Integrand size = 12, antiderivative size = 42 \[ \int x^3 \cot ^{-1}\left (a+b x^4\right ) \, dx=\frac {\left (a+b x^4\right ) \cot ^{-1}\left (a+b x^4\right )}{4 b}+\frac {\log \left (1+\left (a+b x^4\right )^2\right )}{8 b} \] Output:
1/4*(b*x^4+a)*arccot(b*x^4+a)/b+1/8*ln(1+(b*x^4+a)^2)/b
Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int x^3 \cot ^{-1}\left (a+b x^4\right ) \, dx=\frac {2 \left (a+b x^4\right ) \cot ^{-1}\left (a+b x^4\right )+\log \left (1+\left (a+b x^4\right )^2\right )}{8 b} \] Input:
Integrate[x^3*ArcCot[a + b*x^4],x]
Output:
(2*(a + b*x^4)*ArcCot[a + b*x^4] + Log[1 + (a + b*x^4)^2])/(8*b)
Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {7266, 5563, 5346, 240}
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.
\(\displaystyle \int x^3 \cot ^{-1}\left (a+b x^4\right ) \, dx\) |
\(\Big \downarrow \) 7266 |
\(\displaystyle \frac {1}{4} \int \cot ^{-1}\left (b x^4+a\right )dx^4\) |
\(\Big \downarrow \) 5563 |
\(\displaystyle \frac {\int \cot ^{-1}\left (b x^4+a\right )d\left (b x^4+a\right )}{4 b}\) |
\(\Big \downarrow \) 5346 |
\(\displaystyle \frac {\int \frac {b x^4+a}{x^8+1}d\left (b x^4+a\right )+\left (a+b x^4\right ) \cot ^{-1}\left (a+b x^4\right )}{4 b}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {\left (a+b x^4\right ) \cot ^{-1}\left (a+b x^4\right )+\frac {1}{2} \log \left (x^8+1\right )}{4 b}\) |
Input:
Int[x^3*ArcCot[a + b*x^4],x]
Output:
((a + b*x^4)*ArcCot[a + b*x^4] + Log[1 + x^8]/2)/(4*b)
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((a_.) + ArcCot[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCot[c*x^n])^p, x] + Simp[b*c*n*p Int[x^n*((a + b*ArcCot[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0] && (EqQ[n, 1] || EqQ[p, 1])
Int[((a_.) + ArcCot[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[1/d Subst[Int[(a + b*ArcCot[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Time = 0.56 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {\operatorname {arccot}\left (b \,x^{4}+a \right ) \left (b \,x^{4}+a \right )+\frac {\ln \left (1+\left (b \,x^{4}+a \right )^{2}\right )}{2}}{4 b}\) | \(37\) |
default | \(\frac {\operatorname {arccot}\left (b \,x^{4}+a \right ) \left (b \,x^{4}+a \right )+\frac {\ln \left (1+\left (b \,x^{4}+a \right )^{2}\right )}{2}}{4 b}\) | \(37\) |
parallelrisch | \(\frac {2 \,\operatorname {arccot}\left (b \,x^{4}+a \right ) x^{4} b^{2}+2 a \,\operatorname {arccot}\left (b \,x^{4}+a \right ) b +\ln \left (b^{2} x^{8}+2 a b \,x^{4}+a^{2}+1\right ) b}{8 b^{2}}\) | \(57\) |
parts | \(\frac {x^{4} \operatorname {arccot}\left (b \,x^{4}+a \right )}{4}+b \left (\frac {\ln \left (b^{2} x^{8}+2 a b \,x^{4}+a^{2}+1\right )}{8 b^{2}}-\frac {a \arctan \left (\frac {2 b^{2} x^{4}+2 a b}{2 b}\right )}{4 b^{2}}\right )\) | \(68\) |
