\(\int \frac {x^2}{(a+b \sqrt {c+d x})^2} \, dx\) [119]

Optimal result

Integrand size = 19, antiderivative size = 166 \[ \int \frac {x^2}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {\left (3 a^2-2 b^2 c\right ) x}{b^4 d^2}-\frac {8 a \left (a^2-b^2 c\right ) \sqrt {c+d x}}{b^5 d^3}-\frac {4 a (c+d x)^{3/2}}{3 b^3 d^3}+\frac {(c+d x)^2}{2 b^2 d^3}+\frac {2 a \left (a^2-b^2 c\right )^2}{b^6 d^3 \left (a+b \sqrt {c+d x}\right )}+\frac {2 \left (5 a^4-6 a^2 b^2 c+b^4 c^2\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^6 d^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.17 \[ \int \frac {x^2}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {12 a^5-48 a^4 b \sqrt {c+d x}-6 a^3 b^2 (9 c+5 d x)+2 a^2 b^3 \sqrt {c+d x} (29 c+5 d x)+a b^4 \left (43 c^2+26 c d x-5 d^2 x^2\right )+3 b^5 \sqrt {c+d x} \left (-3 c^2-2 c d x+d^2 x^2\right )+12 \left (5 a^4-6 a^2 b^2 c+b^4 c^2\right ) \left (a+b \sqrt {c+d x}\right ) \log \left (a+b \sqrt {c+d x}\right )}{6 b^6 d^3 \left (a+b \sqrt {c+d x}\right )} \] Input:

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

Output:

(12*a^5 - 48*a^4*b*Sqrt[c + d*x] - 6*a^3*b^2*(9*c + 5*d*x) + 2*a^2*b^3*Sqr 
t[c + d*x]*(29*c + 5*d*x) + a*b^4*(43*c^2 + 26*c*d*x - 5*d^2*x^2) + 3*b^5* 
Sqrt[c + d*x]*(-3*c^2 - 2*c*d*x + d^2*x^2) + 12*(5*a^4 - 6*a^2*b^2*c + b^4 
*c^2)*(a + b*Sqrt[c + d*x])*Log[a + b*Sqrt[c + d*x]])/(6*b^6*d^3*(a + b*Sq 
rt[c + d*x]))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {896, 1732, 522, 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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 1732

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

\(\Big \downarrow \) 522

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

rule 896

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff 
icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1)   Subst[Int[Si 
mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; 
 FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
 

rule 1732

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symb 
ol] :> With[{g = Denominator[n]}, Simp[g   Subst[Int[x^(g - 1)*(d + e*x^(g* 
n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p, q} 
, x] && EqQ[n2, 2*n] && FractionQ[n]
 

rule 2009

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (d x +c \right )^{2} b^{3}}{4}+\frac {2 a \left (d x +c \right )^{\frac {3}{2}} b^{2}}{3}+b^{3} c \left (d x +c \right )-\frac {3 a^{2} b \left (d x +c \right )}{2}-4 a c \,b^{2} \sqrt {d x +c}+4 a^{3} \sqrt {d x +c}\right )}{b^{5}}+\frac {2 a \left (b^{4} c^{2}-2 a^{2} b^{2} c +a^{4}\right )}{b^{6} \left (a +b \sqrt {d x +c}\right )}+\frac {2 \left (b^{4} c^{2}-6 a^{2} b^{2} c +5 a^{4}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{b^{6}}}{d^{3}}\)	\(161\)
default	\(\frac {-\frac {2 \left (-\frac {\left (d x +c \right )^{2} b^{3}}{4}+\frac {2 a \left (d x +c \right )^{\frac {3}{2}} b^{2}}{3}+b^{3} c \left (d x +c \right )-\frac {3 a^{2} b \left (d x +c \right )}{2}-4 a c \,b^{2} \sqrt {d x +c}+4 a^{3} \sqrt {d x +c}\right )}{b^{5}}+\frac {2 a \left (b^{4} c^{2}-2 a^{2} b^{2} c +a^{4}\right )}{b^{6} \left (a +b \sqrt {d x +c}\right )}+\frac {2 \left (b^{4} c^{2}-6 a^{2} b^{2} c +5 a^{4}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{b^{6}}}{d^{3}}\)	\(161\)

