\(\int \frac {x^2 (A+B x+C x^2)}{(b x+c x^2)^{3/2}} \, dx\) [1]

Optimal result

Integrand size = 27, antiderivative size = 136 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \left (b B c-A c^2-b^2 C\right ) x}{c^3 \sqrt {b x+c x^2}}+\frac {(4 B c-7 b C) \sqrt {b x+c x^2}}{4 c^3}+\frac {C x \sqrt {b x+c x^2}}{2 c^2}-\frac {\left (12 b B c-8 A c^2-15 b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{7/2}} \] Output:

2*(-A*c^2+B*b*c-C*b^2)*x/c^3/(c*x^2+b*x)^(1/2)+1/4*(4*B*c-7*C*b)*(c*x^2+b* 
x)^(1/2)/c^3+1/2*C*x*(c*x^2+b*x)^(1/2)/c^2-1/4*(-8*A*c^2+12*B*b*c-15*C*b^2 
)*arctanh(c^(1/2)*x/(c*x^2+b*x)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {x^{3/2} \left (\sqrt {c} \sqrt {x} (b+c x) \left (-15 b^2 C+b c (12 B-5 C x)+2 c^2 (-4 A+x (2 B+C x))\right )+2 \left (-12 b B c+8 A c^2+15 b^2 C\right ) (b+c x)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{4 c^{7/2} (x (b+c x))^{3/2}} \] Input:

Integrate[(x^2*(A + B*x + C*x^2))/(b*x + c*x^2)^(3/2),x]
 

Output:

(x^(3/2)*(Sqrt[c]*Sqrt[x]*(b + c*x)*(-15*b^2*C + b*c*(12*B - 5*C*x) + 2*c^ 
2*(-4*A + x*(2*B + C*x))) + 2*(-12*b*B*c + 8*A*c^2 + 15*b^2*C)*(b + c*x)^( 
3/2)*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])]))/(4*c^(7/2)*(x 
*(b + c*x))^(3/2))
 

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.

\(\displaystyle \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (b x+c x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 2165

\(\displaystyle \frac {2 x^2 \text {PolynomialRemainder}\left [A+B x+C x^2,0,x\right ]}{b \sqrt {b x+c x^2}}-\frac {2 \int \frac {b x \text {PolynomialRemainder}\left [C x^2+B x+A,0,x\right ]}{\sqrt {c x^2+b x}}dx}{b^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 x^2 \text {PolynomialRemainder}\left [A+B x+C x^2,0,x\right ]}{b \sqrt {b x+c x^2}}-\frac {2 \int \frac {x \text {PolynomialRemainder}\left [C x^2+B x+A,0,x\right ]}{\sqrt {c x^2+b x}}dx}{b}\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {2 x^2 \text {PolynomialRemainder}\left [A+B x+C x^2,0,x\right ]}{b \sqrt {b x+c x^2}}-\frac {2 \sqrt {x} \sqrt {b+c x} \int \frac {\sqrt {x} \text {PolynomialRemainder}\left [C x^2+B x+A,0,x\right ]}{\sqrt {b+c x}}dx}{b \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 7284

\(\displaystyle \frac {2 x^2 \text {PolynomialRemainder}\left [A+B x+C x^2,0,x\right ]}{b \sqrt {b x+c x^2}}-\frac {4 \sqrt {x} \sqrt {b+c x} \int \frac {x \text {PolynomialRemainder}\left [C x^2+B x+A,0,x\right ]}{\sqrt {b+c x}}d\sqrt {x}}{b \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 7299

\(\displaystyle \frac {2 x^2 \text {PolynomialRemainder}\left [A+B x+C x^2,0,x\right ]}{b \sqrt {b x+c x^2}}-\frac {4 \sqrt {x} \sqrt {b+c x} \int \frac {x \text {PolynomialRemainder}\left [C x^2+B x+A,0,x\right ]}{\sqrt {b+c x}}d\sqrt {x}}{b \sqrt {b x+c x^2}}\)

Input:

Int[(x^2*(A + B*x + C*x^2))/(b*x + c*x^2)^(3/2),x]
 

Output:

