\(\int \frac {2 (3 a q x-2 b p x^3+a p x^5)}{\sqrt [3]{q+p x^4} (b^3 d+c q+3 a b^2 d x^2+(3 a^2 b d+c p) x^4+a^3 d x^6)} \, dx\) [2640]

Optimal result

Integrand size = 76, antiderivative size = 234 \[ \int \frac {2 \left (3 a q x-2 b p x^3+a p x^5\right )}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x^2+\left (3 a^2 b d+c p\right ) x^4+a^3 d x^6\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{c} \sqrt [3]{q+p x^4}}{-2 b \sqrt [3]{d}-2 a \sqrt [3]{d} x^2+\sqrt [3]{c} \sqrt [3]{q+p x^4}}\right )}{c^{2/3} \sqrt [3]{d}}+\frac {\log \left (b \sqrt [3]{d}+a \sqrt [3]{d} x^2+\sqrt [3]{c} \sqrt [3]{q+p x^4}\right )}{c^{2/3} \sqrt [3]{d}}-\frac {\log \left (b^2 d^{2/3}+2 a b d^{2/3} x^2+a^2 d^{2/3} x^4+\left (-b \sqrt [3]{c} \sqrt [3]{d}-a \sqrt [3]{c} \sqrt [3]{d} x^2\right ) \sqrt [3]{q+p x^4}+c^{2/3} \left (q+p x^4\right )^{2/3}\right )}{2 c^{2/3} \sqrt [3]{d}} \]

[Out]

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

Rubi [F]

\[ \int \frac {2 \left (3 a q x-2 b p x^3+a p x^5\right )}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x^2+\left (3 a^2 b d+c p\right ) x^4+a^3 d x^6\right )} \, dx=\int \frac {2 \left (3 a q x-2 b p x^3+a p x^5\right )}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x^2+\left (3 a^2 b d+c p\right ) x^4+a^3 d x^6\right )} \, dx \]

[In]

Int[(2*(3*a*q*x - 2*b*p*x^3 + a*p*x^5))/((q + p*x^4)^(1/3)*(b^3*d + c*q + 3*a*b^2*d*x^2 + (3*a^2*b*d + c*p)*x^
4 + a^3*d*x^6)),x]

[Out]

2*b*p*Defer[Subst][Defer[Int][x/((q + p*x^2)^(1/3)*(-(b^3*d) - c*q - 3*a*b^2*d*x - (3*a^2*b*d + c*p)*x^2 - a^3
*d*x^3)), x], x, x^2] + 3*a*q*Defer[Subst][Defer[Int][1/((q + p*x^2)^(1/3)*(b^3*d + c*q + 3*a*b^2*d*x + (3*a^2
*b*d + c*p)*x^2 + a^3*d*x^3)), x], x, x^2] + a*p*Defer[Subst][Defer[Int][x^2/((q + p*x^2)^(1/3)*(b^3*d + c*q +
3*a*b^2*d*x + (3*a^2*b*d + c*p)*x^2 + a^3*d*x^3)), x], x, x^2]

Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {3 a q x-2 b p x^3+a p x^5}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x^2+\left (3 a^2 b d+c p\right ) x^4+a^3 d x^6\right )} \, dx \\ & = 2 \int \frac {x \left (3 a q-2 b p x^2+a p x^4\right )}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x^2+\left (3 a^2 b d+c p\right ) x^4+a^3 d x^6\right )} \, dx \\ & = \text {Subst}\left (\int \frac {3 a q-2 b p x+a p x^2}{\sqrt [3]{q+p x^2} \left (b^3 d+c q+3 a b^2 d x+\left (3 a^2 b d+c p\right ) x^2+a^3 d x^3\right )} \, dx,x,x^2\right ) \\ & = \text {Subst}\left (\int \left (\frac {2 b p x}{\sqrt [3]{q+p x^2} \left (-b^3 d-c q-3 a b^2 d x-\left (3 a^2 b d+c p\right ) x^2-a^3 d x^3\right )}+\frac {3 a q}{\sqrt [3]{q+p x^2} \left (b^3 d+c q+3 a b^2 d x+\left (3 a^2 b d+c p\right ) x^2+a^3 d x^3\right )}+\frac {a p x^2}{\sqrt [3]{q+p x^2} \left (b^3 d+c q+3 a b^2 d x+\left (3 a^2 b d+c p\right ) x^2+a^3 d x^3\right )}\right ) \, dx,x,x^2\right ) \\ & = (a p) \text {Subst}\left (\int \frac {x^2}{\sqrt [3]{q+p x^2} \left (b^3 d+c q+3 a b^2 d x+\left (3 a^2 b d+c p\right ) x^2+a^3 d x^3\right )} \, dx,x,x^2\right )+(2 b p) \text {Subst}\left (\int \frac {x}{\sqrt [3]{q+p x^2} \left (-b^3 d-c q-3 a b^2 d x-\left (3 a^2 b d+c p\right ) x^2-a^3 d x^3\right )} \, dx,x,x^2\right )+(3 a q) \text {Subst}\left (\int \frac {1}{\sqrt [3]{q+p x^2} \left (b^3 d+c q+3 a b^2 d x+\left (3 a^2 b d+c p\right ) x^2+a^3 d x^3\right )} \, dx,x,x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 8.77 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.88 \[ \int \frac {2 \left (3 a q x-2 b p x^3+a p x^5\right )}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x^2+\left (3 a^2 b d+c p\right ) x^4+a^3 d x^6\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{c} \sqrt [3]{q+p x^4}}{-2 b \sqrt [3]{d}-2 a \sqrt [3]{d} x^2+\sqrt [3]{c} \sqrt [3]{q+p x^4}}\right )+2 \log \left (b \sqrt [3]{d}+a \sqrt [3]{d} x^2+\sqrt [3]{c} \sqrt [3]{q+p x^4}\right )-\log \left (b^2 d^{2/3}+2 a b d^{2/3} x^2+a^2 d^{2/3} x^4-\sqrt [3]{c} \sqrt [3]{d} \left (b+a x^2\right ) \sqrt [3]{q+p x^4}+c^{2/3} \left (q+p x^4\right )^{2/3}\right )}{2 c^{2/3} \sqrt [3]{d}} \]

[In]

Integrate[(2*(3*a*q*x - 2*b*p*x^3 + a*p*x^5))/((q + p*x^4)^(1/3)*(b^3*d + c*q + 3*a*b^2*d*x^2 + (3*a^2*b*d + c
*p)*x^4 + a^3*d*x^6)),x]

[Out]

(2*Sqrt[3]*ArcTan[(Sqrt[3]*c^(1/3)*(q + p*x^4)^(1/3))/(-2*b*d^(1/3) - 2*a*d^(1/3)*x^2 + c^(1/3)*(q + p*x^4)^(1
/3))] + 2*Log[b*d^(1/3) + a*d^(1/3)*x^2 + c^(1/3)*(q + p*x^4)^(1/3)] - Log[b^2*d^(2/3) + 2*a*b*d^(2/3)*x^2 + a
^2*d^(2/3)*x^4 - c^(1/3)*d^(1/3)*(b + a*x^2)*(q + p*x^4)^(1/3) + c^(2/3)*(q + p*x^4)^(2/3)])/(2*c^(2/3)*d^(1/3
))

Maple [F]

\[\int \frac {2 a p \,x^{5}-4 b p \,x^{3}+6 a q x}{\left (p \,x^{4}+q \right )^{\frac {1}{3}} \left (b^{3} d +c q +3 a \,b^{2} d \,x^{2}+\left (3 a^{2} b d +c p \right ) x^{4}+a^{3} d \,x^{6}\right )}d x\]

[In]

int(2*(a*p*x^5-2*b*p*x^3+3*a*q*x)/(p*x^4+q)^(1/3)/(b^3*d+c*q+3*a*b^2*d*x^2+(3*a^2*b*d+c*p)*x^4+a^3*d*x^6),x)

[Out]

int(2*(a*p*x^5-2*b*p*x^3+3*a*q*x)/(p*x^4+q)^(1/3)/(b^3*d+c*q+3*a*b^2*d*x^2+(3*a^2*b*d+c*p)*x^4+a^3*d*x^6),x)

Fricas [F(-1)]

Timed out. \[ \int \frac {2 \left (3 a q x-2 b p x^3+a p x^5\right )}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x^2+\left (3 a^2 b d+c p\right ) x^4+a^3 d x^6\right )} \, dx=\text {Timed out} \]

