\(\int \frac {x^3}{(d+e x) (a+b x+c x^2)^{3/2}} \, dx\) [39]

Optimal result

Integrand size = 25, antiderivative size = 220 \[ \int \frac {x^3}{(d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 \left (a \left (b^2 d-2 a c d-a b e\right )+\left (b^3 d-3 a b c d-a b^2 e+2 a^2 c e\right ) x\right )}{c \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt {a+b x+c x^2}}+\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2} e}-\frac {d^3 \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e \left (c d^2-b d e+a e^2\right )^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.04 \[ \int \frac {x^3}{(d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 \left (-b^3 d x+a b (-b d+3 c d x+b e x)+a^2 (b e+2 c (d-e x))\right )}{c \left (-b^2+4 a c\right ) \left (c d^2+e (-b d+a e)\right ) \sqrt {a+x (b+c x)}}-\frac {2 d^3 \sqrt {-c d^2+b d e-a e^2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )}{e \left (c d^2+e (-b d+a e)\right )^2}-\frac {\log \left (c e \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{c^{3/2} e} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1264, 27, 1269, 1092, 219, 1154, 219}

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.

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

\(\Big \downarrow \) 1264

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1269

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1154

\(\displaystyle \frac {2 \left (x \left (2 a^2 c e-a b^2 e-3 a b c d+b^3 d\right )+a \left (-a b e-2 a c d+b^2 d\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {-\frac {2 c d^3 \left (b^2-4 a c\right ) \int \frac {1}{4 \left (c d^2-b e d+a e^2\right )-\frac {(b d-2 a e+(2 c d-b e) x)^2}{c x^2+b x+a}}d\left (-\frac {b d-2 a e+(2 c d-b e) x}{\sqrt {c x^2+b x+a}}\right )}{e \left (a e^2-b d e+c d^2\right )}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} e}}{c \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 \left (x \left (2 a^2 c e-a b^2 e-3 a b c d+b^3 d\right )+a \left (-a b e-2 a c d+b^2 d\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}-\frac {\frac {c d^3 \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} e}}{c \left (b^2-4 a c\right )}\)

Input:

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

Output:

(2*(a*(b^2*d - 2*a*c*d - a*b*e) + (b^3*d - 3*a*b*c*d - a*b^2*e + 2*a^2*c*e 
)*x))/(c*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*Sqrt[a + b*x + c*x^2]) - (- 
(((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(S 
qrt[c]*e)) + (c*(b^2 - 4*a*c)*d^3*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*(c*d^2 - b*d*e 
+ a*e^2)^(3/2)))/(c*(b^2 - 4*a*c))
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1264

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d + e*x) 
^m*(f + g*x)^n, a + b*x + c*x^2, x], R = Coeff[PolynomialRemainder[(d + e*x 
)^m*(f + g*x)^n, a + b*x + c*x^2, x], x, 0], S = Coeff[PolynomialRemainder[ 
(d + e*x)^m*(f + g*x)^n, a + b*x + c*x^2, x], x, 1]}, Simp[(b*R - 2*a*S + ( 
2*c*R - b*S)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + S 
imp[1/((p + 1)*(b^2 - 4*a*c))   Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*E 
xpandToSum[((p + 1)*(b^2 - 4*a*c)*Q)/(d + e*x)^m - ((2*p + 3)*(2*c*R - b*S) 
)/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[n, 1] 
 && LtQ[p, -1] && ILtQ[m, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(614\) vs. \(2(202)=404\).

Time = 1.49 (sec) , antiderivative size = 615, normalized size of antiderivative = 2.80

method	result	size
default	\(\frac {2 d^{2} \left (2 c x +b \right )}{e^{3} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}}{e}-\frac {d \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{e^{2}}-\frac {d^{3} \left (\frac {e^{2}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {\left (b e -2 c d \right ) e \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {e^{2} \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\right )}{e^{4}}\)	\(615\)

Input:

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

Output:

2*d^2/e^3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+1/e*(-x/c/(c*x^2+b*x+a 
)^(1/2)-1/2*b/c*(-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/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))-d/e^ 
2*(-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)) 
-d^3/e^4*(1/(a*e^2-b*d*e+c*d^2)*e^2/(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a* 
e^2-b*d*e+c*d^2)/e^2)^(1/2)-(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2)*(2*c*(x+d/e) 
+(b*e-2*c*d)/e)/(4*c*(a*e^2-b*d*e+c*d^2)/e^2-(b*e-2*c*d)^2/e^2)/(c*(x+d/e) 
^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)-1/(a*e^2-b*d*e+c*d 
^2)*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)*(c*(x+d/e)^2+(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1180 vs. \(2 (202) = 404\).

Time = 35.40 (sec) , antiderivative size = 4811, normalized size of antiderivative = 21.87 \[ \int \frac {x^3}{(d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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