Integrand size = 26, antiderivative size = 162 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^7} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 x^6 \left (a+b x^3\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^3 \left (a+b x^3\right )}+\frac {b^3 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {3 a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \]
-1/6*a^3*((b*x^3+a)^2)^(1/2)/x^6/(b*x^3+a)-a^2*b*((b*x^3+a)^2)^(1/2)/x^3/( b*x^3+a)+1/3*b^3*x^3*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+3*a*b^2*ln(x)*((b*x^3+a )^2)^(1/2)/(b*x^3+a)
Leaf count is larger than twice the leaf count of optimal. \(612\) vs. \(2(162)=324\).
Time = 0.82 (sec) , antiderivative size = 612, normalized size of antiderivative = 3.78 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^7} \, dx=\frac {4 a^4 \sqrt {a^2}+28 a^3 \sqrt {a^2} b x^3+35 \left (a^2\right )^{3/2} b^2 x^6+3 a \sqrt {a^2} b^3 x^9-8 \sqrt {a^2} b^4 x^{12}-4 a^4 \sqrt {\left (a+b x^3\right )^2}-24 a^3 b x^3 \sqrt {\left (a+b x^3\right )^2}-11 a^2 b^2 x^6 \sqrt {\left (a+b x^3\right )^2}+8 a b^3 x^9 \sqrt {\left (a+b x^3\right )^2}-24 a b^2 x^6 \left (a^2+a b x^3-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right ) \text {arctanh}\left (\frac {b x^3}{\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}}\right )-24 b^2 x^6 \left (\left (a^2\right )^{3/2}+a \sqrt {a^2} b x^3-a^2 \sqrt {\left (a+b x^3\right )^2}\right ) \log \left (x^3\right )+12 \left (a^2\right )^{3/2} b^2 x^6 \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+12 a \sqrt {a^2} b^3 x^9 \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-12 a^2 b^2 x^6 \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+12 \left (a^2\right )^{3/2} b^2 x^6 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+12 a \sqrt {a^2} b^3 x^9 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-12 a^2 b^2 x^6 \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )}{24 x^6 \left (a^2+a b x^3-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )} \]
(4*a^4*Sqrt[a^2] + 28*a^3*Sqrt[a^2]*b*x^3 + 35*(a^2)^(3/2)*b^2*x^6 + 3*a*S qrt[a^2]*b^3*x^9 - 8*Sqrt[a^2]*b^4*x^12 - 4*a^4*Sqrt[(a + b*x^3)^2] - 24*a ^3*b*x^3*Sqrt[(a + b*x^3)^2] - 11*a^2*b^2*x^6*Sqrt[(a + b*x^3)^2] + 8*a*b^ 3*x^9*Sqrt[(a + b*x^3)^2] - 24*a*b^2*x^6*(a^2 + a*b*x^3 - Sqrt[a^2]*Sqrt[( a + b*x^3)^2])*ArcTanh[(b*x^3)/(Sqrt[a^2] - Sqrt[(a + b*x^3)^2])] - 24*b^2 *x^6*((a^2)^(3/2) + a*Sqrt[a^2]*b*x^3 - a^2*Sqrt[(a + b*x^3)^2])*Log[x^3] + 12*(a^2)^(3/2)*b^2*x^6*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2]] + 12 *a*Sqrt[a^2]*b^3*x^9*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2]] - 12*a^2 *b^2*x^6*Sqrt[(a + b*x^3)^2]*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2]] + 12*(a^2)^(3/2)*b^2*x^6*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2]] + 12 *a*Sqrt[a^2]*b^3*x^9*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2]] - 12*a^2 *b^2*x^6*Sqrt[(a + b*x^3)^2]*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2]]) /(24*x^6*(a^2 + a*b*x^3 - Sqrt[a^2]*Sqrt[(a + b*x^3)^2]))
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.44, 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, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^7} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^3 \left (b x^3+a\right )^3}{x^7}dx}{b^3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^3}{x^7}dx}{a+b x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^3}{x^9}dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {a^3}{x^9}+\frac {3 b a^2}{x^6}+\frac {3 b^2 a}{x^3}+b^3\right )dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {a^3}{2 x^6}-\frac {3 a^2 b}{x^3}+3 a b^2 \log \left (x^3\right )+b^3 x^3\right )}{3 \left (a+b x^3\right )}\) |
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-1/2*a^3/x^6 - (3*a^2*b)/x^3 + b^3*x^3 + 3*a*b^2*Log[x^3]))/(3*(a + b*x^3))
3.1.39.3.1 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 [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_)^(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 = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.36
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (-2 b^{3} x^{9}-6 \ln \left (b \,x^{3}\right ) a \,b^{2} x^{6}-2 b^{2} x^{6} a +6 a^{2} b \,x^{3}+a^{3}\right )}{6 x^{6}}\) | \(59\) |
default | \(\frac {{\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}} \left (2 b^{3} x^{9}+18 b^{2} a \ln \left (x \right ) x^{6}-6 a^{2} b \,x^{3}-a^{3}\right )}{6 \left (b \,x^{3}+a \right )^{3} x^{6}}\) | \(60\) |
risch | \(\frac {b^{3} x^{3} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{3 b \,x^{3}+3 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-a^{2} b \,x^{3}-\frac {1}{6} a^{3}\right )}{\left (b \,x^{3}+a \right ) x^{6}}+\frac {3 a \,b^{2} \ln \left (x \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}\) | \(97\) |
Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^7} \, dx=\frac {2 \, b^{3} x^{9} + 18 \, a b^{2} x^{6} \log \left (x\right ) - 6 \, a^{2} b x^{3} - a^{3}}{6 \, x^{6}} \]
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^7} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x^{7}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^7} \, dx=\frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{3} x^{3}}{2 \, a} + \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} a b^{2} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} a b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{2} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{2}}{6 \, a^{2}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b}{6 \, a x^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}}{6 \, a^{2} x^{6}} \]
1/2*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^3*x^3/a + (-1)^(2*b^2*x^3 + 2*a*b)*a *b^2*log(2*b^2*x^3 + 2*a*b) - (-1)^(2*a*b*x^3 + 2*a^2)*a*b^2*log(2*a*b*x/a bs(x) + 2*a^2/(x^2*abs(x))) + 3/2*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^2 + 1/ 6*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^2/a^2 - 1/6*(b^2*x^6 + 2*a*b*x^3 + a ^2)^(3/2)*b/(a*x^3) - 1/6*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)/(a^2*x^6)
Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.53 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^7} \, dx=\frac {1}{3} \, b^{3} x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 3 \, a b^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {9 \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 6 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{6 \, x^{6}} \]
1/3*b^3*x^3*sgn(b*x^3 + a) + 3*a*b^2*log(abs(x))*sgn(b*x^3 + a) - 1/6*(9*a *b^2*x^6*sgn(b*x^3 + a) + 6*a^2*b*x^3*sgn(b*x^3 + a) + a^3*sgn(b*x^3 + a)) /x^6
Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^7} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}}{x^7} \,d x \]