Integrand size = 22, antiderivative size = 202 \[ \int \frac {1}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {\left (a+b x^3\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} \sqrt [3]{b} \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {\left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} \sqrt [3]{b} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{2/3} \sqrt [3]{b} \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:
-1/3*(b*x^3+a)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))*3^(1/2)/a ^(2/3)/b^(1/3)/((b*x^3+a)^2)^(1/2)+1/3*(b*x^3+a)*ln(a^(1/3)+b^(1/3)*x)/a^( 2/3)/b^(1/3)/((b*x^3+a)^2)^(1/2)-1/6*(b*x^3+a)*ln(a^(2/3)-a^(1/3)*b^(1/3)* x+b^(2/3)*x^2)/a^(2/3)/b^(1/3)/((b*x^3+a)^2)^(1/2)
Time = 1.03 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {\left (a+b x^3\right ) \left (2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )\right )}{6 a^{2/3} \sqrt [3]{b} \sqrt {\left (a+b x^3\right )^2}} \] Input:
Integrate[1/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]
Output:
-1/6*((a + b*x^3)*(2*Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 2*Log[a^(1/3) + b^(1/3)*x] + Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^ 2]))/(a^(2/3)*b^(1/3)*Sqrt[(a + b*x^3)^2])
Time = 0.32 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.71, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {1384, 750, 16, 27, 1142, 25, 27, 1082, 217, 1103}
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.
| \(\displaystyle \int \frac {1}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx\) |
| \(\Big \downarrow \) 1384 |
| \(\displaystyle \frac {b \left (a+b x^3\right ) \int \frac {1}{b^2 x^3+a b}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 750 |
| \(\displaystyle \frac {b \left (a+b x^3\right ) \left (\frac {\int \frac {\sqrt [3]{b} \left (2 \sqrt [3]{a}-\sqrt [3]{b} x\right )}{b^{4/3} x^2-\sqrt [3]{a} b x+a^{2/3} b^{2/3}}dx}{3 a^{2/3} b^{2/3}}+\frac {\int \frac {1}{b^{2/3} x+\sqrt [3]{a} \sqrt [3]{b}}dx}{3 a^{2/3} b^{2/3}}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {b \left (a+b x^3\right ) \left (\frac {\int \frac {\sqrt [3]{b} \left (2 \sqrt [3]{a}-\sqrt [3]{b} x\right )}{b^{4/3} x^2-\sqrt [3]{a} b x+a^{2/3} b^{2/3}}dx}{3 a^{2/3} b^{2/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{4/3}}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \left (a+b x^3\right ) \left (\frac {\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{b^{4/3} x^2-\sqrt [3]{a} b x+a^{2/3} b^{2/3}}dx}{3 a^{2/3} \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{4/3}}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {b \left (a+b x^3\right ) \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{4/3} x^2-\sqrt [3]{a} b x+a^{2/3} b^{2/3}}dx-\frac {\int -\frac {b \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{4/3} x^2-\sqrt [3]{a} b x+a^{2/3} b^{2/3}}dx}{2 b}}{3 a^{2/3} \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{4/3}}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b \left (a+b x^3\right ) \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{4/3} x^2-\sqrt [3]{a} b x+a^{2/3} b^{2/3}}dx+\frac {\int \frac {b \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{4/3} x^2-\sqrt [3]{a} b x+a^{2/3} b^{2/3}}dx}{2 b}}{3 a^{2/3} \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{4/3}}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \left (a+b x^3\right ) \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{4/3} x^2-\sqrt [3]{a} b x+a^{2/3} b^{2/3}}dx+\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{4/3} x^2-\sqrt [3]{a} b x+a^{2/3} b^{2/3}}dx}{3 a^{2/3} \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{4/3}}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {b \left (a+b x^3\right ) \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{4/3} x^2-\sqrt [3]{a} b x+a^{2/3} b^{2/3}}dx+\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{b}}{3 a^{2/3} \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{4/3}}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {b \left (a+b x^3\right ) \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{4/3} x^2-\sqrt [3]{a} b x+a^{2/3} b^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b}}{3 a^{2/3} \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{4/3}}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {b \left (a+b x^3\right ) \left (\frac {-\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 b}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b}}{3 a^{2/3} \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{4/3}}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
Input:
Int[1/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]
