Integrand size = 19, antiderivative size = 163 \[ \int \sqrt [3]{c x} \left (a+b x^2\right )^{4/3} \, dx=\frac {a (c x)^{4/3} \sqrt [3]{a+b x^2}}{3 c}+\frac {(c x)^{4/3} \left (a+b x^2\right )^{4/3}}{4 c}-\frac {a^2 \sqrt [3]{c} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+b x^2}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3}}-\frac {a^2 \sqrt [3]{c} \log \left (\sqrt [3]{b} (c x)^{2/3}-c^{2/3} \sqrt [3]{a+b x^2}\right )}{6 b^{2/3}} \]
1/3*a*(c*x)^(4/3)*(b*x^2+a)^(1/3)/c+1/4*(c*x)^(4/3)*(b*x^2+a)^(4/3)/c-1/6* a^2*c^(1/3)*ln(b^(1/3)*(c*x)^(2/3)-c^(2/3)*(b*x^2+a)^(1/3))/b^(2/3)-1/9*a^ 2*c^(1/3)*arctan(1/3*(1+2*b^(1/3)*(c*x)^(2/3)/c^(2/3)/(b*x^2+a)^(1/3))*3^( 1/2))/b^(2/3)*3^(1/2)
Time = 1.27 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.25 \[ \int \sqrt [3]{c x} \left (a+b x^2\right )^{4/3} \, dx=\frac {\sqrt [3]{c x} \left (21 a b^{2/3} x^{4/3} \sqrt [3]{a+b x^2}+9 b^{5/3} x^{10/3} \sqrt [3]{a+b x^2}-4 \sqrt {3} a^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x^{2/3}}{\sqrt [3]{b} x^{2/3}+2 \sqrt [3]{a+b x^2}}\right )-4 a^2 \log \left (-\sqrt [3]{b} x^{2/3}+\sqrt [3]{a+b x^2}\right )+2 a^2 \log \left (b^{2/3} x^{4/3}+\sqrt [3]{b} x^{2/3} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )\right )}{36 b^{2/3} \sqrt [3]{x}} \]
((c*x)^(1/3)*(21*a*b^(2/3)*x^(4/3)*(a + b*x^2)^(1/3) + 9*b^(5/3)*x^(10/3)* (a + b*x^2)^(1/3) - 4*Sqrt[3]*a^2*ArcTan[(Sqrt[3]*b^(1/3)*x^(2/3))/(b^(1/3 )*x^(2/3) + 2*(a + b*x^2)^(1/3))] - 4*a^2*Log[-(b^(1/3)*x^(2/3)) + (a + b* x^2)^(1/3)] + 2*a^2*Log[b^(2/3)*x^(4/3) + b^(1/3)*x^(2/3)*(a + b*x^2)^(1/3 ) + (a + b*x^2)^(2/3)]))/(36*b^(2/3)*x^(1/3))
Time = 0.25 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {248, 248, 266, 807, 853}
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.
\(\displaystyle \int \sqrt [3]{c x} \left (a+b x^2\right )^{4/3} \, dx\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {2}{3} a \int \sqrt [3]{c x} \sqrt [3]{b x^2+a}dx+\frac {(c x)^{4/3} \left (a+b x^2\right )^{4/3}}{4 c}\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {2}{3} a \left (\frac {1}{3} a \int \frac {\sqrt [3]{c x}}{\left (b x^2+a\right )^{2/3}}dx+\frac {(c x)^{4/3} \sqrt [3]{a+b x^2}}{2 c}\right )+\frac {(c x)^{4/3} \left (a+b x^2\right )^{4/3}}{4 c}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2}{3} a \left (\frac {a \int \frac {c x}{\left (b x^2+a\right )^{2/3}}d\sqrt [3]{c x}}{c}+\frac {(c x)^{4/3} \sqrt [3]{a+b x^2}}{2 c}\right )+\frac {(c x)^{4/3} \left (a+b x^2\right )^{4/3}}{4 c}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {2}{3} a \left (\frac {a \int \frac {(c x)^{2/3}}{\left (a+\frac {b x}{c}\right )^{2/3}}d(c x)^{2/3}}{2 c}+\frac {(c x)^{4/3} \sqrt [3]{a+b x^2}}{2 c}\right )+\frac {(c x)^{4/3} \left (a+b x^2\right )^{4/3}}{4 c}\) |
\(\Big \downarrow \) 853 |
\(\displaystyle \frac {2}{3} a \left (\frac {a \left (-\frac {c^{4/3} \arctan \left (\frac {\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+\frac {b x}{c}}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {c^{4/3} \log \left (\frac {\sqrt [3]{b} (c x)^{2/3}}{c^{2/3}}-\sqrt [3]{a+\frac {b x}{c}}\right )}{2 b^{2/3}}\right )}{2 c}+\frac {(c x)^{4/3} \sqrt [3]{a+b x^2}}{2 c}\right )+\frac {(c x)^{4/3} \left (a+b x^2\right )^{4/3}}{4 c}\) |
((c*x)^(4/3)*(a + b*x^2)^(4/3))/(4*c) + (2*a*(((c*x)^(4/3)*(a + b*x^2)^(1/ 3))/(2*c) + (a*(-((c^(4/3)*ArcTan[(1 + (2*b^(1/3)*(c*x)^(2/3))/(c^(2/3)*(a + (b*x)/c)^(1/3)))/Sqrt[3]])/(Sqrt[3]*b^(2/3))) - (c^(4/3)*Log[(b^(1/3)*( c*x)^(2/3))/c^(2/3) - (a + (b*x)/c)^(1/3)])/(2*b^(2/3))))/(2*c)))/3
3.8.56.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim p[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp [Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]
\[\int \left (c x \right )^{\frac {1}{3}} \left (b \,x^{2}+a \right )^{\frac {4}{3}}d x\]
Timed out. \[ \int \sqrt [3]{c x} \left (a+b x^2\right )^{4/3} \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 3.57 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.28 \[ \int \sqrt [3]{c x} \left (a+b x^2\right )^{4/3} \, dx=\frac {a^{\frac {4}{3}} \sqrt [3]{c} x^{\frac {4}{3}} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {5}{3}\right )} \]
a**(4/3)*c**(1/3)*x**(4/3)*gamma(2/3)*hyper((-4/3, 2/3), (5/3,), b*x**2*ex p_polar(I*pi)/a)/(2*gamma(5/3))
\[ \int \sqrt [3]{c x} \left (a+b x^2\right )^{4/3} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {4}{3}} \left (c x\right )^{\frac {1}{3}} \,d x } \]
\[ \int \sqrt [3]{c x} \left (a+b x^2\right )^{4/3} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {4}{3}} \left (c x\right )^{\frac {1}{3}} \,d x } \]
Timed out. \[ \int \sqrt [3]{c x} \left (a+b x^2\right )^{4/3} \, dx=\int {\left (c\,x\right )}^{1/3}\,{\left (b\,x^2+a\right )}^{4/3} \,d x \]