Input:

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

Output:

2/d^3*(-1/b^5*(-1/4*(d*x+c)^2*b^3+2/3*a*(d*x+c)^(3/2)*b^2+b^3*c*(d*x+c)-3/ 
2*a^2*b*(d*x+c)-4*a*c*b^2*(d*x+c)^(1/2)+4*a^3*(d*x+c)^(1/2))+a*(b^4*c^2-2* 
a^2*b^2*c+a^4)/b^6/(a+b*(d*x+c)^(1/2))+1/b^6*(b^4*c^2-6*a^2*b^2*c+5*a^4)*l 
n(a+b*(d*x+c)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.62 \[ \int \frac {x^2}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {3 \, b^{6} d^{3} x^{3} - 9 \, b^{6} c^{3} + 15 \, a^{2} b^{4} c^{2} + 6 \, a^{4} b^{2} c - 12 \, a^{6} - 3 \, {\left (b^{6} c - 5 \, a^{2} b^{4}\right )} d^{2} x^{2} - 3 \, {\left (5 \, b^{6} c^{2} - 14 \, a^{2} b^{4} c + 6 \, a^{4} b^{2}\right )} d x + 12 \, {\left (b^{6} c^{3} - 7 \, a^{2} b^{4} c^{2} + 11 \, a^{4} b^{2} c - 5 \, a^{6} + {\left (b^{6} c^{2} - 6 \, a^{2} b^{4} c + 5 \, a^{4} b^{2}\right )} d x\right )} \log \left (\sqrt {d x + c} b + a\right ) - 4 \, {\left (2 \, a b^{5} d^{2} x^{2} - 13 \, a b^{5} c^{2} + 28 \, a^{3} b^{3} c - 15 \, a^{5} b - 2 \, {\left (4 \, a b^{5} c - 5 \, a^{3} b^{3}\right )} d x\right )} \sqrt {d x + c}}{6 \, {\left (b^{8} d^{4} x + {\left (b^{8} c - a^{2} b^{6}\right )} d^{3}\right )}} \] Input:

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

Output:

1/6*(3*b^6*d^3*x^3 - 9*b^6*c^3 + 15*a^2*b^4*c^2 + 6*a^4*b^2*c - 12*a^6 - 3 
*(b^6*c - 5*a^2*b^4)*d^2*x^2 - 3*(5*b^6*c^2 - 14*a^2*b^4*c + 6*a^4*b^2)*d* 
x + 12*(b^6*c^3 - 7*a^2*b^4*c^2 + 11*a^4*b^2*c - 5*a^6 + (b^6*c^2 - 6*a^2* 
b^4*c + 5*a^4*b^2)*d*x)*log(sqrt(d*x + c)*b + a) - 4*(2*a*b^5*d^2*x^2 - 13 
*a*b^5*c^2 + 28*a^3*b^3*c - 15*a^5*b - 2*(4*a*b^5*c - 5*a^3*b^3)*d*x)*sqrt 
(d*x + c))/(b^8*d^4*x + (b^8*c - a^2*b^6)*d^3)
 

Sympy [A] (verification not implemented)

