Integrand size = 19, antiderivative size = 106 \[ \int \frac {\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx=-\frac {\sqrt {3} \sqrt [3]{c} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+b x^2}}}{\sqrt {3}}\right )}{2 b^{2/3}}-\frac {3 \sqrt [3]{c} \log \left (\sqrt [3]{b} (c x)^{2/3}-c^{2/3} \sqrt [3]{a+b x^2}\right )}{4 b^{2/3}} \] Output:
-1/2*3^(1/2)*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)-3/4*c^(1/3)*ln(b^(1/3)*(c*x)^(2/3)-c^(2/3)*(b*x^2 +a)^(1/3))/b^(2/3)
Time = 0.85 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {\sqrt [3]{c x} \left (-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x^{2/3}}{\sqrt [3]{b} x^{2/3}+2 \sqrt [3]{a+b x^2}}\right )-2 \log \left (-\sqrt [3]{b} x^{2/3}+\sqrt [3]{a+b x^2}\right )+\log \left (b^{2/3} x^{4/3}+\sqrt [3]{b} x^{2/3} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )\right )}{4 b^{2/3} \sqrt [3]{x}} \] Input:
Integrate[(c*x)^(1/3)/(a + b*x^2)^(2/3),x]
Output:
((c*x)^(1/3)*(-2*Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/3)*x^(2/3))/(b^(1/3)*x^(2/3) + 2*(a + b*x^2)^(1/3))] - 2*Log[-(b^(1/3)*x^(2/3)) + (a + b*x^2)^(1/3)] + 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)]))/(4*b^(2/3)*x^(1/3))
Time = 0.21 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.
\(\displaystyle \int \frac {\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {3 \int \frac {c x}{\left (b x^2+a\right )^{2/3}}d\sqrt [3]{c x}}{c}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {3 \int \frac {(c x)^{2/3}}{\left (a+\frac {b x}{c}\right )^{2/3}}d(c x)^{2/3}}{2 c}\) |
\(\Big \downarrow \) 853 |
\(\displaystyle \frac {3 \left (-\frac {c^{4/3} \arctan \left (\frac {\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+\frac {b x}{c}}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {c^{4/3} \log \left (\frac {\sqrt [3]{b} (c x)^{2/3}}{c^{2/3}}-\sqrt [3]{a+\frac {b x}{c}}\right )}{2 b^{2/3}}\right )}{2 c}\) |
Input:
Int[(c*x)^(1/3)/(a + b*x^2)^(2/3),x]
Output:
(3*(-((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)
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]
\[\int \frac {\left (c x \right )^{\frac {1}{3}}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x\]
Input:
int((c*x)^(1/3)/(b*x^2+a)^(2/3),x)
Output:
int((c*x)^(1/3)/(b*x^2+a)^(2/3),x)
Timed out. \[ \int \frac {\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx=\text {Timed out} \] Input:
integrate((c*x)^(1/3)/(b*x^2+a)^(2/3),x, algorithm="fricas")
Output:
Timed out
Result contains complex when optimal does not.
Time = 0.69 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {\sqrt [3]{c} x^{\frac {4}{3}} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {2}{3}} \Gamma \left (\frac {5}{3}\right )} \] Input:
integrate((c*x)**(1/3)/(b*x**2+a)**(2/3),x)
Output:
c**(1/3)*x**(4/3)*gamma(2/3)*hyper((2/3, 2/3), (5/3,), b*x**2*exp_polar(I* pi)/a)/(2*a**(2/3)*gamma(5/3))
\[ \int \frac {\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {\left (c x\right )^{\frac {1}{3}}}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((c*x)^(1/3)/(b*x^2+a)^(2/3),x, algorithm="maxima")
Output:
integrate((c*x)^(1/3)/(b*x^2 + a)^(2/3), x)
\[ \int \frac {\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {\left (c x\right )^{\frac {1}{3}}}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate((c*x)^(1/3)/(b*x^2+a)^(2/3),x, algorithm="giac")
Output:
integrate((c*x)^(1/3)/(b*x^2 + a)^(2/3), x)
Timed out. \[ \int \frac {\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx=\int \frac {{\left (c\,x\right )}^{1/3}}{{\left (b\,x^2+a\right )}^{2/3}} \,d x \] Input:
int((c*x)^(1/3)/(a + b*x^2)^(2/3),x)
Output:
int((c*x)^(1/3)/(a + b*x^2)^(2/3), x)
\[ \int \frac {\sqrt [3]{c x}}{\left (a+b x^2\right )^{2/3}} \, dx=c^{\frac {1}{3}} \left (\int \frac {x^{\frac {1}{3}}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) \] Input:
int((c*x)^(1/3)/(b*x^2+a)^(2/3),x)
Output:
c**(1/3)*int(x**(1/3)/(a + b*x**2)**(2/3),x)