\(\int \frac {x (-4 a+3 x)}{\sqrt [3]{x^2 (-a+x)} (a d-d x+x^4)} \, dx\) [2097]

Optimal result

Integrand size = 35, antiderivative size = 152 \[ \int \frac {x (-4 a+3 x)}{\sqrt [3]{x^2 (-a+x)} \left (a d-d x+x^4\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x^2}{x^2+2 \sqrt [3]{d} \sqrt [3]{-a x^2+x^3}}\right )}{d^{2/3}}+\frac {\log \left (a x^2-a \sqrt [3]{d} \sqrt [3]{-a x^2+x^3}\right )}{d^{2/3}}-\frac {\log \left (a^2 x^4+a^2 \sqrt [3]{d} x^2 \sqrt [3]{-a x^2+x^3}+a^2 d^{2/3} \left (-a x^2+x^3\right )^{2/3}\right )}{2 d^{2/3}} \]

[Out]

-3^(1/2)*arctan(3^(1/2)*x^2/(x^2+2*d^(1/3)*(-a*x^2+x^3)^(1/3)))/d^(2/3)+ln(a*x^2-a*d^(1/3)*(-a*x^2+x^3)^(1/3))
/d^(2/3)-1/2*ln(a^2*x^4+a^2*d^(1/3)*x^2*(-a*x^2+x^3)^(1/3)+a^2*d^(2/3)*(-a*x^2+x^3)^(2/3))/d^(2/3)

Rubi [F]

\[ \int \frac {x (-4 a+3 x)}{\sqrt [3]{x^2 (-a+x)} \left (a d-d x+x^4\right )} \, dx=\int \frac {x (-4 a+3 x)}{\sqrt [3]{x^2 (-a+x)} \left (a d-d x+x^4\right )} \, dx \]

[In]

Int[(x*(-4*a + 3*x))/((x^2*(-a + x))^(1/3)*(a*d - d*x + x^4)),x]

[Out]

(-12*a*x^(2/3)*(-a + x)^(1/3)*Defer[Subst][Defer[Int][x^3/((-a + x^3)^(1/3)*(a*d - d*x^3 + x^12)), x], x, x^(1
/3)])/(-((a - x)*x^2))^(1/3) + (9*x^(2/3)*(-a + x)^(1/3)*Defer[Subst][Defer[Int][x^6/((-a + x^3)^(1/3)*(a*d -
d*x^3 + x^12)), x], x, x^(1/3)])/(-((a - x)*x^2))^(1/3)

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{2/3} \sqrt [3]{-a+x}\right ) \int \frac {\sqrt [3]{x} (-4 a+3 x)}{\sqrt [3]{-a+x} \left (a d-d x+x^4\right )} \, dx}{\sqrt [3]{x^2 (-a+x)}} \\ & = \frac {\left (3 x^{2/3} \sqrt [3]{-a+x}\right ) \text {Subst}\left (\int \frac {x^3 \left (-4 a+3 x^3\right )}{\sqrt [3]{-a+x^3} \left (a d-d x^3+x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2 (-a+x)}} \\ & = \frac {\left (3 x^{2/3} \sqrt [3]{-a+x}\right ) \text {Subst}\left (\int \left (-\frac {4 a x^3}{\sqrt [3]{-a+x^3} \left (a d-d x^3+x^{12}\right )}+\frac {3 x^6}{\sqrt [3]{-a+x^3} \left (a d-d x^3+x^{12}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2 (-a+x)}} \\ & = \frac {\left (9 x^{2/3} \sqrt [3]{-a+x}\right ) \text {Subst}\left (\int \frac {x^6}{\sqrt [3]{-a+x^3} \left (a d-d x^3+x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2 (-a+x)}}-\frac {\left (12 a x^{2/3} \sqrt [3]{-a+x}\right ) \text {Subst}\left (\int \frac {x^3}{\sqrt [3]{-a+x^3} \left (a d-d x^3+x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2 (-a+x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.54 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.71 \[ \int \frac {x (-4 a+3 x)}{\sqrt [3]{x^2 (-a+x)} \left (a d-d x+x^4\right )} \, dx=-\frac {a x^{2/3} \sqrt [3]{-a+x} \text {RootSum}\left [a^3-d \text {$\#$1}^3+3 d \text {$\#$1}^6-3 d \text {$\#$1}^9+d \text {$\#$1}^{12}\&,\frac {-\log \left (\sqrt [3]{x}\right )+\log \left (\sqrt [3]{-a+x}-\sqrt [3]{x} \text {$\#$1}\right )}{-\text {$\#$1}+\text {$\#$1}^4}\&\right ]}{d \sqrt [3]{x^2 (-a+x)}} \]

[In]

Integrate[(x*(-4*a + 3*x))/((x^2*(-a + x))^(1/3)*(a*d - d*x + x^4)),x]

[Out]

