Integrand size = 24, antiderivative size = 145 \[ \int \frac {x^2}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 d^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} \sqrt [3]{b c-a d}} \] Output:
-1/3*arctan(1/3*(1-2*d^(1/3)*(b*x^3+a)^(1/3)/(-a*d+b*c)^(1/3))*3^(1/2))*3^ (1/2)/d^(2/3)/(-a*d+b*c)^(1/3)+1/6*ln(d*x^3+c)/d^(2/3)/(-a*d+b*c)^(1/3)-1/ 2*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(2/3)/(-a*d+b*c)^(1/3)
Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.12 \[ \int \frac {x^2}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {-2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+\log \left ((b c-a d)^{2/3}-\sqrt [3]{d} \sqrt [3]{b c-a d} \sqrt [3]{a+b x^3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{6 d^{2/3} \sqrt [3]{b c-a d}} \] Input:
Integrate[x^2/((a + b*x^3)^(1/3)*(c + d*x^3)),x]
Output:
(-2*Sqrt[3]*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/S qrt[3]] - 2*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] + Log[(b*c - a*d)^(2/3) - d^(1/3)*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3) + d^(2/3)*(a + b*x^3)^(2/3)])/(6*d^(2/3)*(b*c - a*d)^(1/3))
Time = 0.43 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {946, 68, 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.
\(\displaystyle \int \frac {x^2}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 946 |
\(\displaystyle \frac {1}{3} \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx^3\) |
\(\Big \downarrow \) 68 |
\(\displaystyle \frac {1}{3} \left (-\frac {3 \int \frac {1}{\frac {\sqrt [3]{b c-a d}}{\sqrt [3]{d}}+\sqrt [3]{b x^3+a}}d\sqrt [3]{b x^3+a}}{2 d^{2/3} \sqrt [3]{b c-a d}}+\frac {3 \int \frac {1}{x^6+\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} \sqrt [3]{b x^3+a}}{\sqrt [3]{d}}}d\sqrt [3]{b x^3+a}}{2 d}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{3} \left (\frac {3 \int \frac {1}{x^6+\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} \sqrt [3]{b x^3+a}}{\sqrt [3]{d}}}d\sqrt [3]{b x^3+a}}{2 d}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{3} \left (\frac {3 \int \frac {1}{-x^6-3}d\left (1-\frac {2 \sqrt [3]{d} \sqrt [3]{b x^3+a}}{\sqrt [3]{b c-a d}}\right )}{d^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{3} \left (-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{d^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}-\frac {3 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} \sqrt [3]{b c-a d}}\right )\) |
Input:
Int[x^2/((a + b*x^3)^(1/3)*(c + d*x^3)),x]
Output:
(-((Sqrt[3]*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/S qrt[3]])/(d^(2/3)*(b*c - a*d)^(1/3))) + Log[c + d*x^3]/(2*d^(2/3)*(b*c - a *d)^(1/3)) - (3*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*d^( 2/3)*(b*c - a*d)^(1/3)))/3
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ {q = Rt[-(b*c - a*d)/b, 3]}, Simp[Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[3/(2*b*q) Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] / ; FreeQ[{a, b, c, d}, x] && NegQ[(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_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m - n + 1, 0]
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 = 1.10 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.05
method | result | size |
pseudoelliptic | \(\frac {2 \arctan \left (\frac {\sqrt {3}\, \left (2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\right ) \sqrt {3}+2 \ln \left (\left (b \,x^{3}+a \right )^{\frac {1}{3}}-\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}\right )-\ln \left (\left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{d}\right )^{\frac {2}{3}}\right )}{6 d \left (\frac {a d -b c}{d}\right )^{\frac {1}{3}}}\) | \(152\) |
Input:
int(x^2/(b*x^3+a)^(1/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
1/6*(2*arctan(1/3*3^(1/2)*(2*(b*x^3+a)^(1/3)+((a*d-b*c)/d)^(1/3))/((a*d-b* c)/d)^(1/3))*3^(1/2)+2*ln((b*x^3+a)^(1/3)-((a*d-b*c)/d)^(1/3))-ln((b*x^3+a )^(2/3)+((a*d-b*c)/d)^(1/3)*(b*x^3+a)^(1/3)+((a*d-b*c)/d)^(2/3)))/d/((a*d- b*c)/d)^(1/3)
Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (114) = 228\).
