Integrand size = 32, antiderivative size = 160 \[ \int x^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {8 c^3 \left (b c^2-b d^2 x^2\right )^{3/2}}{63 b d^3 (c+d x)^{3/2}}-\frac {2 c^2 \left (b c^2-b d^2 x^2\right )^{3/2}}{21 b d^3 \sqrt {c+d x}}+\frac {4 c \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{21 b d^3}-\frac {2 (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b d^3} \] Output:
-8/63*c^3*(-b*d^2*x^2+b*c^2)^(3/2)/b/d^3/(d*x+c)^(3/2)-2/21*c^2*(-b*d^2*x^ 2+b*c^2)^(3/2)/b/d^3/(d*x+c)^(1/2)+4/21*c*(d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2) ^(3/2)/b/d^3-2/9*(d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(3/2)/b/d^3
Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.47 \[ \int x^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {2 \sqrt {b \left (c^2-d^2 x^2\right )} \left (8 c^4+4 c^3 d x+3 c^2 d^2 x^2-8 c d^3 x^3-7 d^4 x^4\right )}{63 d^3 \sqrt {c+d x}} \] Input:
Integrate[x^2*Sqrt[c + d*x]*Sqrt[b*c^2 - b*d^2*x^2],x]
Output:
(-2*Sqrt[b*(c^2 - d^2*x^2)]*(8*c^4 + 4*c^3*d*x + 3*c^2*d^2*x^2 - 8*c*d^3*x ^3 - 7*d^4*x^4))/(63*d^3*Sqrt[c + d*x])
Time = 0.53 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {574, 581, 27, 672, 458}
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.
\(\displaystyle \int x^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx\) |
\(\Big \downarrow \) 574 |
\(\displaystyle \frac {5}{3} c \int \frac {x^2 \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}dx-\frac {2 x^3 \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 581 |
\(\displaystyle \frac {5}{3} c \left (\frac {2 \int \frac {c (c-6 d x) \sqrt {b c^2-b d^2 x^2}}{2 \sqrt {c+d x}}dx}{7 d^2}-\frac {2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{7 b d^3}\right )-\frac {2 x^3 \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{3} c \left (\frac {c \int \frac {(c-6 d x) \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}dx}{7 d^2}-\frac {2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{7 b d^3}\right )-\frac {2 x^3 \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 672 |
\(\displaystyle \frac {5}{3} c \left (\frac {c \left (\frac {11}{5} c \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}dx+\frac {12 \left (b c^2-b d^2 x^2\right )^{3/2}}{5 b d \sqrt {c+d x}}\right )}{7 d^2}-\frac {2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{7 b d^3}\right )-\frac {2 x^3 \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 458 |
\(\displaystyle \frac {5}{3} c \left (\frac {c \left (\frac {12 \left (b c^2-b d^2 x^2\right )^{3/2}}{5 b d \sqrt {c+d x}}-\frac {22 c \left (b c^2-b d^2 x^2\right )^{3/2}}{15 b d (c+d x)^{3/2}}\right )}{7 d^2}-\frac {2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{7 b d^3}\right )-\frac {2 x^3 \left (b c^2-b d^2 x^2\right )^{3/2}}{9 b (c+d x)^{3/2}}\) |
Input:
Int[x^2*Sqrt[c + d*x]*Sqrt[b*c^2 - b*d^2*x^2],x]
Output:
(-2*x^3*(b*c^2 - b*d^2*x^2)^(3/2))/(9*b*(c + d*x)^(3/2)) + (5*c*((-2*Sqrt[ c + d*x]*(b*c^2 - b*d^2*x^2)^(3/2))/(7*b*d^3) + (c*((-22*c*(b*c^2 - b*d^2* x^2)^(3/2))/(15*b*d*(c + d*x)^(3/2)) + (12*(b*c^2 - b*d^2*x^2)^(3/2))/(5*b *d*Sqrt[c + d*x])))/(7*d^2)))/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c , d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + p, 0]
