Integrand size = 15, antiderivative size = 116 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^3} \, dx=2 b \sqrt [3]{a+b x^2}-\frac {\left (a+b x^2\right )^{4/3}}{2 x^2}-\frac {2 \sqrt [3]{a} b \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^2}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3}}-\frac {2}{3} \sqrt [3]{a} b \log (x)+\sqrt [3]{a} b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \]
2*b*(b*x^2+a)^(1/3)-1/2*(b*x^2+a)^(4/3)/x^2-2/3*a^(1/3)*b*ln(x)+a^(1/3)*b* ln(a^(1/3)-(b*x^2+a)^(1/3))-2/3*a^(1/3)*b*arctan(1/3*(a^(1/3)+2*(b*x^2+a)^ (1/3))/a^(1/3)*3^(1/2))*3^(1/2)
Time = 0.18 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^3} \, dx=\frac {1}{6} \left (-\frac {3 \left (a-3 b x^2\right ) \sqrt [3]{a+b x^2}}{x^2}-4 \sqrt {3} \sqrt [3]{a} b \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+4 \sqrt [3]{a} b \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^2}\right )-2 \sqrt [3]{a} b \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )\right ) \]
((-3*(a - 3*b*x^2)*(a + b*x^2)^(1/3))/x^2 - 4*Sqrt[3]*a^(1/3)*b*ArcTan[(1 + (2*(a + b*x^2)^(1/3))/a^(1/3))/Sqrt[3]] + 4*a^(1/3)*b*Log[-a^(1/3) + (a + b*x^2)^(1/3)] - 2*a^(1/3)*b*Log[a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3)])/6
Time = 0.22 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {243, 51, 60, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{4/3}}{x^3} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{4/3}}{x^4}dx^2\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {4}{3} b \int \frac {\sqrt [3]{b x^2+a}}{x^2}dx^2-\frac {\left (a+b x^2\right )^{4/3}}{x^2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (\frac {4}{3} b \left (a \int \frac {1}{x^2 \left (b x^2+a\right )^{2/3}}dx^2+3 \sqrt [3]{a+b x^2}\right )-\frac {\left (a+b x^2\right )^{4/3}}{x^2}\right )\) |
\(\Big \downarrow \) 69 |
\(\displaystyle \frac {1}{2} \left (\frac {4}{3} b \left (a \left (-\frac {3 \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 a^{2/3}}-\frac {3 \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 \sqrt [3]{a}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )+3 \sqrt [3]{a+b x^2}\right )-\frac {\left (a+b x^2\right )^{4/3}}{x^2}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (\frac {4}{3} b \left (a \left (-\frac {3 \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )+3 \sqrt [3]{a+b x^2}\right )-\frac {\left (a+b x^2\right )^{4/3}}{x^2}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \left (\frac {4}{3} b \left (a \left (\frac {3 \int \frac {1}{-x^4-3}d\left (\frac {2 \sqrt [3]{b x^2+a}}{\sqrt [3]{a}}+1\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )+3 \sqrt [3]{a+b x^2}\right )-\frac {\left (a+b x^2\right )^{4/3}}{x^2}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {4}{3} b \left (a \left (-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 a^{2/3}}-\frac {\log \left (x^2\right )}{2 a^{2/3}}\right )+3 \sqrt [3]{a+b x^2}\right )-\frac {\left (a+b x^2\right )^{4/3}}{x^2}\right )\) |
(-((a + b*x^2)^(4/3)/x^2) + (4*b*(3*(a + b*x^2)^(1/3) + a*(-((Sqrt[3]*ArcT an[(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)/2
3.7.94.3.1 Defintions of rubi rules used
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/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
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 = 2.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.96
method | result | size |
pseudoelliptic | \(\frac {\left (9 b \,x^{2}-3 a \right ) \left (b \,x^{2}+a \right )^{\frac {1}{3}}-2 b \,x^{2} a^{\frac {1}{3}} \left (2 \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b \,x^{2}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}+\ln \left (a^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{2}+a \right )^{\frac {1}{3}}+\left (b \,x^{2}+a \right )^{\frac {2}{3}}\right )-2 \ln \left (\left (b \,x^{2}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )\right )}{6 x^{2}}\) | \(111\) |
1/6*((9*b*x^2-3*a)*(b*x^2+a)^(1/3)-2*b*x^2*a^(1/3)*(2*arctan(1/3*(a^(1/3)+ 2*(b*x^2+a)^(1/3))/a^(1/3)*3^(1/2))*3^(1/2)+ln(a^(2/3)+a^(1/3)*(b*x^2+a)^( 1/3)+(b*x^2+a)^(2/3))-2*ln((b*x^2+a)^(1/3)-a^(1/3))))/x^2
Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^3} \, dx=-\frac {4 \, \sqrt {3} a^{\frac {1}{3}} b x^{2} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + \sqrt {3} a}{3 \, a}\right ) + 2 \, a^{\frac {1}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 4 \, a^{\frac {1}{3}} b x^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - 3 \, {\left (3 \, b x^{2} - a\right )} {\left (b x^{2} + a\right )}^{\frac {1}{3}}}{6 \, x^{2}} \]
-1/6*(4*sqrt(3)*a^(1/3)*b*x^2*arctan(1/3*(2*sqrt(3)*(b*x^2 + a)^(1/3)*a^(2 /3) + sqrt(3)*a)/a) + 2*a^(1/3)*b*x^2*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^ (1/3)*a^(1/3) + a^(2/3)) - 4*a^(1/3)*b*x^2*log((b*x^2 + a)^(1/3) - a^(1/3) ) - 3*(3*b*x^2 - a)*(b*x^2 + a)^(1/3))/x^2
Result contains complex when optimal does not.
