\(\int \frac {x^3}{(a+b x+c x^2)^{3/2} (d+e x+f x^2)} \, dx\) [149]

Optimal result

Integrand size = 30, antiderivative size = 651 \[ \int \frac {x^3}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\frac {2 \left (a \left (b^2 d-a b e-2 a (c d-a f)\right )+\left (b^3 d-a b^2 e+2 a^2 c e-a b (3 c d-a f)\right ) x\right )}{\left (b^2-4 a c\right ) \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {a+b x+c x^2}}+\frac {\left (2 d (b d-a e) f+\left (e-\sqrt {e^2-4 d f}\right ) \left (c d^2-b d e+a \left (e^2-d f\right )\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}}}-\frac {\left (2 d (b d-a e) f+\left (e+\sqrt {e^2-4 d f}\right ) \left (c d^2-b d e+a \left (e^2-d f\right )\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}}} \] Output:

2*(a*(b^2*d-a*b*e-2*a*(-a*f+c*d))+(b^3*d-a*b^2*e+2*a^2*c*e-a*b*(-a*f+3*c*d 
))*x)/(-4*a*c+b^2)/((-a*f+c*d)^2-(-a*e+b*d)*(-b*f+c*e))/(c*x^2+b*x+a)^(1/2 
)+1/2*(2*d*(-a*e+b*d)*f+(e-(-4*d*f+e^2)^(1/2))*(c*d^2-b*d*e+a*(-d*f+e^2))) 
*arctanh(1/4*(4*a*f-b*(e-(-4*d*f+e^2)^(1/2))+2*(b*f-c*(e-(-4*d*f+e^2)^(1/2 
)))*x)*2^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2-(-b*f+c*e)*(-4*d*f+e^2)^(1/2)) 
^(1/2)/(c*x^2+b*x+a)^(1/2))*2^(1/2)/(-4*d*f+e^2)^(1/2)/((-a*f+c*d)^2-(-a*e 
+b*d)*(-b*f+c*e))/(c*e^2-2*c*d*f-b*e*f+2*a*f^2-(-b*f+c*e)*(-4*d*f+e^2)^(1/ 
2))^(1/2)-1/2*(2*d*(-a*e+b*d)*f+(e+(-4*d*f+e^2)^(1/2))*(c*d^2-b*d*e+a*(-d* 
f+e^2)))*arctanh(1/4*(4*a*f-b*(e+(-4*d*f+e^2)^(1/2))+2*(b*f-c*(e+(-4*d*f+e 
^2)^(1/2)))*x)*2^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2+(-b*f+c*e)*(-4*d*f+e^2 
)^(1/2))^(1/2)/(c*x^2+b*x+a)^(1/2))*2^(1/2)/(-4*d*f+e^2)^(1/2)/((-a*f+c*d) 
^2-(-a*e+b*d)*(-b*f+c*e))/(c*e^2-2*c*d*f-b*e*f+2*a*f^2+(-b*f+c*e)*(-4*d*f+ 
e^2)^(1/2))^(1/2)
 

Mathematica [C] (verified)

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

Time = 1.73 (sec) , antiderivative size = 690, normalized size of antiderivative = 1.06 \[ \int \frac {x^3}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\frac {2 \left (2 a^3 f+b^3 d x+a b (b d-3 c d x-b e x)+a^2 (-2 c d-b e+2 c e x+b f x)\right )+\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \text {RootSum}\left [b^2 d-a b e+a^2 f-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {b^2 d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+a c d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 a b d e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+a^2 e^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a^2 d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 b \sqrt {c} d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 a \sqrt {c} d e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-c d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+b d e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a e^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+a d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{2 b \sqrt {c} d-a \sqrt {c} e-4 c d \text {$\#$1}-b e \text {$\#$1}+2 a f \text {$\#$1}+3 \sqrt {c} e \text {$\#$1}^2-2 f \text {$\#$1}^3}\&\right ]}{\left (b^2-4 a c\right ) \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right ) \sqrt {a+x (b+c x)}} \] Input:

Integrate[x^3/((a + b*x + c*x^2)^(3/2)*(d + e*x + f*x^2)),x]
 

Output:

