Integrand size = 26, antiderivative size = 161 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 x^6 \left (a+b x^3\right )}-\frac {a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{x^3 \left (a+b x^3\right )}+\frac {b^3 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \] Output:
-1/9*a^3*((b*x^3+a)^2)^(1/2)/x^9/(b*x^3+a)-1/2*a^2*b*((b*x^3+a)^2)^(1/2)/x ^6/(b*x^3+a)-a*b^2*((b*x^3+a)^2)^(1/2)/x^3/(b*x^3+a)+b^3*((b*x^3+a)^2)^(1/ 2)*ln(x)/(b*x^3+a)
Time = 0.60 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.65 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=\frac {2 a^3 \sqrt {a^2}+9 \left (a^2\right )^{3/2} b x^3+18 a \sqrt {a^2} b^2 x^6-2 a^3 \sqrt {\left (a+b x^3\right )^2}-7 a^2 b x^3 \sqrt {\left (a+b x^3\right )^2}-11 a b^2 x^6 \sqrt {\left (a+b x^3\right )^2}-12 a b^3 x^9 \text {arctanh}\left (\frac {b x^3}{\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}}\right )-12 \sqrt {a^2} b^3 x^9 \log \left (x^3\right )+6 \sqrt {a^2} b^3 x^9 \log \left (a \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )\right )+6 \sqrt {a^2} b^3 x^9 \log \left (a \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )\right )}{36 a x^9} \] Input:
Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^10,x]
Output:
(2*a^3*Sqrt[a^2] + 9*(a^2)^(3/2)*b*x^3 + 18*a*Sqrt[a^2]*b^2*x^6 - 2*a^3*Sq rt[(a + b*x^3)^2] - 7*a^2*b*x^3*Sqrt[(a + b*x^3)^2] - 11*a*b^2*x^6*Sqrt[(a + b*x^3)^2] - 12*a*b^3*x^9*ArcTanh[(b*x^3)/(Sqrt[a^2] - Sqrt[(a + b*x^3)^ 2])] - 12*Sqrt[a^2]*b^3*x^9*Log[x^3] + 6*Sqrt[a^2]*b^3*x^9*Log[a*(Sqrt[a^2 ] - b*x^3 - Sqrt[(a + b*x^3)^2])] + 6*Sqrt[a^2]*b^3*x^9*Log[a*(Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2])])/(36*a*x^9)
Time = 0.23 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.46, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^3 \left (b x^3+a\right )^3}{x^{10}}dx}{b^3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^3}{x^{10}}dx}{a+b x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^3}{x^{12}}dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {a^3}{x^{12}}+\frac {3 b a^2}{x^9}+\frac {3 b^2 a}{x^6}+\frac {b^3}{x^3}\right )dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {a^3}{3 x^9}-\frac {3 a^2 b}{2 x^6}-\frac {3 a b^2}{x^3}+b^3 \log \left (x^3\right )\right )}{3 \left (a+b x^3\right )}\) |
Input:
Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^10,x]
Output:
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-1/3*a^3/x^9 - (3*a^2*b)/(2*x^6) - (3*a* b^2)/x^3 + b^3*Log[x^3]))/(3*(a + b*x^3))
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 = 0.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.32
method | result | size |
pseudoelliptic | \(\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (6 \ln \left (b \,x^{3}\right ) b^{3} x^{9}-18 a \,x^{6} b^{2}-9 a^{2} x^{3} b -2 a^{3}\right )}{18 x^{9}}\) | \(52\) |
default | \(\frac {{\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}} \left (18 b^{3} \ln \left (x \right ) x^{9}-18 a \,x^{6} b^{2}-9 a^{2} x^{3} b -2 a^{3}\right )}{18 \left (b \,x^{3}+a \right )^{3} x^{9}}\) | \(60\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-a \,x^{6} b^{2}-\frac {1}{2} a^{2} x^{3} b -\frac {1}{9} a^{3}\right )}{\left (b \,x^{3}+a \right ) x^{9}}+\frac {b^{3} \sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (x \right )}{b \,x^{3}+a}\) | \(76\) |
Input:
int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x,method=_RETURNVERBOSE)
Output:
1/18*csgn(b*x^3+a)*(6*ln(b*x^3)*b^3*x^9-18*a*x^6*b^2-9*a^2*x^3*b-2*a^3)/x^ 9
Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=\frac {18 \, b^{3} x^{9} \log \left (x\right ) - 18 \, a b^{2} x^{6} - 9 \, a^{2} b x^{3} - 2 \, a^{3}}{18 \, x^{9}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x, algorithm="fricas")
Output:
1/18*(18*b^3*x^9*log(x) - 18*a*b^2*x^6 - 9*a^2*b*x^3 - 2*a^3)/x^9
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x^{10}}\, dx \] Input:
integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x**10,x)
Output:
Integral(((a + b*x**3)**2)**(3/2)/x**10, x)
Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (113) = 226\).
Time = 0.04 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=\frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{4} x^{3}}{6 \, a^{2}} + \frac {1}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} b^{3} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {1}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{3}}{2 \, a} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{3}}{18 \, a^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{2}}{6 \, a^{2} x^{3}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b}{18 \, a^{3} x^{6}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}}{9 \, a^{2} x^{9}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x, algorithm="maxima")
Output:
1/6*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^4*x^3/a^2 + 1/3*(-1)^(2*b^2*x^3 + 2* a*b)*b^3*log(2*b^2*x^3 + 2*a*b) - 1/3*(-1)^(2*a*b*x^3 + 2*a^2)*b^3*log(2*a *b*x/abs(x) + 2*a^2/(x^2*abs(x))) + 1/2*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^ 3/a - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^3/a^3 - 1/6*(b^2*x^6 + 2*a* b*x^3 + a^2)^(3/2)*b^2/(a^2*x^3) + 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)* b/(a^3*x^6) - 1/9*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)/(a^2*x^9)
Time = 0.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.53 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=b^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {11 \, b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 18 \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 9 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 2 \, a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{18 \, x^{9}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x, algorithm="giac")
Output:
b^3*log(abs(x))*sgn(b*x^3 + a) - 1/18*(11*b^3*x^9*sgn(b*x^3 + a) + 18*a*b^ 2*x^6*sgn(b*x^3 + a) + 9*a^2*b*x^3*sgn(b*x^3 + a) + 2*a^3*sgn(b*x^3 + a))/ x^9
Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}}{x^{10}} \,d x \] Input:
int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^10,x)
Output:
int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^10, x)
Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{10}} \, dx=\frac {18 \,\mathrm {log}\left (x \right ) b^{3} x^{9}-2 a^{3}-9 a^{2} b \,x^{3}-18 a \,b^{2} x^{6}}{18 x^{9}} \] Input:
int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^10,x)
Output:
(18*log(x)*b**3*x**9 - 2*a**3 - 9*a**2*b*x**3 - 18*a*b**2*x**6)/(18*x**9)