Integrand size = 15, antiderivative size = 107 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^{2/3}} \, dx=-\frac {\sqrt [3]{a+b x^2}}{2 a x^2}+\frac {b \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^2}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}+\frac {b \log (x)}{3 a^{5/3}}-\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{5/3}} \] Output:
-1/2*(b*x^2+a)^(1/3)/a/x^2+1/3*b*arctan(1/3*(a^(1/3)+2*(b*x^2+a)^(1/3))*3^ (1/2)/a^(1/3))*3^(1/2)/a^(5/3)+1/3*b*ln(x)/a^(5/3)-1/2*b*ln(a^(1/3)-(b*x^2 +a)^(1/3))/a^(5/3)
Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.26 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^{2/3}} \, dx=\frac {-3 a^{2/3} \sqrt [3]{a+b x^2}+2 \sqrt {3} b x^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 b x^2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^2}\right )+b x^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )}{6 a^{5/3} x^2} \] Input:
Integrate[1/(x^3*(a + b*x^2)^(2/3)),x]
Output:
(-3*a^(2/3)*(a + b*x^2)^(1/3) + 2*Sqrt[3]*b*x^2*ArcTan[(1 + (2*(a + b*x^2) ^(1/3))/a^(1/3))/Sqrt[3]] - 2*b*x^2*Log[-a^(1/3) + (a + b*x^2)^(1/3)] + b* x^2*Log[a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3)])/(6*a^(5/ 3)*x^2)
Time = 0.22 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {243, 52, 69, 16, 1082, 217}
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.
\(\displaystyle \int \frac {1}{x^3 \left (a+b x^2\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (b x^2+a\right )^{2/3}}dx^2\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 b \int \frac {1}{x^2 \left (b x^2+a\right )^{2/3}}dx^2}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )\) |
\(\Big \downarrow \) 69 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 b \left (-\frac {3 \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 a^{2/3}}-\frac {3 \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 \sqrt [3]{a}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 b \left (-\frac {3 \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 b \left (\frac {3 \int \frac {1}{-x^4-3}d\left (\frac {2 \sqrt [3]{b x^2+a}}{\sqrt [3]{a}}+1\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 b \left (-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x^2}\right )\) |
Input:
Int[1/(x^3*(a + b*x^2)^(2/3)),x]
Output:
(-((a + b*x^2)^(1/3)/(a*x^2)) - (2*b*(-((Sqrt[3]*ArcTan[(1 + (2*(a + b*x^2 )^(1/3))/a^(1/3))/Sqrt[3]])/a^(2/3)) - Log[x^2]/(2*a^(2/3)) + (3*Log[a^(1/ 3) - (a + b*x^2)^(1/3)])/(2*a^(2/3))))/(3*a))/2
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Simp[3/(2*b*q) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 /3)], x] - Simp[3/(2*b*q^2) Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Time = 0.49 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.04
method | result | size |
pseudoelliptic | \(\frac {2 b \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b \,x^{2}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}\, x^{2}-2 b \ln \left (\left (b \,x^{2}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right ) x^{2}+b \ln \left (a^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{2}+a \right )^{\frac {1}{3}}+\left (b \,x^{2}+a \right )^{\frac {2}{3}}\right ) x^{2}-3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{\frac {2}{3}}}{6 a^{\frac {5}{3}} x^{2}}\) | \(111\) |
Input:
int(1/x^3/(b*x^2+a)^(2/3),x,method=_RETURNVERBOSE)
Output:
1/6*(2*b*arctan(1/3*(a^(1/3)+2*(b*x^2+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)*x ^2-2*b*ln((b*x^2+a)^(1/3)-a^(1/3))*x^2+b*ln(a^(2/3)+a^(1/3)*(b*x^2+a)^(1/3 )+(b*x^2+a)^(2/3))*x^2-3*(b*x^2+a)^(1/3)*a^(2/3))/a^(5/3)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (81) = 162\).
