Integrand size = 24, antiderivative size = 112 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^3} \, dx=-\frac {1}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {4 c}{\left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}-\frac {32 c^2 \log (b+2 c x)}{\left (b^2-4 a c\right )^3 d}+\frac {16 c^2 \log \left (a+b x+c x^2\right )}{\left (b^2-4 a c\right )^3 d} \] Output:
-1/2/(-4*a*c+b^2)/d/(c*x^2+b*x+a)^2+4*c/(-4*a*c+b^2)^2/d/(c*x^2+b*x+a)-32* c^2*ln(2*c*x+b)/(-4*a*c+b^2)^3/d+16*c^2*ln(c*x^2+b*x+a)/(-4*a*c+b^2)^3/d
Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^3} \, dx=-\frac {\frac {\left (b^2-4 a c\right ) \left (b^2-8 b c x-4 c \left (3 a+2 c x^2\right )\right )}{(a+x (b+c x))^2}+64 c^2 \log (b+2 c x)-32 c^2 \log (a+x (b+c x))}{2 \left (b^2-4 a c\right )^3 d} \] Input:
Integrate[1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^3),x]
Output:
-1/2*(((b^2 - 4*a*c)*(b^2 - 8*b*c*x - 4*c*(3*a + 2*c*x^2)))/(a + x*(b + c* x))^2 + 64*c^2*Log[b + 2*c*x] - 32*c^2*Log[a + x*(b + c*x)])/((b^2 - 4*a*c )^3*d)
Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1111, 27, 1111, 1105, 16, 1103}
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.
| \(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^3 (b d+2 c d x)} \, dx\) |
| \(\Big \downarrow \) 1111 |
| \(\displaystyle -\frac {4 c \int \frac {1}{d (b+2 c x) \left (c x^2+b x+a\right )^2}dx}{b^2-4 a c}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {4 c \int \frac {1}{(b+2 c x) \left (c x^2+b x+a\right )^2}dx}{d \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 1111 |
| \(\displaystyle -\frac {4 c \left (-\frac {4 c \int \frac {1}{(b+2 c x) \left (c x^2+b x+a\right )}dx}{b^2-4 a c}-\frac {1}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )}{d \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 1105 |
| \(\displaystyle -\frac {4 c \left (-\frac {4 c \left (\frac {\int \frac {b+2 c x}{c x^2+b x+a}dx}{b^2-4 a c}-\frac {4 c \int \frac {1}{b+2 c x}dx}{b^2-4 a c}\right )}{b^2-4 a c}-\frac {1}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )}{d \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {4 c \left (-\frac {4 c \left (\frac {\int \frac {b+2 c x}{c x^2+b x+a}dx}{b^2-4 a c}-\frac {2 \log (b+2 c x)}{b^2-4 a c}\right )}{b^2-4 a c}-\frac {1}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\right )}{d \left (b^2-4 a c\right )}-\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle -\frac {1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {4 c \left (-\frac {1}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {4 c \left (\frac {\log \left (a+b x+c x^2\right )}{b^2-4 a c}-\frac {2 \log (b+2 c x)}{b^2-4 a c}\right )}{b^2-4 a c}\right )}{d \left (b^2-4 a c\right )}\) |
Input:
Int[1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^3),x]
Output:
