Integrand size = 26, antiderivative size = 76 \[ \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2}{\left (b^2-4 a c\right ) d^2 (b+2 c x) \sqrt {a+b x+c x^2}}-\frac {16 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d^2 (b+2 c x)} \] Output:
-2/(-4*a*c+b^2)/d^2/(2*c*x+b)/(c*x^2+b*x+a)^(1/2)-16*c*(c*x^2+b*x+a)^(1/2) /(-4*a*c+b^2)^2/d^2/(2*c*x+b)
Result contains higher order function than in optimal. Order 3 vs. order 2 in optimal.
Time = 3.06 (sec) , antiderivative size = 463, normalized size of antiderivative = 6.09 \[ \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {\frac {\left (2 a+x (b+2 c x)-2 \sqrt {a} \sqrt {a+x (b+c x)}\right )^3}{a^{3/2} c (b+2 c x) \left (2 a+b x-2 \sqrt {a} \sqrt {a+x (b+c x)}\right ) \left (8 a^2+b^2 x^2+4 a x (2 b+c x)-8 a^{3/2} \sqrt {a+x (b+c x)}-4 \sqrt {a} b x \sqrt {a+x (b+c x)}\right )}+\frac {4 x^3 \left (-\sqrt {a}+\sqrt {a+x (b+c x)}\right )^3}{a^{3/2} \left (2 a^{3/2}+2 \sqrt {a} x (b+c x)-2 a \sqrt {a+x (b+c x)}-b x \sqrt {a+x (b+c x)}\right ) \left (8 a^2+b^2 x^2+4 a x (2 b+c x)-8 a^{3/2} \sqrt {a+x (b+c x)}-4 \sqrt {a} b x \sqrt {a+x (b+c x)}\right )}-\frac {16 \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {b^2-4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{b \left (b^2-4 a c\right )^{3/2}}-\frac {16 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {-b^2+4 a c} x}{\sqrt {a} (b+2 c x)-b \sqrt {a+x (b+c x)}}\right )}{b \left (-b^2+4 a c\right )^{3/2}}}{2 d^2} \] Input:
Integrate[1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(3/2)),x]
Output:
((2*a + x*(b + 2*c*x) - 2*Sqrt[a]*Sqrt[a + x*(b + c*x)])^3/(a^(3/2)*c*(b + 2*c*x)*(2*a + b*x - 2*Sqrt[a]*Sqrt[a + x*(b + c*x)])*(8*a^2 + b^2*x^2 + 4 *a*x*(2*b + c*x) - 8*a^(3/2)*Sqrt[a + x*(b + c*x)] - 4*Sqrt[a]*b*x*Sqrt[a + x*(b + c*x)])) + (4*x^3*(-Sqrt[a] + Sqrt[a + x*(b + c*x)])^3)/(a^(3/2)*( 2*a^(3/2) + 2*Sqrt[a]*x*(b + c*x) - 2*a*Sqrt[a + x*(b + c*x)] - b*x*Sqrt[a + x*(b + c*x)])*(8*a^2 + b^2*x^2 + 4*a*x*(2*b + c*x) - 8*a^(3/2)*Sqrt[a + x*(b + c*x)] - 4*Sqrt[a]*b*x*Sqrt[a + x*(b + c*x)])) - (16*Sqrt[c]*ArcTan [(Sqrt[c]*Sqrt[b^2 - 4*a*c]*x)/(Sqrt[a]*(b + 2*c*x) - b*Sqrt[a + x*(b + c* x)])])/(b*(b^2 - 4*a*c)^(3/2)) - (16*Sqrt[c]*ArcTanh[(Sqrt[c]*Sqrt[-b^2 + 4*a*c]*x)/(Sqrt[a]*(b + 2*c*x) - b*Sqrt[a + x*(b + c*x)])])/(b*(-b^2 + 4*a *c)^(3/2)))/(2*d^2)
Time = 0.21 (sec) , antiderivative size = 76, 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 = {1111, 27, 1106}
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.