Time = 0.07 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.67 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^{2/3}} \, dx=\frac {6 \, \sqrt {\frac {1}{3}} a b x^{2} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {\sqrt {\frac {1}{3}} {\left (\left (-a^{2}\right )^{\frac {1}{3}} a - 2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}}}{a^{2}}\right ) + \left (-a^{2}\right )^{\frac {2}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} a - \left (-a^{2}\right )^{\frac {1}{3}} a + {\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 2 \, \left (-a^{2}\right )^{\frac {2}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} a - \left (-a^{2}\right )^{\frac {2}{3}}\right ) - 3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{2}}{6 \, a^{3} x^{2}} \] Input:
integrate(1/x^3/(b*x^2+a)^(2/3),x, algorithm="fricas")
Output:
1/6*(6*sqrt(1/3)*a*b*x^2*sqrt(-(-a^2)^(1/3))*arctan(-sqrt(1/3)*((-a^2)^(1/ 3)*a - 2*(b*x^2 + a)^(1/3)*(-a^2)^(2/3))*sqrt(-(-a^2)^(1/3))/a^2) + (-a^2) ^(2/3)*b*x^2*log((b*x^2 + a)^(2/3)*a - (-a^2)^(1/3)*a + (b*x^2 + a)^(1/3)* (-a^2)^(2/3)) - 2*(-a^2)^(2/3)*b*x^2*log((b*x^2 + a)^(1/3)*a - (-a^2)^(2/3 )) - 3*(b*x^2 + a)^(1/3)*a^2)/(a^3*x^2)
Result contains complex when optimal does not.
Time = 0.78 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.38 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^{2/3}} \, dx=- \frac {\Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 b^{\frac {2}{3}} x^{\frac {10}{3}} \Gamma \left (\frac {8}{3}\right )} \] Input:
integrate(1/x**3/(b*x**2+a)**(2/3),x)
Output:
-gamma(5/3)*hyper((2/3, 5/3), (8/3,), a*exp_polar(I*pi)/(b*x**2))/(2*b**(2 /3)*x**(10/3)*gamma(8/3))
Time = 0.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^{2/3}} \, dx=\frac {\sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {5}{3}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {1}{3}} b}{2 \, {\left ({\left (b x^{2} + a\right )} a - a^{2}\right )}} + \frac {b \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{6 \, a^{\frac {5}{3}}} - \frac {b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{3 \, a^{\frac {5}{3}}} \] Input:
integrate(1/x^3/(b*x^2+a)^(2/3),x, algorithm="maxima")
Output:
1/3*sqrt(3)*b*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3))/ a^(5/3) - 1/2*(b*x^2 + a)^(1/3)*b/((b*x^2 + a)*a - a^2) + 1/6*b*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/a^(5/3) - 1/3*b*log((b* x^2 + a)^(1/3) - a^(1/3))/a^(5/3)
Time = 0.40 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^{2/3}} \, dx=\frac {1}{6} \, b {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {5}{3}}} + \frac {\log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {5}{3}}} - \frac {2 \, \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {5}{3}}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{a b x^{2}}\right )} \] Input:
integrate(1/x^3/(b*x^2+a)^(2/3),x, algorithm="giac")
Output:
1/6*b*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3 ))/a^(5/3) + log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3))/ a^(5/3) - 2*log(abs((b*x^2 + a)^(1/3) - a^(1/3)))/a^(5/3) - 3*(b*x^2 + a)^ (1/3)/(a*b*x^2))
Time = 0.54 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^3 \left (a+b x^2\right )^{2/3}} \, dx=\frac {\ln \left (\frac {3\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{2\,a^{2/3}}+\frac {3\,b\,{\left (b\,x^2+a\right )}^{1/3}}{a}\right )\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{6\,a^{5/3}}+\frac {\ln \left (\frac {3\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{2\,a^{2/3}}+\frac {3\,b\,{\left (b\,x^2+a\right )}^{1/3}}{a}\right )\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{6\,a^{5/3}}-\frac {b\,\ln \left ({\left (b\,x^2+a\right )}^{1/3}-a^{1/3}\right )}{3\,a^{5/3}}-\frac {{\left (b\,x^2+a\right )}^{1/3}}{2\,a\,x^2} \] Input:
int(1/(x^3*(a + b*x^2)^(2/3)),x)
Output:
(log((3*(b - 3^(1/2)*b*1i))/(2*a^(2/3)) + (3*b*(a + b*x^2)^(1/3))/a)*(b - 3^(1/2)*b*1i))/(6*a^(5/3)) + (log((3*(b + 3^(1/2)*b*1i))/(2*a^(2/3)) + (3* b*(a + b*x^2)^(1/3))/a)*(b + 3^(1/2)*b*1i))/(6*a^(5/3)) - (b*log((a + b*x^ 2)^(1/3) - a^(1/3)))/(3*a^(5/3)) - (a + b*x^2)^(1/3)/(2*a*x^2)
\[ \int \frac {1}{x^3 \left (a+b x^2\right )^{2/3}} \, dx=\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} x^{3}}d x \] Input:
int(1/x^3/(b*x^2+a)^(2/3),x)
Output:
int(1/((a + b*x**2)**(2/3)*x**3),x)