Time = 1.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.40 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^3} \, dx=- \frac {b^{\frac {4}{3}} x^{\frac {2}{3}} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (\frac {2}{3}\right )} \]
-b**(4/3)*x**(2/3)*gamma(-1/3)*hyper((-4/3, -1/3), (2/3,), a*exp_polar(I*p i)/(b*x**2))/(2*gamma(2/3))
Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^3} \, dx=-\frac {2}{3} \, \sqrt {3} a^{\frac {1}{3}} b \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) - \frac {1}{3} \, a^{\frac {1}{3}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + \frac {2}{3} \, a^{\frac {1}{3}} b \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + \frac {3}{2} \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} b - \frac {{\left (b x^{2} + a\right )}^{\frac {1}{3}} a}{2 \, x^{2}} \]
-2/3*sqrt(3)*a^(1/3)*b*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/ a^(1/3)) - 1/3*a^(1/3)*b*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) + 2/3*a^(1/3)*b*log((b*x^2 + a)^(1/3) - a^(1/3)) + 3/2*(b*x^2 + a)^(1/3)*b - 1/2*(b*x^2 + a)^(1/3)*a/x^2
Time = 0.52 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^3} \, dx=-\frac {4 \, \sqrt {3} a^{\frac {1}{3}} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) + 2 \, a^{\frac {1}{3}} b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 4 \, a^{\frac {1}{3}} b^{2} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right ) - 9 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} b^{2} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a b}{x^{2}}}{6 \, b} \]
-1/6*(4*sqrt(3)*a^(1/3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1 /3))/a^(1/3)) + 2*a^(1/3)*b^2*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^ (1/3) + a^(2/3)) - 4*a^(1/3)*b^2*log(abs((b*x^2 + a)^(1/3) - a^(1/3))) - 9 *(b*x^2 + a)^(1/3)*b^2 + 3*(b*x^2 + a)^(1/3)*a*b/x^2)/b
Time = 5.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^3} \, dx=\frac {3\,b\,{\left (b\,x^2+a\right )}^{1/3}}{2}-\frac {a\,{\left (b\,x^2+a\right )}^{1/3}}{2\,x^2}+\frac {2\,a^{1/3}\,b\,\ln \left (6\,a^{4/3}\,b-6\,a\,b\,{\left (b\,x^2+a\right )}^{1/3}\right )}{3}+\frac {a^{1/3}\,b\,\ln \left (6\,a\,b\,{\left (b\,x^2+a\right )}^{1/3}-3\,a^{4/3}\,b\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{3}-\frac {a^{1/3}\,b\,\ln \left (3\,a^{4/3}\,b\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )+6\,a\,b\,{\left (b\,x^2+a\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{3} \]
(3*b*(a + b*x^2)^(1/3))/2 - (a*(a + b*x^2)^(1/3))/(2*x^2) + (2*a^(1/3)*b*l og(6*a^(4/3)*b - 6*a*b*(a + b*x^2)^(1/3)))/3 + (a^(1/3)*b*log(6*a*b*(a + b *x^2)^(1/3) - 3*a^(4/3)*b*(3^(1/2)*1i - 1))*(3^(1/2)*1i - 1))/3 - (a^(1/3) *b*log(3*a^(4/3)*b*(3^(1/2)*1i + 1) + 6*a*b*(a + b*x^2)^(1/3))*(3^(1/2)*1i + 1))/3