3.1816 \(\int \frac{1}{(a c+(b c+a d) x+b d x^2)^2} \, dx\)

Optimal. Leaf size=86 \[ -\frac{a d+b c+2 b d x}{(b c-a d)^2 \left (x (a d+b c)+a c+b d x^2\right )}-\frac{2 b d \log (a+b x)}{(b c-a d)^3}+\frac{2 b d \log (c+d x)}{(b c-a d)^3} \]

[Out]

-((b*c + a*d + 2*b*d*x)/((b*c - a*d)^2*(a*c + (b*c + a*d)*x + b*d*x^2))) - (2*b*d*Log[a + b*x])/(b*c - a*d)^3
+ (2*b*d*Log[c + d*x])/(b*c - a*d)^3

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Rubi [A]  time = 0.0261368, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {614, 616, 31} \[ -\frac{a d+b c+2 b d x}{(b c-a d)^2 \left (x (a d+b c)+a c+b d x^2\right )}-\frac{2 b d \log (a+b x)}{(b c-a d)^3}+\frac{2 b d \log (c+d x)}{(b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c + a*d + 2*b*d*x)/((b*c - a*d)^2*(a*c + (b*c + a*d)*x + b*d*x^2))) - (2*b*d*Log[a + b*x])/(b*c - a*d)^3
+ (2*b*d*Log[c + d*x])/(b*c - a*d)^3

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 31

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

Rubi steps

\begin{align*} \int \frac{1}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx &=-\frac{b c+a d+2 b d x}{(b c-a d)^2 \left (a c+(b c+a d) x+b d x^2\right )}-\frac{(2 b d) \int \frac{1}{a c+(b c+a d) x+b d x^2} \, dx}{(b c-a d)^2}\\ &=-\frac{b c+a d+2 b d x}{(b c-a d)^2 \left (a c+(b c+a d) x+b d x^2\right )}+\frac{\left (2 b^2 d^2\right ) \int \frac{1}{b c+b d x} \, dx}{(b c-a d)^3}-\frac{\left (2 b^2 d^2\right ) \int \frac{1}{a d+b d x} \, dx}{(b c-a d)^3}\\ &=-\frac{b c+a d+2 b d x}{(b c-a d)^2 \left (a c+(b c+a d) x+b d x^2\right )}-\frac{2 b d \log (a+b x)}{(b c-a d)^3}+\frac{2 b d \log (c+d x)}{(b c-a d)^3}\\ \end{align*}

Mathematica [A]  time = 0.0663112, size = 66, normalized size = 0.77 \[ \frac{\frac{b (a d-b c)}{a+b x}+\frac{d (a d-b c)}{c+d x}-2 b d \log (a+b x)+2 b d \log (c+d x)}{(b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*(-(b*c) + a*d))/(a + b*x) + (d*(-(b*c) + a*d))/(c + d*x) - 2*b*d*Log[a + b*x] + 2*b*d*Log[c + d*x])/(b*c -
a*d)^3

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Maple [A]  time = 0.051, size = 82, normalized size = 1. \begin{align*} -{\frac{d}{ \left ( ad-bc \right ) ^{2} \left ( dx+c \right ) }}-2\,{\frac{bd\ln \left ( dx+c \right ) }{ \left ( ad-bc \right ) ^{3}}}-{\frac{b}{ \left ( ad-bc \right ) ^{2} \left ( bx+a \right ) }}+2\,{\frac{bd\ln \left ( bx+a \right ) }{ \left ( ad-bc \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d/(a*d-b*c)^2/(d*x+c)-2*d/(a*d-b*c)^3*b*ln(d*x+c)-b/(a*d-b*c)^2/(b*x+a)+2*d/(a*d-b*c)^3*b*ln(b*x+a)

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Maxima [B]  time = 1.1048, size = 281, normalized size = 3.27 \begin{align*} -\frac{2 \, b d \log \left (b x + a\right )}{b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}} + \frac{2 \, b d \log \left (d x + c\right )}{b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}} - \frac{2 \, b d x + b c + a d}{a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} +{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} +{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b*d*log(b*x + a)/(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3) + 2*b*d*log(d*x + c)/(b^3*c^3 - 3*a*b^
2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3) - (2*b*d*x + b*c + a*d)/(a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d
- 2*a*b^2*c*d^2 + a^2*b*d^3)*x^2 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x)

