\(\int \frac {x^2 (c+d x)}{\sqrt {a x+b x^2}} \, dx\) [126]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\frac {\sqrt {b} x (a+b x) \left (15 a^2 d+4 b^2 x (3 c+2 d x)-2 a b (9 c+5 d x)\right )+6 a^2 (-6 b c+5 a d) \sqrt {x} \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{24 b^{7/2} \sqrt {x (a+b x)}} \] Input:

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

Output:

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

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1134

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 = 0.48 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.63

method	result	size
pseudoelliptic	\(-\frac {5 \left (a^{2} \left (a d -\frac {6 b c}{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )-\left (-\frac {6 a \left (\frac {5 d x}{9}+c \right ) b^{\frac {3}{2}}}{5}+\frac {4 x \left (\frac {2 d x}{3}+c \right ) b^{\frac {5}{2}}}{5}+\sqrt {b}\, a^{2} d \right ) \sqrt {x \left (b x +a \right )}\right )}{8 b^{\frac {7}{2}}}\)	\(80\)
risch	\(\frac {\left (8 b^{2} d \,x^{2}-10 a b d x +12 b^{2} c x +15 a^{2} d -18 a b c \right ) x \left (b x +a \right )}{24 b^{3} \sqrt {x \left (b x +a \right )}}-\frac {a^{2} \left (5 a d -6 b c \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{16 b^{\frac {7}{2}}}\)	\(97\)
default	\(c \left (\frac {x \sqrt {b \,x^{2}+a x}}{2 b}-\frac {3 a \left (\frac {\sqrt {b \,x^{2}+a x}}{b}-\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+d \left (\frac {x^{2} \sqrt {b \,x^{2}+a x}}{3 b}-\frac {5 a \left (\frac {x \sqrt {b \,x^{2}+a x}}{2 b}-\frac {3 a \left (\frac {\sqrt {b \,x^{2}+a x}}{b}-\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )\)	\(172\)

Input:

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

Output:

-5/8/b^(7/2)*(a^2*(a*d-6/5*b*c)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))-(-6/5 
*a*(5/9*d*x+c)*b^(3/2)+4/5*x*(2/3*d*x+c)*b^(5/2)+b^(1/2)*a^2*d)*(x*(b*x+a) 
)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.65 \[ \int \frac {x^2 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\left [-\frac {3 \, {\left (6 \, a^{2} b c - 5 \, a^{3} d\right )} \sqrt {b} \log \left (2 \, b x + a - 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (8 \, b^{3} d x^{2} - 18 \, a b^{2} c + 15 \, a^{2} b d + 2 \, {\left (6 \, b^{3} c - 5 \, a b^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{48 \, b^{4}}, -\frac {3 \, {\left (6 \, a^{2} b c - 5 \, a^{3} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) - {\left (8 \, b^{3} d x^{2} - 18 \, a b^{2} c + 15 \, a^{2} b d + 2 \, {\left (6 \, b^{3} c - 5 \, a b^{2} d\right )} x\right )} \sqrt {b x^{2} + a x}}{24 \, b^{4}}\right ] \] Input:

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

Output:

[-1/48*(3*(6*a^2*b*c - 5*a^3*d)*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a*x 
)*sqrt(b)) - 2*(8*b^3*d*x^2 - 18*a*b^2*c + 15*a^2*b*d + 2*(6*b^3*c - 5*a*b 
^2*d)*x)*sqrt(b*x^2 + a*x))/b^4, -1/24*(3*(6*a^2*b*c - 5*a^3*d)*sqrt(-b)*a 
rctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x + a)) - (8*b^3*d*x^2 - 18*a*b^2*c + 
15*a^2*b*d + 2*(6*b^3*c - 5*a*b^2*d)*x)*sqrt(b*x^2 + a*x))/b^4]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

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

integrate(x**2*(d*x+c)/(b*x**2+a*x)**(1/2),x)
 

Output:

Piecewise((3*a**2*(-5*a*d/(6*b) + c)*Piecewise((log(a + 2*sqrt(b)*sqrt(a*x 
 + b*x**2) + 2*b*x)/sqrt(b), Ne(a**2/b, 0)), ((a/(2*b) + x)*log(a/(2*b) + 
x)/sqrt(b*(a/(2*b) + x)**2), True))/(8*b**2) + sqrt(a*x + b*x**2)*(-3*a*(- 
5*a*d/(6*b) + c)/(4*b**2) + d*x**2/(3*b) + x*(-5*a*d/(6*b) + c)/(2*b)), Ne 
(b, 0)), (2*(c*(a*x)**(5/2)/5 + d*(a*x)**(7/2)/(7*a))/a**3, Ne(a, 0)), (zo 
o*(c*x**3/3 + d*x**4/4), True))
 

Maxima [A] (verification not implemented)

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

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

Output:

1/3*sqrt(b*x^2 + a*x)*d*x^2/b + 1/2*sqrt(b*x^2 + a*x)*c*x/b - 5/12*sqrt(b* 
x^2 + a*x)*a*d*x/b^2 + 3/8*a^2*c*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt( 
b))/b^(5/2) - 5/16*a^3*d*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7 
/2) - 3/4*sqrt(b*x^2 + a*x)*a*c/b^2 + 5/8*sqrt(b*x^2 + a*x)*a^2*d/b^3
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\frac {1}{24} \, \sqrt {b x^{2} + a x} {\left (2 \, {\left (\frac {4 \, d x}{b} + \frac {6 \, b^{2} c - 5 \, a b d}{b^{3}}\right )} x - \frac {3 \, {\left (6 \, a b c - 5 \, a^{2} d\right )}}{b^{3}}\right )} - \frac {{\left (6 \, a^{2} b c - 5 \, a^{3} d\right )} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{16 \, b^{\frac {7}{2}}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

int((x^2*(c + d*x))/(a*x + b*x^2)^(1/2),x)
 

Output:

int((x^2*(c + d*x))/(a*x + b*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.08 \[ \int \frac {x^2 (c+d x)}{\sqrt {a x+b x^2}} \, dx=\frac {15 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b d -18 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} c -10 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} d x +12 \sqrt {x}\, \sqrt {b x +a}\, b^{3} c x +8 \sqrt {x}\, \sqrt {b x +a}\, b^{3} d \,x^{2}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3} d +18 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2} b c}{24 b^{4}} \] Input:

int(x^2*(d*x+c)/(b*x^2+a*x)^(1/2),x)
 

Output:

(15*sqrt(x)*sqrt(a + b*x)*a**2*b*d - 18*sqrt(x)*sqrt(a + b*x)*a*b**2*c - 1 
0*sqrt(x)*sqrt(a + b*x)*a*b**2*d*x + 12*sqrt(x)*sqrt(a + b*x)*b**3*c*x + 8 
*sqrt(x)*sqrt(a + b*x)*b**3*d*x**2 - 15*sqrt(b)*log((sqrt(a + b*x) + sqrt( 
x)*sqrt(b))/sqrt(a))*a**3*d + 18*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt 
(b))/sqrt(a))*a**2*b*c)/(24*b**4)