3.46 \(\int x \tan ^{-1}(a+b x) \, dx\)

Optimal. Leaf size=60 \[ \frac {\left (1-a^2\right ) \tan ^{-1}(a+b x)}{2 b^2}+\frac {a \log \left ((a+b x)^2+1\right )}{2 b^2}+\frac {1}{2} x^2 \tan ^{-1}(a+b x)-\frac {x}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5047, 4862, 702, 635, 203, 260} \[ \frac {\left (1-a^2\right ) \tan ^{-1}(a+b x)}{2 b^2}+\frac {a \log \left ((a+b x)^2+1\right )}{2 b^2}+\frac {1}{2} x^2 \tan ^{-1}(a+b x)-\frac {x}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTan[a + b*x],x]

[Out]

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

Rule 203

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

Rule 260

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 635

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 702

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 4862

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 5047

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*} \int x \tan ^{-1}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right ) \tan ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{2} x^2 \tan ^{-1}(a+b x)-\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^2}{1+x^2} \, dx,x,a+b x\right )\\ &=\frac {1}{2} x^2 \tan ^{-1}(a+b x)-\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{b^2}-\frac {1-a^2+2 a x}{b^2 \left (1+x^2\right )}\right ) \, dx,x,a+b x\right )\\ &=-\frac {x}{2 b}+\frac {1}{2} x^2 \tan ^{-1}(a+b x)+\frac {\operatorname {Subst}\left (\int \frac {1-a^2+2 a x}{1+x^2} \, dx,x,a+b x\right )}{2 b^2}\\ &=-\frac {x}{2 b}+\frac {1}{2} x^2 \tan ^{-1}(a+b x)+\frac {a \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,a+b x\right )}{b^2}+\frac {\left (1-a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,a+b x\right )}{2 b^2}\\ &=-\frac {x}{2 b}+\frac {\left (1-a^2\right ) \tan ^{-1}(a+b x)}{2 b^2}+\frac {1}{2} x^2 \tan ^{-1}(a+b x)+\frac {a \log \left (1+(a+b x)^2\right )}{2 b^2}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 90, normalized size = 1.50 \[ \frac {-i a^2 \log (a+b x+i)+2 b^2 x^2 \tan ^{-1}(a+b x)+2 a \log (a+b x+i)+i (a-i)^2 \log (-a-b x+i)+i \log (a+b x+i)-2 b x}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTan[a + b*x],x]

[Out]

(-2*b*x + 2*b^2*x^2*ArcTan[a + b*x] + I*(-I + a)^2*Log[I - a - b*x] + I*Log[I + a + b*x] + 2*a*Log[I + a + b*x
] - I*a^2*Log[I + a + b*x])/(4*b^2)

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fricas [A]  time = 0.41, size = 52, normalized size = 0.87 \[ -\frac {b x - {\left (b^{2} x^{2} - a^{2} + 1\right )} \arctan \left (b x + a\right ) - a \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(x*arctan(b*x+a),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.04, size = 66, normalized size = 1.10 \[ \frac {x^{2} \arctan \left (b x +a \right )}{2}-\frac {\arctan \left (b x +a \right ) a^{2}}{2 b^{2}}-\frac {x}{2 b}-\frac {a}{2 b^{2}}+\frac {a \ln \left (1+\left (b x +a \right )^{2}\right )}{2 b^{2}}+\frac {\arctan \left (b x +a \right )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctan(b*x+a),x)

[Out]

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

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maxima [A]  time = 0.41, size = 68, normalized size = 1.13 \[ \frac {1}{2} \, x^{2} \arctan \left (b x + a\right ) - \frac {1}{2} \, b {\left (\frac {x}{b^{2}} + \frac {{\left (a^{2} - 1\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{3}} - \frac {a \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.97, size = 61, normalized size = 1.02 \[ \frac {x^2\,\mathrm {atan}\left (a+b\,x\right )}{2}+\frac {\frac {\mathrm {atan}\left (a+b\,x\right )}{2}-\frac {b\,x}{2}-\frac {a^2\,\mathrm {atan}\left (a+b\,x\right )}{2}+\frac {a\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2}}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*atan(a + b*x),x)

[Out]

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

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sympy [A]  time = 0.64, size = 78, normalized size = 1.30 \[ \begin {cases} - \frac {a^{2} \operatorname {atan}{\left (a + b x \right )}}{2 b^{2}} + \frac {a \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{2}} + \frac {x^{2} \operatorname {atan}{\left (a + b x \right )}}{2} - \frac {x}{2 b} + \frac {\operatorname {atan}{\left (a + b x \right )}}{2 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \operatorname {atan}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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