(2*(2*a^3*f + b^3*d*x + a*b*(b*d - 3*c*d*x - b*e*x) + a^2*(-2*c*d - b*e + 
2*c*e*x + b*f*x)) + (b^2 - 4*a*c)*Sqrt[a + x*(b + c*x)]*RootSum[b^2*d - a* 
b*e + a^2*f - 4*b*Sqrt[c]*d*#1 + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^2 + b*e*#1^2 
- 2*a*f*#1^2 - 2*Sqrt[c]*e*#1^3 + f*#1^4 & , (b^2*d^2*Log[-(Sqrt[c]*x) + S 
qrt[a + b*x + c*x^2] - #1] + a*c*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x 
^2] - #1] - 2*a*b*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + a^2 
*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a^2*d*f*Log[-(Sqrt[c 
]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*b*Sqrt[c]*d^2*Log[-(Sqrt[c]*x) + Sq 
rt[a + b*x + c*x^2] - #1]*#1 + 2*a*Sqrt[c]*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + 
 b*x + c*x^2] - #1]*#1 - c*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - 
#1]*#1^2 + b*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 - a*e 
^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + a*d*f*Log[-(Sqrt[ 
c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2)/(2*b*Sqrt[c]*d - a*Sqrt[c]*e - 4 
*c*d*#1 - b*e*#1 + 2*a*f*#1 + 3*Sqrt[c]*e*#1^2 - 2*f*#1^3) & ])/((b^2 - 4* 
a*c)*(c^2*d^2 - b*c*d*e + f*(b^2*d - a*b*e + a^2*f) + a*c*(e^2 - 2*d*f))*S 
qrt[a + x*(b + c*x)])
 

Rubi [A] (warning: unable to verify)

Time = 7.19 (sec) , antiderivative size = 779, normalized size of antiderivative = 1.20, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {7279, 2009}

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.

\(\displaystyle \int \frac {x^3}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {x \left (e^2-d f\right )+d e}{f^2 \left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )}-\frac {e}{f^2 \left (a+b x+c x^2\right )^{3/2}}+\frac {x}{f \left (a+b x+c x^2\right )^{3/2}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (\left (e-\sqrt {e^2-4 d f}\right ) \left (a \left (e^2-d f\right )-b d e+c d^2\right )+2 d f (b d-a e)\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {\left (\left (\sqrt {e^2-4 d f}+e\right ) \left (a \left (e^2-d f\right )-b d e+c d^2\right )+2 d f (b d-a e)\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right ) \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac {2 \left (c x \left (\left (e^2-d f\right ) (a b f-2 a c e+b c d)-d e \left (-c (2 a f+b e)+b^2 f+2 c^2 d\right )\right )-\left (a d f-a e^2+b d e\right ) \left (-c (2 a f+b e)+b^2 f+2 c^2 d\right )+c d e (a b f-2 a c e+b c d)\right )}{f^2 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left ((c d-a f)^2-(b d-a e) (c e-b f)\right )}+\frac {2 e (b+2 c x)}{f^2 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 (2 a+b x)}{f \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\)

Input:

Int[x^3/((a + b*x + c*x^2)^(3/2)*(d + e*x + f*x^2)),x]
 

Output:

(2*(2*a + b*x))/((b^2 - 4*a*c)*f*Sqrt[a + b*x + c*x^2]) + (2*e*(b + 2*c*x) 
)/((b^2 - 4*a*c)*f^2*Sqrt[a + b*x + c*x^2]) + (2*(c*d*e*(b*c*d - 2*a*c*e + 
 a*b*f) - (b*d*e - a*e^2 + a*d*f)*(2*c^2*d + b^2*f - c*(b*e + 2*a*f)) + c* 
((b*c*d - 2*a*c*e + a*b*f)*(e^2 - d*f) - d*e*(2*c^2*d + b^2*f - c*(b*e + 2 
*a*f)))*x))/((b^2 - 4*a*c)*f^2*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*S 
qrt[a + b*x + c*x^2]) + ((2*d*(b*d - a*e)*f + (e - Sqrt[e^2 - 4*d*f])*(c*d 
^2 - b*d*e + a*(e^2 - d*f)))*ArcTanh[(4*a*f - b*(e - Sqrt[e^2 - 4*d*f]) + 
2*(b*f - c*(e - Sqrt[e^2 - 4*d*f]))*x)/(2*Sqrt[2]*Sqrt[c*e^2 - 2*c*d*f - b 
*e*f + 2*a*f^2 - (c*e - b*f)*Sqrt[e^2 - 4*d*f]]*Sqrt[a + b*x + c*x^2])])/( 
Sqrt[2]*Sqrt[e^2 - 4*d*f]*((c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*Sqrt[c 
*e^2 - 2*c*d*f - b*e*f + 2*a*f^2 - (c*e - b*f)*Sqrt[e^2 - 4*d*f]]) - ((2*d 
*(b*d - a*e)*f + (e + Sqrt[e^2 - 4*d*f])*(c*d^2 - b*d*e + a*(e^2 - d*f)))* 
ArcTanh[(4*a*f - b*(e + Sqrt[e^2 - 4*d*f]) + 2*(b*f - c*(e + Sqrt[e^2 - 4* 
d*f]))*x)/(2*Sqrt[2]*Sqrt[c*e^2 - 2*c*d*f - b*e*f + 2*a*f^2 + (c*e - b*f)* 
Sqrt[e^2 - 4*d*f]]*Sqrt[a + b*x + c*x^2])])/(Sqrt[2]*Sqrt[e^2 - 4*d*f]*((c 
*d - a*f)^2 - (b*d - a*e)*(c*e - b*f))*Sqrt[c*e^2 - 2*c*d*f - b*e*f + 2*a* 
f^2 + (c*e - b*f)*Sqrt[e^2 - 4*d*f]])
 