Time = 5.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.14 \[ \int \frac {x^2}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\begin {cases} \frac {2 \left (- \frac {2 a \left (c + d x\right )^{\frac {3}{2}}}{3 b^{3}} - \frac {a \left (a^{2} - b^{2} c\right )^{2} \left (\begin {cases} \frac {\sqrt {c + d x}}{a^{2}} & \text {for}\: b = 0 \\- \frac {1}{b \left (a + b \sqrt {c + d x}\right )} & \text {otherwise} \end {cases}\right )}{b^{5}} + \frac {\left (c + d x\right )^{2}}{4 b^{2}} + \frac {\left (3 a^{2} - 2 b^{2} c\right ) \left (c + d x\right )}{2 b^{4}} + \frac {\left (a^{2} - b^{2} c\right ) \left (5 a^{2} - b^{2} c\right ) \left (\begin {cases} \frac {\sqrt {c + d x}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b \sqrt {c + d x} \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{5}} + \frac {\left (- 4 a^{3} + 4 a b^{2} c\right ) \sqrt {c + d x}}{b^{5}}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {x^{3}}{3 \left (a + b \sqrt {c}\right )^{2}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(-2*a*(c + d*x)**(3/2)/(3*b**3) - a*(a**2 - b**2*c)**2*Piecew 
ise((sqrt(c + d*x)/a**2, Eq(b, 0)), (-1/(b*(a + b*sqrt(c + d*x))), True))/ 
b**5 + (c + d*x)**2/(4*b**2) + (3*a**2 - 2*b**2*c)*(c + d*x)/(2*b**4) + (a 
**2 - b**2*c)*(5*a**2 - b**2*c)*Piecewise((sqrt(c + d*x)/a, Eq(b, 0)), (lo 
g(a + b*sqrt(c + d*x))/b, True))/b**5 + (-4*a**3 + 4*a*b**2*c)*sqrt(c + d* 
x)/b**5)/d**3, Ne(d, 0)), (x**3/(3*(a + b*sqrt(c))**2), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.95 \[ \int \frac {x^2}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {\frac {12 \, {\left (a b^{4} c^{2} - 2 \, a^{3} b^{2} c + a^{5}\right )}}{\sqrt {d x + c} b^{7} + a b^{6}} + \frac {3 \, {\left (d x + c\right )}^{2} b^{3} - 8 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} - 6 \, {\left (2 \, b^{3} c - 3 \, a^{2} b\right )} {\left (d x + c\right )} + 48 \, {\left (a b^{2} c - a^{3}\right )} \sqrt {d x + c}}{b^{5}} + \frac {12 \, {\left (b^{4} c^{2} - 6 \, a^{2} b^{2} c + 5 \, a^{4}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{6}}}{6 \, d^{3}} \] Input:

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

Output:

1/6*(12*(a*b^4*c^2 - 2*a^3*b^2*c + a^5)/(sqrt(d*x + c)*b^7 + a*b^6) + (3*( 
d*x + c)^2*b^3 - 8*(d*x + c)^(3/2)*a*b^2 - 6*(2*b^3*c - 3*a^2*b)*(d*x + c) 
 + 48*(a*b^2*c - a^3)*sqrt(d*x + c))/b^5 + 12*(b^4*c^2 - 6*a^2*b^2*c + 5*a 
^4)*log(sqrt(d*x + c)*b + a)/b^6)/d^3
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.17 \[ \int \frac {x^2}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {\frac {12 \, {\left (b^{4} c^{2} - 6 \, a^{2} b^{2} c + 5 \, a^{4}\right )} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{6} d} + \frac {12 \, {\left (a b^{4} c^{2} - 2 \, a^{3} b^{2} c + a^{5}\right )}}{{\left (\sqrt {d x + c} b + a\right )} b^{6} d} + \frac {3 \, {\left (d x + c\right )}^{2} b^{6} d^{3} - 12 \, {\left (d x + c\right )} b^{6} c d^{3} - 8 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{5} d^{3} + 48 \, \sqrt {d x + c} a b^{5} c d^{3} + 18 \, {\left (d x + c\right )} a^{2} b^{4} d^{3} - 48 \, \sqrt {d x + c} a^{3} b^{3} d^{3}}{b^{8} d^{4}}}{6 \, d^{2}} \] Input:

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

Output:

1/6*(12*(b^4*c^2 - 6*a^2*b^2*c + 5*a^4)*log(abs(sqrt(d*x + c)*b + a))/(b^6 
*d) + 12*(a*b^4*c^2 - 2*a^3*b^2*c + a^5)/((sqrt(d*x + c)*b + a)*b^6*d) + ( 
3*(d*x + c)^2*b^6*d^3 - 12*(d*x + c)*b^6*c*d^3 - 8*(d*x + c)^(3/2)*a*b^5*d 
^3 + 48*sqrt(d*x + c)*a*b^5*c*d^3 + 18*(d*x + c)*a^2*b^4*d^3 - 48*sqrt(d*x 
 + c)*a^3*b^3*d^3)/(b^8*d^4))/d^2
 

Mupad [B] (verification not implemented)

