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

Optimal. Leaf size=78 \[ -\frac {1}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {4 c \log \left (a+b x+c x^2\right )}{d \left (b^2-4 a c\right )^2}+\frac {8 c \log (b+2 c x)}{d \left (b^2-4 a c\right )^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {687, 681, 31, 628} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[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 628

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 681

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 687

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*} \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 {(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*}

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Mathematica [A]  time = 0.06, size = 59, normalized size = 0.76 \begin {gather*} \frac {-\frac {b^2-4 a c}{a+x (b+c x)}-4 c \log (a+x (b+c x))+8 c \log (b+2 c x)}{d \left (b^2-4 a c\right )^2} \end {gather*}

Antiderivative was successfully verified.

[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)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 141, normalized size = 1.81 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.19, size = 108, normalized size = 1.38 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.06, size = 119, normalized size = 1.53 \begin {gather*} \frac {4 a c}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right ) d}-\frac {b^{2}}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right ) d}+\frac {8 c \ln \left (2 c x +b \right )}{\left (4 a c -b^{2}\right )^{2} d}-\frac {4 c \ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right )^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.45, size = 121, normalized size = 1.55 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.58, size = 125, normalized size = 1.60 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 1.63, size = 102, normalized size = 1.31 \begin {gather*} \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} \left (4 a c^{2} d - b^{2} c d\right ) + x \left (4 a b c d - b^{3} d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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