a^3-12 a \left (2-n^2\right )-n \left (8-n^2\right )\right ) (-i a-i b x+1)^{1+\frac{i n}{2}} \, _2F_1\left (\frac{i n}{2}+1,\frac{i n}{2};\frac{i n}{2}+2;\frac{1}{2} (-i a-i b x+1)\right )}{3 b^4 (-n+2 i)}-\frac{(-i a-i b x+1)^{1+\frac{i n}{2}} \left (-18 a^2+2 b x (6 a+n)-10 a n-n^2+6\right ) (i a+i b x+1)^{1-\frac{i n}{2}}}{24 b^4}+\frac{x^2 (-i a-i b x+1)^{1+\frac{i n}{2}} (i a+i b x+1)^{1-\frac{i n}{2}}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTan[a + b*x])*x^3,x]

[Out]

(x^2*(1 - I*a - I*b*x)^(1 + (I/2)*n)*(1 + I*a + I*b*x)^(1 - (I/2)*n))/(4*b^2) - ((1 - I*a - I*b*x)^(1 + (I/2)*
n)*(1 + I*a + I*b*x)^(1 - (I/2)*n)*(6 - 18*a^2 - 10*a*n - n^2 + 2*b*(6*a + n)*x))/(24*b^4) + (2^(-2 - (I/2)*n)
*(24*a^3 + 36*a^2*n - 12*a*(2 - n^2) - n*(8 - 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*b^4*(2*I - n))

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]

Rule 100

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

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]))

Rubi steps

\begin{align*} \int e^{n \tan ^{-1}(a+b x)} x^3 \, dx &=\int x^3 (1-i a-i b x)^{\frac{i n}{2}} (1+i a+i b x)^{-\frac{i n}{2}} \, dx\\ &=\frac{x^2 (1-i a-i b x)^{1+\frac{i n}{2}} (1+i a+i b x)^{1-\frac{i n}{2}}}{4 b^2}+\frac{\int x (1-i a-i b x)^{\frac{i n}{2}} (1+i a+i b x)^{-\frac{i n}{2}} \left (-2 \left (1+a^2\right )-b (6 a+n) x\right ) \, dx}{4 b^2}\\ &=\frac{x^2 (1-i a-i b x)^{1+\frac{i n}{2}} (1+i a+i b x)^{1-\frac{i n}{2}}}{4 b^2}-\frac{(1-i a-i b x)^{1+\frac{i n}{2}} (1+i a+i b x)^{1-\frac{i n}{2}} \left (6-18 a^2-10 a n-n^2+2 b (6 a+n) x\right )}{24 b^4}-\frac{\left (24 a^3+36 a^2 n-12 a \left (2-n^2\right )-n \left (8-n^2\right )\right ) \int (1-i a-i b x)^{\frac{i n}{2}} (1+i a+i b x)^{-\frac{i n}{2}} \, dx}{24 b^3}\\ &=\frac{x^2 (1-i a-i b x)^{1+\frac{i n}{2}} (1+i a+i b x)^{1-\frac{i n}{2}}}{4 b^2}-\frac{(1-i a-i b x)^{1+\frac{i n}{2}} (1+i a+i b x)^{1-\frac{i n}{2}} \left (6-18 a^2-10 a n-n^2+2 b (6 a+n) x\right )}{24 b^4}+\frac{2^{-2-\frac{i n}{2}} \left (24 a^3+36 a^2 n-12 a \left (2-n^2\right )-n \left (8-n^2\right )\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^4 (2 i-n)}\\ \end{align*}

Mathematica [A]  time = 0.300176, size = 272, normalized size = 1.05 \[ \frac{(-i (a+b x+i))^{1+\frac{i n}{2}} \left (b^2 (-n+2 i) x^2 (i a+i b x+1)^{1-\frac{i n}{2}}-2^{3-\frac{i n}{2}} (6 a+n) \, _2F_1\left (\frac{i n}{2}-2,\frac{i n}{2}+1;\frac{i n}{2}+2;-\frac{1}{2} i (a+b x+i)\right )+(1+i a) 2^{3-\frac{i n}{2}} (5 a+n-i) \, _2F_1\left (\frac{i n}{2}-1,\frac{i n}{2}+1;\frac{i n}{2}+2;-\frac{1}{2} i (a+b x+i)\right )+(a-i)^2 2^{1-\frac{i n}{2}} (4 a+n-2 i) \, _2F_1\left (\frac{i n}{2}+1,\frac{i n}{2};\frac{i n}{2}+2;-\frac{1}{2} i (a+b x+i)\right )\right )}{4 b^4 (-n+2 i)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(n*ArcTan[a + b*x])*x^3,x]

[Out]

(((-I)*(I + a + b*x))^(1 + (I/2)*n)*(b^2*(2*I - n)*x^2*(1 + I*a + I*b*x)^(1 - (I/2)*n) - 2^(3 - (I/2)*n)*(6*a
+ n)*Hypergeometric2F1[-2 + (I/2)*n, 1 + (I/2)*n, 2 + (I/2)*n, (-I/2)*(I + a + b*x)] + 2^(3 - (I/2)*n)*(1 + I*
a)*(-I + 5*a + n)*Hypergeometric2F1[-1 + (I/2)*n, 1 + (I/2)*n, 2 + (I/2)*n, (-I/2)*(I + a + b*x)] + 2^(1 - (I/
2)*n)*(-I + a)^2*(-2*I + 4*a + n)*Hypergeometric2F1[1 + (I/2)*n, (I/2)*n, 2 + (I/2)*n, (-I/2)*(I + a + b*x)]))
/(4*b^4*(2*I - n))

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Maple [F]  time = 0.197, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n\arctan \left ( bx+a \right ) }}{x}^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctan(b*x+a))*x^3,x)

[Out]

int(exp(n*arctan(b*x+a))*x^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} e^{\left (n \arctan \left (b x + a\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(b*x+a))*x^3,x, algorithm="maxima")

[Out]

integrate(x^3*e^(n*arctan(b*x + a)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{3} e^{\left (n \arctan \left (b x + a\right )\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(b*x+a))*x^3,x, algorithm="fricas")

[Out]

integral(x^3*e^(n*arctan(b*x + a)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} e^{n \operatorname{atan}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atan(b*x+a))*x**3,x)

[Out]

Integral(x**3*exp(n*atan(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} e^{\left (n \arctan \left (b x + a\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctan(b*x+a))*x^3,x, algorithm="giac")

[Out]

integrate(x^3*e^(n*arctan(b*x + a)), x)