Integrand size = 26, antiderivative size = 59 \[ \int (b d+2 c d x)^3 \sqrt {a+b x+c x^2} \, dx=\frac {4}{15} \left (b^2-4 a c\right ) d^3 \left (a+b x+c x^2\right )^{3/2}+\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2} \] Output:
4/15*(-4*a*c+b^2)*d^3*(c*x^2+b*x+a)^(3/2)+2/5*d^3*(2*c*x+b)^2*(c*x^2+b*x+a )^(3/2)
Time = 0.89 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.36 \[ \int (b d+2 c d x)^3 \sqrt {a+b x+c x^2} \, dx=-\frac {2}{15} d^3 \left (-5 a b^2+8 a^2 c-5 b^3 x-4 a b c x-17 b^2 c x^2-4 a c^2 x^2-24 b c^2 x^3-12 c^3 x^4\right ) \sqrt {a+x (b+c x)} \] Input:
Integrate[(b*d + 2*c*d*x)^3*Sqrt[a + b*x + c*x^2],x]
Output:
(-2*d^3*(-5*a*b^2 + 8*a^2*c - 5*b^3*x - 4*a*b*c*x - 17*b^2*c*x^2 - 4*a*c^2 *x^2 - 24*b*c^2*x^3 - 12*c^3*x^4)*Sqrt[a + x*(b + c*x)])/15
Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1116, 27, 1104}
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.
\(\displaystyle \int \sqrt {a+b x+c x^2} (b d+2 c d x)^3 \, dx\) |
\(\Big \downarrow \) 1116 |
\(\displaystyle \frac {2}{5} d^2 \left (b^2-4 a c\right ) \int d (b+2 c x) \sqrt {c x^2+b x+a}dx+\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{5} d^3 \left (b^2-4 a c\right ) \int (b+2 c x) \sqrt {c x^2+b x+a}dx+\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}\) |
\(\Big \downarrow \) 1104 |
\(\displaystyle \frac {4}{15} d^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}+\frac {2}{5} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{3/2}\) |
Input:
Int[(b*d + 2*c*d*x)^3*Sqrt[a + b*x + c*x^2],x]
Output:
(4*(b^2 - 4*a*c)*d^3*(a + b*x + c*x^2)^(3/2))/15 + (2*d^3*(b + 2*c*x)^2*(a + b*x + c*x^2)^(3/2))/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol ] :> Simp[d*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || OddQ[m])
Time = 0.87 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69
method | result | size |
gosper | \(-\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (-12 c^{2} x^{2}-12 c b x +8 a c -5 b^{2}\right ) d^{3}}{15}\) | \(41\) |
pseudoelliptic | \(-\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (-12 c^{2} x^{2}-12 c b x +8 a c -5 b^{2}\right ) d^{3}}{15}\) | \(41\) |
orering | \(-\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (-12 c^{2} x^{2}-12 c b x +8 a c -5 b^{2}\right ) \left (2 c d x +b d \right )^{3}}{15 \left (2 c x +b \right )^{3}}\) | \(57\) |
trager | \(d^{3} \left (\frac {8}{5} c^{3} x^{4}+\frac {16}{5} b \,c^{2} x^{3}+\frac {8}{15} a \,c^{2} x^{2}+\frac {34}{15} b^{2} c \,x^{2}+\frac {8}{15} a b c x +\frac {2}{3} b^{3} x -\frac {16}{15} a^{2} c +\frac {2}{3} a \,b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\) | \(77\) |
risch | \(-\frac {2 d^{3} \left (-12 c^{3} x^{4}-24 b \,c^{2} x^{3}-4 a \,c^{2} x^{2}-17 b^{2} c \,x^{2}-4 a b c x -5 b^{3} x +8 a^{2} c -5 a \,b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{15}\) | \(78\) |
default | \(d^{3} \left (b^{3} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )+8 c^{3} \left (\frac {x^{2} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{5 c}-\frac {7 b \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )}{10 c}-\frac {2 a \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{5 c}\right )+12 b \,c^{2} \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )+6 b^{2} c \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )\right )\) | \(666\) |
Input:
int((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2/15*(c*x^2+b*x+a)^(3/2)*(-12*c^2*x^2-12*b*c*x+8*a*c-5*b^2)*d^3
Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.54 \[ \int (b d+2 c d x)^3 \sqrt {a+b x+c x^2} \, dx=\frac {2}{15} \, {\left (12 \, c^{3} d^{3} x^{4} + 24 \, b c^{2} d^{3} x^{3} + {\left (17 \, b^{2} c + 4 \, a c^{2}\right )} d^{3} x^{2} + {\left (5 \, b^{3} + 4 \, a b c\right )} d^{3} x + {\left (5 \, a b^{2} - 8 \, a^{2} c\right )} d^{3}\right )} \sqrt {c x^{2} + b x + a} \] Input:
integrate((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
Output:
2/15*(12*c^3*d^3*x^4 + 24*b*c^2*d^3*x^3 + (17*b^2*c + 4*a*c^2)*d^3*x^2 + ( 5*b^3 + 4*a*b*c)*d^3*x + (5*a*b^2 - 8*a^2*c)*d^3)*sqrt(c*x^2 + b*x + a)
Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (58) = 116\).