Defintions of rubi rules used

rule 2009

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

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2082\) vs. \(2(602)=1204\).

Time = 2.77 (sec) , antiderivative size = 2083, normalized size of antiderivative = 3.20

method	result	size
default	\(\text {Expression too large to display}\)	\(2083\)

Input:

int(x^3/(c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
 

Output:

1/f*(-1/c/(c*x^2+b*x+a)^(1/2)-b/c*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2 
))-1/2/f^3*(d*f*(-4*d*f+e^2)^(1/2)-e^2*(-4*d*f+e^2)^(1/2)+3*d*e*f-e^3)/(-4 
*d*f+e^2)^(1/2)*(2/(-f*b*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2 
-b*e*f-2*d*f*c+c*e^2)*f^2/(c*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2+1/f*(-c*(- 
4*d*f+e^2)^(1/2)+f*b-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*(-f*b*(-4*d 
*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(1/ 
2)-2*f*(-c*(-4*d*f+e^2)^(1/2)+f*b-c*e)/(-f*b*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^ 
2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)*(2*c*(x+1/2*(e+(-4*d*f+e^2)^(1/2 
))/f)+1/f*(-c*(-4*d*f+e^2)^(1/2)+f*b-c*e))/(2*c*(-f*b*(-4*d*f+e^2)^(1/2)+( 
-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2-1/f^2*(-c*(-4*d*f+e 
^2)^(1/2)+f*b-c*e)^2)/(c*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2+1/f*(-c*(-4*d* 
f+e^2)^(1/2)+f*b-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*(-f*b*(-4*d*f+e 
^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(1/2)-2 
/(-f*b*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e 
^2)*f^2*2^(1/2)/((-f*b*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b 
*e*f-2*d*f*c+c*e^2)/f^2)^(1/2)*ln(((-f*b*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^( 
1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2+1/f*(-c*(-4*d*f+e^2)^(1/2)+f*b-c 
*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^(1/2)*((-f*b*(-4*d*f+e^2)^(1/2) 
+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(1/2)*(4*c*(x+1/ 
2*(e+(-4*d*f+e^2)^(1/2))/f)^2+4/f*(-c*(-4*d*f+e^2)^(1/2)+f*b-c*e)*(x+1/...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(x^3/(c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {x^3}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\int \frac {x^{3}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}} \left (d + e x + f x^{2}\right )}\, dx \] Input:

integrate(x**3/(c*x**2+b*x+a)**(3/2)/(f*x**2+e*x+d),x)
 

Output:

Integral(x**3/((a + b*x + c*x**2)**(3/2)*(d + e*x + f*x**2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^3/(c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d),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(4*d*f-e^2>0)', see `assume?` for 
 more deta
 

Giac [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(x^3/(c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d),x, algorithm="giac")
 

Output:

Timed out
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\int \frac {x^3}{{\left (c\,x^2+b\,x+a\right )}^{3/2}\,\left (f\,x^2+e\,x+d\right )} \,d x \] Input:

int(x^3/((a + b*x + c*x^2)^(3/2)*(d + e*x + f*x^2)),x)
 

Output:

int(x^3/((a + b*x + c*x^2)^(3/2)*(d + e*x + f*x^2)), x)
 

Reduce [F]

\[ \int \frac {x^3}{\left (a+b x+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx=\int \frac {x^{3}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (f \,x^{2}+e x +d \right )}d x \] Input:

int(x^3/(c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d),x)
 

Output:

int(x^3/(c*x^2+b*x+a)^(3/2)/(f*x^2+e*x+d),x)