$Aborted
 

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 2165

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, a*e + c*d*x, x], R = Pol 
ynomialRemainder[Pq, a*e + c*d*x, x]}, Simp[R*(2*c*d - b*e)*(d + e*x)^m*((a 
 + b*x + c*x^2)^(p + 1)/(e*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b 
^2 - 4*a*c))   Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[ 
d*e*(p + 1)*(b^2 - 4*a*c)*Qx - R*(2*c*d - b*e)*(m + 2*p + 2), x], x], x]] / 
; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c 
*d^2 - b*d*e + a*e^2, 0] && ILtQ[p + 1/2, 0] && GtQ[m, 0]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 7284

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.74

method	result	size
pseudoelliptic	\(\frac {3 b x \left (-\frac {5 C x}{12}+B \right ) c^{\frac {3}{2}}-2 \left (-\frac {1}{4} C \,x^{2}-\frac {1}{2} B x +A \right ) x \,c^{\frac {5}{2}}-\frac {15 C \sqrt {c}\, b^{2} x}{4}+2 \left (\frac {15}{8} b^{2} C -\frac {3}{2} B b c +A \,c^{2}\right ) \sqrt {x \left (c x +b \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{c^{\frac {7}{2}} \sqrt {x \left (c x +b \right )}}\)	\(101\)
risch	\(\frac {\left (2 C x c +4 B c -7 C b \right ) x \left (c x +b \right )}{4 c^{3} \sqrt {x \left (c x +b \right )}}+\frac {8 A \,c^{\frac {3}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )+\frac {15 b^{2} C \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}-12 B b \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )-\frac {16 \left (A \,c^{2}-B b c +b^{2} C \right ) \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{c \left (\frac {b}{c}+x \right )}}{8 c^{3}}\)	\(190\)
default	\(A \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )+B \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )+C \left (\frac {x^{3}}{2 c \sqrt {c \,x^{2}+b x}}-\frac {5 b \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )}{4 c}\right )\)	\(363\)

Input:

int(x^2*(C*x^2+B*x+A)/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/(x*(c*x+b))^(1/2)/c^(7/2)*(3/2*b*x*(-5/12*C*x+B)*c^(3/2)-(-1/4*C*x^2-1/2 
*B*x+A)*x*c^(5/2)-15/8*C*c^(1/2)*b^2*x+(15/8*b^2*C-3/2*B*b*c+A*c^2)*(x*(c* 
x+b))^(1/2)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.20 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (b x+c x^2\right )^{3/2}} \, dx=\left [\frac {{\left (15 \, C b^{3} - 12 \, B b^{2} c + 8 \, A b c^{2} + {\left (15 \, C b^{2} c - 12 \, B b c^{2} + 8 \, A c^{3}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, C c^{3} x^{2} - 15 \, C b^{2} c + 12 \, B b c^{2} - 8 \, A c^{3} - {\left (5 \, C b c^{2} - 4 \, B c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{8 \, {\left (c^{5} x + b c^{4}\right )}}, -\frac {{\left (15 \, C b^{3} - 12 \, B b^{2} c + 8 \, A b c^{2} + {\left (15 \, C b^{2} c - 12 \, B b c^{2} + 8 \, A c^{3}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) - {\left (2 \, C c^{3} x^{2} - 15 \, C b^{2} c + 12 \, B b c^{2} - 8 \, A c^{3} - {\left (5 \, C b c^{2} - 4 \, B c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{4 \, {\left (c^{5} x + b c^{4}\right )}}\right ] \] Input:

integrate(x^2*(C*x^2+B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")
 

Output:

[1/8*((15*C*b^3 - 12*B*b^2*c + 8*A*b*c^2 + (15*C*b^2*c - 12*B*b*c^2 + 8*A* 
c^3)*x)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(2*C*c^3* 
x^2 - 15*C*b^2*c + 12*B*b*c^2 - 8*A*c^3 - (5*C*b*c^2 - 4*B*c^3)*x)*sqrt(c* 
x^2 + b*x))/(c^5*x + b*c^4), -1/4*((15*C*b^3 - 12*B*b^2*c + 8*A*b*c^2 + (1 
5*C*b^2*c - 12*B*b*c^2 + 8*A*c^3)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqr 
t(-c)/(c*x + b)) - (2*C*c^3*x^2 - 15*C*b^2*c + 12*B*b*c^2 - 8*A*c^3 - (5*C 
*b*c^2 - 4*B*c^3)*x)*sqrt(c*x^2 + b*x))/(c^5*x + b*c^4)]
                                                                                    