Time = 0.10 (sec) , antiderivative size = 216, normalized size of antiderivative = 3.66 \[ \int (b d+2 c d x)^3 \sqrt {a+b x+c x^2} \, dx=- \frac {16 a^{2} c d^{3} \sqrt {a + b x + c x^{2}}}{15} + \frac {2 a b^{2} d^{3} \sqrt {a + b x + c x^{2}}}{3} + \frac {8 a b c d^{3} x \sqrt {a + b x + c x^{2}}}{15} + \frac {8 a c^{2} d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{15} + \frac {2 b^{3} d^{3} x \sqrt {a + b x + c x^{2}}}{3} + \frac {34 b^{2} c d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{15} + \frac {16 b c^{2} d^{3} x^{3} \sqrt {a + b x + c x^{2}}}{5} + \frac {8 c^{3} d^{3} x^{4} \sqrt {a + b x + c x^{2}}}{5} \] Input:
integrate((2*c*d*x+b*d)**3*(c*x**2+b*x+a)**(1/2),x)
Output:
-16*a**2*c*d**3*sqrt(a + b*x + c*x**2)/15 + 2*a*b**2*d**3*sqrt(a + b*x + c *x**2)/3 + 8*a*b*c*d**3*x*sqrt(a + b*x + c*x**2)/15 + 8*a*c**2*d**3*x**2*s qrt(a + b*x + c*x**2)/15 + 2*b**3*d**3*x*sqrt(a + b*x + c*x**2)/3 + 34*b** 2*c*d**3*x**2*sqrt(a + b*x + c*x**2)/15 + 16*b*c**2*d**3*x**3*sqrt(a + b*x + c*x**2)/5 + 8*c**3*d**3*x**4*sqrt(a + b*x + c*x**2)/5
Exception generated. \[ \int (b d+2 c d x)^3 \sqrt {a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.98 \[ \int (b d+2 c d x)^3 \sqrt {a+b x+c x^2} \, dx=\frac {2}{3} \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} b^{2} d^{3} + \frac {8}{5} \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} c d^{3} - \frac {8}{3} \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} a c d^{3} \] Input:
integrate((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
Output:
2/3*(c*x^2 + b*x + a)^(3/2)*b^2*d^3 + 8/5*(c*x^2 + b*x + a)^(5/2)*c*d^3 - 8/3*(c*x^2 + b*x + a)^(3/2)*a*c*d^3
Time = 5.38 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.51 \[ \int (b d+2 c d x)^3 \sqrt {a+b x+c x^2} \, dx=\sqrt {c\,x^2+b\,x+a}\,\left (\frac {8\,c^3\,d^3\,x^4}{5}-\frac {2\,a\,d^3\,\left (8\,a\,c-5\,b^2\right )}{15}+\frac {2\,c\,d^3\,x^2\,\left (17\,b^2+4\,a\,c\right )}{15}+\frac {16\,b\,c^2\,d^3\,x^3}{5}+\frac {2\,b\,d^3\,x\,\left (5\,b^2+4\,a\,c\right )}{15}\right ) \] Input:
int((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(1/2),x)
Output:
(a + b*x + c*x^2)^(1/2)*((8*c^3*d^3*x^4)/5 - (2*a*d^3*(8*a*c - 5*b^2))/15 + (2*c*d^3*x^2*(4*a*c + 17*b^2))/15 + (16*b*c^2*d^3*x^3)/5 + (2*b*d^3*x*(4 *a*c + 5*b^2))/15)
Time = 0.20 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29 \[ \int (b d+2 c d x)^3 \sqrt {a+b x+c x^2} \, dx=\frac {2 \sqrt {c \,x^{2}+b x +a}\, d^{3} \left (12 c^{3} x^{4}+24 b \,c^{2} x^{3}+4 a \,c^{2} x^{2}+17 b^{2} c \,x^{2}+4 a b c x +5 b^{3} x -8 a^{2} c +5 a \,b^{2}\right )}{15} \] Input:
int((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^(1/2),x)
Output:
(2*sqrt(a + b*x + c*x**2)*d**3*( - 8*a**2*c + 5*a*b**2 + 4*a*b*c*x + 4*a*c **2*x**2 + 5*b**3*x + 17*b**2*c*x**2 + 24*b*c**2*x**3 + 12*c**3*x**4))/15