Integrand size = 26, antiderivative size = 147 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {1}{3 a^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{6 a \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {\left (a+b x^3\right ) \log (x)}{a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:
1/3/a^2/((b*x^3+a)^2)^(1/2)+1/6/a/(b*x^3+a)/((b*x^3+a)^2)^(1/2)+(b*x^3+a)* ln(x)/a^3/((b*x^3+a)^2)^(1/2)-1/3*(b*x^3+a)*ln(b*x^3+a)/a^3/((b*x^3+a)^2)^ (1/2)
Time = 1.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {a \left (3 a+2 b x^3\right )+6 \left (a+b x^3\right )^2 \log (x)-2 \left (a+b x^3\right )^2 \log \left (a+b x^3\right )}{6 a^3 \left (a+b x^3\right ) \sqrt {\left (a+b x^3\right )^2}} \] Input:
Integrate[1/(x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)),x]
Output:
(a*(3*a + 2*b*x^3) + 6*(a + b*x^3)^2*Log[x] - 2*(a + b*x^3)^2*Log[a + b*x^ 3])/(6*a^3*(a + b*x^3)*Sqrt[(a + b*x^3)^2])
Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.57, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 27, 798, 54, 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.
\(\displaystyle \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^3\right ) \int \frac {1}{b^3 x \left (b x^3+a\right )^3}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x \left (b x^3+a\right )^3}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x^3 \left (b x^3+a\right )^3}dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \left (-\frac {b}{a^3 \left (b x^3+a\right )}-\frac {b}{a^2 \left (b x^3+a\right )^2}-\frac {b}{a \left (b x^3+a\right )^3}+\frac {1}{a^3 x^3}\right )dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^3\right ) \left (-\frac {\log \left (a+b x^3\right )}{a^3}+\frac {\log \left (x^3\right )}{a^3}+\frac {1}{a^2 \left (a+b x^3\right )}+\frac {1}{2 a \left (a+b x^3\right )^2}\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
Input:
Int[1/(x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)),x]
Output:
((a + b*x^3)*(1/(2*a*(a + b*x^3)^2) + 1/(a^2*(a + b*x^3)) + Log[x^3]/a^3 - Log[a + b*x^3]/a^3))/(3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])
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_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
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 = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.48
method | result | size |
pseudoelliptic | \(\frac {\left (-\ln \left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{2}+\ln \left (b \,x^{3}\right ) \left (b \,x^{3}+a \right )^{2}+a \,x^{3} b +\frac {3 a^{2}}{2}\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{3 \left (b \,x^{3}+a \right )^{2} a^{3}}\) | \(70\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (\frac {b \,x^{3}}{3 a^{2}}+\frac {1}{2 a}\right )}{\left (b \,x^{3}+a \right )^{3}}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a^{3}}-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (b \,x^{3}+a \right )}{3 \left (b \,x^{3}+a \right ) a^{3}}\) | \(97\) |
default | \(-\frac {\left (2 b^{2} \ln \left (b \,x^{3}+a \right ) x^{6}-6 b^{2} \ln \left (x \right ) x^{6}+4 \ln \left (b \,x^{3}+a \right ) a b \,x^{3}-12 \ln \left (x \right ) a b \,x^{3}-2 a \,x^{3} b +2 a^{2} \ln \left (b \,x^{3}+a \right )-6 \ln \left (x \right ) a^{2}-3 a^{2}\right ) \left (b \,x^{3}+a \right )}{6 a^{3} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(107\) |
Input:
int(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/3*(-ln(b*x^3+a)*(b*x^3+a)^2+ln(b*x^3)*(b*x^3+a)^2+a*x^3*b+3/2*a^2)*csgn( b*x^3+a)/(b*x^3+a)^2/a^3
Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {2 \, a b x^{3} + 3 \, a^{2} - 2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )} \log \left (b x^{3} + a\right ) + 6 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )} \log \left (x\right )}{6 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{3} + a^{5}\right )}} \] Input:
integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")
Output:
1/6*(2*a*b*x^3 + 3*a^2 - 2*(b^2*x^6 + 2*a*b*x^3 + a^2)*log(b*x^3 + a) + 6* (b^2*x^6 + 2*a*b*x^3 + a^2)*log(x))/(a^3*b^2*x^6 + 2*a^4*b*x^3 + a^5)
\[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\int \frac {1}{x \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/x/(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)
Output:
Integral(1/(x*((a + b*x**3)**2)**(3/2)), x)
Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right )}{3 \, a^{3}} + \frac {1}{3 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{2}} + \frac {1}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} a b^{2}} \] Input:
integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")
Output:
-1/3*(-1)^(2*a*b*x^3 + 2*a^2)*log(2*a*b*x/abs(x) + 2*a^2/(x^2*abs(x)))/a^3 + 1/3/(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^2) + 1/6/((x^3 + a/b)^2*a*b^2)
Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {\log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{3} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {\log \left ({\left | x \right |}\right )}{a^{3} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {3 \, b^{2} x^{6} + 8 \, a b x^{3} + 6 \, a^{2}}{6 \, {\left (b x^{3} + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x^{3} + a\right )} \] Input:
integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")
Output:
-1/3*log(abs(b*x^3 + a))/(a^3*sgn(b*x^3 + a)) + log(abs(x))/(a^3*sgn(b*x^3 + a)) + 1/6*(3*b^2*x^6 + 8*a*b*x^3 + 6*a^2)/((b*x^3 + a)^2*a^3*sgn(b*x^3 + a))
Timed out. \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\int \frac {1}{x\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}} \,d x \] Input:
int(1/(x*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)),x)
Output:
int(1/(x*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)), x)
Time = 0.20 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.35 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {-2 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2}-4 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a b \,x^{3}-2 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b^{2} x^{6}-2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2}-4 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a b \,x^{3}-2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b^{2} x^{6}+6 \,\mathrm {log}\left (x \right ) a^{2}+12 \,\mathrm {log}\left (x \right ) a b \,x^{3}+6 \,\mathrm {log}\left (x \right ) b^{2} x^{6}+2 a^{2}-b^{2} x^{6}}{6 a^{3} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )} \] Input:
int(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x)
Output:
( - 2*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2 - 4*log(a** (2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b*x**3 - 2*log(a**(2/3) - b **(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**2*x**6 - 2*log(a**(1/3) + b**(1/3)* x)*a**2 - 4*log(a**(1/3) + b**(1/3)*x)*a*b*x**3 - 2*log(a**(1/3) + b**(1/3 )*x)*b**2*x**6 + 6*log(x)*a**2 + 12*log(x)*a*b*x**3 + 6*log(x)*b**2*x**6 + 2*a**2 - b**2*x**6)/(6*a**3*(a**2 + 2*a*b*x**3 + b**2*x**6))