\(\int x^2 \arctan (a+b x) \, dx\) [45]

Optimal result

Integrand size = 10, antiderivative size = 79 \[ \int x^2 \arctan (a+b x) \, dx=\frac {a x}{b^2}-\frac {(a+b x)^2}{6 b^3}-\frac {a \left (3-a^2\right ) \arctan (a+b x)}{3 b^3}+\frac {1}{3} x^3 \arctan (a+b x)+\frac {\left (1-3 a^2\right ) \log \left (1+(a+b x)^2\right )}{6 b^3} \]

[Out]

a*x/b^2-1/6*(b*x+a)^2/b^3-1/3*a*(-a^2+3)*arctan(b*x+a)/b^3+1/3*x^3*arctan(b*x+a)+1/6*(-3*a^2+1)*ln(1+(b*x+a)^2
)/b^3

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5155, 4972, 716, 649, 209, 266} \[ \int x^2 \arctan (a+b x) \, dx=-\frac {a \left (3-a^2\right ) \arctan (a+b x)}{3 b^3}+\frac {\left (1-3 a^2\right ) \log \left ((a+b x)^2+1\right )}{6 b^3}+\frac {1}{3} x^3 \arctan (a+b x)-\frac {(a+b x)^2}{6 b^3}+\frac {a x}{b^2} \]

[In]

Int[x^2*ArcTan[a + b*x],x]

[Out]

(a*x)/b^2 - (a + b*x)^2/(6*b^3) - (a*(3 - a^2)*ArcTan[a + b*x])/(3*b^3) + (x^3*ArcTan[a + b*x])/3 + ((1 - 3*a^
2)*Log[1 + (a + b*x)^2])/(6*b^3)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 716

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 4972

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((a + b*
ArcTan[c*x])/(e*(q + 1))), x] - Dist[b*(c/(e*(q + 1))), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 5155

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right )^2 \arctan (x) \, dx,x,a+b x\right )}{b} \\ & = \frac {1}{3} x^3 \arctan (a+b x)-\frac {1}{3} \text {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^3}{1+x^2} \, dx,x,a+b x\right ) \\ & = \frac {1}{3} x^3 \arctan (a+b x)-\frac {1}{3} \text {Subst}\left (\int \left (-\frac {3 a}{b^3}+\frac {x}{b^3}+\frac {a \left (3-a^2\right )-\left (1-3 a^2\right ) x}{b^3 \left (1+x^2\right )}\right ) \, dx,x,a+b x\right ) \\ & = \frac {a x}{b^2}-\frac {(a+b x)^2}{6 b^3}+\frac {1}{3} x^3 \arctan (a+b x)-\frac {\text {Subst}\left (\int \frac {a \left (3-a^2\right )-\left (1-3 a^2\right ) x}{1+x^2} \, dx,x,a+b x\right )}{3 b^3} \\ & = \frac {a x}{b^2}-\frac {(a+b x)^2}{6 b^3}+\frac {1}{3} x^3 \arctan (a+b x)+\frac {\left (1-3 a^2\right ) \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,a+b x\right )}{3 b^3}-\frac {\left (a \left (3-a^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,a+b x\right )}{3 b^3} \\ & = \frac {a x}{b^2}-\frac {(a+b x)^2}{6 b^3}-\frac {a \left (3-a^2\right ) \arctan (a+b x)}{3 b^3}+\frac {1}{3} x^3 \arctan (a+b x)+\frac {\left (1-3 a^2\right ) \log \left (1+(a+b x)^2\right )}{6 b^3} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.05 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.44 \[ \int x^2 \arctan (a+b x) \, dx=\frac {\frac {1}{3} b \left (-\frac {a}{b}+\frac {a+b x}{b}\right )^3 \arctan (a+b x)-\frac {1}{3} b \left (-\frac {3 a x}{b^2}+\frac {(a+b x)^2}{2 b^3}-\frac {(1+i a)^3 \log (i-a-b x)}{2 b^3}-\frac {(1-i a)^3 \log (i+a+b x)}{2 b^3}\right )}{b} \]

[In]

Integrate[x^2*ArcTan[a + b*x],x]

[Out]

((b*(-(a/b) + (a + b*x)/b)^3*ArcTan[a + b*x])/3 - (b*((-3*a*x)/b^2 + (a + b*x)^2/(2*b^3) - ((1 + I*a)^3*Log[I
- a - b*x])/(2*b^3) - ((1 - I*a)^3*Log[I + a + b*x])/(2*b^3)))/3)/b

