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

Optimal. Leaf size=106 \[ \frac{\left (1-6 a^2\right ) x}{4 b^3}-\frac{a \left (1-a^2\right ) \log \left ((a+b x)^2+1\right )}{2 b^4}-\frac{\left (a^4-6 a^2+1\right ) \tan ^{-1}(a+b x)}{4 b^4}-\frac{(a+b x)^3}{12 b^4}+\frac{a (a+b x)^2}{2 b^4}+\frac{1}{4} x^4 \tan ^{-1}(a+b x) \]

[Out]

((1 - 6*a^2)*x)/(4*b^3) + (a*(a + b*x)^2)/(2*b^4) - (a + b*x)^3/(12*b^4) - ((1 - 6*a^2 + a^4)*ArcTan[a + b*x])
/(4*b^4) + (x^4*ArcTan[a + b*x])/4 - (a*(1 - a^2)*Log[1 + (a + b*x)^2])/(2*b^4)

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Rubi [A]  time = 0.110634, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5047, 4862, 702, 635, 203, 260} \[ \frac{\left (1-6 a^2\right ) x}{4 b^3}-\frac{a \left (1-a^2\right ) \log \left ((a+b x)^2+1\right )}{2 b^4}-\frac{\left (a^4-6 a^2+1\right ) \tan ^{-1}(a+b x)}{4 b^4}-\frac{(a+b x)^3}{12 b^4}+\frac{a (a+b x)^2}{2 b^4}+\frac{1}{4} x^4 \tan ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTan[a + b*x],x]

[Out]

((1 - 6*a^2)*x)/(4*b^3) + (a*(a + b*x)^2)/(2*b^4) - (a + b*x)^3/(12*b^4) - ((1 - 6*a^2 + a^4)*ArcTan[a + b*x])
/(4*b^4) + (x^4*ArcTan[a + b*x])/4 - (a*(1 - a^2)*Log[1 + (a + b*x)^2])/(2*b^4)

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]

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 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 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 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]

Rubi steps

\begin{align*} \int x^3 \tan ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^3 \tan ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{4} x^4 \tan ^{-1}(a+b x)-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^4}{1+x^2} \, dx,x,a+b x\right )\\ &=\frac{1}{4} x^4 \tan ^{-1}(a+b x)-\frac{1}{4} \operatorname{Subst}\left (\int \left (-\frac{1-6 a^2}{b^4}-\frac{4 a x}{b^4}+\frac{x^2}{b^4}+\frac{1-6 a^2+a^4+4 a \left (1-a^2\right ) x}{b^4 \left (1+x^2\right )}\right ) \, dx,x,a+b x\right )\\ &=\frac{\left (1-6 a^2\right ) x}{4 b^3}+\frac{a (a+b x)^2}{2 b^4}-\frac{(a+b x)^3}{12 b^4}+\frac{1}{4} x^4 \tan ^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int \frac{1-6 a^2+a^4+4 a \left (1-a^2\right ) x}{1+x^2} \, dx,x,a+b x\right )}{4 b^4}\\ &=\frac{\left (1-6 a^2\right ) x}{4 b^3}+\frac{a (a+b x)^2}{2 b^4}-\frac{(a+b x)^3}{12 b^4}+\frac{1}{4} x^4 \tan ^{-1}(a+b x)-\frac{\left (a \left (1-a^2\right )\right ) \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,a+b x\right )}{b^4}-\frac{\left (1-6 a^2+a^4\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,a+b x\right )}{4 b^4}\\ &=\frac{\left (1-6 a^2\right ) x}{4 b^3}+\frac{a (a+b x)^2}{2 b^4}-\frac{(a+b x)^3}{12 b^4}-\frac{\left (1-6 a^2+a^4\right ) \tan ^{-1}(a+b x)}{4 b^4}+\frac{1}{4} x^4 \tan ^{-1}(a+b x)-\frac{a \left (1-a^2\right ) \log \left (1+(a+b x)^2\right )}{2 b^4}\\ \end{align*}

Mathematica [C]  time = 0.0699706, size = 95, normalized size = 0.9 \[ \frac{6 \left (1-6 a^2\right ) b x+6 b^4 x^4 \tan ^{-1}(a+b x)-2 (a+b x)^3+12 a (a+b x)^2+3 i (a-i)^4 \log (-a-b x+i)-3 i (a+i)^4 \log (a+b x+i)}{24 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcTan[a + b*x],x]

[Out]

(6*(1 - 6*a^2)*b*x + 12*a*(a + b*x)^2 - 2*(a + b*x)^3 + 6*b^4*x^4*ArcTan[a + b*x] + (3*I)*(-I + a)^4*Log[I - a
- b*x] - (3*I)*(I + a)^4*Log[I + a + b*x])/(24*b^4)

