Integrand size = 24, antiderivative size = 78 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=-\frac {1}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}+\frac {8 c \log (b+2 c x)}{\left (b^2-4 a c\right )^2 d}-\frac {4 c \log \left (a+b x+c x^2\right )}{\left (b^2-4 a c\right )^2 d} \] Output:
-1/(-4*a*c+b^2)/d/(c*x^2+b*x+a)+8*c*ln(2*c*x+b)/(-4*a*c+b^2)^2/d-4*c*ln(c* x^2+b*x+a)/(-4*a*c+b^2)^2/d
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=\frac {-\frac {b^2-4 a c}{a+x (b+c x)}+8 c \log (b+2 c x)-4 c \log (a+x (b+c x))}{\left (b^2-4 a c\right )^2 d} \] Input:
Integrate[1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^2),x]
Output:
(-((b^2 - 4*a*c)/(a + x*(b + c*x))) + 8*c*Log[b + 2*c*x] - 4*c*Log[a + x*( b + c*x)])/((b^2 - 4*a*c)^2*d)
Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1111, 27, 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.
\(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^2 (b d+2 c d x)} \, dx\) |
\(\Big \downarrow \) 1111 |
\(\displaystyle -\frac {4 c \int \frac {1}{d (b+2 c x) \left (c x^2+b x+a\right )}dx}{b^2-4 a c}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 c \int \frac {1}{(b+2 c x) \left (c x^2+b x+a\right )}dx}{d \left (b^2-4 a c\right )}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
\(\Big \downarrow \) 1105 |
\(\displaystyle -\frac {4 c \left (\frac {\int \frac {b+2 c x}{c x^2+b x+a}dx}{b^2-4 a c}-\frac {4 c \int \frac {1}{b+2 c x}dx}{b^2-4 a c}\right )}{d \left (b^2-4 a c\right )}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle -\frac {4 c \left (\frac {\int \frac {b+2 c x}{c x^2+b x+a}dx}{b^2-4 a c}-\frac {2 \log (b+2 c x)}{b^2-4 a c}\right )}{d \left (b^2-4 a c\right )}-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle -\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {4 c \left (\frac {\log \left (a+b x+c x^2\right )}{b^2-4 a c}-\frac {2 \log (b+2 c x)}{b^2-4 a c}\right )}{d \left (b^2-4 a c\right )}\) |
Input:
Int[1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^2),x]
Output:
-(1/((b^2 - 4*a*c)*d*(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)*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 = 0.98 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {\frac {8 c \ln \left (2 c x +b \right )}{\left (4 a c -b^{2}\right )^{2}}-\frac {\frac {-4 a c +b^{2}}{c \,x^{2}+b x +a}+4 c \ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right )^{2}}}{d}\) | \(78\) |
norman | \(\frac {1}{d \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )}+\frac {8 c \ln \left (2 c x +b \right )}{d \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}-\frac {4 c \ln \left (c \,x^{2}+b x +a \right )}{d \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}\) | \(102\) |
risch | \(\frac {1}{d \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )}+\frac {8 c \ln \left (2 c x +b \right )}{d \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}-\frac {4 c \ln \left (c \,x^{2}+b x +a \right )}{d \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}\) | \(102\) |
parallelrisch | \(\frac {-b^{2} c +4 a \,c^{2}+8 \ln \left (\frac {b}{2}+c x \right ) x b \,c^{2}-4 \ln \left (c \,x^{2}+b x +a \right ) x b \,c^{2}-4 \ln \left (c \,x^{2}+b x +a \right ) x^{2} c^{3}+8 \ln \left (\frac {b}{2}+c x \right ) a \,c^{2}-4 \ln \left (c \,x^{2}+b x +a \right ) a \,c^{2}+8 \ln \left (\frac {b}{2}+c x \right ) x^{2} c^{3}}{\left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \left (c \,x^{2}+b x +a \right ) c d}\) | \(153\) |
Input:
int(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(8*c/(4*a*c-b^2)^2*ln(2*c*x+b)-1/(4*a*c-b^2)^2*((-4*a*c+b^2)/(c*x^2+b* x+a)+4*c*ln(c*x^2+b*x+a)))
Time = 0.09 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.81 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=-\frac {b^{2} - 4 \, a c + 4 \, {\left (c^{2} x^{2} + b c x + a c\right )} \log \left (c x^{2} + b x + a\right ) - 8 \, {\left (c^{2} x^{2} + b c x + a c\right )} \log \left (2 \, c x + b\right )}{{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d x + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} d} \] Input:
integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^2,x, algorithm="fricas")
Output:
-(b^2 - 4*a*c + 4*(c^2*x^2 + b*c*x + a*c)*log(c*x^2 + b*x + a) - 8*(c^2*x^ 2 + b*c*x + a*c)*log(2*c*x + b))/((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d*x^2 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*d*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^ 2)*d)
Time = 0.80 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.31 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=\frac {8 c \log {\left (\frac {b}{2 c} + x \right )}}{d \left (4 a c - b^{2}\right )^{2}} - \frac {4 c \log {\left (\frac {a}{c} + \frac {b x}{c} + x^{2} \right )}}{d \left (4 a c - b^{2}\right )^{2}} + \frac {1}{4 a^{2} c d - a b^{2} d + x^{2} \cdot \left (4 a c^{2} d - b^{2} c d\right ) + x \left (4 a b c d - b^{3} d\right )} \] Input:
integrate(1/(2*c*d*x+b*d)/(c*x**2+b*x+a)**2,x)
Output:
8*c*log(b/(2*c) + x)/(d*(4*a*c - b**2)**2) - 4*c*log(a/c + b*x/c + x**2)/( d*(4*a*c - b**2)**2) + 1/(4*a**2*c*d - a*b**2*d + x**2*(4*a*c**2*d - b**2* c*d) + x*(4*a*b*c*d - b**3*d))
Time = 0.04 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.55 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=-\frac {4 \, c \log \left (c x^{2} + b x + a\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d} + \frac {8 \, c \log \left (2 \, c x + b\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d} - \frac {1}{{\left (b^{2} c - 4 \, a c^{2}\right )} d x^{2} + {\left (b^{3} - 4 \, a b c\right )} d x + {\left (a b^{2} - 4 \, a^{2} c\right )} d} \] Input:
integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^2,x, algorithm="maxima")
Output:
-4*c*log(c*x^2 + b*x + a)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*d) + 8*c*log(2*c *x + b)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*d) - 1/((b^2*c - 4*a*c^2)*d*x^2 + (b^3 - 4*a*b*c)*d*x + (a*b^2 - 4*a^2*c)*d)
Time = 0.13 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.38 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=\frac {8 \, c^{2} \log \left ({\left | 2 \, c x + b \right |}\right )}{b^{4} c d - 8 \, a b^{2} c^{2} d + 16 \, a^{2} c^{3} d} - \frac {4 \, c \log \left (c x^{2} + b x + a\right )}{b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} - \frac {1}{{\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )} d} \] Input:
integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^2,x, algorithm="giac")
Output:
8*c^2*log(abs(2*c*x + b))/(b^4*c*d - 8*a*b^2*c^2*d + 16*a^2*c^3*d) - 4*c*l og(c*x^2 + b*x + a)/(b^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d) - 1/((c*x^2 + b*x + a)*(b^2 - 4*a*c)*d)
Time = 5.75 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.60 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=\frac {8\,c\,\ln \left (b+2\,c\,x\right )}{16\,d\,a^2\,c^2-8\,d\,a\,b^2\,c+d\,b^4}-\frac {4\,c\,\ln \left (c\,x^2+b\,x+a\right )}{16\,d\,a^2\,c^2-8\,d\,a\,b^2\,c+d\,b^4}-\frac {1}{-4\,d\,a^2\,c+d\,a\,b^2-4\,d\,a\,b\,c\,x-4\,d\,a\,c^2\,x^2+d\,b^3\,x+d\,b^2\,c\,x^2} \] Input:
int(1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^2),x)
Output:
(8*c*log(b + 2*c*x))/(b^4*d + 16*a^2*c^2*d - 8*a*b^2*c*d) - (4*c*log(a + b *x + c*x^2))/(b^4*d + 16*a^2*c^2*d - 8*a*b^2*c*d) - 1/(a*b^2*d - 4*a^2*c*d + b^3*d*x - 4*a*c^2*d*x^2 + b^2*c*d*x^2 - 4*a*b*c*d*x)
Time = 0.19 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.32 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=\frac {-4 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a c -4 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) b c x -4 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) c^{2} x^{2}+8 \,\mathrm {log}\left (2 c x +b \right ) a c +8 \,\mathrm {log}\left (2 c x +b \right ) b c x +8 \,\mathrm {log}\left (2 c x +b \right ) c^{2} x^{2}+4 a c -b^{2}}{d \left (16 a^{2} c^{3} x^{2}-8 a \,b^{2} c^{2} x^{2}+b^{4} c \,x^{2}+16 a^{2} b \,c^{2} x -8 a \,b^{3} c x +b^{5} x +16 a^{3} c^{2}-8 a^{2} b^{2} c +a \,b^{4}\right )} \] Input:
int(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^2,x)
Output:
( - 4*log(a + b*x + c*x**2)*a*c - 4*log(a + b*x + c*x**2)*b*c*x - 4*log(a + b*x + c*x**2)*c**2*x**2 + 8*log(b + 2*c*x)*a*c + 8*log(b + 2*c*x)*b*c*x + 8*log(b + 2*c*x)*c**2*x**2 + 4*a*c - b**2)/(d*(16*a**3*c**2 - 8*a**2*b** 2*c + 16*a**2*b*c**2*x + 16*a**2*c**3*x**2 + a*b**4 - 8*a*b**3*c*x - 8*a*b **2*c**2*x**2 + b**5*x + b**4*c*x**2))