Integrand size = 22, antiderivative size = 288 \[ \int \frac {x^2}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{2 a}+\frac {\sqrt {3} \sqrt [3]{b} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a-b} x}{\sqrt [3]{a-b} x-2 \sqrt [3]{b} \sqrt [3]{x+x^3}}\right )}{2 a \sqrt [3]{a-b}}-\frac {\log \left (-x+\sqrt [3]{x+x^3}\right )}{2 a}-\frac {\sqrt [3]{b} \log \left (\sqrt [3]{a-b} x+\sqrt [3]{b} \sqrt [3]{x+x^3}\right )}{2 a \sqrt [3]{a-b}}+\frac {\log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right )}{4 a}+\frac {\sqrt [3]{b} \log \left ((a-b)^{2/3} x^2-\sqrt [3]{a-b} \sqrt [3]{b} x \sqrt [3]{x+x^3}+b^{2/3} \left (x+x^3\right )^{2/3}\right )}{4 a \sqrt [3]{a-b}} \] Output:
1/2*3^(1/2)*arctan(3^(1/2)*x/(x+2*(x^3+x)^(1/3)))/a+1/2*3^(1/2)*b^(1/3)*ar ctan(3^(1/2)*(a-b)^(1/3)*x/((a-b)^(1/3)*x-2*b^(1/3)*(x^3+x)^(1/3)))/a/(a-b )^(1/3)-1/2*ln(-x+(x^3+x)^(1/3))/a-1/2*b^(1/3)*ln((a-b)^(1/3)*x+b^(1/3)*(x ^3+x)^(1/3))/a/(a-b)^(1/3)+1/4*ln(x^2+x*(x^3+x)^(1/3)+(x^3+x)^(2/3))/a+1/4 *b^(1/3)*ln((a-b)^(2/3)*x^2-(a-b)^(1/3)*b^(1/3)*x*(x^3+x)^(1/3)+b^(2/3)*(x ^3+x)^(2/3))/a/(a-b)^(1/3)
Time = 8.36 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.16 \[ \int \frac {x^2}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=\frac {\sqrt [3]{x} \sqrt [3]{1+x^2} \left (2 \sqrt {3} \sqrt [3]{a-b} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )+2 \sqrt {3} \sqrt [3]{b} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a-b} x^{2/3}}{\sqrt [3]{a-b} x^{2/3}-2 \sqrt [3]{b} \sqrt [3]{1+x^2}}\right )-2 \sqrt [3]{a-b} \log \left (a \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )\right )-2 \sqrt [3]{b} \log \left (\sqrt [3]{a-b} x^{2/3}+\sqrt [3]{b} \sqrt [3]{1+x^2}\right )+\sqrt [3]{a-b} \log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )+\sqrt [3]{b} \log \left ((a-b)^{2/3} x^{4/3}-\sqrt [3]{a-b} \sqrt [3]{b} x^{2/3} \sqrt [3]{1+x^2}+b^{2/3} \left (1+x^2\right )^{2/3}\right )\right )}{4 a \sqrt [3]{a-b} \sqrt [3]{x+x^3}} \] Input:
Integrate[x^2/((b + a*x^2)*(x + x^3)^(1/3)),x]
Output:
(x^(1/3)*(1 + x^2)^(1/3)*(2*Sqrt[3]*(a - b)^(1/3)*ArcTan[(Sqrt[3]*x^(2/3)) /(x^(2/3) + 2*(1 + x^2)^(1/3))] + 2*Sqrt[3]*b^(1/3)*ArcTan[(Sqrt[3]*(a - b )^(1/3)*x^(2/3))/((a - b)^(1/3)*x^(2/3) - 2*b^(1/3)*(1 + x^2)^(1/3))] - 2* (a - b)^(1/3)*Log[a*(-x^(2/3) + (1 + x^2)^(1/3))] - 2*b^(1/3)*Log[(a - b)^ (1/3)*x^(2/3) + b^(1/3)*(1 + x^2)^(1/3)] + (a - b)^(1/3)*Log[x^(4/3) + x^( 2/3)*(1 + x^2)^(1/3) + (1 + x^2)^(2/3)] + b^(1/3)*Log[(a - b)^(2/3)*x^(4/3 ) - (a - b)^(1/3)*b^(1/3)*x^(2/3)*(1 + x^2)^(1/3) + b^(2/3)*(1 + x^2)^(2/3 )]))/(4*a*(a - b)^(1/3)*(x + x^3)^(1/3))
Time = 0.36 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.77, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1948, 368, 965, 983, 769, 901}
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.
