Integrand size = 14, antiderivative size = 52 \[ \int (a+b x)^2 \cot ^{-1}(a+b x) \, dx=\frac {(a+b x)^2}{6 b}+\frac {(a+b x)^3 \cot ^{-1}(a+b x)}{3 b}-\frac {\log \left (1+(a+b x)^2\right )}{6 b} \] Output:
1/6*(b*x+a)^2/b+1/3*(b*x+a)^3*arccot(b*x+a)/b-1/6*ln(1+(b*x+a)^2)/b
Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int (a+b x)^2 \cot ^{-1}(a+b x) \, dx=\frac {(a+b x)^2+2 (a+b x)^3 \cot ^{-1}(a+b x)-\log \left (1+(a+b x)^2\right )}{6 b} \] Input:
Integrate[(a + b*x)^2*ArcCot[a + b*x],x]
Output:
((a + b*x)^2 + 2*(a + b*x)^3*ArcCot[a + b*x] - Log[1 + (a + b*x)^2])/(6*b)
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5567, 5362, 243, 49, 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.
\(\displaystyle \int (a+b x)^2 \cot ^{-1}(a+b x) \, dx\) |
\(\Big \downarrow \) 5567 |
\(\displaystyle \frac {\int (a+b x)^2 \cot ^{-1}(a+b x)d(a+b x)}{b}\) |
\(\Big \downarrow \) 5362 |
\(\displaystyle \frac {\frac {1}{3} \int \frac {(a+b x)^3}{(a+b x)^2+1}d(a+b x)+\frac {1}{3} (a+b x)^3 \cot ^{-1}(a+b x)}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {1}{6} \int \frac {(a+b x)^2}{(a+b x)^2+1}d(a+b x)^2+\frac {1}{3} (a+b x)^3 \cot ^{-1}(a+b x)}{b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\frac {1}{6} \int \left (1+\frac {1}{-a-b x-1}\right )d(a+b x)^2+\frac {1}{3} (a+b x)^3 \cot ^{-1}(a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{6} \left ((a+b x)^2-\log (a+b x+1)\right )+\frac {1}{3} (a+b x)^3 \cot ^{-1}(a+b x)}{b}\) |
Input:
Int[(a + b*x)^2*ArcCot[a + b*x],x]
Output:
(((a + b*x)^3*ArcCot[a + b*x])/3 + ((a + b*x)^2 - Log[1 + a + b*x])/6)/b
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcCot[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCot[c*x^n])^p/(m + 1)), x] + Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcCot[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] & & IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[(f*(x/d))^m*(a + b*ArcCot[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0] && IGtQ[p, 0]
Time = 0.37 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(\frac {\frac {\operatorname {arccot}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}+\frac {\left (b x +a \right )^{2}}{6}-\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{6}}{b}\) | \(42\) |
default | \(\frac {\frac {\operatorname {arccot}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}+\frac {\left (b x +a \right )^{2}}{6}-\frac {\ln \left (1+\left (b x +a \right )^{2}\right )}{6}}{b}\) | \(42\) |
parts | \(\frac {\operatorname {arccot}\left (b x +a \right ) b^{2} x^{3}}{3}+\operatorname {arccot}\left (b x +a \right ) b a \,x^{2}+\operatorname {arccot}\left (b x +a \right ) a^{2} x +\frac {\operatorname {arccot}\left (b x +a \right ) a^{3}}{3 b}+\frac {x^{2} b}{6}+\frac {a x}{3}-\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{6 b}\) | \(86\) |
parallelrisch | \(-\frac {-2 b^{4} \operatorname {arccot}\left (b x +a \right ) x^{3}-6 a \,b^{3} x^{2} \operatorname {arccot}\left (b x +a \right )-6 x \,\operatorname {arccot}\left (b x +a \right ) a^{2} b^{2}-x^{2} b^{3}-2 \,\operatorname {arccot}\left (b x +a \right ) a^{3} b -2 x a \,b^{2}+5 a^{2} b +\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) b +b}{6 b^{2}}\) | \(105\) |
risch | \(\frac {i \left (b x +a \right )^{3} \ln \left (1+i \left (b x +a \right )\right )}{6 b}-\frac {i b^{2} x^{3} \ln \left (1-i \left (b x +a \right )\right )}{6}+\frac {\pi \,b^{2} x^{3}}{6}-\frac {i b a \,x^{2} \ln \left (1-i \left (b x +a \right )\right )}{2}+\frac {b \pi a \,x^{2}}{2}-\frac {i a^{2} x \ln \left (1-i \left (b x +a \right )\right )}{2}+\frac {x \,a^{2} \pi }{2}-\frac {i a^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{12 b}-\frac {\arctan \left (b x +a \right ) a^{3}}{6 b}+\frac {x^{2} b}{6}+\frac {a x}{3}-\frac {\ln \left (-b^{2} x^{2}-2 a b x -a^{2}-1\right )}{6 b}\) | \(184\) |
Input:
int((b*x+a)^2*arccot(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/b*(1/3*arccot(b*x+a)*(b*x+a)^3+1/6*(b*x+a)^2-1/6*ln(1+(b*x+a)^2))
Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.56 \[ \int (a+b x)^2 \cot ^{-1}(a+b x) \, dx=\frac {b^{2} x^{2} - 2 \, a^{3} \arctan \left (b x + a\right ) + 2 \, a b x + 2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x\right )} \operatorname {arccot}\left (b x + a\right ) - \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{6 \, b} \] Input:
integrate((b*x+a)^2*arccot(b*x+a),x, algorithm="fricas")
Output:
1/6*(b^2*x^2 - 2*a^3*arctan(b*x + a) + 2*a*b*x + 2*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x)*arccot(b*x + a) - log(b^2*x^2 + 2*a*b*x + a^2 + 1))/b
Result contains complex when optimal does not.