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.29

method	result	size
parallelrisch	\(-\frac {-2 \arctan \left (b x +a \right ) x^{3} b^{3}+b^{2} x^{2}-2 \arctan \left (b x +a \right ) a^{3}+3 a^{2} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )-4 a b x +6 a \arctan \left (b x +a \right )+7 a^{2}-1-\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{6 b^{3}}\)	\(102\)
derivativedivides	\(\frac {-\frac {\arctan \left (b x +a \right ) a^{3}}{3}+\arctan \left (b x +a \right ) a^{2} \left (b x +a \right )-\arctan \left (b x +a \right ) a \left (b x +a \right )^{2}+\frac {\arctan \left (b x +a \right ) \left (b x +a \right )^{3}}{3}+\left (b x +a \right ) a -\frac {\left (b x +a \right )^{2}}{6}+\frac {\left (-3 a^{2}+1\right ) \ln \left (1+\left (b x +a \right )^{2}\right )}{6}+\frac {\left (a^{3}-3 a \right ) \arctan \left (b x +a \right )}{3}}{b^{3}}\)	\(113\)
default	\(\frac {-\frac {\arctan \left (b x +a \right ) a^{3}}{3}+\arctan \left (b x +a \right ) a^{2} \left (b x +a \right )-\arctan \left (b x +a \right ) a \left (b x +a \right )^{2}+\frac {\arctan \left (b x +a \right ) \left (b x +a \right )^{3}}{3}+\left (b x +a \right ) a -\frac {\left (b x +a \right )^{2}}{6}+\frac {\left (-3 a^{2}+1\right ) \ln \left (1+\left (b x +a \right )^{2}\right )}{6}+\frac {\left (a^{3}-3 a \right ) \arctan \left (b x +a \right )}{3}}{b^{3}}\)	\(113\)
parts	\(\frac {x^{3} \arctan \left (b x +a \right )}{3}-\frac {b \left (-\frac {-\frac {1}{2} x^{2} b +2 a x}{b^{3}}+\frac {\frac {\left (3 a^{2} b -b \right ) \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}+\frac {\left (2 a^{3}+2 a -\frac {\left (3 a^{2} b -b \right ) a}{b}\right ) \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}}{b^{3}}\right )}{3}\)	\(117\)
risch	\(-\frac {i x^{3} \ln \left (1+i \left (b x +a \right )\right )}{6}+\frac {i x^{3} \ln \left (1-i \left (b x +a \right )\right )}{6}+\frac {a^{3} \arctan \left (b x +a \right )}{3 b^{3}}-\frac {x^{2}}{6 b}-\frac {a^{2} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{3}}+\frac {2 a x}{3 b^{2}}-\frac {a \arctan \left (b x +a \right )}{b^{3}}+\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{6 b^{3}}\)	\(126\)

[In]

int(x^2*arctan(b*x+a),x,method=_RETURNVERBOSE)

[Out]

-1/6*(-2*arctan(b*x+a)*x^3*b^3+b^2*x^2-2*arctan(b*x+a)*a^3+3*a^2*ln(b^2*x^2+2*a*b*x+a^2+1)-4*a*b*x+6*a*arctan(
b*x+a)+7*a^2-1-ln(b^2*x^2+2*a*b*x+a^2+1))/b^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int x^2 \arctan (a+b x) \, dx=-\frac {b^{2} x^{2} - 4 \, a b x - 2 \, {\left (b^{3} x^{3} + a^{3} - 3 \, a\right )} \arctan \left (b x + a\right ) + {\left (3 \, a^{2} - 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{6 \, b^{3}} \]

[In]

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

[Out]

-1/6*(b^2*x^2 - 4*a*b*x - 2*(b^3*x^3 + a^3 - 3*a)*arctan(b*x + a) + (3*a^2 - 1)*log(b^2*x^2 + 2*a*b*x + a^2 +
1))/b^3

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.48 \[ \int x^2 \arctan (a+b x) \, dx=\begin {cases} \frac {a^{3} \operatorname {atan}{\left (a + b x \right )}}{3 b^{3}} - \frac {a^{2} \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{3}} + \frac {2 a x}{3 b^{2}} - \frac {a \operatorname {atan}{\left (a + b x \right )}}{b^{3}} + \frac {x^{3} \operatorname {atan}{\left (a + b x \right )}}{3} - \frac {x^{2}}{6 b} + \frac {\log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{6 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \operatorname {atan}{\left (a \right )}}{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**3*atan(a + b*x)/(3*b**3) - a**2*log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(2*b**3) + 2*a*x/(3*b**2) -
a*atan(a + b*x)/b**3 + x**3*atan(a + b*x)/3 - x**2/(6*b) + log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(6*b**3), Ne(b,
0)), (x**3*atan(a)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.08 \[ \int x^2 \arctan (a+b x) \, dx=\frac {1}{3} \, x^{3} \arctan \left (b x + a\right ) - \frac {1}{6} \, b {\left (\frac {b x^{2} - 4 \, a x}{b^{3}} - \frac {2 \, {\left (a^{3} - 3 \, a\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{4}} + \frac {{\left (3 \, a^{2} - 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{4}}\right )} \]

[In]

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

[Out]

1/3*x^3*arctan(b*x + a) - 1/6*b*((b*x^2 - 4*a*x)/b^3 - 2*(a^3 - 3*a)*arctan((b^2*x + a*b)/b)/b^4 + (3*a^2 - 1)
*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/b^4)

Giac [F]

\[ \int x^2 \arctan (a+b x) \, dx=\int { x^{2} \arctan \left (b x + a\right ) \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.29 \[ \int x^2 \arctan (a+b x) \, dx=\frac {\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{6\,b^3}+\frac {x^3\,\mathrm {atan}\left (a+b\,x\right )}{3}-\frac {x^2}{6\,b}-\frac {a^2\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2\,b^3}+\frac {a^3\,\mathrm {atan}\left (a+b\,x\right )}{3\,b^3}-\frac {a\,\mathrm {atan}\left (a+b\,x\right )}{b^3}+\frac {2\,a\,x}{3\,b^2} \]

[In]

int(x^2*atan(a + b*x),x)

[Out]

log(a^2 + b^2*x^2 + 2*a*b*x + 1)/(6*b^3) + (x^3*atan(a + b*x))/3 - x^2/(6*b) - (a^2*log(a^2 + b^2*x^2 + 2*a*b*
x + 1))/(2*b^3) + (a^3*atan(a + b*x))/(3*b^3) - (a*atan(a + b*x))/b^3 + (2*a*x)/(3*b^2)