\(\int \frac {x^3 (A+B x)}{(d+e x) \sqrt {a+b x+c x^2}} \, dx\) [13]

Optimal result

Integrand size = 30, antiderivative size = 362 \[ \int \frac {x^3 (A+B x)}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=-\frac {\left (6 A c e (4 c d+3 b e)-B \left (24 c^2 d^2+15 b^2 e^2+2 c e (9 b d-8 a e)\right )\right ) \sqrt {a+b x+c x^2}}{24 c^3 e^3}-\frac {(6 B c d+5 b B e-6 A c e) x \sqrt {a+b x+c x^2}}{12 c^2 e^2}+\frac {B x^2 \sqrt {a+b x+c x^2}}{3 c e}+\frac {\left (2 A c e \left (8 c^2 d^2+3 b^2 e^2+4 c e (b d-a e)\right )-B \left (16 c^3 d^3+5 b^3 e^3+6 b c e^2 (b d-2 a e)+8 c^2 d e (b d-a e)\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{7/2} e^4}+\frac {d^3 (B d-A e) \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^4 \sqrt {c d^2-b d e+a e^2}} \] Output:

-1/24*(6*A*c*e*(3*b*e+4*c*d)-B*(24*c^2*d^2+15*b^2*e^2+2*c*e*(-8*a*e+9*b*d) 
))*(c*x^2+b*x+a)^(1/2)/c^3/e^3-1/12*(-6*A*c*e+5*B*b*e+6*B*c*d)*x*(c*x^2+b* 
x+a)^(1/2)/c^2/e^2+1/3*B*x^2*(c*x^2+b*x+a)^(1/2)/c/e+1/16*(2*A*c*e*(8*c^2* 
d^2+3*b^2*e^2+4*c*e*(-a*e+b*d))-B*(16*c^3*d^3+5*b^3*e^3+6*b*c*e^2*(-2*a*e+ 
b*d)+8*c^2*d*e*(-a*e+b*d)))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1 
/2))/c^(7/2)/e^4+d^3*(-A*e+B*d)*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a* 
e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/e^4/(a*e^2-b*d*e+c*d^2)^(1/2)
 

Mathematica [A] (verified)

Time = 2.78 (sec) , antiderivative size = 321, normalized size of antiderivative = 0.89 \[ \int \frac {x^3 (A+B x)}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\frac {\frac {2 e \sqrt {a+x (b+c x)} \left (6 A c e (-4 c d-3 b e+2 c e x)+B \left (15 b^2 e^2-2 c e (-9 b d+8 a e+5 b e x)+4 c^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )\right )}{c^3}+\frac {96 d^3 (B d-A e) \sqrt {-c d^2+b d e-a e^2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )}{c d^2+e (-b d+a e)}+\frac {3 \left (-2 A c e \left (8 c^2 d^2+3 b^2 e^2+4 c e (b d-a e)\right )+B \left (16 c^3 d^3+5 b^3 e^3+6 b c e^2 (b d-2 a e)+8 c^2 d e (b d-a e)\right )\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{c^{7/2}}}{48 e^4} \] Input:

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

Output:

((2*e*Sqrt[a + x*(b + c*x)]*(6*A*c*e*(-4*c*d - 3*b*e + 2*c*e*x) + B*(15*b^ 
2*e^2 - 2*c*e*(-9*b*d + 8*a*e + 5*b*e*x) + 4*c^2*(6*d^2 - 3*d*e*x + 2*e^2* 
x^2))))/c^3 + (96*d^3*(B*d - A*e)*Sqrt[-(c*d^2) + b*d*e - a*e^2]*ArcTan[(S 
qrt[c]*(d + e*x) - e*Sqrt[a + x*(b + c*x)])/Sqrt[-(c*d^2) + e*(b*d - a*e)] 
])/(c*d^2 + e*(-(b*d) + a*e)) + (3*(-2*A*c*e*(8*c^2*d^2 + 3*b^2*e^2 + 4*c* 
e*(b*d - a*e)) + B*(16*c^3*d^3 + 5*b^3*e^3 + 6*b*c*e^2*(b*d - 2*a*e) + 8*c 
^2*d*e*(b*d - a*e)))*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/c^( 
7/2))/(48*e^4)
 

Rubi [A] (verified)

Time = 1.46 (sec) , antiderivative size = 475, normalized size of antiderivative = 1.31, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2153, 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.

\(\displaystyle \int \frac {x^3 (A+B x)}{(d+e x) \sqrt {a+b x+c x^2}} \, dx\)

\(\Big \downarrow \) 2153

\(\displaystyle \int \left (\frac {d^3 (B d-A e)}{e^4 (d+e x) \sqrt {a+b x+c x^2}}-\frac {d^2 (B d-A e)}{e^4 \sqrt {a+b x+c x^2}}+\frac {d x (B d-A e)}{e^3 \sqrt {a+b x+c x^2}}+\frac {x^2 (A e-B d)}{e^2 \sqrt {a+b x+c x^2}}+\frac {B x^3}{e \sqrt {a+b x+c x^2}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\left (3 b^2-4 a c\right ) (B d-A e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2} e^2}-\frac {b d (B d-A e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} e^3}-\frac {d^2 (B d-A e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} e^4}+\frac {d^3 (B d-A e) \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^4 \sqrt {a e^2-b d e+c d^2}}+\frac {3 b \sqrt {a+b x+c x^2} (B d-A e)}{4 c^2 e^2}+\frac {d \sqrt {a+b x+c x^2} (B d-A e)}{c e^3}-\frac {x \sqrt {a+b x+c x^2} (B d-A e)}{2 c e^2}-\frac {b B \left (5 b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{7/2} e}+\frac {B \left (-16 a c+15 b^2-10 b c x\right ) \sqrt {a+b x+c x^2}}{24 c^3 e}+\frac {B x^2 \sqrt {a+b x+c x^2}}{3 c e}\)

