Integrand size = 26, antiderivative size = 277 \[ \int \frac {x^6}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {a x \left (a+b x^3\right )}{b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {x^4 \left (a+b x^3\right )}{4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a^{4/3} \left (a+b x^3\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{7/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^{4/3} \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{7/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a^{4/3} \left (a+b x^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{7/3} \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:
-a*x*(b*x^3+a)/b^2/((b*x^3+a)^2)^(1/2)+1/4*x^4*(b*x^3+a)/b/((b*x^3+a)^2)^( 1/2)-1/3*a^(4/3)*(b*x^3+a)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3 ))*3^(1/2)/b^(7/3)/((b*x^3+a)^2)^(1/2)+1/3*a^(4/3)*(b*x^3+a)*ln(a^(1/3)+b^ (1/3)*x)/b^(7/3)/((b*x^3+a)^2)^(1/2)-1/6*a^(4/3)*(b*x^3+a)*ln(a^(2/3)-a^(1 /3)*b^(1/3)*x+b^(2/3)*x^2)/b^(7/3)/((b*x^3+a)^2)^(1/2)
Time = 1.08 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.51 \[ \int \frac {x^6}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\left (a+b x^3\right ) \left (-12 a \sqrt [3]{b} x+3 b^{4/3} x^4-4 \sqrt {3} a^{4/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+4 a^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-2 a^{4/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )\right )}{12 b^{7/3} \sqrt {\left (a+b x^3\right )^2}} \] Input:
Integrate[x^6/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]
Output:
((a + b*x^3)*(-12*a*b^(1/3)*x + 3*b^(4/3)*x^4 - 4*Sqrt[3]*a^(4/3)*ArcTan[( 1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] + 4*a^(4/3)*Log[a^(1/3) + b^(1/3)*x] - 2*a^(4/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]))/(12*b^(7/3)*Sq rt[(a + b*x^3)^2])
Time = 0.32 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.58, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1384, 27, 831, 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.
\(\displaystyle \int \frac {x^6}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b \left (a+b x^3\right ) \int \frac {x^6}{b \left (b x^3+a\right )}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {x^6}{b x^3+a}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 831 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \left (\frac {x^3}{b}+\frac {a^2}{b^2 \left (b x^3+a\right )}-\frac {a}{b^2}\right )dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^3\right ) \left (-\frac {a^{4/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{7/3}}-\frac {a^{4/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{7/3}}+\frac {a^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{7/3}}-\frac {a x}{b^2}+\frac {x^4}{4 b}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
Input:
Int[x^6/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]
Output:
((a + b*x^3)*(-((a*x)/b^2) + x^4/(4*b) - (a^(4/3)*ArcTan[(a^(1/3) - 2*b^(1 /3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(7/3)) + (a^(4/3)*Log[a^(1/3) + b^(1 /3)*x])/(3*b^(7/3)) - (a^(4/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x ^2])/(6*b^(7/3))))/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[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x ^m, a + b*x^n, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && Gt Q[m, 2*n - 1]
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 = 3.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.31
method | result | size |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (\frac {1}{4} b \,x^{4}-x a \right )}{\left (b \,x^{3}+a \right ) b^{2}}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{3 \left (b \,x^{3}+a \right ) b^{3}}\) | \(86\) |
default | \(\frac {\left (b \,x^{3}+a \right ) \left (3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2} x^{4}-12 \left (\frac {a}{b}\right )^{\frac {2}{3}} a b x -4 \arctan \left (\frac {\sqrt {3}\, \left (-2 x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) \sqrt {3}\, a^{2}+4 \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right ) a^{2}-2 \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right ) a^{2}\right )}{12 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}}\) | \(133\) |
