Integrand size = 26, antiderivative size = 188 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {2 b}{3 a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b}{6 a^2 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a+b x^3}{3 a^3 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {3 b \left (a+b x^3\right ) \log (x)}{a^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {b \left (a+b x^3\right ) \log \left (a+b x^3\right )}{a^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \]
-2/3*b/a^3/((b*x^3+a)^2)^(1/2)-1/6*b/a^2/(b*x^3+a)/((b*x^3+a)^2)^(1/2)+1/3 *(-b*x^3-a)/a^3/x^3/((b*x^3+a)^2)^(1/2)-3*b*(b*x^3+a)*ln(x)/a^4/((b*x^3+a) ^2)^(1/2)+b*(b*x^3+a)*ln(b*x^3+a)/a^4/((b*x^3+a)^2)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(901\) vs. \(2(188)=376\).
Time = 1.33 (sec) , antiderivative size = 901, normalized size of antiderivative = 4.79 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {-2 a^6-5 a^5 b x^3+2 a^4 b^2 x^6+4 a^3 b^3 x^9-a b^5 x^{15}+2 a^4 \sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}+3 a^3 \sqrt {a^2} b x^3 \sqrt {\left (a+b x^3\right )^2}-5 \left (a^2\right )^{3/2} b^2 x^6 \sqrt {\left (a+b x^3\right )^2}+a \sqrt {a^2} b^3 x^9 \sqrt {\left (a+b x^3\right )^2}-\sqrt {a^2} b^4 x^{12} \sqrt {\left (a+b x^3\right )^2}+6 b x^3 \left (\left (a^2\right )^{3/2} b^2 x^6+a^4 \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )+a^3 b x^3 \left (2 \sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )\right ) \text {arctanh}\left (\frac {b x^3}{\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}}\right )-6 b x^3 \left (a^5+2 a^4 b x^3-\left (a^2\right )^{3/2} b x^3 \sqrt {\left (a+b x^3\right )^2}+a^3 \left (b^2 x^6-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )\right ) \log \left (x^3\right )+3 a^5 b x^3 \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+6 a^4 b^2 x^6 \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+3 a^3 b^3 x^9 \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-3 a^3 \sqrt {a^2} b x^3 \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-3 \left (a^2\right )^{3/2} b^2 x^6 \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+3 a^5 b x^3 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+6 a^4 b^2 x^6 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+3 a^3 b^3 x^9 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-3 a^3 \sqrt {a^2} b x^3 \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-3 \left (a^2\right )^{3/2} b^2 x^6 \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )}{3 a^4 \sqrt {a^2} x^3 \left (a^2+a b x^3-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )^2} \]
-1/3*(-2*a^6 - 5*a^5*b*x^3 + 2*a^4*b^2*x^6 + 4*a^3*b^3*x^9 - a*b^5*x^15 + 2*a^4*Sqrt[a^2]*Sqrt[(a + b*x^3)^2] + 3*a^3*Sqrt[a^2]*b*x^3*Sqrt[(a + b*x^ 3)^2] - 5*(a^2)^(3/2)*b^2*x^6*Sqrt[(a + b*x^3)^2] + a*Sqrt[a^2]*b^3*x^9*Sq rt[(a + b*x^3)^2] - Sqrt[a^2]*b^4*x^12*Sqrt[(a + b*x^3)^2] + 6*b*x^3*((a^2 )^(3/2)*b^2*x^6 + a^4*(Sqrt[a^2] - Sqrt[(a + b*x^3)^2]) + a^3*b*x^3*(2*Sqr t[a^2] - Sqrt[(a + b*x^3)^2]))*ArcTanh[(b*x^3)/(Sqrt[a^2] - Sqrt[(a + b*x^ 3)^2])] - 6*b*x^3*(a^5 + 2*a^4*b*x^3 - (a^2)^(3/2)*b*x^3*Sqrt[(a + b*x^3)^ 2] + a^3*(b^2*x^6 - Sqrt[a^2]*Sqrt[(a + b*x^3)^2]))*Log[x^3] + 3*a^5*b*x^3 *Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2]] + 6*a^4*b^2*x^6*Log[Sqrt[a^2 ] - b*x^3 - Sqrt[(a + b*x^3)^2]] + 3*a^3*b^3*x^9*Log[Sqrt[a^2] - b*x^3 - S qrt[(a + b*x^3)^2]] - 3*a^3*Sqrt[a^2]*b*x^3*Sqrt[(a + b*x^3)^2]*Log[Sqrt[a ^2] - b*x^3 - Sqrt[(a + b*x^3)^2]] - 3*(a^2)^(3/2)*b^2*x^6*Sqrt[(a + b*x^3 )^2]*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2]] + 3*a^5*b*x^3*Log[Sqrt[a ^2] + b*x^3 - Sqrt[(a + b*x^3)^2]] + 6*a^4*b^2*x^6*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2]] + 3*a^3*b^3*x^9*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b* x^3)^2]] - 3*a^3*Sqrt[a^2]*b*x^3*Sqrt[(a + b*x^3)^2]*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2]] - 3*(a^2)^(3/2)*b^2*x^6*Sqrt[(a + b*x^3)^2]*Log[Sq rt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2]])/(a^4*Sqrt[a^2]*x^3*(a^2 + a*b*x^3 - Sqrt[a^2]*Sqrt[(a + b*x^3)^2])^2)
Time = 0.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.52, 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.