\(\displaystyle \int \frac {x^2}{\sqrt [3]{x^3+x} \left (a x^2+b\right )} \, dx\) |
\(\Big \downarrow \) 1948 |
\(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{x^2+1} \int \frac {x^{5/3}}{\sqrt [3]{x^2+1} \left (a x^2+b\right )}dx}{\sqrt [3]{x^3+x}}\) |
\(\Big \downarrow \) 368 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^2+1} \int \frac {x^{7/3}}{\sqrt [3]{x^2+1} \left (a x^2+b\right )}d\sqrt [3]{x}}{\sqrt [3]{x^3+x}}\) |
\(\Big \downarrow \) 965 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^2+1} \int \frac {x}{\sqrt [3]{x+1} (b+a x)}dx^{2/3}}{2 \sqrt [3]{x^3+x}}\) |
\(\Big \downarrow \) 983 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^2+1} \left (\frac {\int \frac {1}{\sqrt [3]{x+1}}dx^{2/3}}{a}-\frac {b \int \frac {1}{\sqrt [3]{x+1} (b+a x)}dx^{2/3}}{a}\right )}{2 \sqrt [3]{x^3+x}}\) |
\(\Big \downarrow \) 769 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^2+1} \left (\frac {\frac {\arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (\sqrt [3]{x+1}-x^{2/3}\right )}{a}-\frac {b \int \frac {1}{\sqrt [3]{x+1} (b+a x)}dx^{2/3}}{a}\right )}{2 \sqrt [3]{x^3+x}}\) |
\(\Big \downarrow \) 901 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{x^2+1} \left (\frac {\frac {\arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (\sqrt [3]{x+1}-x^{2/3}\right )}{a}-\frac {b \left (-\frac {\arctan \left (\frac {1-\frac {2 x^{2/3} \sqrt [3]{a-b}}{\sqrt [3]{b} \sqrt [3]{x+1}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3} \sqrt [3]{a-b}}+\frac {\log \left (-\frac {x^{2/3} \sqrt [3]{a-b}}{\sqrt [3]{b}}-\sqrt [3]{x+1}\right )}{2 b^{2/3} \sqrt [3]{a-b}}-\frac {\log (a x+b)}{6 b^{2/3} \sqrt [3]{a-b}}\right )}{a}\right )}{2 \sqrt [3]{x^3+x}}\) |
Input:
Int[x^2/((b + a*x^2)*(x + x^3)^(1/3)),x]
Output:
(3*x^(1/3)*(1 + x^2)^(1/3)*(-((b*(-(ArcTan[(1 - (2*(a - b)^(1/3)*x^(2/3))/ (b^(1/3)*(1 + x)^(1/3)))/Sqrt[3]]/(Sqrt[3]*(a - b)^(1/3)*b^(2/3))) - Log[b + a*x]/(6*(a - b)^(1/3)*b^(2/3)) + Log[-(((a - b)^(1/3)*x^(2/3))/b^(1/3)) - (1 + x)^(1/3)]/(2*(a - b)^(1/3)*b^(2/3))))/a) + (ArcTan[(1 + (2*x^(2/3) )/(1 + x)^(1/3))/Sqrt[3]]/Sqrt[3] - Log[-x^(2/3) + (1 + x)^(1/3)]/2)/a))/( 2*(x + x^3)^(1/3))
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m ] && IntegerQ[p]
Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* (x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]
Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S qrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /; Free Q[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]
Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_)^(n_))^(q_.))/((a_) + (b_.)*(x_)^( n_)), x_Symbol] :> Simp[e^n/b Int[(e*x)^(m - n)*(c + d*x^n)^q, x], x] - S imp[a*(e^n/b) Int[(e*x)^(m - n)*((c + d*x^n)^q/(a + b*x^n)), x], x] /; Fr eeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1] && IntBinomialQ[a, b, c, d, e, m, n, -1, q, x]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Simp[e^IntPart[m]*(e*x)^FracPart[m]*( (a*x^j + b*x^(j + n))^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x ^n)^FracPart[p])) Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p, q}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && !(EqQ[n, 1] && EqQ[j, 1])
Time = 0.77 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.90
method | result | size |
pseudoelliptic | \(-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right ) \left (\frac {a -b}{b}\right )^{\frac {1}{3}}+\ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}-x}{x}\right ) \left (\frac {a -b}{b}\right )^{\frac {1}{3}}-\frac {\ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +x^{2}}{x^{2}}\right ) \left (\frac {a -b}{b}\right )^{\frac {1}{3}}}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a -b}{b}\right )^{\frac {1}{3}} x -2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}\right )}{3 \left (\frac {a -b}{b}\right )^{\frac {1}{3}} x}\right )+\ln \left (\frac {\left (\frac {a -b}{b}\right )^{\frac {1}{3}} x +{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {\left (\frac {a -b}{b}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a -b}{b}\right )^{\frac {1}{3}} {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}}{x^{2}}\right )}{2}}{2 \left (\frac {a -b}{b}\right )^{\frac {1}{3}} a}\) | \(259\) |