risch | \(\frac {i x^{4} \ln \left (1+i \left (b \,x^{4}+a \right )\right )}{8}-\frac {i x^{4} \ln \left (1-i \left (b \,x^{4}+a \right )\right )}{8}+\frac {\pi \,x^{4}}{8}-\frac {a \arctan \left (\frac {b \,x^{4}}{a^{2}+1}+\frac {a^{2} b \,x^{4}}{a^{2}+1}+\frac {a^{3}}{a^{2}+1}+\frac {a}{a^{2}+1}\right )}{4 b}+\frac {a \arctan \left (a \right )}{4 b}+\frac {\ln \left (a^{2} b^{2} x^{8}+b^{2} x^{8}+2 a^{3} b \,x^{4}+2 a b \,x^{4}+a^{4}+2 a^{2}+1\right )}{8 b}\) | \(158\) |
Input:
int(x^3*arccot(b*x^4+a),x,method=_RETURNVERBOSE)
Output:
1/4/b*(arccot(b*x^4+a)*(b*x^4+a)+1/2*ln(1+(b*x^4+a)^2))
Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.21 \[ \int x^3 \cot ^{-1}\left (a+b x^4\right ) \, dx=\frac {2 \, b x^{4} \operatorname {arccot}\left (b x^{4} + a\right ) - 2 \, a \arctan \left (b x^{4} + a\right ) + \log \left (b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1\right )}{8 \, b} \] Input:
integrate(x^3*arccot(b*x^4+a),x, algorithm="fricas")
Output:
1/8*(2*b*x^4*arccot(b*x^4 + a) - 2*a*arctan(b*x^4 + a) + log(b^2*x^8 + 2*a *b*x^4 + a^2 + 1))/b
Time = 0.64 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.43 \[ \int x^3 \cot ^{-1}\left (a+b x^4\right ) \, dx=\begin {cases} \frac {a \operatorname {acot}{\left (a + b x^{4} \right )}}{4 b} + \frac {x^{4} \operatorname {acot}{\left (a + b x^{4} \right )}}{4} + \frac {\log {\left (a^{2} + 2 a b x^{4} + b^{2} x^{8} + 1 \right )}}{8 b} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {acot}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*acot(b*x**4+a),x)
Output:
Piecewise((a*acot(a + b*x**4)/(4*b) + x**4*acot(a + b*x**4)/4 + log(a**2 + 2*a*b*x**4 + b**2*x**8 + 1)/(8*b), Ne(b, 0)), (x**4*acot(a)/4, True))
Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.83 \[ \int x^3 \cot ^{-1}\left (a+b x^4\right ) \, dx=\frac {2 \, {\left (b x^{4} + a\right )} \operatorname {arccot}\left (b x^{4} + a\right ) + \log \left ({\left (b x^{4} + a\right )}^{2} + 1\right )}{8 \, b} \] Input:
integrate(x^3*arccot(b*x^4+a),x, algorithm="maxima")
Output:
1/8*(2*(b*x^4 + a)*arccot(b*x^4 + a) + log((b*x^4 + a)^2 + 1))/b
Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (38) = 76\).
Time = 0.17 (sec) , antiderivative size = 127, normalized size of antiderivative = 3.02 \[ \int x^3 \cot ^{-1}\left (a+b x^4\right ) \, dx=-\frac {\arctan \left (\frac {1}{b x^{4} + a}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right )^{2} + \log \left (\frac {16 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right )^{2}}{\tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right )^{4} + 2 \, \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right )^{2} + 1}\right ) \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right ) - \arctan \left (\frac {1}{b x^{4} + a}\right )}{8 \, b \tan \left (\frac {1}{2} \, \arctan \left (\frac {1}{b x^{4} + a}\right )\right )} \] Input:
integrate(x^3*arccot(b*x^4+a),x, algorithm="giac")
Output:
-1/8*(arctan(1/(b*x^4 + a))*tan(1/2*arctan(1/(b*x^4 + a)))^2 + log(16*tan( 1/2*arctan(1/(b*x^4 + a)))^2/(tan(1/2*arctan(1/(b*x^4 + a)))^4 + 2*tan(1/2 *arctan(1/(b*x^4 + a)))^2 + 1))*tan(1/2*arctan(1/(b*x^4 + a))) - arctan(1/ (b*x^4 + a)))/(b*tan(1/2*arctan(1/(b*x^4 + a))))
Time = 0.99 (sec) , antiderivative size = 230, normalized size of antiderivative = 5.48 \[ \int x^3 \cot ^{-1}\left (a+b x^4\right ) \, dx=\frac {\ln \left (a^2+2\,a\,b\,x^4+b^2\,x^8+1\right )}{8\,b}+\frac {x^4\,\mathrm {acot}\left (b\,x^4+a\right )}{4}-\frac {a\,\mathrm {atan}\left (\frac {a}{a^6+3\,a^4+3\,a^2+1}+\frac {3\,a^3}{a^6+3\,a^4+3\,a^2+1}+\frac {3\,a^5}{a^6+3\,a^4+3\,a^2+1}+\frac {a^7}{a^6+3\,a^4+3\,a^2+1}+\frac {b\,x^4}{a^6+3\,a^4+3\,a^2+1}+\frac {3\,a^2\,b\,x^4}{a^6+3\,a^4+3\,a^2+1}+\frac {3\,a^4\,b\,x^4}{a^6+3\,a^4+3\,a^2+1}+\frac {a^6\,b\,x^4}{a^6+3\,a^4+3\,a^2+1}\right )}{4\,b} \] Input:
int(x^3*acot(a + b*x^4),x)
Output:
log(a^2 + b^2*x^8 + 2*a*b*x^4 + 1)/(8*b) + (x^4*acot(a + b*x^4))/4 - (a*at an(a/(3*a^2 + 3*a^4 + a^6 + 1) + (3*a^3)/(3*a^2 + 3*a^4 + a^6 + 1) + (3*a^ 5)/(3*a^2 + 3*a^4 + a^6 + 1) + a^7/(3*a^2 + 3*a^4 + a^6 + 1) + (b*x^4)/(3* a^2 + 3*a^4 + a^6 + 1) + (3*a^2*b*x^4)/(3*a^2 + 3*a^4 + a^6 + 1) + (3*a^4* b*x^4)/(3*a^2 + 3*a^4 + a^6 + 1) + (a^6*b*x^4)/(3*a^2 + 3*a^4 + a^6 + 1))) /(4*b)
Time = 0.24 (sec) , antiderivative size = 215, normalized size of antiderivative = 5.12 \[ \int x^3 \cot ^{-1}\left (a+b x^4\right ) \, dx=\frac {2 \mathit {acot} \left (b \,x^{4}+a \right ) a +2 \mathit {acot} \left (b \,x^{4}+a \right ) b \,x^{4}+\mathrm {log}\left (\left (a^{2}+1\right )^{\frac {1}{4}}-b^{\frac {1}{4}} \sqrt {2 \left (a^{2}+1\right )^{\frac {1}{4}}-\sqrt {\sqrt {a^{2}+1}-a}\, \sqrt {2}}\, x +\sqrt {b}\, x^{2}\right )+\mathrm {log}\left (\left (a^{2}+1\right )^{\frac {1}{4}}-b^{\frac {1}{4}} \sqrt {2 \left (a^{2}+1\right )^{\frac {1}{4}}+\sqrt {\sqrt {a^{2}+1}-a}\, \sqrt {2}}\, x +\sqrt {b}\, x^{2}\right )+\mathrm {log}\left (\left (a^{2}+1\right )^{\frac {1}{4}}+b^{\frac {1}{4}} \sqrt {2 \left (a^{2}+1\right )^{\frac {1}{4}}-\sqrt {\sqrt {a^{2}+1}-a}\, \sqrt {2}}\, x +\sqrt {b}\, x^{2}\right )+\mathrm {log}\left (\left (a^{2}+1\right )^{\frac {1}{4}}+b^{\frac {1}{4}} \sqrt {2 \left (a^{2}+1\right )^{\frac {1}{4}}+\sqrt {\sqrt {a^{2}+1}-a}\, \sqrt {2}}\, x +\sqrt {b}\, x^{2}\right )}{8 b} \] Input:
int(x^3*acot(b*x^4+a),x)
Output:
(2*acot(a + b*x**4)*a + 2*acot(a + b*x**4)*b*x**4 + log((a**2 + 1)**(1/4) - b**(1/4)*sqrt(2*(a**2 + 1)**(1/4) - sqrt(sqrt(a**2 + 1) - a)*sqrt(2))*x + sqrt(b)*x**2) + log((a**2 + 1)**(1/4) - b**(1/4)*sqrt(2*(a**2 + 1)**(1/4 ) + sqrt(sqrt(a**2 + 1) - a)*sqrt(2))*x + sqrt(b)*x**2) + log((a**2 + 1)** (1/4) + b**(1/4)*sqrt(2*(a**2 + 1)**(1/4) - sqrt(sqrt(a**2 + 1) - a)*sqrt( 2))*x + sqrt(b)*x**2) + log((a**2 + 1)**(1/4) + b**(1/4)*sqrt(2*(a**2 + 1) **(1/4) + sqrt(sqrt(a**2 + 1) - a)*sqrt(2))*x + sqrt(b)*x**2))/(8*b)