Integrand size = 19, antiderivative size = 151 \[ \int \frac {x^2}{a+b \sqrt {c+d x}} \, dx=-\frac {a \left (a^2-2 b^2 c\right ) x}{b^4 d^2}+\frac {2 \left (a^2-b^2 c\right )^2 \sqrt {c+d x}}{b^5 d^3}+\frac {2 \left (a^2-2 b^2 c\right ) (c+d x)^{3/2}}{3 b^3 d^3}-\frac {a (c+d x)^2}{2 b^2 d^3}+\frac {2 (c+d x)^{5/2}}{5 b d^3}-\frac {2 a \left (a^2-b^2 c\right )^2 \log \left (a+b \sqrt {c+d x}\right )}{b^6 d^3} \] Output:
-a*(-2*b^2*c+a^2)*x/b^4/d^2+2*(-b^2*c+a^2)^2*(d*x+c)^(1/2)/b^5/d^3+2/3*(-2 *b^2*c+a^2)*(d*x+c)^(3/2)/b^3/d^3-1/2*a*(d*x+c)^2/b^2/d^3+2/5*(d*x+c)^(5/2 )/b/d^3-2*a*(-b^2*c+a^2)^2*ln(a+b*(d*x+c)^(1/2))/b^6/d^3
Time = 0.17 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{a+b \sqrt {c+d x}} \, dx=\frac {b \left (60 a^4 \sqrt {c+d x}-20 a^2 b^2 (5 c-d x) \sqrt {c+d x}-30 a^3 b (c+d x)+15 a b^3 \left (3 c^2+2 c d x-d^2 x^2\right )+4 b^4 \sqrt {c+d x} \left (8 c^2-4 c d x+3 d^2 x^2\right )\right )-60 a \left (a^2-b^2 c\right )^2 \log \left (a+b \sqrt {c+d x}\right )}{30 b^6 d^3} \] Input:
Integrate[x^2/(a + b*Sqrt[c + d*x]),x]
Output:
(b*(60*a^4*Sqrt[c + d*x] - 20*a^2*b^2*(5*c - d*x)*Sqrt[c + d*x] - 30*a^3*b *(c + d*x) + 15*a*b^3*(3*c^2 + 2*c*d*x - d^2*x^2) + 4*b^4*Sqrt[c + d*x]*(8 *c^2 - 4*c*d*x + 3*d^2*x^2)) - 60*a*(a^2 - b^2*c)^2*Log[a + b*Sqrt[c + d*x ]])/(30*b^6*d^3)
Time = 0.55 (sec) , antiderivative size = 143, 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.
\(\displaystyle \int \frac {x^2}{a+b \sqrt {c+d x}} \, dx\) |
\(\Big \downarrow \) 896 |
\(\displaystyle \frac {\int \frac {d^2 x^2}{a+b \sqrt {c+d x}}d(c+d x)}{d^3}\) |
\(\Big \downarrow \) 1732 |
\(\displaystyle \frac {2 \int \frac {d^2 x^2 \sqrt {c+d x}}{a+b \sqrt {c+d x}}d\sqrt {c+d x}}{d^3}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {2 \int \left (-\frac {a \left (a^2-b^2 c\right )^2}{b^5 \left (a+b \sqrt {c+d x}\right )}+\frac {\left (b^2 c-a^2\right )^2}{b^5}+\frac {(c+d x)^2}{b}-\frac {a (c+d x)^{3/2}}{b^2}-\frac {\left (2 b^2 c-a^2\right ) (c+d x)}{b^3}-\frac {a \left (a^2-2 b^2 c\right ) \sqrt {c+d x}}{b^4}\right )d\sqrt {c+d x}}{d^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (-\frac {a \left (a^2-b^2 c\right )^2 \log \left (a+b \sqrt {c+d x}\right )}{b^6}+\frac {\left (a^2-b^2 c\right )^2 \sqrt {c+d x}}{b^5}-\frac {a \left (a^2-2 b^2 c\right ) (c+d x)}{2 b^4}+\frac {\left (a^2-2 b^2 c\right ) (c+d x)^{3/2}}{3 b^3}-\frac {a (c+d x)^2}{4 b^2}+\frac {(c+d x)^{5/2}}{5 b}\right )}{d^3}\) |
Input:
Int[x^2/(a + b*Sqrt[c + d*x]),x]
Output:
(2*(((a^2 - b^2*c)^2*Sqrt[c + d*x])/b^5 - (a*(a^2 - 2*b^2*c)*(c + d*x))/(2 *b^4) + ((a^2 - 2*b^2*c)*(c + d*x)^(3/2))/(3*b^3) - (a*(c + d*x)^2)/(4*b^2 ) + (c + d*x)^(5/2)/(5*b) - (a*(a^2 - b^2*c)^2*Log[a + b*Sqrt[c + d*x]])/b ^6))/d^3