-((a*x^(2/3)*(-a + x)^(1/3)*RootSum[a^3 - d*#1^3 + 3*d*#1^6 - 3*d*#1^9 + d*#1^12 & , (-Log[x^(1/3)] + Log[(-a
+ x)^(1/3) - x^(1/3)*#1])/(-#1 + #1^4) & ])/(d*(x^2*(-a + x))^(1/3)))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.46

method	result	size
pseudoelliptic	\(-\frac {a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,\textit {\_Z}^{12}-3 d \,\textit {\_Z}^{9}+3 d \,\textit {\_Z}^{6}-d \,\textit {\_Z}^{3}+a^{3}\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (-\left (a -x \right ) x^{2}\right )^{\frac {1}{3}}}{x}\right )}{\textit {\_R}^{4}-\textit {\_R}}\right )}{d}\)	\(70\)

[In]

int(x*(-4*a+3*x)/(x^2*(-a+x))^(1/3)/(x^4+a*d-d*x),x,method=_RETURNVERBOSE)

[Out]

-a*sum(ln((-_R*x+(-(a-x)*x^2)^(1/3))/x)/(_R^4-_R),_R=RootOf(_Z^12*d-3*_Z^9*d+3*_Z^6*d-_Z^3*d+a^3))/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.03 \[ \int \frac {x (-4 a+3 x)}{\sqrt [3]{x^2 (-a+x)} \left (a d-d x+x^4\right )} \, dx=\frac {2 \, \sqrt {3} {\left (d^{2}\right )}^{\frac {1}{6}} d \arctan \left (\frac {\sqrt {3} {\left ({\left (d^{2}\right )}^{\frac {1}{3}} x^{2} + 2 \, {\left (-a x^{2} + x^{3}\right )}^{\frac {1}{3}} d\right )} {\left (d^{2}\right )}^{\frac {1}{6}}}{3 \, d x^{2}}\right ) + 2 \, {\left (d^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (d^{2}\right )}^{\frac {1}{3}} x^{2} - {\left (-a x^{2} + x^{3}\right )}^{\frac {1}{3}} d}{x^{2}}\right ) - {\left (d^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (d^{2}\right )}^{\frac {2}{3}} x^{4} + {\left (-a x^{2} + x^{3}\right )}^{\frac {1}{3}} {\left (d^{2}\right )}^{\frac {1}{3}} d x^{2} + {\left (-a x^{2} + x^{3}\right )}^{\frac {2}{3}} d^{2}}{x^{4}}\right )}{2 \, d^{2}} \]

[In]

integrate(x*(-4*a+3*x)/(x^2*(-a+x))^(1/3)/(x^4+a*d-d*x),x, algorithm="fricas")

[Out]

1/2*(2*sqrt(3)*(d^2)^(1/6)*d*arctan(1/3*sqrt(3)*((d^2)^(1/3)*x^2 + 2*(-a*x^2 + x^3)^(1/3)*d)*(d^2)^(1/6)/(d*x^
2)) + 2*(d^2)^(2/3)*log(((d^2)^(1/3)*x^2 - (-a*x^2 + x^3)^(1/3)*d)/x^2) - (d^2)^(2/3)*log(((d^2)^(2/3)*x^4 + (
-a*x^2 + x^3)^(1/3)*(d^2)^(1/3)*d*x^2 + (-a*x^2 + x^3)^(2/3)*d^2)/x^4))/d^2

Sympy [F(-1)]

Timed out. \[ \int \frac {x (-4 a+3 x)}{\sqrt [3]{x^2 (-a+x)} \left (a d-d x+x^4\right )} \, dx=\text {Timed out} \]

[In]

integrate(x*(-4*a+3*x)/(x**2*(-a+x))**(1/3)/(x**4+a*d-d*x),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x (-4 a+3 x)}{\sqrt [3]{x^2 (-a+x)} \left (a d-d x+x^4\right )} \, dx=\int { -\frac {{\left (4 \, a - 3 \, x\right )} x}{{\left (x^{4} + a d - d x\right )} \left (-{\left (a - x\right )} x^{2}\right )^{\frac {1}{3}}} \,d x } \]

[In]

integrate(x*(-4*a+3*x)/(x^2*(-a+x))^(1/3)/(x^4+a*d-d*x),x, algorithm="maxima")

[Out]

-integrate((4*a - 3*x)*x/((x^4 + a*d - d*x)*(-(a - x)*x^2)^(1/3)), x)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.01 \[ \int \frac {x (-4 a+3 x)}{\sqrt [3]{x^2 (-a+x)} \left (a d-d x+x^4\right )} \, dx=0 \]

[In]

integrate(x*(-4*a+3*x)/(x^2*(-a+x))^(1/3)/(x^4+a*d-d*x),x, algorithm="giac")

[Out]

0

Mupad [F(-1)]

Timed out. \[ \int \frac {x (-4 a+3 x)}{\sqrt [3]{x^2 (-a+x)} \left (a d-d x+x^4\right )} \, dx=\int -\frac {x\,\left (4\,a-3\,x\right )}{{\left (-x^2\,\left (a-x\right )\right )}^{1/3}\,\left (x^4-d\,x+a\,d\right )} \,d x \]

[In]

int(-(x*(4*a - 3*x))/((-x^2*(a - x))^(1/3)*(a*d - d*x + x^4)),x)

[Out]

int(-(x*(4*a - 3*x))/((-x^2*(a - x))^(1/3)*(a*d - d*x + x^4)), x)