\(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^2} \, dx\) |
\(\Big \downarrow \) 1111 |
\(\displaystyle -\frac {8 c \int \frac {1}{d^2 (b+2 c x)^2 \sqrt {c x^2+b x+a}}dx}{b^2-4 a c}-\frac {2}{d^2 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {8 c \int \frac {1}{(b+2 c x)^2 \sqrt {c x^2+b x+a}}dx}{d^2 \left (b^2-4 a c\right )}-\frac {2}{d^2 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 1106 |
\(\displaystyle -\frac {16 c \sqrt {a+b x+c x^2}}{d^2 \left (b^2-4 a c\right )^2 (b+2 c x)}-\frac {2}{d^2 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a+b x+c x^2}}\) |
Input:
Int[1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(3/2)),x]
Output:
-2/((b^2 - 4*a*c)*d^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2]) - (16*c*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)^2*d^2*(b + 2*c*x))
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_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*c*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)* (b^2 - 4*a*c))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*c*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)* (b^2 - 4*a*c))), x] - Simp[2*c*e*((m + 2*p + 3)/(e*(p + 1)*(b^2 - 4*a*c))) Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e , m}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] && !G tQ[m, 1] && RationalQ[m] && IntegerQ[2*p]
Time = 1.22 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.89
method | result | size |
gosper | \(-\frac {2 \left (8 c^{2} x^{2}+8 c b x +4 a c +b^{2}\right )}{\left (2 c x +b \right ) d^{2} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \sqrt {c \,x^{2}+b x +a}}\) | \(68\) |
orering | \(-\frac {2 \left (8 c^{2} x^{2}+8 c b x +4 a c +b^{2}\right ) \left (2 c x +b \right )}{\left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \sqrt {c \,x^{2}+b x +a}\, \left (2 c d x +b d \right )^{2}}\) | \(74\) |
trager | \(-\frac {2 \left (8 c^{2} x^{2}+8 c b x +4 a c +b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{d^{2} \left (4 a c -b^{2}\right )^{2} \left (2 x^{3} c^{2}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right )}\) | \(82\) |
default | \(\frac {-\frac {4 c}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right ) \sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}}-\frac {32 c^{3} \left (x +\frac {b}{2 c}\right )}{\left (4 a c -b^{2}\right )^{2} \sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}}}{4 d^{2} c^{2}}\) | \(120\) |
Input:
int(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2*(8*c^2*x^2+8*b*c*x+4*a*c+b^2)/(2*c*x+b)/d^2/(16*a^2*c^2-8*a*b^2*c+b^4)/ (c*x^2+b*x+a)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (72) = 144\).
Time = 0.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.04 \[ \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d^{2} x^{3} + 3 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d^{2} x^{2} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} d^{2} x + {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} d^{2}} \] Input:
integrate(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")
Output:
-2*(8*c^2*x^2 + 8*b*c*x + b^2 + 4*a*c)*sqrt(c*x^2 + b*x + a)/(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^2*x^3 + 3*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3 )*d^2*x^2 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d^2*x + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d^2)
\[ \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {\int \frac {1}{a b^{2} \sqrt {a + b x + c x^{2}} + 4 a b c x \sqrt {a + b x + c x^{2}} + 4 a c^{2} x^{2} \sqrt {a + b x + c x^{2}} + b^{3} x \sqrt {a + b x + c x^{2}} + 5 b^{2} c x^{2} \sqrt {a + b x + c x^{2}} + 8 b c^{2} x^{3} \sqrt {a + b x + c x^{2}} + 4 c^{3} x^{4} \sqrt {a + b x + c x^{2}}}\, dx}{d^{2}} \] Input:
integrate(1/(2*c*d*x+b*d)**2/(c*x**2+b*x+a)**(3/2),x)
Output:
Integral(1/(a*b**2*sqrt(a + b*x + c*x**2) + 4*a*b*c*x*sqrt(a + b*x + c*x** 2) + 4*a*c**2*x**2*sqrt(a + b*x + c*x**2) + b**3*x*sqrt(a + b*x + c*x**2) + 5*b**2*c*x**2*sqrt(a + b*x + c*x**2) + 8*b*c**2*x**3*sqrt(a + b*x + c*x* *2) + 4*c**3*x**4*sqrt(a + b*x + c*x**2)), x)/d**2
Exception generated. \[ \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(3/2),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
Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (72) = 144\).
