Integrand size = 26, antiderivative size = 52 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {16 c d^3}{3 \sqrt {a+b x+c x^2}} \] Output:
-2/3*d^3*(2*c*x+b)^2/(c*x^2+b*x+a)^(3/2)-16/3*c*d^3/(c*x^2+b*x+a)^(1/2)
Time = 0.84 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d^3 \left (b^2+12 b c x+4 c \left (2 a+3 c x^2\right )\right )}{3 (a+x (b+c x))^{3/2}} \] Input:
Integrate[(b*d + 2*c*d*x)^3/(a + b*x + c*x^2)^(5/2),x]
Output:
(-2*d^3*(b^2 + 12*b*c*x + 4*c*(2*a + 3*c*x^2)))/(3*(a + x*(b + c*x))^(3/2) )
Time = 0.20 (sec) , antiderivative size = 52, 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 = {1110, 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.
\(\displaystyle \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1110 |
\(\displaystyle \frac {8}{3} c d^2 \int \frac {d (b+2 c x)}{\left (c x^2+b x+a\right )^{3/2}}dx-\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {8}{3} c d^3 \int \frac {b+2 c x}{\left (c x^2+b x+a\right )^{3/2}}dx-\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1104 |
\(\displaystyle -\frac {16 c d^3}{3 \sqrt {a+b x+c x^2}}-\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}}\) |
Input:
Int[(b*d + 2*c*d*x)^3/(a + b*x + c*x^2)^(5/2),x]
Output:
(-2*d^3*(b + 2*c*x)^2)/(3*(a + b*x + c*x^2)^(3/2)) - (16*c*d^3)/(3*Sqrt[a + b*x + c*x^2])
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_Sy mbol] :> Simp[d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), x] - Simp[d*e*((m - 1)/(b*(p + 1))) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^ 2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0] && N eQ[m + 2*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
Time = 0.93 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.75
method | result | size |
gosper | \(-\frac {2 d^{3} \left (12 c^{2} x^{2}+12 c b x +8 a c +b^{2}\right )}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\) | \(39\) |
trager | \(-\frac {2 d^{3} \left (12 c^{2} x^{2}+12 c b x +8 a c +b^{2}\right )}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\) | \(39\) |
pseudoelliptic | \(-\frac {2 d^{3} \left (12 c^{2} x^{2}+12 c b x +8 a c +b^{2}\right )}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\) | \(39\) |
orering | \(-\frac {2 \left (12 c^{2} x^{2}+12 c b x +8 a c +b^{2}\right ) \left (2 c d x +b d \right )^{3}}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} \left (2 c x +b \right )^{3}}\) | \(55\) |
default | \(d^{3} \left (b^{3} \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )+8 c^{3} \left (-\frac {x^{2}}{c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {b \left (-\frac {x}{2 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{4 c}+\frac {a \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{2 c}+\frac {2 a \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{c}\right )+12 b \,c^{2} \left (-\frac {x}{2 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{4 c}+\frac {a \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )+6 b^{2} c \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )\right )\) | \(680\) |
Input:
int((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*d^3*(12*c^2*x^2+12*b*c*x+8*a*c+b^2)/(c*x^2+b*x+a)^(3/2)
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.60 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (12 \, c^{2} d^{3} x^{2} + 12 \, b c d^{3} x + {\left (b^{2} + 8 \, a c\right )} d^{3}\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \] Input:
integrate((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")
Output:
-2/3*(12*c^2*d^3*x^2 + 12*b*c*d^3*x + (b^2 + 8*a*c)*d^3)*sqrt(c*x^2 + b*x + a)/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)
Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (51) = 102\).
Time = 0.47 (sec) , antiderivative size = 320, normalized size of antiderivative = 6.15 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\begin {cases} - \frac {16 a c d^{3}}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} - \frac {2 b^{2} d^{3}}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} - \frac {24 b c d^{3} x}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} - \frac {24 c^{2} d^{3} x^{2}}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} & \text {for}\: a \neq - x \left (b + c x\right ) \\\tilde {\infty } b^{3} d^{3} x + \tilde {\infty } b^{2} c d^{3} x^{2} + \tilde {\infty } b c^{2} d^{3} x^{3} + \tilde {\infty } c^{3} d^{3} x^{4} & \text {otherwise} \end {cases} \] Input:
integrate((2*c*d*x+b*d)**3/(c*x**2+b*x+a)**(5/2),x)
Output:
Piecewise((-16*a*c*d**3/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x + c*x**2)) - 2*b**2*d**3/(3*a*sqrt(a + b* x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x + c*x** 2)) - 24*b*c*d**3*x/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x **2) + 3*c*x**2*sqrt(a + b*x + c*x**2)) - 24*c**2*d**3*x**2/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x + c*x **2)), Ne(a, -x*(b + c*x))), (zoo*b**3*d**3*x + zoo*b**2*c*d**3*x**2 + zoo *b*c**2*d**3*x**3 + zoo*c**3*d**3*x**4, True))
Exception generated. \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),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 = 0.39 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.29 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (b^{2} d^{7} - 4 \, a c d^{7} + 12 \, {\left (a d^{2} + {\left (c d x^{2} + b d x\right )} d\right )} c d^{5}\right )}}{3 \, {\left (a d^{2} + {\left (c d x^{2} + b d x\right )} d\right )}^{\frac {3}{2}} {\left | d \right |}} \] Input:
integrate((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")
Output:
-2/3*(b^2*d^7 - 4*a*c*d^7 + 12*(a*d^2 + (c*d*x^2 + b*d*x)*d)*c*d^5)/((a*d^ 2 + (c*d*x^2 + b*d*x)*d)^(3/2)*abs(d))
Time = 6.37 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.19 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2\,b^2\,d^3+24\,c\,d^3\,\left (c\,x^2+b\,x+a\right )-8\,a\,c\,d^3}{\sqrt {c\,x^2+b\,x+a}\,\left (3\,c\,x^2+3\,b\,x+3\,a\right )} \] Input:
int((b*d + 2*c*d*x)^3/(a + b*x + c*x^2)^(5/2),x)
Output:
-(2*b^2*d^3 + 24*c*d^3*(a + b*x + c*x^2) - 8*a*c*d^3)/((a + b*x + c*x^2)^( 1/2)*(3*a + 3*b*x + 3*c*x^2))
Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.58 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {c \,x^{2}+b x +a}\, d^{3} \left (-12 c^{2} x^{2}-12 b c x -8 a c -b^{2}\right )}{3 c^{2} x^{4}+6 b c \,x^{3}+6 a c \,x^{2}+3 b^{2} x^{2}+6 a b x +3 a^{2}} \] Input:
int((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x)
Output:
(2*sqrt(a + b*x + c*x**2)*d**3*( - 8*a*c - b**2 - 12*b*c*x - 12*c**2*x**2) )/(3*(a**2 + 2*a*b*x + 2*a*c*x**2 + b**2*x**2 + 2*b*c*x**3 + c**2*x**4))