Integrand size = 17, antiderivative size = 89 \[ \int x \left (a+b \sqrt {c+d x}\right )^2 \, dx=-\frac {a^2 c x}{d}-\frac {4 a b c (c+d x)^{3/2}}{3 d^2}+\frac {\left (a^2-b^2 c\right ) (c+d x)^2}{2 d^2}+\frac {4 a b (c+d x)^{5/2}}{5 d^2}+\frac {b^2 (c+d x)^3}{3 d^2} \] Output:
-a^2*c*x/d-4/3*a*b*c*(d*x+c)^(3/2)/d^2+1/2*(-b^2*c+a^2)*(d*x+c)^2/d^2+4/5* a*b*(d*x+c)^(5/2)/d^2+1/3*b^2*(d*x+c)^3/d^2
Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76 \[ \int x \left (a+b \sqrt {c+d x}\right )^2 \, dx=\frac {(c+d x) \left (-15 a^2 (c-d x)-8 a b (2 c-3 d x) \sqrt {c+d x}+5 b^2 \left (-c^2+c d x+2 d^2 x^2\right )\right )}{30 d^2} \] Input:
Integrate[x*(a + b*Sqrt[c + d*x])^2,x]
Output:
((c + d*x)*(-15*a^2*(c - d*x) - 8*a*b*(2*c - 3*d*x)*Sqrt[c + d*x] + 5*b^2* (-c^2 + c*d*x + 2*d^2*x^2)))/(30*d^2)
Time = 0.41 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {896, 25, 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.
\(\displaystyle \int x \left (a+b \sqrt {c+d x}\right )^2 \, dx\) |
\(\Big \downarrow \) 896 |
\(\displaystyle \frac {\int d x \left (a+b \sqrt {c+d x}\right )^2d(c+d x)}{d^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int -d x \left (a+b \sqrt {c+d x}\right )^2d(c+d x)}{d^2}\) |
\(\Big \downarrow \) 1732 |
\(\displaystyle -\frac {2 \int -d x \sqrt {c+d x} \left (a+b \sqrt {c+d x}\right )^2d\sqrt {c+d x}}{d^2}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {2 \int \left (-b^2 (c+d x)^{5/2}-2 a b (c+d x)^2-\left (a^2-b^2 c\right ) (c+d x)^{3/2}+2 a b c (c+d x)+a^2 c \sqrt {c+d x}\right )d\sqrt {c+d x}}{d^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (-\frac {1}{4} \left (a^2-b^2 c\right ) (c+d x)^2+\frac {1}{2} a^2 c (c+d x)-\frac {2}{5} a b (c+d x)^{5/2}+\frac {2}{3} a b c (c+d x)^{3/2}-\frac {1}{6} b^2 (c+d x)^3\right )}{d^2}\) |
Input:
Int[x*(a + b*Sqrt[c + d*x])^2,x]
Output:
(-2*((a^2*c*(c + d*x))/2 + (2*a*b*c*(c + d*x)^(3/2))/3 - ((a^2 - b^2*c)*(c + d*x)^2)/4 - (2*a*b*(c + d*x)^(5/2))/5 - (b^2*(c + d*x)^3)/6))/d^2
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 = 0.37 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.61
method | result | size |
default | \(b^{2} \left (\frac {1}{3} d \,x^{3}+\frac {1}{2} c \,x^{2}\right )+\frac {4 a b \left (\frac {\left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {c \left (d x +c \right )^{\frac {3}{2}}}{3}\right )}{d^{2}}+\frac {a^{2} x^{2}}{2}\) | \(54\) |
trager | \(\frac {\left (2 b^{2} d x +3 b^{2} c +3 a^{2}\right ) x^{2}}{6}-\frac {4 a b \left (-3 d^{2} x^{2}-c d x +2 c^{2}\right ) \sqrt {d x +c}}{15 d^{2}}\) | \(59\) |
derivativedivides | \(\frac {\frac {b^{2} \left (d x +c \right )^{3}}{3}+\frac {4 a b \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {\left (-b^{2} c +a^{2}\right ) \left (d x +c \right )^{2}}{2}-\frac {4 a b c \left (d x +c \right )^{\frac {3}{2}}}{3}-c \,a^{2} \left (d x +c \right )}{d^{2}}\) | \(72\) |
orering | \(-\frac {\left (18 b^{2} d^{3} x^{3}+23 b^{2} c \,d^{2} x^{2}-21 a^{2} d^{2} x^{2}-14 b^{2} c^{2} d x +2 a^{2} c d x -16 c^{3} b^{2}+16 a^{2} c^{2}\right ) \left (a +b \sqrt {d x +c}\right )^{2}}{30 d^{2} \left (-b^{2} d x -b^{2} c +a^{2}\right )}+\frac {\left (2 b^{2} d^{2} x^{2}+b^{2} c x d -3 a^{2} d x -4 b^{2} c^{2}+4 a^{2} c \right ) \left (d x +c \right ) \left (\left (a +b \sqrt {d x +c}\right )^{2}+\frac {x \left (a +b \sqrt {d x +c}\right ) b d}{\sqrt {d x +c}}\right )}{15 d^{2} \left (-b^{2} d x -b^{2} c +a^{2}\right )}\) | \(213\) |
Input:
int(x*(a+b*(d*x+c)^(1/2))^2,x,method=_RETURNVERBOSE)
Output:
b^2*(1/3*d*x^3+1/2*c*x^2)+4*a*b/d^2*(1/5*(d*x+c)^(5/2)-1/3*c*(d*x+c)^(3/2) )+1/2*a^2*x^2
Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.75 \[ \int x \left (a+b \sqrt {c+d x}\right )^2 \, dx=\frac {10 \, b^{2} d^{3} x^{3} + 15 \, {\left (b^{2} c + a^{2}\right )} d^{2} x^{2} + 8 \, {\left (3 \, a b d^{2} x^{2} + a b c d x - 2 \, a b c^{2}\right )} \sqrt {d x + c}}{30 \, d^{2}} \] Input:
integrate(x*(a+b*(d*x+c)^(1/2))^2,x, algorithm="fricas")
Output:
1/30*(10*b^2*d^3*x^3 + 15*(b^2*c + a^2)*d^2*x^2 + 8*(3*a*b*d^2*x^2 + a*b*c *d*x - 2*a*b*c^2)*sqrt(d*x + c))/d^2
Time = 0.53 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99 \[ \int x \left (a+b \sqrt {c+d x}\right )^2 \, dx=\begin {cases} \frac {2 \left (- \frac {a^{2} c \left (c + d x\right )}{2} - \frac {2 a b c \left (c + d x\right )^{\frac {3}{2}}}{3} + \frac {2 a b \left (c + d x\right )^{\frac {5}{2}}}{5} + \frac {b^{2} \left (c + d x\right )^{3}}{6} + \frac {\left (a^{2} - b^{2} c\right ) \left (c + d x\right )^{2}}{4}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{2} \left (a + b \sqrt {c}\right )^{2}}{2} & \text {otherwise} \end {cases} \] Input:
integrate(x*(a+b*(d*x+c)**(1/2))**2,x)
Output:
Piecewise((2*(-a**2*c*(c + d*x)/2 - 2*a*b*c*(c + d*x)**(3/2)/3 + 2*a*b*(c + d*x)**(5/2)/5 + b**2*(c + d*x)**3/6 + (a**2 - b**2*c)*(c + d*x)**2/4)/d* *2, Ne(d, 0)), (x**2*(a + b*sqrt(c))**2/2, True))
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.81 \[ \int x \left (a+b \sqrt {c+d x}\right )^2 \, dx=\frac {10 \, {\left (d x + c\right )}^{3} b^{2} + 24 \, {\left (d x + c\right )}^{\frac {5}{2}} a b - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} a b c - 30 \, {\left (d x + c\right )} a^{2} c - 15 \, {\left (b^{2} c - a^{2}\right )} {\left (d x + c\right )}^{2}}{30 \, d^{2}} \] Input:
integrate(x*(a+b*(d*x+c)^(1/2))^2,x, algorithm="maxima")
Output:
1/30*(10*(d*x + c)^3*b^2 + 24*(d*x + c)^(5/2)*a*b - 40*(d*x + c)^(3/2)*a*b *c - 30*(d*x + c)*a^2*c - 15*(b^2*c - a^2)*(d*x + c)^2)/d^2
Time = 0.16 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.47 \[ \int x \left (a+b \sqrt {c+d x}\right )^2 \, dx=\frac {10 \, b^{2} d^{2} x^{3} + \frac {40 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a b c}{d} + \frac {15 \, {\left ({\left (d x + c\right )}^{2} - 2 \, {\left (d x + c\right )} c\right )} b^{2} c}{d} + \frac {15 \, {\left ({\left (d x + c\right )}^{2} - 2 \, {\left (d x + c\right )} c\right )} a^{2}}{d} + \frac {8 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a b}{d}}{30 \, d} \] Input:
integrate(x*(a+b*(d*x+c)^(1/2))^2,x, algorithm="giac")
Output:
1/30*(10*b^2*d^2*x^3 + 40*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a*b*c/d + 15*((d*x + c)^2 - 2*(d*x + c)*c)*b^2*c/d + 15*((d*x + c)^2 - 2*(d*x + c)*c )*a^2/d + 8*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c ^2)*a*b/d)/d
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int x \left (a+b \sqrt {c+d x}\right )^2 \, dx=\frac {b^2\,{\left (c+d\,x\right )}^3}{3\,d^2}-\frac {\left (2\,b^2\,c-2\,a^2\right )\,{\left (c+d\,x\right )}^2}{4\,d^2}+\frac {4\,a\,b\,{\left (c+d\,x\right )}^{5/2}}{5\,d^2}-\frac {a^2\,c\,x}{d}-\frac {4\,a\,b\,c\,{\left (c+d\,x\right )}^{3/2}}{3\,d^2} \] Input:
int(x*(a + b*(c + d*x)^(1/2))^2,x)
Output:
(b^2*(c + d*x)^3)/(3*d^2) - ((2*b^2*c - 2*a^2)*(c + d*x)^2)/(4*d^2) + (4*a *b*(c + d*x)^(5/2))/(5*d^2) - (a^2*c*x)/d - (4*a*b*c*(c + d*x)^(3/2))/(3*d ^2)
Time = 0.23 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92 \[ \int x \left (a+b \sqrt {c+d x}\right )^2 \, dx=\frac {-16 \sqrt {d x +c}\, a b \,c^{2}+8 \sqrt {d x +c}\, a b c d x +24 \sqrt {d x +c}\, a b \,d^{2} x^{2}+15 a^{2} d^{2} x^{2}+15 b^{2} c \,d^{2} x^{2}+10 b^{2} d^{3} x^{3}}{30 d^{2}} \] Input:
int(x*(a+b*(d*x+c)^(1/2))^2,x)
Output:
( - 16*sqrt(c + d*x)*a*b*c**2 + 8*sqrt(c + d*x)*a*b*c*d*x + 24*sqrt(c + d* x)*a*b*d**2*x**2 + 15*a**2*d**2*x**2 + 15*b**2*c*d**2*x**2 + 10*b**2*d**3* x**3)/(30*d**2)