Time = 0.12 (sec) , antiderivative size = 592, normalized size of antiderivative = 4.08 \[ \int \frac {x^2}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx =\text {Too large to display} \] Input:
integrate(x^2/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")
Output:
[1/6*(3*sqrt(1/3)*(b*c*d - a*d^2)*sqrt((-b*c*d^2 + a*d^3)^(1/3)/(b*c - a*d ))*log((2*b*d^2*x^3 - b*c*d + 3*a*d^2 + 3*sqrt(1/3)*(2*(-b*c*d^2 + a*d^3)^ (2/3)*(b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(b*c*d - a*d^2) + (-b*c*d^2 + a*d^3)^(1/3)*(b*c - a*d))*sqrt((-b*c*d^2 + a*d^3)^(1/3)/(b*c - a*d)) - 3*( -b*c*d^2 + a*d^3)^(2/3)*(b*x^3 + a)^(1/3))/(d*x^3 + c)) + (-b*c*d^2 + a*d^ 3)^(2/3)*log((b*x^3 + a)^(2/3)*d^2 + (-b*c*d^2 + a*d^3)^(1/3)*(b*x^3 + a)^ (1/3)*d + (-b*c*d^2 + a*d^3)^(2/3)) - 2*(-b*c*d^2 + a*d^3)^(2/3)*log((b*x^ 3 + a)^(1/3)*d - (-b*c*d^2 + a*d^3)^(1/3)))/(b*c*d^2 - a*d^3), 1/6*(6*sqrt (1/3)*(b*c*d - a*d^2)*sqrt(-(-b*c*d^2 + a*d^3)^(1/3)/(b*c - a*d))*arctan(s qrt(1/3)*(2*(b*x^3 + a)^(1/3)*d + (-b*c*d^2 + a*d^3)^(1/3))*sqrt(-(-b*c*d^ 2 + a*d^3)^(1/3)/(b*c - a*d))/d) + (-b*c*d^2 + a*d^3)^(2/3)*log((b*x^3 + a )^(2/3)*d^2 + (-b*c*d^2 + a*d^3)^(1/3)*(b*x^3 + a)^(1/3)*d + (-b*c*d^2 + a *d^3)^(2/3)) - 2*(-b*c*d^2 + a*d^3)^(2/3)*log((b*x^3 + a)^(1/3)*d - (-b*c* d^2 + a*d^3)^(1/3)))/(b*c*d^2 - a*d^3)]
\[ \int \frac {x^2}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\int \frac {x^{2}}{\sqrt [3]{a + b x^{3}} \left (c + d x^{3}\right )}\, dx \] Input:
integrate(x**2/(b*x**3+a)**(1/3)/(d*x**3+c),x)
Output:
Integral(x**2/((a + b*x**3)**(1/3)*(c + d*x**3)), x)
Exception generated. \[ \int \frac {x^2}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^2/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.14 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.56 \[ \int \frac {x^2}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=-\frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b c d^{2} - \sqrt {3} a d^{3}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b c d^{2} - a d^{3}\right )}} - \frac {\left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b c - a d\right )}} \] Input:
integrate(x^2/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")
Output:
-(-b*c*d^2 + a*d^3)^(2/3)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d)^(1/3))/(sqrt(3)*b*c*d^2 - sqrt(3)*a*d^3 ) + 1/6*(-b*c*d^2 + a*d^3)^(2/3)*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3) *(-(b*c - a*d)/d)^(1/3) + (-(b*c - a*d)/d)^(2/3))/(b*c*d^2 - a*d^3) - 1/3* (-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)) )/(b*c - a*d)
Time = 3.59 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.43 \[ \int \frac {x^2}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\frac {\ln \left (d\,{\left (b\,x^3+a\right )}^{1/3}-\frac {9\,a\,d^3-9\,b\,c\,d^2}{9\,d^{4/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )}{3\,d^{2/3}\,{\left (a\,d-b\,c\right )}^{1/3}}+\frac {\ln \left (d\,{\left (b\,x^3+a\right )}^{1/3}-\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{36\,d^{4/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,d^{2/3}\,{\left (a\,d-b\,c\right )}^{1/3}}-\frac {\ln \left (d\,{\left (b\,x^3+a\right )}^{1/3}-\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{36\,d^{4/3}\,{\left (a\,d-b\,c\right )}^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,d^{2/3}\,{\left (a\,d-b\,c\right )}^{1/3}} \] Input:
int(x^2/((a + b*x^3)^(1/3)*(c + d*x^3)),x)
Output:
log(d*(a + b*x^3)^(1/3) - (9*a*d^3 - 9*b*c*d^2)/(9*d^(4/3)*(a*d - b*c)^(2/ 3)))/(3*d^(2/3)*(a*d - b*c)^(1/3)) + (log(d*(a + b*x^3)^(1/3) - ((3^(1/2)* 1i - 1)^2*(9*a*d^3 - 9*b*c*d^2))/(36*d^(4/3)*(a*d - b*c)^(2/3)))*(3^(1/2)* 1i - 1))/(6*d^(2/3)*(a*d - b*c)^(1/3)) - (log(d*(a + b*x^3)^(1/3) - ((3^(1 /2)*1i + 1)^2*(9*a*d^3 - 9*b*c*d^2))/(36*d^(4/3)*(a*d - b*c)^(2/3)))*(3^(1 /2)*1i + 1))/(6*d^(2/3)*(a*d - b*c)^(1/3))
\[ \int \frac {x^2}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx=\int \frac {x^{2}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} c +\left (b \,x^{3}+a \right )^{\frac {1}{3}} d \,x^{3}}d x \] Input:
int(x^2/(b*x^3+a)^(1/3)/(d*x^3+c),x)
Output:
int(x**2/((a + b*x**3)**(1/3)*c + (a + b*x**3)**(1/3)*d*x**3),x)