\(\int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx\) [144]

Optimal result

Integrand size = 22, antiderivative size = 127 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {b (5 b B-6 A c) \sqrt {b x+c x^2}}{8 c^3}-\frac {(5 b B-6 A c) x \sqrt {b x+c x^2}}{12 c^2}+\frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {b^2 (5 b B-6 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}} \] Output:

1/8*b*(-6*A*c+5*B*b)*(c*x^2+b*x)^(1/2)/c^3-1/12*(-6*A*c+5*B*b)*x*(c*x^2+b* 
x)^(1/2)/c^2+1/3*B*x^2*(c*x^2+b*x)^(1/2)/c-1/8*b^2*(-6*A*c+5*B*b)*arctanh( 
c^(1/2)*x/(c*x^2+b*x)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c} x (b+c x) \left (15 b^2 B+4 c^2 x (3 A+2 B x)-2 b c (9 A+5 B x)\right )+6 b^2 (5 b B-6 A c) \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{24 c^{7/2} \sqrt {x (b+c x)}} \] Input:

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

Output:

(Sqrt[c]*x*(b + c*x)*(15*b^2*B + 4*c^2*x*(3*A + 2*B*x) - 2*b*c*(9*A + 5*B* 
x)) + 6*b^2*(5*b*B - 6*A*c)*Sqrt[x]*Sqrt[b + c*x]*ArcTanh[(Sqrt[c]*Sqrt[x] 
)/(Sqrt[b] - Sqrt[b + c*x])])/(24*c^(7/2)*Sqrt[x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1221, 1134, 1160, 1091, 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.

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

\(\Big \downarrow \) 1221

\(\displaystyle \frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {(5 b B-6 A c) \int \frac {x^2}{\sqrt {c x^2+b x}}dx}{6 c}\)

\(\Big \downarrow \) 1134

\(\displaystyle \frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {(5 b B-6 A c) \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \int \frac {x}{\sqrt {c x^2+b x}}dx}{4 c}\right )}{6 c}\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {(5 b B-6 A c) \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \int \frac {1}{\sqrt {c x^2+b x}}dx}{2 c}\right )}{4 c}\right )}{6 c}\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {(5 b B-6 A c) \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{c}\right )}{4 c}\right )}{6 c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {B x^2 \sqrt {b x+c x^2}}{3 c}-\frac {(5 b B-6 A c) \left (\frac {x \sqrt {b x+c x^2}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x+c x^2}}{c}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}}\right )}{4 c}\right )}{6 c}\)

Input:

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

Output:

(B*x^2*Sqrt[b*x + c*x^2])/(3*c) - ((5*b*B - 6*A*c)*((x*Sqrt[b*x + c*x^2])/ 
(2*c) - (3*b*(Sqrt[b*x + c*x^2]/c - (b*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^ 
2]])/c^(3/2)))/(4*c)))/(6*c)
 

Defintions of rubi rules used

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 1091

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

rule 1134

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1)))   Int[(d + e*x)^ 
(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[ 
c*d^2 - b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 
*p]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

rule 1221

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

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(\frac {\left (\frac {3}{2} A \,b^{2} c -\frac {5}{4} B \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\left (-\frac {3 \left (\frac {5 B x}{9}+A \right ) b \,c^{\frac {3}{2}}}{2}+x \left (\frac {2 B x}{3}+A \right ) c^{\frac {5}{2}}+\frac {5 B \sqrt {c}\, b^{2}}{4}\right ) \sqrt {x \left (c x +b \right )}}{2 c^{\frac {7}{2}}}\)	\(82\)
risch	\(-\frac {\left (-8 B \,c^{2} x^{2}-12 A \,c^{2} x +10 B b c x +18 A b c -15 B \,b^{2}\right ) x \left (c x +b \right )}{24 c^{3} \sqrt {x \left (c x +b \right )}}+\frac {b^{2} \left (6 A c -5 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {7}{2}}}\)	\(97\)
default	\(A \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )+B \left (\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )\)	\(172\)

Input:

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

Output:

1/2*((3/2*A*b^2*c-5/4*B*b^3)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(-3/2*(5 
/9*B*x+A)*b*c^(3/2)+x*(2/3*B*x+A)*c^(5/2)+5/4*B*c^(1/2)*b^2)*(x*(c*x+b))^( 
1/2))/c^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.64 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\left [-\frac {3 \, {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{2} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) + {\left (8 \, B c^{3} x^{2} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \, {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c^{4}}\right ] \] Input:

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

Output:

[-1/48*(3*(5*B*b^3 - 6*A*b^2*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x 
)*sqrt(c)) - 2*(8*B*c^3*x^2 + 15*B*b^2*c - 18*A*b*c^2 - 2*(5*B*b*c^2 - 6*A 
*c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/24*(3*(5*B*b^3 - 6*A*b^2*c)*sqrt(-c)*ar 
ctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x + b)) + (8*B*c^3*x^2 + 15*B*b^2*c - 1 
8*A*b*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x)*sqrt(c*x^2 + b*x))/c^4]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.42 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\begin {cases} \frac {3 b^{2} \left (A - \frac {5 B b}{6 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 c^{2}} + \sqrt {b x + c x^{2}} \left (\frac {B x^{2}}{3 c} - \frac {3 b \left (A - \frac {5 B b}{6 c}\right )}{4 c^{2}} + \frac {x \left (A - \frac {5 B b}{6 c}\right )}{2 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {5}{2}}}{5} + \frac {B \left (b x\right )^{\frac {7}{2}}}{7 b}\right )}{b^{3}} & \text {for}\: b \neq 0 \\\tilde {\infty } \left (\frac {A x^{3}}{3} + \frac {B x^{4}}{4}\right ) & \text {otherwise} \end {cases} \] Input:

integrate(x**2*(B*x+A)/(c*x**2+b*x)**(1/2),x)
 

Output:

Piecewise((3*b**2*(A - 5*B*b/(6*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x 
+ c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x)*log(b/(2*c) + x 
)/sqrt(c*(b/(2*c) + x)**2), True))/(8*c**2) + sqrt(b*x + c*x**2)*(B*x**2/( 
3*c) - 3*b*(A - 5*B*b/(6*c))/(4*c**2) + x*(A - 5*B*b/(6*c))/(2*c)), Ne(c, 
0)), (2*(A*(b*x)**(5/2)/5 + B*(b*x)**(7/2)/(7*b))/b**3, Ne(b, 0)), (zoo*(A 
*x**3/3 + B*x**4/4), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.26 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x} B x^{2}}{3 \, c} - \frac {5 \, \sqrt {c x^{2} + b x} B b x}{12 \, c^{2}} + \frac {\sqrt {c x^{2} + b x} A x}{2 \, c} - \frac {5 \, B b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} + \frac {3 \, A b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {5}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} B b^{2}}{8 \, c^{3}} - \frac {3 \, \sqrt {c x^{2} + b x} A b}{4 \, c^{2}} \] Input:

integrate(x^2*(B*x+A)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")
 

Output:

1/3*sqrt(c*x^2 + b*x)*B*x^2/c - 5/12*sqrt(c*x^2 + b*x)*B*b*x/c^2 + 1/2*sqr 
t(c*x^2 + b*x)*A*x/c - 5/16*B*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt 
(c))/c^(7/2) + 3/8*A*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5 
/2) + 5/8*sqrt(c*x^2 + b*x)*B*b^2/c^3 - 3/4*sqrt(c*x^2 + b*x)*A*b/c^2
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (\frac {4 \, B x}{c} - \frac {5 \, B b c - 6 \, A c^{2}}{c^{3}}\right )} x + \frac {3 \, {\left (5 \, B b^{2} - 6 \, A b c\right )}}{c^{3}}\right )} + \frac {{\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {7}{2}}} \] Input:

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

Output:

1/24*sqrt(c*x^2 + b*x)*(2*(4*B*x/c - (5*B*b*c - 6*A*c^2)/c^3)*x + 3*(5*B*b 
^2 - 6*A*b*c)/c^3) + 1/16*(5*B*b^3 - 6*A*b^2*c)*log(abs(2*(sqrt(c)*x - sqr 
t(c*x^2 + b*x))*sqrt(c) + b))/c^(7/2)
 

Mupad [F(-1)]

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

int((x^2*(A + B*x))/(b*x + c*x^2)^(1/2),x)
 

Output:

int((x^2*(A + B*x))/(b*x + c*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.07 \[ \int \frac {x^2 (A+B x)}{\sqrt {b x+c x^2}} \, dx=\frac {-18 \sqrt {x}\, \sqrt {c x +b}\, a b \,c^{2}+12 \sqrt {x}\, \sqrt {c x +b}\, a \,c^{3} x +15 \sqrt {x}\, \sqrt {c x +b}\, b^{3} c -10 \sqrt {x}\, \sqrt {c x +b}\, b^{2} c^{2} x +8 \sqrt {x}\, \sqrt {c x +b}\, b \,c^{3} x^{2}+18 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) a \,b^{2} c -15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{4}}{24 c^{4}} \] Input:

int(x^2*(B*x+A)/(c*x^2+b*x)^(1/2),x)
 

Output:

( - 18*sqrt(x)*sqrt(b + c*x)*a*b*c**2 + 12*sqrt(x)*sqrt(b + c*x)*a*c**3*x 
+ 15*sqrt(x)*sqrt(b + c*x)*b**3*c - 10*sqrt(x)*sqrt(b + c*x)*b**2*c**2*x + 
 8*sqrt(x)*sqrt(b + c*x)*b*c**3*x**2 + 18*sqrt(c)*log((sqrt(b + c*x) + sqr 
t(x)*sqrt(c))/sqrt(b))*a*b**2*c - 15*sqrt(c)*log((sqrt(b + c*x) + sqrt(x)* 
sqrt(c))/sqrt(b))*b**4)/(24*c**4)