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

3.27.40.1 Optimal result

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

output

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)
 

3.27.40.2 Mathematica [A] (verified)

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

input

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]
 

output

(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))
 

3.27.40.3 Rubi [F]

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.

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2028

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

\(\Big \downarrow \) 7266

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

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]
 

output

$Aborted
 

3.27.40.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2028

Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.) + (c_.)*(x_)^(t_.))^(p_.), 
x_Symbol] :> Int[x^(p*r)*(a + b*x^(s - r) + c*x^(t - r))^p*Fx, x] /; FreeQ[ 
{a, b, c, r, s, t}, x] && IntegerQ[p] && PosQ[s - r] && PosQ[t - r] &&  !(E 
qQ[p, 1] && EqQ[u, 1])
 

rule 7266

Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1)   Subst[Int[SubstFor[x^(m 
+ 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function 
OfQ[x^(m + 1), u, x]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.27.40.4 Maple [F]

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

input

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)
 

output

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)
 

3.27.40.5 Fricas [F(-1)]

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

input

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")
 

output

Timed out
 

3.27.40.6 Sympy [F(-1)]

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

input

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)
 

output

Timed out
 

3.27.40.7 Maxima [F]

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

input

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")
 

output

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)
 

3.27.40.8 Giac [F]

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

input

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")
 

output

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)
 

3.27.40.9 Mupad [F(-1)]

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

input

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)
 

output

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)