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Fricas [B]  time = 1.61919, size = 486, normalized size = 5.65 \begin{align*} -\frac{b^{2} c^{2} - a^{2} d^{2} + 2 \,{\left (b^{2} c d - a b d^{2}\right )} x + 2 \,{\left (b^{2} d^{2} x^{2} + a b c d +{\left (b^{2} c d + a b d^{2}\right )} x\right )} \log \left (b x + a\right ) - 2 \,{\left (b^{2} d^{2} x^{2} + a b c d +{\left (b^{2} c d + a b d^{2}\right )} x\right )} \log \left (d x + c\right )}{a b^{3} c^{4} - 3 \, a^{2} b^{2} c^{3} d + 3 \, a^{3} b c^{2} d^{2} - a^{4} c d^{3} +{\left (b^{4} c^{3} d - 3 \, a b^{3} c^{2} d^{2} + 3 \, a^{2} b^{2} c d^{3} - a^{3} b d^{4}\right )} x^{2} +{\left (b^{4} c^{4} - 2 \, a b^{3} c^{3} d + 2 \, a^{3} b c d^{3} - a^{4} d^{4}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*c^2 - a^2*d^2 + 2*(b^2*c*d - a*b*d^2)*x + 2*(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)*log(b*x + a)
- 2*(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)*log(d*x + c))/(a*b^3*c^4 - 3*a^2*b^2*c^3*d + 3*a^3*b*c^2*
d^2 - a^4*c*d^3 + (b^4*c^3*d - 3*a*b^3*c^2*d^2 + 3*a^2*b^2*c*d^3 - a^3*b*d^4)*x^2 + (b^4*c^4 - 2*a*b^3*c^3*d +
2*a^3*b*c*d^3 - a^4*d^4)*x)

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Sympy [B]  time = 1.93193, size = 405, normalized size = 4.71 \begin{align*} - \frac{2 b d \log{\left (x + \frac{- \frac{2 a^{4} b d^{5}}{\left (a d - b c\right )^{3}} + \frac{8 a^{3} b^{2} c d^{4}}{\left (a d - b c\right )^{3}} - \frac{12 a^{2} b^{3} c^{2} d^{3}}{\left (a d - b c\right )^{3}} + \frac{8 a b^{4} c^{3} d^{2}}{\left (a d - b c\right )^{3}} + 2 a b d^{2} - \frac{2 b^{5} c^{4} d}{\left (a d - b c\right )^{3}} + 2 b^{2} c d}{4 b^{2} d^{2}} \right )}}{\left (a d - b c\right )^{3}} + \frac{2 b d \log{\left (x + \frac{\frac{2 a^{4} b d^{5}}{\left (a d - b c\right )^{3}} - \frac{8 a^{3} b^{2} c d^{4}}{\left (a d - b c\right )^{3}} + \frac{12 a^{2} b^{3} c^{2} d^{3}}{\left (a d - b c\right )^{3}} - \frac{8 a b^{4} c^{3} d^{2}}{\left (a d - b c\right )^{3}} + 2 a b d^{2} + \frac{2 b^{5} c^{4} d}{\left (a d - b c\right )^{3}} + 2 b^{2} c d}{4 b^{2} d^{2}} \right )}}{\left (a d - b c\right )^{3}} - \frac{a d + b c + 2 b d x}{a^{3} c d^{2} - 2 a^{2} b c^{2} d + a b^{2} c^{3} + x^{2} \left (a^{2} b d^{3} - 2 a b^{2} c d^{2} + b^{3} c^{2} d\right ) + x \left (a^{3} d^{3} - a^{2} b c d^{2} - a b^{2} c^{2} d + b^{3} c^{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b*d*log(x + (-2*a**4*b*d**5/(a*d - b*c)**3 + 8*a**3*b**2*c*d**4/(a*d - b*c)**3 - 12*a**2*b**3*c**2*d**3/(a*
d - b*c)**3 + 8*a*b**4*c**3*d**2/(a*d - b*c)**3 + 2*a*b*d**2 - 2*b**5*c**4*d/(a*d - b*c)**3 + 2*b**2*c*d)/(4*b
**2*d**2))/(a*d - b*c)**3 + 2*b*d*log(x + (2*a**4*b*d**5/(a*d - b*c)**3 - 8*a**3*b**2*c*d**4/(a*d - b*c)**3 +
12*a**2*b**3*c**2*d**3/(a*d - b*c)**3 - 8*a*b**4*c**3*d**2/(a*d - b*c)**3 + 2*a*b*d**2 + 2*b**5*c**4*d/(a*d -
b*c)**3 + 2*b**2*c*d)/(4*b**2*d**2))/(a*d - b*c)**3 - (a*d + b*c + 2*b*d*x)/(a**3*c*d**2 - 2*a**2*b*c**2*d + a
*b**2*c**3 + x**2*(a**2*b*d**3 - 2*a*b**2*c*d**2 + b**3*c**2*d) + x*(a**3*d**3 - a**2*b*c*d**2 - a*b**2*c**2*d
+ b**3*c**3))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError