3.28.82 \(\int \frac {(b+a x)^2 (-2 a q+3 b p x^2+a p x^3)}{\sqrt {q+p x^3} (b^4 c+d q^2+4 a b^3 c x+6 a^2 b^2 c x^2+(4 a^3 b c+2 d p q) x^3+a^4 c x^4+d p^2 x^6)} \, dx\) [2782]

3.28.82.1 Optimal result

Integrand size = 103, antiderivative size = 267 \[ \int \frac {(b+a x)^2 \left (-2 a q+3 b p x^2+a p x^3\right )}{\sqrt {q+p x^3} \left (b^4 c+d q^2+4 a b^3 c x+6 a^2 b^2 c x^2+\left (4 a^3 b c+2 d p q\right ) x^3+a^4 c x^4+d p^2 x^6\right )} \, dx=\frac {\arctan \left (\frac {b \sqrt [4]{c}+a \sqrt [4]{c} x}{b \sqrt [4]{c}+a \sqrt [4]{c} x-\sqrt {2} \sqrt [4]{d} \sqrt {q+p x^3}}\right )}{\sqrt {2} c^{3/4} \sqrt [4]{d}}-\frac {\arctan \left (\frac {b \sqrt [4]{c}+a \sqrt [4]{c} x}{b \sqrt [4]{c}+a \sqrt [4]{c} x+\sqrt {2} \sqrt [4]{d} \sqrt {q+p x^3}}\right )}{\sqrt {2} c^{3/4} \sqrt [4]{d}}+\frac {\text {arctanh}\left (\frac {\left (\sqrt {2} b \sqrt [4]{c} \sqrt [4]{d}+\sqrt {2} a \sqrt [4]{c} \sqrt [4]{d} x\right ) \sqrt {q+p x^3}}{b^2 \sqrt {c}+\sqrt {d} q+2 a b \sqrt {c} x+a^2 \sqrt {c} x^2+\sqrt {d} p x^3}\right )}{\sqrt {2} c^{3/4} \sqrt [4]{d}} \]

output

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

3.28.82.2 Mathematica [C] (warning: unable to verify)

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

Time = 16.87 (sec) , antiderivative size = 52633, normalized size of antiderivative = 197.13 \[ \int \frac {(b+a x)^2 \left (-2 a q+3 b p x^2+a p x^3\right )}{\sqrt {q+p x^3} \left (b^4 c+d q^2+4 a b^3 c x+6 a^2 b^2 c x^2+\left (4 a^3 b c+2 d p q\right ) x^3+a^4 c x^4+d p^2 x^6\right )} \, dx=\text {Result too large to show} \]

input

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

output

Result too large to show
 

3.28.82.3 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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {5 a^2 b p x^4}{\sqrt {p x^3+q} \left (a^4 c x^4+4 a^3 b c x^3 \left (\frac {d p q}{2 a^3 b c}+1\right )+6 a^2 b^2 c x^2+4 a b^3 c x+b^4 c \left (\frac {d q^2}{b^4 c}+1\right )+d p^2 x^6\right )}+\frac {7 a b^2 p x^3}{\sqrt {p x^3+q} \left (a^4 c x^4+4 a^3 b c x^3 \left (\frac {d p q}{2 a^3 b c}+1\right )+6 a^2 b^2 c x^2+4 a b^3 c x+b^4 c \left (\frac {d q^2}{b^4 c}+1\right )+d p^2 x^6\right )}+\frac {2 a^3 q x^2 \left (1-\frac {3 b^3 p}{2 a^3 q}\right )}{\sqrt {p x^3+q} \left (-a^4 c x^4-4 a^3 b c x^3 \left (\frac {d p q}{2 a^3 b c}+1\right )-6 a^2 b^2 c x^2-4 a b^3 c x-b^4 c \left (\frac {d q^2}{b^4 c}+1\right )-d p^2 x^6\right )}+\frac {4 a^2 b q x}{\sqrt {p x^3+q} \left (-a^4 c x^4-4 a^3 b c x^3 \left (\frac {d p q}{2 a^3 b c}+1\right )-6 a^2 b^2 c x^2-4 a b^3 c x-b^4 c \left (\frac {d q^2}{b^4 c}+1\right )-d p^2 x^6\right )}+\frac {2 a b^2 q}{\sqrt {p x^3+q} \left (-a^4 c x^4-4 a^3 b c x^3 \left (\frac {d p q}{2 a^3 b c}+1\right )-6 a^2 b^2 c x^2-4 a b^3 c x-b^4 c \left (\frac {d q^2}{b^4 c}+1\right )-d p^2 x^6\right )}+\frac {a^3 p x^5}{\sqrt {p x^3+q} \left (a^4 c x^4+4 a^3 b c x^3 \left (\frac {d p q}{2 a^3 b c}+1\right )+6 a^2 b^2 c x^2+4 a b^3 c x+b^4 c \left (\frac {d q^2}{b^4 c}+1\right )+d p^2 x^6\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 a^2 b q \int \frac {x}{\sqrt {p x^3+q} \left (-d p^2 x^6-a^4 c x^4-4 a^3 b c \left (\frac {d p q}{2 a^3 b c}+1\right ) x^3-6 a^2 b^2 c x^2-4 a b^3 c x-b^4 c \left (\frac {d q^2}{b^4 c}+1\right )\right )}dx+5 a^2 b p \int \frac {x^4}{\sqrt {p x^3+q} \left (d p^2 x^6+a^4 c x^4+4 a^3 b c \left (\frac {d p q}{2 a^3 b c}+1\right ) x^3+6 a^2 b^2 c x^2+4 a b^3 c x+b^4 c \left (\frac {d q^2}{b^4 c}+1\right )\right )}dx+2 a b^2 q \int \frac {1}{\sqrt {p x^3+q} \left (-d p^2 x^6-a^4 c x^4-4 a^3 b c \left (\frac {d p q}{2 a^3 b c}+1\right ) x^3-6 a^2 b^2 c x^2-4 a b^3 c x-b^4 c \left (\frac {d q^2}{b^4 c}+1\right )\right )}dx+7 a b^2 p \int \frac {x^3}{\sqrt {p x^3+q} \left (d p^2 x^6+a^4 c x^4+4 a^3 b c \left (\frac {d p q}{2 a^3 b c}+1\right ) x^3+6 a^2 b^2 c x^2+4 a b^3 c x+b^4 c \left (\frac {d q^2}{b^4 c}+1\right )\right )}dx-\left (3 b^3 p-2 a^3 q\right ) \int \frac {x^2}{\sqrt {p x^3+q} \left (-d p^2 x^6-a^4 c x^4-4 a^3 b c \left (\frac {d p q}{2 a^3 b c}+1\right ) x^3-6 a^2 b^2 c x^2-4 a b^3 c x-b^4 c \left (\frac {d q^2}{b^4 c}+1\right )\right )}dx+a^3 p \int \frac {x^5}{\sqrt {p x^3+q} \left (d p^2 x^6+a^4 c x^4+4 a^3 b c \left (\frac {d p q}{2 a^3 b c}+1\right ) x^3+6 a^2 b^2 c x^2+4 a b^3 c x+b^4 c \left (\frac {d q^2}{b^4 c}+1\right )\right )}dx\)

input

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

output

$Aborted
 

3.28.82.3.1 Defintions of rubi rules used

rule 2009

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

rule 7292

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

rule 7293

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

3.28.82.4 Maple [C] (warning: unable to verify)

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

Time = 2.63 (sec) , antiderivative size = 7274, normalized size of antiderivative = 27.24

method	result	size
default	\(\text {Expression too large to display}\)	\(7274\)
elliptic	\(\text {Expression too large to display}\)	\(7274\)

input

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

output

result too large to display
 

3.28.82.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.28.82.6 Sympy [F]

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

input

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

output

Integral((a*x + b)**2*(a*p*x**3 - 2*a*q + 3*b*p*x**2)/(sqrt(p*x**3 + q)*(a 
**4*c*x**4 + 4*a**3*b*c*x**3 + 6*a**2*b**2*c*x**2 + 4*a*b**3*c*x + b**4*c 
+ d*p**2*x**6 + 2*d*p*q*x**3 + d*q**2)), x)
 

3.28.82.7 Maxima [F]

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

input

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

output

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

3.28.82.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.28.82.9 Mupad [F(-1)]

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

input

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

output

\text{Hanged}