Integrand size = 26, antiderivative size = 315 \[ \int \frac {b+a x^2}{\left (d+c x^2\right ) \sqrt [3]{x+x^3}} \, dx=\frac {\sqrt {3} a \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{2 c}-\frac {\sqrt {3} (b c-a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{c-d} x}{\sqrt [3]{c-d} x-2 \sqrt [3]{d} \sqrt [3]{x+x^3}}\right )}{2 c \sqrt [3]{c-d} d^{2/3}}-\frac {a \log \left (-x+\sqrt [3]{x+x^3}\right )}{2 c}+\frac {(b c-a d) \log \left (\sqrt [3]{c-d} x+\sqrt [3]{d} \sqrt [3]{x+x^3}\right )}{2 c \sqrt [3]{c-d} d^{2/3}}+\frac {a \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right )}{4 c}+\frac {(-b c+a d) \log \left ((c-d)^{2/3} x^2-\sqrt [3]{c-d} \sqrt [3]{d} x \sqrt [3]{x+x^3}+d^{2/3} \left (x+x^3\right )^{2/3}\right )}{4 c \sqrt [3]{c-d} d^{2/3}} \]
1/2*3^(1/2)*a*arctan(3^(1/2)*x/(x+2*(x^3+x)^(1/3)))/c-1/2*3^(1/2)*(-a*d+b* c)*arctan(3^(1/2)*(c-d)^(1/3)*x/((c-d)^(1/3)*x-2*d^(1/3)*(x^3+x)^(1/3)))/c /(c-d)^(1/3)/d^(2/3)-1/2*a*ln(-x+(x^3+x)^(1/3))/c+1/2*(-a*d+b*c)*ln((c-d)^ (1/3)*x+d^(1/3)*(x^3+x)^(1/3))/c/(c-d)^(1/3)/d^(2/3)+1/4*a*ln(x^2+x*(x^3+x )^(1/3)+(x^3+x)^(2/3))/c+1/4*(a*d-b*c)*ln((c-d)^(2/3)*x^2-(c-d)^(1/3)*d^(1 /3)*x*(x^3+x)^(1/3)+d^(2/3)*(x^3+x)^(2/3))/c/(c-d)^(1/3)/d^(2/3)
Time = 10.37 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.04 \[ \int \frac {b+a x^2}{\left (d+c x^2\right ) \sqrt [3]{x+x^3}} \, dx=\frac {\sqrt [3]{x} \sqrt [3]{1+x^2} \left (2 \sqrt {3} a \arctan \left (\frac {1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )-\frac {2 \sqrt {3} (b c-a d) \arctan \left (\frac {1-\frac {2 \sqrt [3]{c-d} x^{2/3}}{\sqrt [3]{d} \sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{\sqrt [3]{c-d} d^{2/3}}-2 a \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )+a \log \left (1+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )+\frac {2 (b c-a d) \log \left (\sqrt [3]{d}+\frac {\sqrt [3]{c-d} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{\sqrt [3]{c-d} d^{2/3}}-\frac {(b c-a d) \log \left (d^{2/3}+\frac {(c-d)^{2/3} x^{4/3}}{\left (1+x^2\right )^{2/3}}-\frac {\sqrt [3]{c-d} \sqrt [3]{d} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{\sqrt [3]{c-d} d^{2/3}}\right )}{4 c \sqrt [3]{x+x^3}} \]
(x^(1/3)*(1 + x^2)^(1/3)*(2*Sqrt[3]*a*ArcTan[(1 + (2*x^(2/3))/(1 + x^2)^(1 /3))/Sqrt[3]] - (2*Sqrt[3]*(b*c - a*d)*ArcTan[(1 - (2*(c - d)^(1/3)*x^(2/3 ))/(d^(1/3)*(1 + x^2)^(1/3)))/Sqrt[3]])/((c - d)^(1/3)*d^(2/3)) - 2*a*Log[ 1 - x^(2/3)/(1 + x^2)^(1/3)] + a*Log[1 + x^(4/3)/(1 + x^2)^(2/3) + x^(2/3) /(1 + x^2)^(1/3)] + (2*(b*c - a*d)*Log[d^(1/3) + ((c - d)^(1/3)*x^(2/3))/( 1 + x^2)^(1/3)])/((c - d)^(1/3)*d^(2/3)) - ((b*c - a*d)*Log[d^(2/3) + ((c - d)^(2/3)*x^(4/3))/(1 + x^2)^(2/3) - ((c - d)^(1/3)*d^(1/3)*x^(2/3))/(1 + x^2)^(1/3)])/((c - d)^(1/3)*d^(2/3))))/(4*c*(x + x^3)^(1/3))
Time = 0.56 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2467, 446, 2009}
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.