[In]

integrate(2*(a*p*x^5-2*b*p*x^3+3*a*q*x)/(p*x^4+q)^(1/3)/(b^3*d+c*q+3*a*b^2*d*x^2+(3*a^2*b*d+c*p)*x^4+a^3*d*x^6
),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {2 \left (3 a q x-2 b p x^3+a p x^5\right )}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x^2+\left (3 a^2 b d+c p\right ) x^4+a^3 d x^6\right )} \, dx=\text {Timed out} \]

[In]

integrate(2*(a*p*x**5-2*b*p*x**3+3*a*q*x)/(p*x**4+q)**(1/3)/(b**3*d+c*q+3*a*b**2*d*x**2+(3*a**2*b*d+c*p)*x**4+
a**3*d*x**6),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {2 \left (3 a q x-2 b p x^3+a p x^5\right )}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x^2+\left (3 a^2 b d+c p\right ) x^4+a^3 d x^6\right )} \, dx=\int { \frac {2 \, {\left (a p x^{5} - 2 \, b p x^{3} + 3 \, a q x\right )}}{{\left (a^{3} d x^{6} + 3 \, a b^{2} d x^{2} + {\left (3 \, a^{2} b d + c p\right )} x^{4} + b^{3} d + c q\right )} {\left (p x^{4} + q\right )}^{\frac {1}{3}}} \,d x } \]

[In]

integrate(2*(a*p*x^5-2*b*p*x^3+3*a*q*x)/(p*x^4+q)^(1/3)/(b^3*d+c*q+3*a*b^2*d*x^2+(3*a^2*b*d+c*p)*x^4+a^3*d*x^6
),x, algorithm="maxima")

[Out]

2*integrate((a*p*x^5 - 2*b*p*x^3 + 3*a*q*x)/((a^3*d*x^6 + 3*a*b^2*d*x^2 + (3*a^2*b*d + c*p)*x^4 + b^3*d + c*q)
*(p*x^4 + q)^(1/3)), x)

Giac [F]

\[ \int \frac {2 \left (3 a q x-2 b p x^3+a p x^5\right )}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x^2+\left (3 a^2 b d+c p\right ) x^4+a^3 d x^6\right )} \, dx=\int { \frac {2 \, {\left (a p x^{5} - 2 \, b p x^{3} + 3 \, a q x\right )}}{{\left (a^{3} d x^{6} + 3 \, a b^{2} d x^{2} + {\left (3 \, a^{2} b d + c p\right )} x^{4} + b^{3} d + c q\right )} {\left (p x^{4} + q\right )}^{\frac {1}{3}}} \,d x } \]

[In]

integrate(2*(a*p*x^5-2*b*p*x^3+3*a*q*x)/(p*x^4+q)^(1/3)/(b^3*d+c*q+3*a*b^2*d*x^2+(3*a^2*b*d+c*p)*x^4+a^3*d*x^6
),x, algorithm="giac")

[Out]

integrate(2*(a*p*x^5 - 2*b*p*x^3 + 3*a*q*x)/((a^3*d*x^6 + 3*a*b^2*d*x^2 + (3*a^2*b*d + c*p)*x^4 + b^3*d + c*q)
*(p*x^4 + q)^(1/3)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {2 \left (3 a q x-2 b p x^3+a p x^5\right )}{\sqrt [3]{q+p x^4} \left (b^3 d+c q+3 a b^2 d x^2+\left (3 a^2 b d+c p\right ) x^4+a^3 d x^6\right )} \, dx=\int \frac {2\,a\,p\,x^5-4\,b\,p\,x^3+6\,a\,q\,x}{{\left (p\,x^4+q\right )}^{1/3}\,\left (c\,q+b^3\,d+x^4\,\left (3\,b\,d\,a^2+c\,p\right )+a^3\,d\,x^6+3\,a\,b^2\,d\,x^2\right )} \,d x \]

[In]

int((2*a*p*x^5 - 4*b*p*x^3 + 6*a*q*x)/((q + p*x^4)^(1/3)*(c*q + b^3*d + x^4*(c*p + 3*a^2*b*d) + a^3*d*x^6 + 3*
a*b^2*d*x^2)),x)

[Out]

int((2*a*p*x^5 - 4*b*p*x^3 + 6*a*q*x)/((q + p*x^4)^(1/3)*(c*q + b^3*d + x^4*(c*p + 3*a^2*b*d) + a^3*d*x^6 + 3*
a*b^2*d*x^2)), x)