-1/2*1/((b^2 - 4*a*c)*d*(a + b*x + c*x^2)^2) - (4*c*(-(1/((b^2 - 4*a*c)*(a + b*x + c*x^2))) - (4*c*((-2*Log[b + 2*c*x])/(b^2 - 4*a*c) + Log[a + b*x + c*x^2]/(b^2 - 4*a*c)))/(b^2 - 4*a*c)))/((b^2 - 4*a*c)*d)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, 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_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[1/(((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Simp[-4*b*(c/(d*(b^2 - 4*a*c))) Int[1/(b + 2*c*x), x], x] + Simp[b^2/( d^2*(b^2 - 4*a*c)) Int[(d + e*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b , c, d, e}, 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*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 = 1.00 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.14
| method | result | size |
| default | \(\frac {\frac {32 c^{2} \ln \left (2 c x +b \right )}{\left (4 a c -b^{2}\right )^{3}}-\frac {\frac {-4 c^{2} \left (4 a c -b^{2}\right ) x^{2}-4 b c \left (4 a c -b^{2}\right ) x -24 a^{2} c^{2}+8 c a \,b^{2}-\frac {b^{4}}{2}}{\left (c \,x^{2}+b x +a \right )^{2}}+16 c^{2} \ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right )^{3}}}{d}\) | \(128\) |
| risch | \(\frac {\frac {4 c^{2} x^{2}}{16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}}+\frac {4 b c x}{16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}}+\frac {12 a c -b^{2}}{32 a^{2} c^{2}-16 c a \,b^{2}+2 b^{4}}}{d \left (c \,x^{2}+b x +a \right )^{2}}+\frac {32 c^{2} \ln \left (2 c x +b \right )}{d \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}-\frac {16 c^{2} \ln \left (c \,x^{2}+b x +a \right )}{d \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}\) | \(209\) |
| norman | \(\frac {\frac {4 c^{2} x^{2}}{d \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}+\frac {12 c^{3} a -b^{2} c^{2}}{2 c^{2} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) d}+\frac {4 c b x}{d \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}}{\left (c \,x^{2}+b x +a \right )^{2}}+\frac {32 c^{2} \ln \left (2 c x +b \right )}{d \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}-\frac {16 c^{2} \ln \left (c \,x^{2}+b x +a \right )}{d \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}\) | \(223\) |
| parallelrisch | \(\frac {b^{4} c^{2}-16 a \,b^{2} c^{3}+48 a^{2} c^{4}+32 a b \,c^{4} x +128 \ln \left (\frac {b}{2}+c x \right ) x^{3} b \,c^{5}-64 \ln \left (c \,x^{2}+b x +a \right ) x^{3} b \,c^{5}+128 \ln \left (\frac {b}{2}+c x \right ) x^{2} a \,c^{5}+64 \ln \left (\frac {b}{2}+c x \right ) x^{2} b^{2} c^{4}-64 \ln \left (c \,x^{2}+b x +a \right ) x^{2} a \,c^{5}-32 \ln \left (c \,x^{2}+b x +a \right ) x^{2} b^{2} c^{4}-8 b^{3} c^{3} x -8 b^{2} c^{4} x^{2}+128 \ln \left (\frac {b}{2}+c x \right ) x a b \,c^{4}-64 \ln \left (c \,x^{2}+b x +a \right ) x a b \,c^{4}+32 a \,c^{5} x^{2}-32 \ln \left (c \,x^{2}+b x +a \right ) x^{4} c^{6}+64 \ln \left (\frac {b}{2}+c x \right ) a^{2} c^{4}-32 \ln \left (c \,x^{2}+b x +a \right ) a^{2} c^{4}+64 \ln \left (\frac {b}{2}+c x \right ) x^{4} c^{6}}{2 \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \left (c \,x^{2}+b x +a \right )^{2} c^{2} d}\) | \(337\) |
Input:
int(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(32*c^2/(4*a*c-b^2)^3*ln(2*c*x+b)-1/(4*a*c-b^2)^3*((-4*c^2*(4*a*c-b^2) *x^2-4*b*c*(4*a*c-b^2)*x-24*a^2*c^2+8*c*a*b^2-1/2*b^4)/(c*x^2+b*x+a)^2+16* c^2*ln(c*x^2+b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (110) = 220\).