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]
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]
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]
Time = 0.04 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (\frac {\left (d x +c \right )^{\frac {5}{2}} b^{4}}{5}-\frac {a \left (d x +c \right )^{2} b^{3}}{4}-\frac {2 b^{4} c \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {a^{2} b^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+a \,b^{3} c \left (d x +c \right )+b^{4} c^{2} \sqrt {d x +c}-\frac {a^{3} b \left (d x +c \right )}{2}-2 a^{2} b^{2} c \sqrt {d x +c}+a^{4} \sqrt {d x +c}\right )}{b^{5}}-\frac {2 a \left (b^{4} c^{2}-2 a^{2} b^{2} c +a^{4}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{b^{6}}}{d^{3}}\) | \(166\) |
default | \(\frac {\frac {2 \left (\frac {\left (d x +c \right )^{\frac {5}{2}} b^{4}}{5}-\frac {a \left (d x +c \right )^{2} b^{3}}{4}-\frac {2 b^{4} c \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {a^{2} b^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+a \,b^{3} c \left (d x +c \right )+b^{4} c^{2} \sqrt {d x +c}-\frac {a^{3} b \left (d x +c \right )}{2}-2 a^{2} b^{2} c \sqrt {d x +c}+a^{4} \sqrt {d x +c}\right )}{b^{5}}-\frac {2 a \left (b^{4} c^{2}-2 a^{2} b^{2} c +a^{4}\right ) \ln \left (a +b \sqrt {d x +c}\right )}{b^{6}}}{d^{3}}\) | \(166\) |
Input:
int(x^2/(a+b*(d*x+c)^(1/2)),x,method=_RETURNVERBOSE)
Output:
2/d^3*(1/b^5*(1/5*(d*x+c)^(5/2)*b^4-1/4*a*(d*x+c)^2*b^3-2/3*b^4*c*(d*x+c)^ (3/2)+1/3*a^2*b^2*(d*x+c)^(3/2)+a*b^3*c*(d*x+c)+b^4*c^2*(d*x+c)^(1/2)-1/2* a^3*b*(d*x+c)-2*a^2*b^2*c*(d*x+c)^(1/2)+a^4*(d*x+c)^(1/2))-a*(b^4*c^2-2*a^ 2*b^2*c+a^4)/b^6*ln(a+b*(d*x+c)^(1/2)))
Time = 0.10 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.91 \[ \int \frac {x^2}{a+b \sqrt {c+d x}} \, dx=-\frac {15 \, a b^{4} d^{2} x^{2} - 30 \, {\left (a b^{4} c - a^{3} b^{2}\right )} d x + 60 \, {\left (a b^{4} c^{2} - 2 \, a^{3} b^{2} c + a^{5}\right )} \log \left (\sqrt {d x + c} b + a\right ) - 4 \, {\left (3 \, b^{5} d^{2} x^{2} + 8 \, b^{5} c^{2} - 25 \, a^{2} b^{3} c + 15 \, a^{4} b - {\left (4 \, b^{5} c - 5 \, a^{2} b^{3}\right )} d x\right )} \sqrt {d x + c}}{30 \, b^{6} d^{3}} \] Input:
integrate(x^2/(a+b*(d*x+c)^(1/2)),x, algorithm="fricas")
Output:
-1/30*(15*a*b^4*d^2*x^2 - 30*(a*b^4*c - a^3*b^2)*d*x + 60*(a*b^4*c^2 - 2*a ^3*b^2*c + a^5)*log(sqrt(d*x + c)*b + a) - 4*(3*b^5*d^2*x^2 + 8*b^5*c^2 - 25*a^2*b^3*c + 15*a^4*b - (4*b^5*c - 5*a^2*b^3)*d*x)*sqrt(d*x + c))/(b^6*d ^3)
Time = 1.37 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.11 \[ \int \frac {x^2}{a+b \sqrt {c+d x}} \, dx=\begin {cases} \frac {2 \left (- \frac {a \left (c + d x\right )^{2}}{4 b^{2}} - \frac {a \left (a^{2} - b^{2} c\right )^{2} \left (\begin {cases} \frac {\sqrt {c + d x}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b \sqrt {c + d x} \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{5}} + \frac {\left (c + d x\right )^{\frac {5}{2}}}{5 b} + \frac {\left (a^{2} - 2 b^{2} c\right ) \left (c + d x\right )^{\frac {3}{2}}}{3 b^{3}} + \frac {\left (- a^{3} + 2 a b^{2} c\right ) \left (c + d x\right )}{2 b^{4}} + \frac {\sqrt {c + d x} \left (a^{4} - 2 a^{2} b^{2} c + b^{4} c^{2}\right )}{b^{5}}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {x^{3}}{3 \left (a + b \sqrt {c}\right )} & \text {otherwise} \end {cases} \] Input:
integrate(x**2/(a+b*(d*x+c)**(1/2)),x)
Output:
Piecewise((2*(-a*(c + d*x)**2/(4*b**2) - a*(a**2 - b**2*c)**2*Piecewise((s qrt(c + d*x)/a, Eq(b, 0)), (log(a + b*sqrt(c + d*x))/b, True))/b**5 + (c + d*x)**(5/2)/(5*b) + (a**2 - 2*b**2*c)*(c + d*x)**(3/2)/(3*b**3) + (-a**3 + 2*a*b**2*c)*(c + d*x)/(2*b**4) + sqrt(c + d*x)*(a**4 - 2*a**2*b**2*c + b **4*c**2)/b**5)/d**3, Ne(d, 0)), (x**3/(3*(a + b*sqrt(c))), True))
Time = 0.03 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.98 \[ \int \frac {x^2}{a+b \sqrt {c+d x}} \, dx=\frac {\frac {12 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} - 15 \, {\left (d x + c\right )}^{2} a b^{3} - 20 \, {\left (2 \, b^{4} c - a^{2} b^{2}\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 30 \, {\left (2 \, a b^{3} c - a^{3} b\right )} {\left (d x + c\right )} + 60 \, {\left (b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}\right )} \sqrt {d x + c}}{b^{5}} - \frac {60 \, {\left (a b^{4} c^{2} - 2 \, a^{3} b^{2} c + a^{5}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{6}}}{30 \, d^{3}} \] Input:
integrate(x^2/(a+b*(d*x+c)^(1/2)),x, algorithm="maxima")
Output:
1/30*((12*(d*x + c)^(5/2)*b^4 - 15*(d*x + c)^2*a*b^3 - 20*(2*b^4*c - a^2*b ^2)*(d*x + c)^(3/2) + 30*(2*a*b^3*c - a^3*b)*(d*x + c) + 60*(b^4*c^2 - 2*a ^2*b^2*c + a^4)*sqrt(d*x + c))/b^5 - 60*(a*b^4*c^2 - 2*a^3*b^2*c + a^5)*lo g(sqrt(d*x + c)*b + a)/b^6)/d^3
Time = 0.11 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.34 \[ \int \frac {x^2}{a+b \sqrt {c+d x}} \, dx=-\frac {\frac {60 \, {\left (a b^{4} c^{2} - 2 \, a^{3} b^{2} c + a^{5}\right )} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{6} d} - \frac {12 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} d^{4} - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c d^{4} + 60 \, \sqrt {d x + c} b^{4} c^{2} d^{4} - 15 \, {\left (d x + c\right )}^{2} a b^{3} d^{4} + 60 \, {\left (d x + c\right )} a b^{3} c d^{4} + 20 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} b^{2} d^{4} - 120 \, \sqrt {d x + c} a^{2} b^{2} c d^{4} - 30 \, {\left (d x + c\right )} a^{3} b d^{4} + 60 \, \sqrt {d x + c} a^{4} d^{4}}{b^{5} d^{5}}}{30 \, d^{2}} \] Input:
integrate(x^2/(a+b*(d*x+c)^(1/2)),x, algorithm="giac")
Output:
-1/30*(60*(a*b^4*c^2 - 2*a^3*b^2*c + a^5)*log(abs(sqrt(d*x + c)*b + a))/(b ^6*d) - (12*(d*x + c)^(5/2)*b^4*d^4 - 40*(d*x + c)^(3/2)*b^4*c*d^4 + 60*sq rt(d*x + c)*b^4*c^2*d^4 - 15*(d*x + c)^2*a*b^3*d^4 + 60*(d*x + c)*a*b^3*c* d^4 + 20*(d*x + c)^(3/2)*a^2*b^2*d^4 - 120*sqrt(d*x + c)*a^2*b^2*c*d^4 - 3 0*(d*x + c)*a^3*b*d^4 + 60*sqrt(d*x + c)*a^4*d^4)/(b^5*d^5))/d^2
Time = 9.46 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.22 \[ \int \frac {x^2}{a+b \sqrt {c+d x}} \, dx=\frac {2\,{\left (c+d\,x\right )}^{5/2}}{5\,b\,d^3}-\left (\frac {a^2\,\left (\frac {4\,c}{b\,d^3}-\frac {2\,a^2}{b^3\,d^3}\right )}{b^2}-\frac {2\,c^2}{b\,d^3}\right )\,\sqrt {c+d\,x}-\left (\frac {4\,c}{3\,b\,d^3}-\frac {2\,a^2}{3\,b^3\,d^3}\right )\,{\left (c+d\,x\right )}^{3/2}-\frac {\ln \left (a+b\,\sqrt {c+d\,x}\right )\,\left (2\,a^5-4\,a^3\,b^2\,c+2\,a\,b^4\,c^2\right )}{b^6\,d^3}-\frac {a\,{\left (c+d\,x\right )}^2}{2\,b^2\,d^3}+\frac {a\,d\,x\,\left (\frac {4\,c}{b\,d^3}-\frac {2\,a^2}{b^3\,d^3}\right )}{2\,b} \] Input:
int(x^2/(a + b*(c + d*x)^(1/2)),x)
Output:
(2*(c + d*x)^(5/2))/(5*b*d^3) - ((a^2*((4*c)/(b*d^3) - (2*a^2)/(b^3*d^3))) /b^2 - (2*c^2)/(b*d^3))*(c + d*x)^(1/2) - ((4*c)/(3*b*d^3) - (2*a^2)/(3*b^ 3*d^3))*(c + d*x)^(3/2) - (log(a + b*(c + d*x)^(1/2))*(2*a^5 - 4*a^3*b^2*c + 2*a*b^4*c^2))/(b^6*d^3) - (a*(c + d*x)^2)/(2*b^2*d^3) + (a*d*x*((4*c)/( b*d^3) - (2*a^2)/(b^3*d^3)))/(2*b)
Time = 0.25 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.34 \[ \int \frac {x^2}{a+b \sqrt {c+d x}} \, dx=\frac {60 \sqrt {d x +c}\, a^{4} b -100 \sqrt {d x +c}\, a^{2} b^{3} c +20 \sqrt {d x +c}\, a^{2} b^{3} d x +32 \sqrt {d x +c}\, b^{5} c^{2}-16 \sqrt {d x +c}\, b^{5} c d x +12 \sqrt {d x +c}\, b^{5} d^{2} x^{2}-60 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a^{5}+120 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a^{3} b^{2} c -60 \,\mathrm {log}\left (\sqrt {d x +c}\, b +a \right ) a \,b^{4} c^{2}-30 a^{3} b^{2} c -30 a^{3} b^{2} d x +45 a \,b^{4} c^{2}+30 a \,b^{4} c d x -15 a \,b^{4} d^{2} x^{2}}{30 b^{6} d^{3}} \] Input:
int(x^2/(a+b*(d*x+c)^(1/2)),x)
Output:
(60*sqrt(c + d*x)*a**4*b - 100*sqrt(c + d*x)*a**2*b**3*c + 20*sqrt(c + d*x )*a**2*b**3*d*x + 32*sqrt(c + d*x)*b**5*c**2 - 16*sqrt(c + d*x)*b**5*c*d*x + 12*sqrt(c + d*x)*b**5*d**2*x**2 - 60*log(sqrt(c + d*x)*b + a)*a**5 + 12 0*log(sqrt(c + d*x)*b + a)*a**3*b**2*c - 60*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 + 45*a*b**4*c**2 + 30*a*b**4*c* d*x - 15*a*b**4*d**2*x**2)/(30*b**6*d**3)