3.1.44 \(\int x^3 \text {ArcTan}(a+b x) \, dx\) [44]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5155, 4972, 716, 649, 209, 266} \begin {gather*} -\frac {a \left (1-a^2\right ) \log \left ((a+b x)^2+1\right )}{2 b^4}+\frac {\left (1-6 a^2\right ) x}{4 b^3}-\frac {\left (a^4-6 a^2+1\right ) \text {ArcTan}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \text {ArcTan}(a+b x)-\frac {(a+b x)^3}{12 b^4}+\frac {a (a+b x)^2}{2 b^4} \end {gather*}

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 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*} \int x^3 \tan ^{-1}(a+b x) \, dx &=\frac {\text {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} \text {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} \text {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 {\text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.05, size = 95, normalized size = 0.90 \begin {gather*} \frac {6 \left (1-6 a^2\right ) b x+12 a (a+b x)^2-2 (a+b x)^3+6 b^4 x^4 \text {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} \end {gather*}

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 = 0.06, size = 157, normalized size = 1.48

method	result	size
derivativedivides	\(\frac {\frac {\arctan \left (b x +a \right ) a^{4}}{4}-\arctan \left (b x +a \right ) a^{3} \left (b x +a \right )+\frac {3 \arctan \left (b x +a \right ) a^{2} \left (b x +a \right )^{2}}{2}-\arctan \left (b x +a \right ) a \left (b x +a \right )^{3}+\frac {\arctan \left (b x +a \right ) \left (b x +a \right )^{4}}{4}-\frac {3 a^{2} \left (b x +a \right )}{2}+\frac {\left (b x +a \right )^{2} a}{2}-\frac {\left (b x +a \right )^{3}}{12}+\frac {b x}{4}+\frac {a}{4}-\frac {\left (-4 a^{3}+4 a \right ) \ln \left (1+\left (b x +a \right )^{2}\right )}{8}-\frac {\left (a^{4}-6 a^{2}+1\right ) \arctan \left (b x +a \right )}{4}}{b^{4}}\)	\(157\)
default	\(\frac {\frac {\arctan \left (b x +a \right ) a^{4}}{4}-\arctan \left (b x +a \right ) a^{3} \left (b x +a \right )+\frac {3 \arctan \left (b x +a \right ) a^{2} \left (b x +a \right )^{2}}{2}-\arctan \left (b x +a \right ) a \left (b x +a \right )^{3}+\frac {\arctan \left (b x +a \right ) \left (b x +a \right )^{4}}{4}-\frac {3 a^{2} \left (b x +a \right )}{2}+\frac {\left (b x +a \right )^{2} a}{2}-\frac {\left (b x +a \right )^{3}}{12}+\frac {b x}{4}+\frac {a}{4}-\frac {\left (-4 a^{3}+4 a \right ) \ln \left (1+\left (b x +a \right )^{2}\right )}{8}-\frac {\left (a^{4}-6 a^{2}+1\right ) \arctan \left (b x +a \right )}{4}}{b^{4}}\)	\(157\)
risch	\(-\frac {i x^{4} \ln \left (1+i \left (b x +a \right )\right )}{8}+\frac {i x^{4} \ln \left (1-i \left (b x +a \right )\right )}{8}-\frac {x^{3}}{12 b}-\frac {a^{4} \arctan \left (b x +a \right )}{4 b^{4}}+\frac {a \,x^{2}}{4 b^{2}}+\frac {a^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{4}}-\frac {3 x \,a^{2}}{4 b^{3}}+\frac {3 a^{2} \arctan \left (b x +a \right )}{2 b^{4}}-\frac {a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{4}}+\frac {x}{4 b^{3}}-\frac {\arctan \left (b x +a \right )}{4 b^{4}}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^4*(1/4*arctan(b*x+a)*a^4-arctan(b*x+a)*a^3*(b*x+a)+3/2*arctan(b*x+a)*a^2*(b*x+a)^2-arctan(b*x+a)*a*(b*x+a)
^3+1/4*arctan(b*x+a)*(b*x+a)^4-3/2*a^2*(b*x+a)+1/2*(b*x+a)^2*a-1/12*(b*x+a)^3+1/4*b*x+1/4*a-1/8*(-4*a^3+4*a)*l
n(1+(b*x+a)^2)-1/4*(a^4-6*a^2+1)*arctan(b*x+a))

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Maxima [A]
time = 0.50, size = 104, normalized size = 0.98 \begin {gather*} \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 {gather*}

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 = 3.19, size = 87, normalized size = 0.82 \begin {gather*} -\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 {gather*}

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 = 0.42, size = 155, normalized size = 1.46 \begin {gather*} \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 {gather*}

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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [B]
time = 0.59, size = 133, normalized size = 1.25 \begin {gather*} \frac {x^4\,\mathrm {atan}\left (a+b\,x\right )}{4}-\frac {\mathrm {atan}\left (a+b\,x\right )}{4\,b^4}+\frac {x}{4\,b^3}-\frac {x^3}{12\,b}+\frac {a^3\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2\,b^4}+\frac {3\,a^2\,\mathrm {atan}\left (a+b\,x\right )}{2\,b^4}-\frac {a^4\,\mathrm {atan}\left (a+b\,x\right )}{4\,b^4}+\frac {a\,x^2}{4\,b^2}-\frac {3\,a^2\,x}{4\,b^3}-\frac {a\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2\,b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*atan(a + b*x),x)

[Out]

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

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