Int[((e_.)*(x_))^(n_)*((c_) + (d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d^2*(e*x)^(n + 1)*(c + d*x)^(m - 2)*((a + b*x^2)^(p + 1)/ (b*e*(n + p + 2))), x] + Simp[c*((2*n + p + 3)/(n + p + 2)) Int[(e*x)^n*( c + d*x)^(m - 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x ] && EqQ[b*c^2 + a*d^2, 0] && EqQ[m + p - 1, 0] && !LtQ[n, -1] && IntegerQ [2*p]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 2*p + 1))), x] + Simp[1/(d^m*(m + n + 2*p + 1)) Int[(c + d*x)^n*(a + b*x^ 2)^p*ExpandToSum[d^m*(m + n + 2*p + 1)*x^m - (m + n + 2*p + 1)*(c + d*x)^m + c*(c + d*x)^(m - 2)*(c*(m + n - 1) + c*(m + n + 2*p + 1) + 2*d*(m + n + p )*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] & & IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[2*p] || ILtQ[m + n, 0] )
Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)) Int[(d + e*x )^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^ 2 + a*e^2, 0] && NeQ[m + 2*p + 2, 0]
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.41
method | result | size |
default | \(-\frac {2 \sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (-d x +c \right ) \left (7 d^{3} x^{3}+15 c \,d^{2} x^{2}+12 c^{2} d x +8 c^{3}\right )}{63 \sqrt {d x +c}\, d^{3}}\) | \(65\) |
gosper | \(-\frac {2 \left (-d x +c \right ) \left (7 d^{3} x^{3}+15 c \,d^{2} x^{2}+12 c^{2} d x +8 c^{3}\right ) \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}{63 d^{3} \sqrt {d x +c}}\) | \(66\) |
orering | \(-\frac {2 \left (-d x +c \right ) \left (7 d^{3} x^{3}+15 c \,d^{2} x^{2}+12 c^{2} d x +8 c^{3}\right ) \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}{63 d^{3} \sqrt {d x +c}}\) | \(66\) |
risch | \(-\frac {2 \sqrt {-\frac {b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b \left (-7 d^{4} x^{4}-8 c \,d^{3} x^{3}+3 d^{2} c^{2} x^{2}+4 c^{3} d x +8 c^{4}\right ) \left (-d x +c \right )}{63 \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}\, d^{3} \sqrt {-b \left (d x -c \right )}}\) | \(116\) |
Input:
int(x^2*(d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/63/(d*x+c)^(1/2)*(b*(-d^2*x^2+c^2))^(1/2)*(-d*x+c)*(7*d^3*x^3+15*c*d^2* x^2+12*c^2*d*x+8*c^3)/d^3
Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.50 \[ \int x^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=\frac {2 \, {\left (7 \, d^{4} x^{4} + 8 \, c d^{3} x^{3} - 3 \, c^{2} d^{2} x^{2} - 4 \, c^{3} d x - 8 \, c^{4}\right )} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c}}{63 \, {\left (d^{4} x + c d^{3}\right )}} \] Input:
integrate(x^2*(d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2),x, algorithm="fricas" )
Output:
2/63*(7*d^4*x^4 + 8*c*d^3*x^3 - 3*c^2*d^2*x^2 - 4*c^3*d*x - 8*c^4)*sqrt(-b *d^2*x^2 + b*c^2)*sqrt(d*x + c)/(d^4*x + c*d^3)
\[ \int x^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=\int x^{2} \sqrt {- b \left (- c + d x\right ) \left (c + d x\right )} \sqrt {c + d x}\, dx \] Input:
integrate(x**2*(d*x+c)**(1/2)*(-b*d**2*x**2+b*c**2)**(1/2),x)
Output:
Integral(x**2*sqrt(-b*(-c + d*x)*(c + d*x))*sqrt(c + d*x), x)
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.52 \[ \int x^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=\frac {2 \, {\left (7 \, \sqrt {b} d^{4} x^{4} + 8 \, \sqrt {b} c d^{3} x^{3} - 3 \, \sqrt {b} c^{2} d^{2} x^{2} - 4 \, \sqrt {b} c^{3} d x - 8 \, \sqrt {b} c^{4}\right )} {\left (d x + c\right )} \sqrt {-d x + c}}{63 \, {\left (d^{4} x + c d^{3}\right )}} \] Input:
integrate(x^2*(d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2),x, algorithm="maxima" )
Output:
2/63*(7*sqrt(b)*d^4*x^4 + 8*sqrt(b)*c*d^3*x^3 - 3*sqrt(b)*c^2*d^2*x^2 - 4* sqrt(b)*c^3*d*x - 8*sqrt(b)*c^4)*(d*x + c)*sqrt(-d*x + c)/(d^4*x + c*d^3)
Time = 0.13 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.23 \[ \int x^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {2 \, {\left (\frac {21 \, {\left (15 \, \sqrt {-b d x + b c} b^{2} c^{2} - 10 \, {\left (-b d x + b c\right )}^{\frac {3}{2}} b c + 3 \, {\left (b d x - b c\right )}^{2} \sqrt {-b d x + b c}\right )} c^{2}}{b^{2}} - \frac {315 \, \sqrt {-b d x + b c} b^{4} c^{4} - 420 \, {\left (-b d x + b c\right )}^{\frac {3}{2}} b^{3} c^{3} + 378 \, {\left (b d x - b c\right )}^{2} \sqrt {-b d x + b c} b^{2} c^{2} + 180 \, {\left (b d x - b c\right )}^{3} \sqrt {-b d x + b c} b c + 35 \, {\left (b d x - b c\right )}^{4} \sqrt {-b d x + b c}}{b^{4}}\right )}}{315 \, d^{3}} \] Input:
integrate(x^2*(d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2),x, algorithm="giac")
Output:
-2/315*(21*(15*sqrt(-b*d*x + b*c)*b^2*c^2 - 10*(-b*d*x + b*c)^(3/2)*b*c + 3*(b*d*x - b*c)^2*sqrt(-b*d*x + b*c))*c^2/b^2 - (315*sqrt(-b*d*x + b*c)*b^ 4*c^4 - 420*(-b*d*x + b*c)^(3/2)*b^3*c^3 + 378*(b*d*x - b*c)^2*sqrt(-b*d*x + b*c)*b^2*c^2 + 180*(b*d*x - b*c)^3*sqrt(-b*d*x + b*c)*b*c + 35*(b*d*x - b*c)^4*sqrt(-b*d*x + b*c))/b^4)/d^3
Time = 7.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.66 \[ \int x^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {\sqrt {b\,c^2-b\,d^2\,x^2}\,\left (\frac {16\,c^4\,\sqrt {c+d\,x}}{63\,d^4}-\frac {2\,x^4\,\sqrt {c+d\,x}}{9}+\frac {2\,c^2\,x^2\,\sqrt {c+d\,x}}{21\,d^2}-\frac {16\,c\,x^3\,\sqrt {c+d\,x}}{63\,d}+\frac {8\,c^3\,x\,\sqrt {c+d\,x}}{63\,d^3}\right )}{x+\frac {c}{d}} \] Input:
int(x^2*(b*c^2 - b*d^2*x^2)^(1/2)*(c + d*x)^(1/2),x)
Output:
-((b*c^2 - b*d^2*x^2)^(1/2)*((16*c^4*(c + d*x)^(1/2))/(63*d^4) - (2*x^4*(c + d*x)^(1/2))/9 + (2*c^2*x^2*(c + d*x)^(1/2))/(21*d^2) - (16*c*x^3*(c + d *x)^(1/2))/(63*d) + (8*c^3*x*(c + d*x)^(1/2))/(63*d^3)))/(x + c/d)
Time = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.34 \[ \int x^2 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=\frac {2 \sqrt {b}\, \sqrt {-d x +c}\, \left (7 d^{4} x^{4}+8 c \,d^{3} x^{3}-3 c^{2} d^{2} x^{2}-4 c^{3} d x -8 c^{4}\right )}{63 d^{3}} \] Input:
int(x^2*(d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2),x)
Output:
(2*sqrt(b)*sqrt(c - d*x)*( - 8*c**4 - 4*c**3*d*x - 3*c**2*d**2*x**2 + 8*c* d**3*x**3 + 7*d**4*x**4))/(63*d**3)