| \(\displaystyle \int \frac {a x^2+b}{\sqrt [3]{x^3+x} \left (c x^2+d\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{x^2+1} \int \frac {a x^2+b}{\sqrt [3]{x} \sqrt [3]{x^2+1} \left (c x^2+d\right )}dx}{\sqrt [3]{x^3+x}}\) |
| \(\Big \downarrow \) 446 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{x^2+1} \int \left (\frac {a}{c \sqrt [3]{x} \sqrt [3]{x^2+1}}+\frac {b c-a d}{c \sqrt [3]{x} \sqrt [3]{x^2+1} \left (c x^2+d\right )}\right )dx}{\sqrt [3]{x^3+x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{x^2+1} \left (-\frac {\sqrt {3} (b c-a d) \arctan \left (\frac {1-\frac {2 x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{d} \sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{2 c d^{2/3} \sqrt [3]{c-d}}+\frac {\sqrt {3} a \arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{2 c}-\frac {(b c-a d) \log \left (c x^2+d\right )}{4 c d^{2/3} \sqrt [3]{c-d}}+\frac {3 (b c-a d) \log \left (x^{2/3} \sqrt [3]{c-d}+\sqrt [3]{d} \sqrt [3]{x^2+1}\right )}{4 c d^{2/3} \sqrt [3]{c-d}}-\frac {3 a \log \left (x^{2/3}-\sqrt [3]{x^2+1}\right )}{4 c}\right )}{\sqrt [3]{x^3+x}}\) |
(x^(1/3)*(1 + x^2)^(1/3)*((Sqrt[3]*a*ArcTan[(1 + (2*x^(2/3))/(1 + x^2)^(1/ 3))/Sqrt[3]])/(2*c) - (Sqrt[3]*(b*c - a*d)*ArcTan[(1 - (2*(c - d)^(1/3)*x^ (2/3))/(d^(1/3)*(1 + x^2)^(1/3)))/Sqrt[3]])/(2*c*(c - d)^(1/3)*d^(2/3)) - ((b*c - a*d)*Log[d + c*x^2])/(4*c*(c - d)^(1/3)*d^(2/3)) - (3*a*Log[x^(2/3 ) - (1 + x^2)^(1/3)])/(4*c) + (3*(b*c - a*d)*Log[(c - d)^(1/3)*x^(2/3) + d ^(1/3)*(1 + x^2)^(1/3)])/(4*c*(c - d)^(1/3)*d^(2/3))))/(x + x^3)^(1/3)
3.29.89.3.1 Defintions of rubi rules used
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((e_) + (f_.)*(x_)^2))/( (c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^2)^ p*((e + f*x^2)/(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 0.97 (sec) , antiderivative size = 288, normalized size of antiderivative = 0.91
| method | result | size |
| pseudoelliptic | \(\frac {\ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}+{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +x^{2}}{x^{2}}\right ) a d \left (\frac {c -d}{d}\right )^{\frac {1}{3}}+\left (-2 a d +2 b c \right ) \ln \left (\frac {\left (\frac {c -d}{d}\right )^{\frac {1}{3}} x +{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}}{x}\right )-2 \sqrt {3}\, \arctan \left (\frac {\left (2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right ) a d \left (\frac {c -d}{d}\right )^{\frac {1}{3}}-2 a \ln \left (\frac {{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}-x}{x}\right ) d \left (\frac {c -d}{d}\right )^{\frac {1}{3}}+\left (a d -b c \right ) \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {c -d}{d}\right )^{\frac {1}{3}} x -2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}\right )}{3 \left (\frac {c -d}{d}\right )^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {\left (\frac {c -d}{d}\right )^{\frac {2}{3}} x^{2}-\left (\frac {c -d}{d}\right )^{\frac {1}{3}} {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}}{x^{2}}\right )\right )}{4 \left (\frac {c -d}{d}\right )^{\frac {1}{3}} d c}\) | \(288\) |