Input:
int(x^2/(a*x^2+b)/(x^3+x)^(1/3),x,method=_RETURNVERBOSE)
Output:
-1/2*(3^(1/2)*arctan(1/3*(2*(x*(x^2+1))^(1/3)+x)*3^(1/2)/x)*((a-b)/b)^(1/3 )+ln(((x*(x^2+1))^(1/3)-x)/x)*((a-b)/b)^(1/3)-1/2*ln(((x*(x^2+1))^(2/3)+(x *(x^2+1))^(1/3)*x+x^2)/x^2)*((a-b)/b)^(1/3)+3^(1/2)*arctan(1/3*3^(1/2)*((( a-b)/b)^(1/3)*x-2*(x*(x^2+1))^(1/3))/((a-b)/b)^(1/3)/x)+ln((((a-b)/b)^(1/3 )*x+(x*(x^2+1))^(1/3))/x)-1/2*ln((((a-b)/b)^(2/3)*x^2-((a-b)/b)^(1/3)*(x*( x^2+1))^(1/3)*x+(x*(x^2+1))^(2/3))/x^2))/((a-b)/b)^(1/3)/a
Timed out. \[ \int \frac {x^2}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=\text {Timed out} \] Input:
integrate(x^2/(a*x^2+b)/(x^3+x)^(1/3),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {x^2}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=\int \frac {x^{2}}{\sqrt [3]{x \left (x^{2} + 1\right )} \left (a x^{2} + b\right )}\, dx \] Input:
integrate(x**2/(a*x**2+b)/(x**3+x)**(1/3),x)
Output:
Integral(x**2/((x*(x**2 + 1))**(1/3)*(a*x**2 + b)), x)
\[ \int \frac {x^2}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=\int { \frac {x^{2}}{{\left (a x^{2} + b\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate(x^2/(a*x^2+b)/(x^3+x)^(1/3),x, algorithm="maxima")
Output:
integrate(x^2/((a*x^2 + b)*(x^3 + x)^(1/3)), x)
Time = 0.88 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=-\frac {b \left (-\frac {a - b}{b}\right )^{\frac {2}{3}} \log \left ({\left | -\left (-\frac {a - b}{b}\right )^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \right |}\right )}{2 \, {\left (a^{2} - a b\right )}} - \frac {3 \, {\left (-a b^{2} + b^{3}\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a - b}{b}\right )^{\frac {1}{3}} + 2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a - b}{b}\right )^{\frac {1}{3}}}\right )}{2 \, {\left (\sqrt {3} a^{2} b - \sqrt {3} a b^{2}\right )}} + \frac {{\left (-a b^{2} + b^{3}\right )}^{\frac {2}{3}} \log \left (\left (-\frac {a - b}{b}\right )^{\frac {2}{3}} + \left (-\frac {a - b}{b}\right )^{\frac {1}{3}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}}\right )}{4 \, {\left (a^{2} b - a b^{2}\right )}} - \frac {\sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right )}{2 \, a} + \frac {\log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}{4 \, a} - \frac {\log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right )}{2 \, a} \] Input:
integrate(x^2/(a*x^2+b)/(x^3+x)^(1/3),x, algorithm="giac")
Output:
-1/2*b*(-(a - b)/b)^(2/3)*log(abs(-(-(a - b)/b)^(1/3) + (1/x^2 + 1)^(1/3)) )/(a^2 - a*b) - 3/2*(-a*b^2 + b^3)^(2/3)*arctan(1/3*sqrt(3)*((-(a - b)/b)^ (1/3) + 2*(1/x^2 + 1)^(1/3))/(-(a - b)/b)^(1/3))/(sqrt(3)*a^2*b - sqrt(3)* a*b^2) + 1/4*(-a*b^2 + b^3)^(2/3)*log((-(a - b)/b)^(2/3) + (-(a - b)/b)^(1 /3)*(1/x^2 + 1)^(1/3) + (1/x^2 + 1)^(2/3))/(a^2*b - a*b^2) - 1/2*sqrt(3)*a rctan(1/3*sqrt(3)*(2*(1/x^2 + 1)^(1/3) + 1))/a + 1/4*log((1/x^2 + 1)^(2/3) + (1/x^2 + 1)^(1/3) + 1)/a - 1/2*log(abs((1/x^2 + 1)^(1/3) - 1))/a
Timed out. \[ \int \frac {x^2}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=\int \frac {x^2}{\left (a\,x^2+b\right )\,{\left (x^3+x\right )}^{1/3}} \,d x \] Input:
int(x^2/((b + a*x^2)*(x + x^3)^(1/3)),x)
Output:
int(x^2/((b + a*x^2)*(x + x^3)^(1/3)), x)
\[ \int \frac {x^2}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=\int \frac {x^{2}}{x^{\frac {7}{3}} \left (x^{2}+1\right )^{\frac {1}{3}} a +x^{\frac {1}{3}} \left (x^{2}+1\right )^{\frac {1}{3}} b}d x \] Input:
int(x^2/(a*x^2+b)/(x^3+x)^(1/3),x)
Output:
int(x**2/(x**(1/3)*(x**2 + 1)**(1/3)*a*x**2 + x**(1/3)*(x**2 + 1)**(1/3)*b ),x)