Input:
int(x^6/((b*x^3+a)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
((b*x^3+a)^2)^(1/2)/(b*x^3+a)*(1/4*b*x^4-x*a)/b^2+1/3*((b*x^3+a)^2)^(1/2)/ (b*x^3+a)/b^3*a^2*sum(1/_R^2*ln(x-_R),_R=RootOf(_Z^3*b+a))
Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.39 \[ \int \frac {x^6}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {3 \, b x^{4} + 4 \, \sqrt {3} a \left (\frac {a}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (\frac {a}{b}\right )^{\frac {2}{3}} - \sqrt {3} a}{3 \, a}\right ) - 2 \, a \left (\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right ) + 4 \, a \left (\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right ) - 12 \, a x}{12 \, b^{2}} \] Input:
integrate(x^6/((b*x^3+a)^2)^(1/2),x, algorithm="fricas")
Output:
1/12*(3*b*x^4 + 4*sqrt(3)*a*(a/b)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*(a/b)^(2 /3) - sqrt(3)*a)/a) - 2*a*(a/b)^(1/3)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3 )) + 4*a*(a/b)^(1/3)*log(x + (a/b)^(1/3)) - 12*a*x)/b^2
\[ \int \frac {x^6}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\int \frac {x^{6}}{\sqrt {\left (a + b x^{3}\right )^{2}}}\, dx \] Input:
integrate(x**6/((b*x**3+a)**2)**(1/2),x)
Output:
Integral(x**6/sqrt((a + b*x**3)**2), x)
Time = 0.13 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.44 \[ \int \frac {x^6}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {a^{2} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {a^{2} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {b x^{4} - 4 \, a x}{4 \, b^{2}} \] Input:
integrate(x^6/((b*x^3+a)^2)^(1/2),x, algorithm="maxima")
Output:
1/3*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^3*( a/b)^(2/3)) - 1/6*a^2*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^3*(a/b)^(2 /3)) + 1/3*a^2*log(x + (a/b)^(1/3))/(b^3*(a/b)^(2/3)) + 1/4*(b*x^4 - 4*a*x )/b^2
Time = 0.12 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.61 \[ \int \frac {x^6}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {a \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right )}{3 \, b^{2}} + \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} a \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right ) \mathrm {sgn}\left (b x^{3} + a\right )}{3 \, b^{3}} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} a \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right ) \mathrm {sgn}\left (b x^{3} + a\right )}{6 \, b^{3}} + \frac {b^{3} x^{4} \mathrm {sgn}\left (b x^{3} + a\right ) - 4 \, a b^{2} x \mathrm {sgn}\left (b x^{3} + a\right )}{4 \, b^{4}} \] Input:
integrate(x^6/((b*x^3+a)^2)^(1/2),x, algorithm="giac")
Output:
-1/3*a*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))*sgn(b*x^3 + a)/b^2 + 1/3*sq rt(3)*(-a*b^2)^(1/3)*a*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3 ))*sgn(b*x^3 + a)/b^3 + 1/6*(-a*b^2)^(1/3)*a*log(x^2 + x*(-a/b)^(1/3) + (- a/b)^(2/3))*sgn(b*x^3 + a)/b^3 + 1/4*(b^3*x^4*sgn(b*x^3 + a) - 4*a*b^2*x*s gn(b*x^3 + a))/b^4
Timed out. \[ \int \frac {x^6}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\int \frac {x^6}{\sqrt {{\left (b\,x^3+a\right )}^2}} \,d x \] Input:
int(x^6/((a + b*x^3)^2)^(1/2),x)
Output:
int(x^6/((a + b*x^3)^2)^(1/2), x)
Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.32 \[ \int \frac {x^6}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {-4 a^{\frac {4}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right )-2 a^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right )+4 a^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right )-12 b^{\frac {1}{3}} a x +3 b^{\frac {4}{3}} x^{4}}{12 b^{\frac {7}{3}}} \] Input:
int(x^6/((b*x^3+a)^2)^(1/2),x)
Output:
( - 4*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))* a - 2*a**(1/3)*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a + 4*a **(1/3)*log(a**(1/3) + b**(1/3)*x)*a - 12*b**(1/3)*a*x + 3*b**(1/3)*b*x**4 )/(12*b**(1/3)*b**2)