Time = 9.23 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.19 \[ \int \frac {x^2}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\left (\frac {4\,a^3}{b^5\,d^3}+\frac {2\,a\,\left (\frac {4\,c}{b^2\,d^3}-\frac {6\,a^2}{b^4\,d^3}\right )}{b}\right )\,\sqrt {c+d\,x}+\frac {2\,\left (a^5-2\,a^3\,b^2\,c+a\,b^4\,c^2\right )}{b\,\left (b^6\,d^3\,\sqrt {c+d\,x}+a\,b^5\,d^3\right )}+\frac {{\left (c+d\,x\right )}^2}{2\,b^2\,d^3}-d\,x\,\left (\frac {2\,c}{b^2\,d^3}-\frac {3\,a^2}{b^4\,d^3}\right )-\frac {4\,a\,{\left (c+d\,x\right )}^{3/2}}{3\,b^3\,d^3}+\frac {\ln \left (a+b\,\sqrt {c+d\,x}\right )\,\left (10\,a^4-12\,a^2\,b^2\,c+2\,b^4\,c^2\right )}{b^6\,d^3} \] Input:

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

Output:

((4*a^3)/(b^5*d^3) + (2*a*((4*c)/(b^2*d^3) - (6*a^2)/(b^4*d^3)))/b)*(c + d 
*x)^(1/2) + (2*(a^5 - 2*a^3*b^2*c + a*b^4*c^2))/(b*(b^6*d^3*(c + d*x)^(1/2 
) + a*b^5*d^3)) + (c + d*x)^2/(2*b^2*d^3) - d*x*((2*c)/(b^2*d^3) - (3*a^2) 
/(b^4*d^3)) - (4*a*(c + d*x)^(3/2))/(3*b^3*d^3) + (log(a + b*(c + d*x)^(1/ 
2))*(10*a^4 + 2*b^4*c^2 - 12*a^2*b^2*c))/(b^6*d^3)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.73 \[ \int \frac {x^2}{\left (a+b \sqrt {c+d x}\right )^2} \, dx=\frac {60 \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a^{4} b -72 \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a^{2} b^{3} c +12 \sqrt {d x +c}\, \mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) b^{5} c^{2}-60 \sqrt {d x +c}\, a^{4} b +82 \sqrt {d x +c}\, a^{2} b^{3} c +10 \sqrt {d x +c}\, a^{2} b^{3} d x -21 \sqrt {d x +c}\, b^{5} c^{2}-6 \sqrt {d x +c}\, b^{5} c d x +3 \sqrt {d x +c}\, b^{5} d^{2} x^{2}+60 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a^{5}-72 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a^{3} b^{2} c +12 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a \,b^{4} c^{2}-30 a^{3} b^{2} c -30 a^{3} b^{2} d x +31 a \,b^{4} c^{2}+26 a \,b^{4} c d x -5 a \,b^{4} d^{2} x^{2}}{6 b^{6} d^{3} \left (\sqrt {d x +c}\, b +a \right )} \] Input:

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

Output:

(60*sqrt(c + d*x)*log(sqrt(c + d*x)*b + a)*a**4*b - 72*sqrt(c + d*x)*log(s 
qrt(c + d*x)*b + a)*a**2*b**3*c + 12*sqrt(c + d*x)*log(sqrt(c + d*x)*b + a 
)*b**5*c**2 - 60*sqrt(c + d*x)*a**4*b + 82*sqrt(c + d*x)*a**2*b**3*c + 10* 
sqrt(c + d*x)*a**2*b**3*d*x - 21*sqrt(c + d*x)*b**5*c**2 - 6*sqrt(c + d*x) 
*b**5*c*d*x + 3*sqrt(c + d*x)*b**5*d**2*x**2 + 60*log(sqrt(c + d*x)*b + a) 
*a**5 - 72*log(sqrt(c + d*x)*b + a)*a**3*b**2*c + 12*log(sqrt(c + d*x)*b + 
 a)*a*b**4*c**2 - 30*a**3*b**2*c - 30*a**3*b**2*d*x + 31*a*b**4*c**2 + 26* 
a*b**4*c*d*x - 5*a*b**4*d**2*x**2)/(6*b**6*d**3*(sqrt(c + d*x)*b + a))