Output:
(b*(a + b*x^3)*(Log[a^(1/3) + b^(1/3)*x]/(3*a^(2/3)*b^(4/3)) + (-((Sqrt[3] *ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b) - Log[a^(2/3) - a^(1/3)*b ^(1/3)*x + b^(2/3)*x^2]/(2*b))/(3*a^(2/3)*b^(1/3))))/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2) Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.23
| method | result | size |
| risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{3 \left (b \,x^{3}+a \right ) b}\) | \(47\) |
| default | \(\frac {\left (b \,x^{3}+a \right ) \left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-2 x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )+2 \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )-\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )\right )}{6 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\) | \(97\) |
Input:
int(1/((b*x^3+a)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/3*((b*x^3+a)^2)^(1/2)/(b*x^3+a)/b*sum(1/_R^2*ln(x-_R),_R=RootOf(_Z^3*b+a ))
Time = 0.09 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.48 \[ \int \frac {1}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\left [\frac {3 \, \sqrt {\frac {1}{3}} a b \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x^{3} - 3 \, \left (a^{2} b\right )^{\frac {1}{3}} a x - a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{2} + \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{b x^{3} + a}\right ) - \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) + 2 \, \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{6 \, a^{2} b}, \frac {6 \, \sqrt {\frac {1}{3}} a b \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) - \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) + 2 \, \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{6 \, a^{2} b}\right ] \] Input:
integrate(1/((b*x^3+a)^2)^(1/2),x, algorithm="fricas")
Output:
[1/6*(3*sqrt(1/3)*a*b*sqrt(-(a^2*b)^(1/3)/b)*log((2*a*b*x^3 - 3*(a^2*b)^(1 /3)*a*x - a^2 + 3*sqrt(1/3)*(2*a*b*x^2 + (a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a )*sqrt(-(a^2*b)^(1/3)/b))/(b*x^3 + a)) - (a^2*b)^(2/3)*log(a*b*x^2 - (a^2* b)^(2/3)*x + (a^2*b)^(1/3)*a) + 2*(a^2*b)^(2/3)*log(a*b*x + (a^2*b)^(2/3)) )/(a^2*b), 1/6*(6*sqrt(1/3)*a*b*sqrt((a^2*b)^(1/3)/b)*arctan(sqrt(1/3)*(2* (a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*sqrt((a^2*b)^(1/3)/b)/a^2) - (a^2*b)^(2 /3)*log(a*b*x^2 - (a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) + 2*(a^2*b)^(2/3)*log (a*b*x + (a^2*b)^(2/3)))/(a^2*b)]
\[ \int \frac {1}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\int \frac {1}{\sqrt {\left (a + b x^{3}\right )^{2}}}\, dx \] Input:
integrate(1/((b*x**3+a)**2)**(1/2),x)
Output:
Integral(1/sqrt((a + b*x**3)**2), x)
Time = 0.12 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.49 \[ \int \frac {1}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {\log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {\log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \] Input:
integrate(1/((b*x^3+a)^2)^(1/2),x, algorithm="maxima")
Output:
1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b*(a/b)^( 2/3)) - 1/6*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b*(a/b)^(2/3)) + 1/3*l og(x + (a/b)^(1/3))/(b*(a/b)^(2/3))
Time = 0.12 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.60 \[ \int \frac {1}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {1}{6} \, {\left (\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a} - \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b}\right )} \mathrm {sgn}\left (b x^{3} + a\right ) \] Input:
integrate(1/((b*x^3+a)^2)^(1/2),x, algorithm="giac")
Output:
-1/6*(2*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a - 2*sqrt(3)*(-a*b^2)^(1/ 3)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a*b) - (-a*b^2)^ (1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a*b))*sgn(b*x^3 + a)
Timed out. \[ \int \frac {1}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\int \frac {1}{\sqrt {{\left (b\,x^3+a\right )}^2}} \,d x \] Input:
int(1/((a + b*x^3)^2)^(1/2),x)
Output:
int(1/((a + b*x^3)^2)^(1/2), x)
Time = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.33 \[ \int \frac {1}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {-2 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right )-\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right )+2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right )}{6 a^{\frac {2}{3}} b^{\frac {1}{3}}} \] Input:
int(1/((b*x^3+a)^2)^(1/2),x)
Output:
(a**(1/3)*( - 2*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3))) - log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2) + 2*log(a**(1/3) + b**(1/3)*x)))/(6*b**(1/3)*a)