Time = 5.45 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.90 \[ \int (a+b x)^2 \cot ^{-1}(a+b x) \, dx=\begin {cases} \frac {a^{3} \operatorname {acot}{\left (a + b x \right )}}{3 b} + a^{2} x \operatorname {acot}{\left (a + b x \right )} + a b x^{2} \operatorname {acot}{\left (a + b x \right )} + \frac {a x}{3} + \frac {b^{2} x^{3} \operatorname {acot}{\left (a + b x \right )}}{3} + \frac {b x^{2}}{6} - \frac {\log {\left (\frac {a}{b} + x - \frac {i}{b} \right )}}{3 b} - \frac {i \operatorname {acot}{\left (a + b x \right )}}{3 b} & \text {for}\: b \neq 0 \\a^{2} x \operatorname {acot}{\left (a \right )} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**2*acot(b*x+a),x)
Output:
Piecewise((a**3*acot(a + b*x)/(3*b) + a**2*x*acot(a + b*x) + a*b*x**2*acot (a + b*x) + a*x/3 + b**2*x**3*acot(a + b*x)/3 + b*x**2/6 - log(a/b + x - I /b)/(3*b) - I*acot(a + b*x)/(3*b), Ne(b, 0)), (a**2*x*acot(a), True))
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (46) = 92\).
Time = 0.14 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.79 \[ \int (a+b x)^2 \cot ^{-1}(a+b x) \, dx=-\frac {1}{6} \, {\left (\frac {2 \, a^{3} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{2}} - \frac {b x^{2} + 2 \, a x}{b} + \frac {\log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{2}}\right )} b + \frac {1}{3} \, {\left (b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x\right )} \operatorname {arccot}\left (b x + a\right ) \] Input:
integrate((b*x+a)^2*arccot(b*x+a),x, algorithm="maxima")
Output:
-1/6*(2*a^3*arctan((b^2*x + a*b)/b)/b^2 - (b*x^2 + 2*a*x)/b + log(b^2*x^2 + 2*a*b*x + a^2 + 1)/b^2)*b + 1/3*(b^2*x^3 + 3*a*b*x^2 + 3*a^2*x)*arccot(b *x + a)
Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (46) = 92\).
Time = 0.23 (sec) , antiderivative size = 203, normalized size of antiderivative = 3.90 \[ \int (a+b x)^2 \cot ^{-1}(a+b x) \, dx=-\frac {\arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{6} - 3 \, \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} - \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{5} - 4 \, \log \left (\frac {16 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} + 3 \, \arctan \left (\frac {1}{b x + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{2} - 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3} - \arctan \left (\frac {1}{b x + a}\right ) - \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )}{24 \, b \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x + a}\right )\right )^{3}} \] Input:
integrate((b*x+a)^2*arccot(b*x+a),x, algorithm="giac")
Output:
-1/24*(arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))^6 - 3*arctan(1/(b* x + a))*tan(1/2*arctan(1/(b*x + a)))^4 - tan(1/2*arctan(1/(b*x + a)))^5 - 4*log(16*tan(1/2*arctan(1/(b*x + a)))^2/(tan(1/2*arctan(1/(b*x + a)))^4 + 2*tan(1/2*arctan(1/(b*x + a)))^2 + 1))*tan(1/2*arctan(1/(b*x + a)))^3 + 3* arctan(1/(b*x + a))*tan(1/2*arctan(1/(b*x + a)))^2 - 2*tan(1/2*arctan(1/(b *x + a)))^3 - arctan(1/(b*x + a)) - tan(1/2*arctan(1/(b*x + a))))/(b*tan(1 /2*arctan(1/(b*x + a)))^3)
Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.63 \[ \int (a+b x)^2 \cot ^{-1}(a+b x) \, dx=\frac {a\,x}{3}-\frac {\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{6\,b}+\frac {b\,x^2}{6}-\frac {a^3\,\mathrm {atan}\left (a+b\,x\right )}{3\,b}+\frac {b^2\,x^3\,\mathrm {acot}\left (a+b\,x\right )}{3}+a^2\,x\,\mathrm {acot}\left (a+b\,x\right )+a\,b\,x^2\,\mathrm {acot}\left (a+b\,x\right ) \] Input:
int(acot(a + b*x)*(a + b*x)^2,x)
Output:
(a*x)/3 - log(a^2 + b^2*x^2 + 2*a*b*x + 1)/(6*b) + (b*x^2)/6 - (a^3*atan(a + b*x))/(3*b) + (b^2*x^3*acot(a + b*x))/3 + a^2*x*acot(a + b*x) + a*b*x^2 *acot(a + b*x)
Time = 0.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.75 \[ \int (a+b x)^2 \cot ^{-1}(a+b x) \, dx=\frac {2 \mathit {acot} \left (b x +a \right ) a^{3}+6 \mathit {acot} \left (b x +a \right ) a^{2} b x +6 \mathit {acot} \left (b x +a \right ) a \,b^{2} x^{2}+2 \mathit {acot} \left (b x +a \right ) b^{3} x^{3}-\mathrm {log}\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )+2 a b x +b^{2} x^{2}}{6 b} \] Input:
int((b*x+a)^2*acot(b*x+a),x)
Output:
(2*acot(a + b*x)*a**3 + 6*acot(a + b*x)*a**2*b*x + 6*acot(a + b*x)*a*b**2* x**2 + 2*acot(a + b*x)*b**3*x**3 - log(a**2 + 2*a*b*x + b**2*x**2 + 1) + 2 *a*b*x + b**2*x**2)/(6*b)