Time = 0.08 (sec) , antiderivative size = 386, normalized size of antiderivative = 3.45 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^3} \, dx=-\frac {b^{4} - 16 \, a b^{2} c + 48 \, a^{2} c^{2} - 8 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} - 8 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x - 32 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{3} + 2 \, a b c^{2} x + a^{2} c^{2} + {\left (b^{2} c^{2} + 2 \, a c^{3}\right )} x^{2}\right )} \log \left (c x^{2} + b x + a\right ) + 64 \, {\left (c^{4} x^{4} + 2 \, b c^{3} x^{3} + 2 \, a b c^{2} x + a^{2} c^{2} + {\left (b^{2} c^{2} + 2 \, a c^{3}\right )} x^{2}\right )} \log \left (2 \, c x + b\right )}{2 \, {\left ({\left (b^{6} c^{2} - 12 \, a b^{4} c^{3} + 48 \, a^{2} b^{2} c^{4} - 64 \, a^{3} c^{5}\right )} d x^{4} + 2 \, {\left (b^{7} c - 12 \, a b^{5} c^{2} + 48 \, a^{2} b^{3} c^{3} - 64 \, a^{3} b c^{4}\right )} d x^{3} + {\left (b^{8} - 10 \, a b^{6} c + 24 \, a^{2} b^{4} c^{2} + 32 \, a^{3} b^{2} c^{3} - 128 \, a^{4} c^{4}\right )} d x^{2} + 2 \, {\left (a b^{7} - 12 \, a^{2} b^{5} c + 48 \, a^{3} b^{3} c^{2} - 64 \, a^{4} b c^{3}\right )} d x + {\left (a^{2} b^{6} - 12 \, a^{3} b^{4} c + 48 \, a^{4} b^{2} c^{2} - 64 \, a^{5} c^{3}\right )} d\right )}} \] Input:
integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^3,x, algorithm="fricas")
Output:
-1/2*(b^4 - 16*a*b^2*c + 48*a^2*c^2 - 8*(b^2*c^2 - 4*a*c^3)*x^2 - 8*(b^3*c - 4*a*b*c^2)*x - 32*(c^4*x^4 + 2*b*c^3*x^3 + 2*a*b*c^2*x + a^2*c^2 + (b^2 *c^2 + 2*a*c^3)*x^2)*log(c*x^2 + b*x + a) + 64*(c^4*x^4 + 2*b*c^3*x^3 + 2* a*b*c^2*x + a^2*c^2 + (b^2*c^2 + 2*a*c^3)*x^2)*log(2*c*x + b))/((b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*d*x^4 + 2*(b^7*c - 12*a*b^5*c ^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*d*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4 *c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*d*x^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48* a^3*b^3*c^2 - 64*a^4*b*c^3)*d*x + (a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*d)
Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (104) = 208\).
Time = 1.44 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.20 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^3} \, dx=\frac {32 c^{2} \log {\left (\frac {b}{2 c} + x \right )}}{d \left (4 a c - b^{2}\right )^{3}} - \frac {16 c^{2} \log {\left (\frac {a}{c} + \frac {b x}{c} + x^{2} \right )}}{d \left (4 a c - b^{2}\right )^{3}} + \frac {12 a c - b^{2} + 8 b c x + 8 c^{2} x^{2}}{32 a^{4} c^{2} d - 16 a^{3} b^{2} c d + 2 a^{2} b^{4} d + x^{4} \cdot \left (32 a^{2} c^{4} d - 16 a b^{2} c^{3} d + 2 b^{4} c^{2} d\right ) + x^{3} \cdot \left (64 a^{2} b c^{3} d - 32 a b^{3} c^{2} d + 4 b^{5} c d\right ) + x^{2} \cdot \left (64 a^{3} c^{3} d - 12 a b^{4} c d + 2 b^{6} d\right ) + x \left (64 a^{3} b c^{2} d - 32 a^{2} b^{3} c d + 4 a b^{5} d\right )} \] Input:
integrate(1/(2*c*d*x+b*d)/(c*x**2+b*x+a)**3,x)
Output:
32*c**2*log(b/(2*c) + x)/(d*(4*a*c - b**2)**3) - 16*c**2*log(a/c + b*x/c + x**2)/(d*(4*a*c - b**2)**3) + (12*a*c - b**2 + 8*b*c*x + 8*c**2*x**2)/(32 *a**4*c**2*d - 16*a**3*b**2*c*d + 2*a**2*b**4*d + x**4*(32*a**2*c**4*d - 1 6*a*b**2*c**3*d + 2*b**4*c**2*d) + x**3*(64*a**2*b*c**3*d - 32*a*b**3*c**2 *d + 4*b**5*c*d) + x**2*(64*a**3*c**3*d - 12*a*b**4*c*d + 2*b**6*d) + x*(6 4*a**3*b*c**2*d - 32*a**2*b**3*c*d + 4*a*b**5*d))
Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (110) = 220\).