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Maple [A]  time = 0.039, size = 132, normalized size = 1.3 \begin{align*}{\frac{{x}^{4}\arctan \left ( bx+a \right ) }{4}}-{\frac{\arctan \left ( bx+a \right ){a}^{4}}{4\,{b}^{4}}}-{\frac{{x}^{3}}{12\,b}}+{\frac{a{x}^{2}}{4\,{b}^{2}}}-{\frac{3\,{a}^{2}x}{4\,{b}^{3}}}-{\frac{13\,{a}^{3}}{12\,{b}^{4}}}+{\frac{x}{4\,{b}^{3}}}+{\frac{a}{4\,{b}^{4}}}+{\frac{\ln \left ( 1+ \left ( bx+a \right ) ^{2} \right ){a}^{3}}{2\,{b}^{4}}}-{\frac{\ln \left ( 1+ \left ( bx+a \right ) ^{2} \right ) a}{2\,{b}^{4}}}+{\frac{3\,\arctan \left ( bx+a \right ){a}^{2}}{2\,{b}^{4}}}-{\frac{\arctan \left ( bx+a \right ) }{4\,{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arctan(b*x+a)-1/4/b^4*arctan(b*x+a)*a^4-1/12/b*x^3+1/4/b^2*x^2*a-3/4/b^3*x*a^2-13/12/b^4*a^3+1/4/b^3*x
+1/4/b^4*a+1/2/b^4*ln(1+(b*x+a)^2)*a^3-1/2/b^4*ln(1+(b*x+a)^2)*a+3/2/b^4*arctan(b*x+a)*a^2-1/4/b^4*arctan(b*x+
a)

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Maxima [A]  time = 1.50389, size = 140, normalized size = 1.32 \begin{align*} \frac{1}{4} \, x^{4} \arctan \left (b x + a\right ) - \frac{1}{12} \, b{\left (\frac{b^{2} x^{3} - 3 \, a b x^{2} + 3 \,{\left (3 \, a^{2} - 1\right )} x}{b^{4}} + \frac{3 \,{\left (a^{4} - 6 \, a^{2} + 1\right )} \arctan \left (\frac{b^{2} x + a b}{b}\right )}{b^{5}} - \frac{6 \,{\left (a^{3} - a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{5}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arctan(b*x + a) - 1/12*b*((b^2*x^3 - 3*a*b*x^2 + 3*(3*a^2 - 1)*x)/b^4 + 3*(a^4 - 6*a^2 + 1)*arctan((b^
2*x + a*b)/b)/b^5 - 6*(a^3 - a)*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/b^5)

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Fricas [A]  time = 1.68514, size = 203, normalized size = 1.92 \begin{align*} -\frac{b^{3} x^{3} - 3 \, a b^{2} x^{2} + 3 \,{\left (3 \, a^{2} - 1\right )} b x - 3 \,{\left (b^{4} x^{4} - a^{4} + 6 \, a^{2} - 1\right )} \arctan \left (b x + a\right ) - 6 \,{\left (a^{3} - a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{12 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(b^3*x^3 - 3*a*b^2*x^2 + 3*(3*a^2 - 1)*b*x - 3*(b^4*x^4 - a^4 + 6*a^2 - 1)*arctan(b*x + a) - 6*(a^3 - a)
*log(b^2*x^2 + 2*a*b*x + a^2 + 1))/b^4

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Sympy [A]  time = 2.36794, size = 155, normalized size = 1.46 \begin{align*} \begin{cases} - \frac{a^{4} \operatorname{atan}{\left (a + b x \right )}}{4 b^{4}} + \frac{a^{3} \log{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{4}} - \frac{3 a^{2} x}{4 b^{3}} + \frac{3 a^{2} \operatorname{atan}{\left (a + b x \right )}}{2 b^{4}} + \frac{a x^{2}}{4 b^{2}} - \frac{a \log{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{4}} + \frac{x^{4} \operatorname{atan}{\left (a + b x \right )}}{4} - \frac{x^{3}}{12 b} + \frac{x}{4 b^{3}} - \frac{\operatorname{atan}{\left (a + b x \right )}}{4 b^{4}} & \text{for}\: b \neq 0 \\\frac{x^{4} \operatorname{atan}{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**4*atan(a + b*x)/(4*b**4) + a**3*log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(2*b**4) - 3*a**2*x/(4*b**3
) + 3*a**2*atan(a + b*x)/(2*b**4) + a*x**2/(4*b**2) - a*log(a**2 + 2*a*b*x + b**2*x**2 + 1)/(2*b**4) + x**4*at
an(a + b*x)/4 - x**3/(12*b) + x/(4*b**3) - atan(a + b*x)/(4*b**4), Ne(b, 0)), (x**4*atan(a)/4, True))

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Giac [A]  time = 1.10813, size = 139, normalized size = 1.31 \begin{align*} \frac{1}{4} \, x^{4} \arctan \left (b x + a\right ) - \frac{1}{12} \, b{\left (\frac{3 \,{\left (a^{4} - 6 \, a^{2} + 1\right )} \arctan \left (b x + a\right )}{b^{5}} - \frac{6 \,{\left (a^{3} - a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{5}} + \frac{b^{4} x^{3} - 3 \, a b^{3} x^{2} + 9 \, a^{2} b^{2} x - 3 \, b^{2} x}{b^{6}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arctan(b*x + a) - 1/12*b*(3*(a^4 - 6*a^2 + 1)*arctan(b*x + a)/b^5 - 6*(a^3 - a)*log(b^2*x^2 + 2*a*b*x
+ a^2 + 1)/b^5 + (b^4*x^3 - 3*a*b^3*x^2 + 9*a^2*b^2*x - 3*b^2*x)/b^6)