\(\int \frac {1}{(b d+2 c d x) (a+b x+c x^2)^2} \, dx\) [1174]

Optimal result

Integrand size = 24, antiderivative size = 78 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=-\frac {1}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}+\frac {8 c \log (b+2 c x)}{\left (b^2-4 a c\right )^2 d}-\frac {4 c \log \left (a+b x+c x^2\right )}{\left (b^2-4 a c\right )^2 d} \]

[Out]

-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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {701, 695, 31, 642} \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=-\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {4 c \log \left (a+b x+c x^2\right )}{d \left (b^2-4 a c\right )^2}+\frac {8 c \log (b+2 c x)}{d \left (b^2-4 a c\right )^2} \]

[In]

Int[1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^2),x]

[Out]

-(1/((b^2 - 4*a*c)*d*(a + b*x + c*x^2))) + (8*c*Log[b + 2*c*x])/((b^2 - 4*a*c)^2*d) - (4*c*Log[a + b*x + c*x^2
])/((b^2 - 4*a*c)^2*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 695

Int[1/(((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[-4*b*(c/(d*(b^2 - 4*a*c))),
Int[1/(b + 2*c*x), x], x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0]

Rule 701

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> 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] - Dist[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] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && RationalQ[m] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}-\frac {(4 c) \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )} \, dx}{b^2-4 a c} \\ & = -\frac {1}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}-\frac {(4 c) \int \frac {b d+2 c d x}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2 d^2}+\frac {\left (16 c^2\right ) \int \frac {1}{b+2 c x} \, dx}{\left (b^2-4 a c\right )^2 d} \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=\frac {-\frac {b^2-4 a c}{a+x (b+c x)}+8 c \log (b+2 c x)-4 c \log (a+x (b+c x))}{\left (b^2-4 a c\right )^2 d} \]

[In]

Integrate[1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^2),x]

[Out]

(-((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)

Maple [A] (verified)

Time = 2.61 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {-\frac {\frac {-4 a c +b^{2}}{c \,x^{2}+b x +a}+4 c \ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right )^{2}}+\frac {8 c \ln \left (2 c x +b \right )}{\left (4 a c -b^{2}\right )^{2}}}{d}\)	\(78\)
norman	\(\frac {1}{d \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )}+\frac {8 c \ln \left (2 c x +b \right )}{d \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {4 c \ln \left (c \,x^{2}+b x +a \right )}{d \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}\)	\(102\)
risch	\(\frac {1}{d \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )}+\frac {8 c \ln \left (2 c x +b \right )}{d \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {4 c \ln \left (c \,x^{2}+b x +a \right )}{d \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}\)	\(102\)
parallelrisch	\(\frac {-b^{2} c +4 c^{2} a +8 \ln \left (\frac {b}{2}+c x \right ) x^{2} c^{3}-4 \ln \left (c \,x^{2}+b x +a \right ) x^{2} c^{3}+8 \ln \left (\frac {b}{2}+c x \right ) a \,c^{2}+8 \ln \left (\frac {b}{2}+c x \right ) x b \,c^{2}-4 \ln \left (c \,x^{2}+b x +a \right ) x b \,c^{2}-4 \ln \left (c \,x^{2}+b x +a \right ) a \,c^{2}}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \left (c \,x^{2}+b x +a \right ) c d}\)	\(153\)

[In]

int(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-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))+8*c/(4*a*c-b^2)^2*ln(2*c*x+b))

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.81 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=-\frac {b^{2} - 4 \, a c + 4 \, {\left (c^{2} x^{2} + b c x + a c\right )} \log \left (c x^{2} + b x + a\right ) - 8 \, {\left (c^{2} x^{2} + b c x + a c\right )} \log \left (2 \, c x + b\right )}{{\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d x^{2} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d x + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} d} \]

[In]

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-(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)

Sympy [A] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.31 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=\frac {8 c \log {\left (\frac {b}{2 c} + x \right )}}{d \left (4 a c - b^{2}\right )^{2}} - \frac {4 c \log {\left (\frac {a}{c} + \frac {b x}{c} + x^{2} \right )}}{d \left (4 a c - b^{2}\right )^{2}} + \frac {1}{4 a^{2} c d - a b^{2} d + x^{2} \cdot \left (4 a c^{2} d - b^{2} c d\right ) + x \left (4 a b c d - b^{3} d\right )} \]

[In]

integrate(1/(2*c*d*x+b*d)/(c*x**2+b*x+a)**2,x)

[Out]

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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.55 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=-\frac {4 \, c \log \left (c x^{2} + b x + a\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d} + \frac {8 \, c \log \left (2 \, c x + b\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d} - \frac {1}{{\left (b^{2} c - 4 \, a c^{2}\right )} d x^{2} + {\left (b^{3} - 4 \, a b c\right )} d x + {\left (a b^{2} - 4 \, a^{2} c\right )} d} \]

[In]

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

-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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.38 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=\frac {8 \, c^{2} \log \left ({\left | 2 \, c x + b \right |}\right )}{b^{4} c d - 8 \, a b^{2} c^{2} d + 16 \, a^{2} c^{3} d} - \frac {4 \, c \log \left (c x^{2} + b x + a\right )}{b^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} - \frac {1}{{\left (c x^{2} + b x + a\right )} {\left (b^{2} - 4 \, a c\right )} d} \]

[In]

integrate(1/(2*c*d*x+b*d)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

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*log(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)

Mupad [B] (verification not implemented)

Time = 9.70 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.60 \[ \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx=\frac {8\,c\,\ln \left (b+2\,c\,x\right )}{16\,d\,a^2\,c^2-8\,d\,a\,b^2\,c+d\,b^4}-\frac {4\,c\,\ln \left (c\,x^2+b\,x+a\right )}{16\,d\,a^2\,c^2-8\,d\,a\,b^2\,c+d\,b^4}-\frac {1}{-4\,d\,a^2\,c+d\,a\,b^2-4\,d\,a\,b\,c\,x-4\,d\,a\,c^2\,x^2+d\,b^3\,x+d\,b^2\,c\,x^2} \]

[In]

int(1/((b*d + 2*c*d*x)*(a + b*x + c*x^2)^2),x)

[Out]

(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)