1/4/((c-d)/d)^(1/3)*(ln(((x*(x^2+1))^(2/3)+(x*(x^2+1))^(1/3)*x+x^2)/x^2)*a *d*((c-d)/d)^(1/3)+(-2*a*d+2*b*c)*ln((((c-d)/d)^(1/3)*x+(x*(x^2+1))^(1/3)) /x)-2*3^(1/2)*arctan(1/3*(2*(x*(x^2+1))^(1/3)+x)*3^(1/2)/x)*a*d*((c-d)/d)^ (1/3)-2*a*ln(((x*(x^2+1))^(1/3)-x)/x)*d*((c-d)/d)^(1/3)+(a*d-b*c)*(-2*arct an(1/3*3^(1/2)*(((c-d)/d)^(1/3)*x-2*(x*(x^2+1))^(1/3))/((c-d)/d)^(1/3)/x)* 3^(1/2)+ln((((c-d)/d)^(2/3)*x^2-((c-d)/d)^(1/3)*(x*(x^2+1))^(1/3)*x+(x*(x^ 2+1))^(2/3))/x^2)))/d/c
Timed out. \[ \int \frac {b+a x^2}{\left (d+c x^2\right ) \sqrt [3]{x+x^3}} \, dx=\text {Timed out} \]
\[ \int \frac {b+a x^2}{\left (d+c x^2\right ) \sqrt [3]{x+x^3}} \, dx=\int \frac {a x^{2} + b}{\sqrt [3]{x \left (x^{2} + 1\right )} \left (c x^{2} + d\right )}\, dx \]
\[ \int \frac {b+a x^2}{\left (d+c x^2\right ) \sqrt [3]{x+x^3}} \, dx=\int { \frac {a x^{2} + b}{{\left (c x^{2} + d\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}}} \,d x } \]
Time = 0.35 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.90 \[ \int \frac {b+a x^2}{\left (d+c x^2\right ) \sqrt [3]{x+x^3}} \, dx=-\frac {\sqrt {3} a \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right )}{2 \, c} + \frac {{\left (b c \left (-\frac {c - d}{d}\right )^{\frac {1}{3}} - a d \left (-\frac {c - d}{d}\right )^{\frac {1}{3}}\right )} \left (-\frac {c - d}{d}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {c - d}{d}\right )^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \right |}\right )}{2 \, {\left (c^{2} - c d\right )}} + \frac {a \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}{4 \, c} - \frac {a \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right )}{2 \, c} - \frac {{\left (\sqrt {3} b c - \sqrt {3} a d\right )} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {c - d}{d}\right )^{\frac {1}{3}} + 2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c - d}{d}\right )^{\frac {1}{3}}}\right )}{2 \, {\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} c} + \frac {{\left (b c - a d\right )} \log \left (\left (-\frac {c - d}{d}\right )^{\frac {2}{3}} + \left (-\frac {c - d}{d}\right )^{\frac {1}{3}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}}\right )}{4 \, {\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} c} \]
-1/2*sqrt(3)*a*arctan(1/3*sqrt(3)*(2*(1/x^2 + 1)^(1/3) + 1))/c + 1/2*(b*c* (-(c - d)/d)^(1/3) - a*d*(-(c - d)/d)^(1/3))*(-(c - d)/d)^(1/3)*log(abs(-( -(c - d)/d)^(1/3) + (1/x^2 + 1)^(1/3)))/(c^2 - c*d) + 1/4*a*log((1/x^2 + 1 )^(2/3) + (1/x^2 + 1)^(1/3) + 1)/c - 1/2*a*log(abs((1/x^2 + 1)^(1/3) - 1)) /c - 1/2*(sqrt(3)*b*c - sqrt(3)*a*d)*arctan(1/3*sqrt(3)*((-(c - d)/d)^(1/3 ) + 2*(1/x^2 + 1)^(1/3))/(-(c - d)/d)^(1/3))/((-c*d^2 + d^3)^(1/3)*c) + 1/ 4*(b*c - a*d)*log((-(c - d)/d)^(2/3) + (-(c - d)/d)^(1/3)*(1/x^2 + 1)^(1/3 ) + (1/x^2 + 1)^(2/3))/((-c*d^2 + d^3)^(1/3)*c)
Timed out. \[ \int \frac {b+a x^2}{\left (d+c x^2\right ) \sqrt [3]{x+x^3}} \, dx=\int \frac {a\,x^2+b}{\left (c\,x^2+d\right )\,{\left (x^3+x\right )}^{1/3}} \,d x \]