Input:

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

Output:

(d*(B*d - A*e)*Sqrt[a + b*x + c*x^2])/(c*e^3) + (3*b*(B*d - A*e)*Sqrt[a + 
b*x + c*x^2])/(4*c^2*e^2) - ((B*d - A*e)*x*Sqrt[a + b*x + c*x^2])/(2*c*e^2 
) + (B*x^2*Sqrt[a + b*x + c*x^2])/(3*c*e) + (B*(15*b^2 - 16*a*c - 10*b*c*x 
)*Sqrt[a + b*x + c*x^2])/(24*c^3*e) - (b*B*(5*b^2 - 12*a*c)*ArcTanh[(b + 2 
*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(7/2)*e) - (d^2*(B*d - A*e 
)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(Sqrt[c]*e^4) - 
(b*d*(B*d - A*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/( 
2*c^(3/2)*e^3) - ((3*b^2 - 4*a*c)*(B*d - A*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[ 
c]*Sqrt[a + b*x + c*x^2])])/(8*c^(5/2)*e^2) + (d^3*(B*d - A*e)*ArcTanh[(b* 
d - 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + 
 c*x^2])])/(e^4*Sqrt[c*d^2 - b*d*e + a*e^2])
 

Defintions of rubi rules used

rule 2009

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

rule 2153

Int[(Px_)*((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b 
_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(d + e* 
x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, n, p}, x] && PolyQ[Px, x] && (IntegerQ[p] || (IntegerQ[2*p] && IntegerQ 
[m] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] && IGtQ[n, 0])
 

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.17

method	result	size
risch	\(-\frac {\left (-8 B \,e^{2} c^{2} x^{2}-12 A \,c^{2} e^{2} x +10 B b c \,e^{2} x +12 B \,c^{2} d e x +18 A b c \,e^{2}+24 A \,c^{2} d e +16 B a c \,e^{2}-15 B \,b^{2} e^{2}-18 B b c d e -24 B \,c^{2} d^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{24 c^{3} e^{3}}-\frac {\frac {\left (8 A a \,c^{2} e^{3}-6 A \,b^{2} c \,e^{3}-8 A b \,c^{2} d \,e^{2}-16 A \,c^{3} d^{2} e -12 B a b c \,e^{3}-8 B a \,c^{2} d \,e^{2}+5 B \,b^{3} e^{3}+6 B \,b^{2} c d \,e^{2}+8 B b \,c^{2} d^{2} e +16 B \,c^{3} d^{3}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{e \sqrt {c}}-\frac {16 d^{3} \left (A e -B d \right ) c^{3} \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}}{16 c^{3} e^{3}}\)	\(425\)
default	\(\frac {e^{2} \left (A e -B d \right ) \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )-d e \left (A e -B d \right ) \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+\frac {A \,d^{2} e \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+B \,e^{3} \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )-\frac {B \,d^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}}{e^{4}}+\frac {d^{3} \left (A e -B d \right ) \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{5} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\)	\(607\)

Input:

int(x^3*(B*x+A)/(e*x+d)/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(-8*B*c^2*e^2*x^2-12*A*c^2*e^2*x+10*B*b*c*e^2*x+12*B*c^2*d*e*x+18*A* 
b*c*e^2+24*A*c^2*d*e+16*B*a*c*e^2-15*B*b^2*e^2-18*B*b*c*d*e-24*B*c^2*d^2)* 
(c*x^2+b*x+a)^(1/2)/c^3/e^3-1/16/c^3/e^3*((8*A*a*c^2*e^3-6*A*b^2*c*e^3-8*A 
*b*c^2*d*e^2-16*A*c^3*d^2*e-12*B*a*b*c*e^3-8*B*a*c^2*d*e^2+5*B*b^3*e^3+6*B 
*b^2*c*d*e^2+8*B*b*c^2*d^2*e+16*B*c^3*d^3)/e*ln((1/2*b+c*x)/c^(1/2)+(c*x^2 
+b*x+a)^(1/2))/c^(1/2)-16*d^3*(A*e-B*d)*c^3/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^ 
(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+ 
c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e 
^2)^(1/2))/(x+d/e)))
 

Fricas [F(-1)]

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

integrate(x^3*(B*x+A)/(e*x+d)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate(x**3*(B*x+A)/(e*x+d)/(c*x**2+b*x+a)**(1/2),x)
 

Output:

Integral(x**3*(A + B*x)/((d + e*x)*sqrt(a + b*x + c*x**2)), x)
 

Maxima [F(-2)]

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

integrate(x^3*(B*x+A)/(e*x+d)/(c*x^2+b*x+a)^(1/2),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((b/e-(2*c*d)/e^2)^2>0)', see `as 
sume?` for
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 (A+B x)}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^3*(B*x+A)/(e*x+d)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

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

int((x^3*(A + B*x))/((d + e*x)*(a + b*x + c*x^2)^(1/2)),x)
 

Output:

int((x^3*(A + B*x))/((d + e*x)*(a + b*x + c*x^2)^(1/2)), x)
 

Reduce [F]

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

int(x^3*(B*x+A)/(e*x+d)/(c*x^2+b*x+a)^(1/2),x)
 

Output:

int(x^3*(B*x+A)/(e*x+d)/(c*x^2+b*x+a)^(1/2),x)