                                                                                    
 

Sympy [F]

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

integrate(x**2*(C*x**2+B*x+A)/(c*x**2+b*x)**(3/2),x)
 

Output:

Integral(x**2*(A + B*x + C*x**2)/(x*(b + c*x))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.54 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {C x^{3}}{2 \, \sqrt {c x^{2} + b x} c} - \frac {5 \, C b x^{2}}{4 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {B x^{2}}{\sqrt {c x^{2} + b x} c} - \frac {15 \, C b^{2} x}{4 \, \sqrt {c x^{2} + b x} c^{3}} + \frac {3 \, B b x}{\sqrt {c x^{2} + b x} c^{2}} - \frac {2 \, A x}{\sqrt {c x^{2} + b x} c} + \frac {15 \, C b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {7}{2}}} - \frac {3 \, B b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} + \frac {A \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} \] Input:

integrate(x^2*(C*x^2+B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")
 

Output:

1/2*C*x^3/(sqrt(c*x^2 + b*x)*c) - 5/4*C*b*x^2/(sqrt(c*x^2 + b*x)*c^2) + B* 
x^2/(sqrt(c*x^2 + b*x)*c) - 15/4*C*b^2*x/(sqrt(c*x^2 + b*x)*c^3) + 3*B*b*x 
/(sqrt(c*x^2 + b*x)*c^2) - 2*A*x/(sqrt(c*x^2 + b*x)*c) + 15/8*C*b^2*log(2* 
c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) - 3/2*B*b*log(2*c*x + b + 2 
*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) + A*log(2*c*x + b + 2*sqrt(c*x^2 + b*x 
)*sqrt(c))/c^(3/2)
 

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.09 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (\frac {2 \, C x}{c^{2}} - \frac {7 \, C b c^{5} - 4 \, B c^{6}}{c^{8}}\right )} - \frac {{\left (15 \, C b^{2} - 12 \, B b c + 8 \, A c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {7}{2}}} - \frac {2 \, {\left (C b^{3} \sqrt {c} - B b^{2} c^{\frac {3}{2}} + A b c^{\frac {5}{2}}\right )}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b\right )} c^{4}} \] Input:

integrate(x^2*(C*x^2+B*x+A)/(c*x^2+b*x)^(3/2),x, algorithm="giac")
 

Output:

1/4*sqrt(c*x^2 + b*x)*(2*C*x/c^2 - (7*C*b*c^5 - 4*B*c^6)/c^8) - 1/8*(15*C* 
b^2 - 12*B*b*c + 8*A*c^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c 
) + b))/c^(7/2) - 2*(C*b^3*sqrt(c) - B*b^2*c^(3/2) + A*b*c^(5/2))/(((sqrt( 
c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b)*c^4)
 

Mupad [F(-1)]

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

int((x^2*(A + B*x + C*x^2))/(b*x + c*x^2)^(3/2),x)
 

Output:

int((x^2*(A + B*x + C*x^2))/(b*x + c*x^2)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {8 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) a c +3 \sqrt {c}\, \sqrt {c x +b}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{2}-8 \sqrt {c}\, \sqrt {c x +b}\, a c -\sqrt {c}\, \sqrt {c x +b}\, b^{2}-8 \sqrt {x}\, a \,c^{2}-3 \sqrt {x}\, b^{2} c -\sqrt {x}\, b \,c^{2} x +2 \sqrt {x}\, c^{3} x^{2}}{4 \sqrt {c x +b}\, c^{3}} \] Input:

int(x^2*(C*x^2+B*x+A)/(c*x^2+b*x)^(3/2),x)
 

Output:

(8*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b))*a* 
c + 3*sqrt(c)*sqrt(b + c*x)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b)) 
*b**2 - 8*sqrt(c)*sqrt(b + c*x)*a*c - sqrt(c)*sqrt(b + c*x)*b**2 - 8*sqrt( 
x)*a*c**2 - 3*sqrt(x)*b**2*c - sqrt(x)*b*c**2*x + 2*sqrt(x)*c**3*x**2)/(4* 
sqrt(b + c*x)*c**3)