\(\displaystyle \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^3\right ) \int \frac {1}{b^3 x^4 \left (b x^3+a\right )^3}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x^4 \left (b x^3+a\right )^3}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x^6 \left (b x^3+a\right )^3}dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \left (\frac {3 b^2}{a^4 \left (b x^3+a\right )}+\frac {2 b^2}{a^3 \left (b x^3+a\right )^2}+\frac {b^2}{a^2 \left (b x^3+a\right )^3}-\frac {3 b}{a^4 x^3}+\frac {1}{a^3 x^6}\right )dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^3\right ) \left (-\frac {3 b \log \left (x^3\right )}{a^4}+\frac {3 b \log \left (a+b x^3\right )}{a^4}-\frac {2 b}{a^3 \left (a+b x^3\right )}-\frac {1}{a^3 x^3}-\frac {b}{2 a^2 \left (a+b x^3\right )^2}\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
((a + b*x^3)*(-(1/(a^3*x^3)) - b/(2*a^2*(a + b*x^3)^2) - (2*b)/(a^3*(a + b *x^3)) - (3*b*Log[x^3])/a^4 + (3*b*Log[a + b*x^3])/a^4))/(3*Sqrt[a^2 + 2*a *b*x^3 + b^2*x^6])
3.2.6.3.1 Defintions of rubi rules used
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 = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.49
method | result | size |
pseudoelliptic | \(-\frac {\left (-3 b \,x^{3} \left (b \,x^{3}+a \right )^{2} \ln \left (b \,x^{3}+a \right )+3 b \,x^{3} \left (b \,x^{3}+a \right )^{2} \ln \left (b \,x^{3}\right )+a \left (3 b^{2} x^{6}+\frac {9}{2} a b \,x^{3}+a^{2}\right )\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{3 \left (b \,x^{3}+a \right )^{2} a^{4} x^{3}}\) | \(92\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {b^{2} x^{6}}{a^{3}}-\frac {3 b \,x^{3}}{2 a^{2}}-\frac {1}{3 a}\right )}{\left (b \,x^{3}+a \right )^{3} x^{3}}-\frac {3 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a^{4}}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \ln \left (-b \,x^{3}-a \right )}{\left (b \,x^{3}+a \right ) a^{4}}\) | \(116\) |
default | \(-\frac {\left (18 b^{3} \ln \left (x \right ) x^{9}-6 \ln \left (b \,x^{3}+a \right ) b^{3} x^{9}+36 b^{2} a \ln \left (x \right ) x^{6}-12 \ln \left (b \,x^{3}+a \right ) a \,b^{2} x^{6}+6 b^{2} x^{6} a +18 a^{2} b \ln \left (x \right ) x^{3}-6 \ln \left (b \,x^{3}+a \right ) a^{2} b \,x^{3}+9 a^{2} b \,x^{3}+2 a^{3}\right ) \left (b \,x^{3}+a \right )}{6 x^{3} a^{4} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(133\) |
-1/3*(-3*b*x^3*(b*x^3+a)^2*ln(b*x^3+a)+3*b*x^3*(b*x^3+a)^2*ln(b*x^3)+a*(3* b^2*x^6+9/2*a*b*x^3+a^2))*csgn(b*x^3+a)/(b*x^3+a)^2/a^4/x^3
Time = 0.25 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {6 \, a b^{2} x^{6} + 9 \, a^{2} b x^{3} + 2 \, a^{3} - 6 \, {\left (b^{3} x^{9} + 2 \, a b^{2} x^{6} + a^{2} b x^{3}\right )} \log \left (b x^{3} + a\right ) + 18 \, {\left (b^{3} x^{9} + 2 \, a b^{2} x^{6} + a^{2} b x^{3}\right )} \log \left (x\right )}{6 \, {\left (a^{4} b^{2} x^{9} + 2 \, a^{5} b x^{6} + a^{6} x^{3}\right )}} \]
-1/6*(6*a*b^2*x^6 + 9*a^2*b*x^3 + 2*a^3 - 6*(b^3*x^9 + 2*a*b^2*x^6 + a^2*b *x^3)*log(b*x^3 + a) + 18*(b^3*x^9 + 2*a*b^2*x^6 + a^2*b*x^3)*log(x))/(a^4 *b^2*x^9 + 2*a^5*b*x^6 + a^6*x^3)
\[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\int \frac {1}{x^{4} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {\left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right )}{a^{4}} - \frac {b}{\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{3}} - \frac {1}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} a^{2} b} - \frac {1}{3 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{2} x^{3}} \]
(-1)^(2*a*b*x^3 + 2*a^2)*b*log(2*a*b*x/abs(x) + 2*a^2/(x^2*abs(x)))/a^4 - b/(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^3) - 1/6/((x^3 + a/b)^2*a^2*b) - 1/3/ (sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^2*x^3)
Time = 0.30 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {b \log \left ({\left | b x^{3} + a \right |}\right )}{a^{4} \mathrm {sgn}\left (b x^{3} + a\right )} - \frac {3 \, b \log \left ({\left | x \right |}\right )}{a^{4} \mathrm {sgn}\left (b x^{3} + a\right )} - \frac {9 \, b^{3} x^{6} + 22 \, a b^{2} x^{3} + 14 \, a^{2} b}{6 \, {\left (b x^{3} + a\right )}^{2} a^{4} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {3 \, b x^{3} - a}{3 \, a^{4} x^{3} \mathrm {sgn}\left (b x^{3} + a\right )} \]
b*log(abs(b*x^3 + a))/(a^4*sgn(b*x^3 + a)) - 3*b*log(abs(x))/(a^4*sgn(b*x^ 3 + a)) - 1/6*(9*b^3*x^6 + 22*a*b^2*x^3 + 14*a^2*b)/((b*x^3 + a)^2*a^4*sgn (b*x^3 + a)) + 1/3*(3*b*x^3 - a)/(a^4*x^3*sgn(b*x^3 + a))
Timed out. \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\int \frac {1}{x^4\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}} \,d x \]