Time = 0.04 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.38 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^3} \, dx=\frac {16 \, c^{2} \log \left (c x^{2} + b x + a\right )}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d} - \frac {32 \, c^{2} \log \left (2 \, c x + b\right )}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d} + \frac {8 \, c^{2} x^{2} + 8 \, b c x - b^{2} + 12 \, a c}{2 \, {\left ({\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d x^{4} + 2 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d x^{3} + {\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} d x^{2} + 2 \, {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} d x + {\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2}\right )} d\right )}} \] Input:
integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^3,x, algorithm="maxima")
Output:
16*c^2*log(c*x^2 + b*x + a)/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c ^3)*d) - 32*c^2*log(2*c*x + b)/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^ 3*c^3)*d) + 1/2*(8*c^2*x^2 + 8*b*c*x - b^2 + 12*a*c)/((b^4*c^2 - 8*a*b^2*c ^3 + 16*a^2*c^4)*d*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d*x^3 + (b ^6 - 6*a*b^4*c + 32*a^3*c^3)*d*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2 )*d*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*d)
Time = 0.11 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.68 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^3} \, dx=-\frac {32 \, c^{3} \log \left ({\left | 2 \, c x + b \right |}\right )}{b^{6} c d - 12 \, a b^{4} c^{2} d + 48 \, a^{2} b^{2} c^{3} d - 64 \, a^{3} c^{4} d} + \frac {16 \, c^{2} \log \left (c x^{2} + b x + a\right )}{b^{6} d - 12 \, a b^{4} c d + 48 \, a^{2} b^{2} c^{2} d - 64 \, a^{3} c^{3} d} - \frac {b^{4} - 16 \, a b^{2} c + 48 \, a^{2} c^{2} - 8 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} - 8 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x}{2 \, {\left (c x^{2} + b x + a\right )}^{2} {\left (b^{2} - 4 \, a c\right )}^{3} d} \] Input:
integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^3,x, algorithm="giac")
Output:
-32*c^3*log(abs(2*c*x + b))/(b^6*c*d - 12*a*b^4*c^2*d + 48*a^2*b^2*c^3*d - 64*a^3*c^4*d) + 16*c^2*log(c*x^2 + b*x + a)/(b^6*d - 12*a*b^4*c*d + 48*a^ 2*b^2*c^2*d - 64*a^3*c^3*d) - 1/2*(b^4 - 16*a*b^2*c + 48*a^2*c^2 - 8*(b^2* c^2 - 4*a*c^3)*x^2 - 8*(b^3*c - 4*a*b*c^2)*x)/((c*x^2 + b*x + a)^2*(b^2 - 4*a*c)^3*d)
Time = 5.68 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.13 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^3} \, dx=\frac {\frac {12\,a\,c-b^2}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {4\,c^2\,x^2}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {4\,b\,c\,x}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}}{a^2\,d+x^2\,\left (d\,b^2+2\,a\,c\,d\right )+c^2\,d\,x^4+2\,b\,c\,d\,x^3+2\,a\,b\,d\,x}-\frac {32\,c^2\,\ln \left (b+2\,c\,x\right )}{-64\,d\,a^3\,c^3+48\,d\,a^2\,b^2\,c^2-12\,d\,a\,b^4\,c+d\,b^6}+\frac {16\,c^2\,\ln \left (c\,x^2+b\,x+a\right )}{-64\,d\,a^3\,c^3+48\,d\,a^2\,b^2\,c^2-12\,d\,a\,b^4\,c+d\,b^6} \] Input:
int(1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^3),x)
Output:
((12*a*c - b^2)/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (4*c^2*x^2)/(b^4 + 16 *a^2*c^2 - 8*a*b^2*c) + (4*b*c*x)/(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(a^2*d + x^2*(b^2*d + 2*a*c*d) + c^2*d*x^4 + 2*b*c*d*x^3 + 2*a*b*d*x) - (32*c^2*lo g(b + 2*c*x))/(b^6*d - 64*a^3*c^3*d + 48*a^2*b^2*c^2*d - 12*a*b^4*c*d) + ( 16*c^2*log(a + b*x + c*x^2))/(b^6*d - 64*a^3*c^3*d + 48*a^2*b^2*c^2*d - 12 *a*b^4*c*d)
Time = 0.19 (sec) , antiderivative size = 503, normalized size of antiderivative = 4.49 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^3} \, dx=\frac {-32 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a^{2} c^{2}-64 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a b \,c^{2} x -64 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a \,c^{3} x^{2}-32 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) b^{2} c^{2} x^{2}-64 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) b \,c^{3} x^{3}-32 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) c^{4} x^{4}+64 \,\mathrm {log}\left (2 c x +b \right ) a^{2} c^{2}+128 \,\mathrm {log}\left (2 c x +b \right ) a b \,c^{2} x +128 \,\mathrm {log}\left (2 c x +b \right ) a \,c^{3} x^{2}+64 \,\mathrm {log}\left (2 c x +b \right ) b^{2} c^{2} x^{2}+128 \,\mathrm {log}\left (2 c x +b \right ) b \,c^{3} x^{3}+64 \,\mathrm {log}\left (2 c x +b \right ) c^{4} x^{4}+48 a^{2} c^{2}-16 a \,b^{2} c +32 a b \,c^{2} x +32 a \,c^{3} x^{2}+b^{4}-8 b^{3} c x -8 b^{2} c^{2} x^{2}}{2 d \left (64 a^{3} c^{5} x^{4}-48 a^{2} b^{2} c^{4} x^{4}+12 a \,b^{4} c^{3} x^{4}-b^{6} c^{2} x^{4}+128 a^{3} b \,c^{4} x^{3}-96 a^{2} b^{3} c^{3} x^{3}+24 a \,b^{5} c^{2} x^{3}-2 b^{7} c \,x^{3}+128 a^{4} c^{4} x^{2}-32 a^{3} b^{2} c^{3} x^{2}-24 a^{2} b^{4} c^{2} x^{2}+10 a \,b^{6} c \,x^{2}-b^{8} x^{2}+128 a^{4} b \,c^{3} x -96 a^{3} b^{3} c^{2} x +24 a^{2} b^{5} c x -2 a \,b^{7} x +64 a^{5} c^{3}-48 a^{4} b^{2} c^{2}+12 a^{3} b^{4} c -a^{2} b^{6}\right )} \] Input:
int(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^3,x)
Output:
( - 32*log(a + b*x + c*x**2)*a**2*c**2 - 64*log(a + b*x + c*x**2)*a*b*c**2 *x - 64*log(a + b*x + c*x**2)*a*c**3*x**2 - 32*log(a + b*x + c*x**2)*b**2* c**2*x**2 - 64*log(a + b*x + c*x**2)*b*c**3*x**3 - 32*log(a + b*x + c*x**2 )*c**4*x**4 + 64*log(b + 2*c*x)*a**2*c**2 + 128*log(b + 2*c*x)*a*b*c**2*x + 128*log(b + 2*c*x)*a*c**3*x**2 + 64*log(b + 2*c*x)*b**2*c**2*x**2 + 128* log(b + 2*c*x)*b*c**3*x**3 + 64*log(b + 2*c*x)*c**4*x**4 + 48*a**2*c**2 - 16*a*b**2*c + 32*a*b*c**2*x + 32*a*c**3*x**2 + b**4 - 8*b**3*c*x - 8*b**2* c**2*x**2)/(2*d*(64*a**5*c**3 - 48*a**4*b**2*c**2 + 128*a**4*b*c**3*x + 12 8*a**4*c**4*x**2 + 12*a**3*b**4*c - 96*a**3*b**3*c**2*x - 32*a**3*b**2*c** 3*x**2 + 128*a**3*b*c**4*x**3 + 64*a**3*c**5*x**4 - a**2*b**6 + 24*a**2*b* *5*c*x - 24*a**2*b**4*c**2*x**2 - 96*a**2*b**3*c**3*x**3 - 48*a**2*b**2*c* *4*x**4 - 2*a*b**7*x + 10*a*b**6*c*x**2 + 24*a*b**5*c**2*x**3 + 12*a*b**4* c**3*x**4 - b**8*x**2 - 2*b**7*c*x**3 - b**6*c**2*x**4))