Integrand size = 10, antiderivative size = 106 \[ \int x^3 \arctan (a+b x) \, dx=\frac {\left (1-6 a^2\right ) x}{4 b^3}+\frac {a (a+b x)^2}{2 b^4}-\frac {(a+b x)^3}{12 b^4}-\frac {\left (1-6 a^2+a^4\right ) \arctan (a+b x)}{4 b^4}+\frac {1}{4} x^4 \arctan (a+b x)-\frac {a \left (1-a^2\right ) \log \left (1+(a+b x)^2\right )}{2 b^4} \] Output:
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
Result contains complex when optimal does not.
Time = 0.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90 \[ \int x^3 \arctan (a+b x) \, dx=\frac {6 \left (1-6 a^2\right ) b x+12 a (a+b x)^2-2 (a+b x)^3+6 b^4 x^4 \arctan (a+b x)+3 i (-i+a)^4 \log (i-a-b x)-3 i (i+a)^4 \log (i+a+b x)}{24 b^4} \] Input:
Integrate[x^3*ArcTan[a + b*x],x]
Output:
(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)
Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5570, 25, 27, 5387, 478, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \arctan (a+b x) \, dx\) |
\(\Big \downarrow \) 5570 |
\(\displaystyle \frac {\int x^3 \arctan (a+b x)d(a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int -x^3 \arctan (a+b x)d(a+b x)}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int -b^3 x^3 \arctan (a+b x)d(a+b x)}{b^4}\) |
\(\Big \downarrow \) 5387 |
\(\displaystyle -\frac {\frac {1}{4} \int \frac {b^4 x^4}{(a+b x)^2+1}d(a+b x)-\frac {1}{4} b^4 x^4 \arctan (a+b x)}{b^4}\) |
\(\Big \downarrow \) 478 |
\(\displaystyle -\frac {\frac {1}{4} \int \left (6 a^2-4 (a+b x) a+(a+b x)^2+\frac {a^4-6 a^2+4 \left (1-a^2\right ) (a+b x) a+1}{(a+b x)^2+1}-1\right )d(a+b x)-\frac {1}{4} b^4 x^4 \arctan (a+b x)}{b^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{4} \left (-\left (1-6 a^2\right ) (a+b x)+2 a \left (1-a^2\right ) \log \left ((a+b x)^2+1\right )+\left (a^4-6 a^2+1\right ) \arctan (a+b x)+\frac {1}{3} (a+b x)^3-2 a (a+b x)^2\right )-\frac {1}{4} b^4 x^4 \arctan (a+b x)}{b^4}\) |
Input:
Int[x^3*ArcTan[a + b*x],x]
Output:
-((-1/4*(b^4*x^4*ArcTan[a + b*x]) + (-((1 - 6*a^2)*(a + b*x)) - 2*a*(a + b *x)^2 + (a + b*x)^3/3 + (1 - 6*a^2 + a^4)*ArcTan[a + b*x] + 2*a*(1 - a^2)* Log[1 + (a + b*x)^2])/4)/b^4)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[Expand Integrand[(c + d*x)^n/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ [n, 1]
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] - Simp[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]
Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A rcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && I GtQ[p, 0]
Time = 0.39 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.24
method | result | size |
parallelrisch | \(\frac {3 \arctan \left (b x +a \right ) x^{4} b^{4}-b^{3} x^{3}+3 a \,b^{2} x^{2}-3 \arctan \left (b x +a \right ) a^{4}+6 a^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )-9 a^{2} b x +18 \arctan \left (b x +a \right ) a^{2}+15 a^{3}-6 a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )+3 b x -3 \arctan \left (b x +a \right )-9 a}{12 b^{4}}\) | \(131\) |
parts | \(\frac {x^{4} \arctan \left (b x +a \right )}{4}-\frac {b \left (\frac {\frac {1}{3} b^{2} x^{3}-a b \,x^{2}+3 a^{2} x -x}{b^{4}}+\frac {\frac {\left (-4 a^{3} b +4 a b \right ) \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}+\frac {\left (-3 a^{4}-2 a^{2}+1-\frac {\left (-4 a^{3} b +4 a b \right ) a}{b}\right ) \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}}{b^{4}}\right )}{4}\) | \(135\) |
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 | \(-\frac {i x^{4} \ln \left (1+i \left (b x +a \right )\right )}{8}+\frac {i x^{4} \ln \left (1-i \left (b x +a \right )\right )}{8}-\frac {x^{3}}{12 b}-\frac {a^{4} \arctan \left (b x +a \right )}{4 b^{4}}+\frac {a \,x^{2}}{4 b^{2}}+\frac {a^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{4}}-\frac {3 a^{2} x}{4 b^{3}}+\frac {3 a^{2} \arctan \left (b x +a \right )}{2 b^{4}}-\frac {a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{4}}+\frac {x}{4 b^{3}}-\frac {\arctan \left (b x +a \right )}{4 b^{4}}\) | \(157\) |
Input:
int(x^3*arctan(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/12*(3*arctan(b*x+a)*x^4*b^4-b^3*x^3+3*a*b^2*x^2-3*arctan(b*x+a)*a^4+6*a^ 3*ln(b^2*x^2+2*a*b*x+a^2+1)-9*a^2*b*x+18*arctan(b*x+a)*a^2+15*a^3-6*a*ln(b ^2*x^2+2*a*b*x+a^2+1)+3*b*x-3*arctan(b*x+a)-9*a)/b^4
Time = 0.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.82 \[ \int x^3 \arctan (a+b x) \, dx=-\frac {b^{3} x^{3} - 3 \, a b^{2} x^{2} + 3 \, {\left (3 \, a^{2} - 1\right )} b x - 3 \, {\left (b^{4} x^{4} - a^{4} + 6 \, a^{2} - 1\right )} \arctan \left (b x + a\right ) - 6 \, {\left (a^{3} - a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{12 \, b^{4}} \] Input:
integrate(x^3*arctan(b*x+a),x, algorithm="fricas")
Output:
-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
Time = 0.80 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.46 \[ \int x^3 \arctan (a+b x) \, dx=\begin {cases} - \frac {a^{4} \operatorname {atan}{\left (a + b x \right )}}{4 b^{4}} + \frac {a^{3} \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{4}} - \frac {3 a^{2} x}{4 b^{3}} + \frac {3 a^{2} \operatorname {atan}{\left (a + b x \right )}}{2 b^{4}} + \frac {a x^{2}}{4 b^{2}} - \frac {a \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{4}} + \frac {x^{4} \operatorname {atan}{\left (a + b x \right )}}{4} - \frac {x^{3}}{12 b} + \frac {x}{4 b^{3}} - \frac {\operatorname {atan}{\left (a + b x \right )}}{4 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {atan}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*atan(b*x+a),x)
Output:
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*atan (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))
Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.98 \[ \int x^3 \arctan (a+b x) \, dx=\frac {1}{4} \, x^{4} \arctan \left (b x + a\right ) - \frac {1}{12} \, b {\left (\frac {b^{2} x^{3} - 3 \, a b x^{2} + 3 \, {\left (3 \, a^{2} - 1\right )} x}{b^{4}} + \frac {3 \, {\left (a^{4} - 6 \, a^{2} + 1\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{5}} - \frac {6 \, {\left (a^{3} - a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{5}}\right )} \] Input:
integrate(x^3*arctan(b*x+a),x, algorithm="maxima")
Output:
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)
Time = 0.12 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.97 \[ \int x^3 \arctan (a+b x) \, dx=\frac {1}{4} \, x^{4} \arctan \left (b x + a\right ) - \frac {1}{12} \, b {\left (\frac {3 \, {\left (a^{4} - 6 \, a^{2} + 1\right )} \arctan \left (b x + a\right )}{b^{5}} - \frac {6 \, {\left (a^{3} - a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{5}} + \frac {b^{4} x^{3} - 3 \, a b^{3} x^{2} + 9 \, a^{2} b^{2} x - 3 \, b^{2} x}{b^{6}}\right )} \] Input:
integrate(x^3*arctan(b*x+a),x, algorithm="giac")
Output:
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)
Time = 0.98 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.25 \[ \int x^3 \arctan (a+b x) \, dx=\frac {x^4\,\mathrm {atan}\left (a+b\,x\right )}{4}-\frac {\mathrm {atan}\left (a+b\,x\right )}{4\,b^4}+\frac {x}{4\,b^3}-\frac {x^3}{12\,b}+\frac {a^3\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2\,b^4}+\frac {3\,a^2\,\mathrm {atan}\left (a+b\,x\right )}{2\,b^4}-\frac {a^4\,\mathrm {atan}\left (a+b\,x\right )}{4\,b^4}+\frac {a\,x^2}{4\,b^2}-\frac {3\,a^2\,x}{4\,b^3}-\frac {a\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2\,b^4} \] Input:
int(x^3*atan(a + b*x),x)
Output:
(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)
Time = 0.21 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.15 \[ \int x^3 \arctan (a+b x) \, dx=\frac {-3 \mathit {atan} \left (b x +a \right ) a^{4}+18 \mathit {atan} \left (b x +a \right ) a^{2}+3 \mathit {atan} \left (b x +a \right ) b^{4} x^{4}-3 \mathit {atan} \left (b x +a \right )+6 \,\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}-6 \,\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a -9 a^{2} b x +3 a \,b^{2} x^{2}-b^{3} x^{3}+3 b x}{12 b^{4}} \] Input:
int(x^3*atan(b*x+a),x)
Output:
( - 3*atan(a + b*x)*a**4 + 18*atan(a + b*x)*a**2 + 3*atan(a + b*x)*b**4*x* *4 - 3*atan(a + b*x) + 6*log(a**2 + 2*a*b*x + b**2*x**2 + 1)*a**3 - 6*log( a**2 + 2*a*b*x + b**2*x**2 + 1)*a - 9*a**2*b*x + 3*a*b**2*x**2 - b**3*x**3 + 3*b*x)/(12*b**4)