Time = 0.38 (sec) , antiderivative size = 266, normalized size of antiderivative = 3.50 \[ \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {4 \, {\left (\frac {c^{2} d^{4} {\left (\frac {\sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c} c}{b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right ) - 4 \, a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )} + \frac {c^{2}}{{\left (b^{2} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right ) - 4 \, a c \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )\right )} \sqrt {-\frac {b^{2} c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac {4 \, a c^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + c}}\right )} {\left | c \right |}}{b^{2} c^{3} d^{4} - 4 \, a c^{4} d^{4}} - \frac {2 \, \sqrt {c} {\left | c \right |} \mathrm {sgn}\left (\frac {1}{2 \, c d x + b d}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}\right )}}{d^{2} {\left | c \right |}} \] Input:
integrate(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")
Output:
-4*(c^2*d^4*(sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b* d)^2 + c)*c/(b^2*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 4*a*c*sgn(1/(2*c*d *x + b*d))*sgn(c)*sgn(d)) + c^2/((b^2*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d) - 4*a*c*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d))*sqrt(-b^2*c*d^2/(2*c*d*x + b*d)^2 + 4*a*c^2*d^2/(2*c*d*x + b*d)^2 + c)))*abs(c)/(b^2*c^3*d^4 - 4*a*c^ 4*d^4) - 2*sqrt(c)*abs(c)*sgn(1/(2*c*d*x + b*d))*sgn(c)*sgn(d)/(b^4 - 8*a* b^2*c + 16*a^2*c^2))/(d^2*abs(c))
Time = 5.89 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.47 \[ \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {\left (\frac {2\,\left (4\,c^2+\frac {64\,b^2\,c^5}{32\,a\,c^4-8\,b^2\,c^3}\right )}{d^2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}+\frac {512\,c^7\,x^2}{d^2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )\,\left (32\,a\,c^4-8\,b^2\,c^3\right )}+\frac {512\,b\,c^6\,x}{d^2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )\,\left (32\,a\,c^4-8\,b^2\,c^3\right )}\right )\,\sqrt {c\,x^2+b\,x+a}}{x\,\left (b^2+2\,a\,c\right )+a\,b+2\,c^2\,x^3+3\,b\,c\,x^2} \] Input:
int(1/((b*d + 2*c*d*x)^2*(a + b*x + c*x^2)^(3/2)),x)
Output:
-(((2*(4*c^2 + (64*b^2*c^5)/(32*a*c^4 - 8*b^2*c^3)))/(d^2*(16*a*c^3 - 4*b^ 2*c^2)) + (512*c^7*x^2)/(d^2*(16*a*c^3 - 4*b^2*c^2)*(32*a*c^4 - 8*b^2*c^3) ) + (512*b*c^6*x)/(d^2*(16*a*c^3 - 4*b^2*c^2)*(32*a*c^4 - 8*b^2*c^3)))*(a + b*x + c*x^2)^(1/2))/(x*(2*a*c + b^2) + a*b + 2*c^2*x^3 + 3*b*c*x^2)
Time = 0.19 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.99 \[ \int \frac {1}{(b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {-8 \sqrt {c \,x^{2}+b x +a}\, a c -2 \sqrt {c \,x^{2}+b x +a}\, b^{2}-16 \sqrt {c \,x^{2}+b x +a}\, b c x -16 \sqrt {c \,x^{2}+b x +a}\, c^{2} x^{2}-8 \sqrt {c}\, a b -16 \sqrt {c}\, a c x -8 \sqrt {c}\, b^{2} x -24 \sqrt {c}\, b c \,x^{2}-16 \sqrt {c}\, c^{2} x^{3}}{d^{2} \left (32 a^{2} c^{4} x^{3}-16 a \,b^{2} c^{3} x^{3}+2 b^{4} c^{2} x^{3}+48 a^{2} b \,c^{3} x^{2}-24 a \,b^{3} c^{2} x^{2}+3 b^{5} c \,x^{2}+32 a^{3} c^{3} x -6 a \,b^{4} c x +b^{6} x +16 a^{3} b \,c^{2}-8 a^{2} b^{3} c +a \,b^{5}\right )} \] Input:
int(1/(2*c*d*x+b*d)^2/(c*x^2+b*x+a)^(3/2),x)
Output:
(2*( - 4*sqrt(a + b*x + c*x**2)*a*c - sqrt(a + b*x + c*x**2)*b**2 - 8*sqrt (a + b*x + c*x**2)*b*c*x - 8*sqrt(a + b*x + c*x**2)*c**2*x**2 - 4*sqrt(c)* a*b - 8*sqrt(c)*a*c*x - 4*sqrt(c)*b**2*x - 12*sqrt(c)*b*c*x**2 - 8*sqrt(c) *c**2*x**3))/(d**2*(16*a**3*b*c**2 + 32*a**3*c**3*x - 8*a**2*b**3*c + 48*a **2*b*c**3*x**2 + 32*a**2*c**4*x**3 + a*b**5 - 6*a*b**4*c*x - 24*a*b**3*c* *2*x**2 - 16*a*b**2*c**3*x**3 + b**6*x + 3*b**5*c*x**2 + 2*b**4*c**2*x**3) )