### 3.1806 $$\int \frac{1}{(a+b x) (a c+(b c+a d) x+b d x^2)} \, dx$$

Optimal. Leaf size=57 $-\frac{1}{(a+b x) (b c-a d)}-\frac{d \log (a+b x)}{(b c-a d)^2}+\frac{d \log (c+d x)}{(b c-a d)^2}$

[Out]

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

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Rubi [A]  time = 0.0355253, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.069, Rules used = {626, 44} $-\frac{1}{(a+b x) (b c-a d)}-\frac{d \log (a+b x)}{(b c-a d)^2}+\frac{d \log (c+d x)}{(b c-a d)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0238165, size = 53, normalized size = 0.93 $\frac{d (a+b x) \log (c+d x)-d (a+b x) \log (a+b x)+a d-b c}{(a+b x) (b c-a d)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 57, normalized size = 1. \begin{align*}{\frac{d\ln \left ( dx+c \right ) }{ \left ( ad-bc \right ) ^{2}}}+{\frac{1}{ \left ( ad-bc \right ) \left ( bx+a \right ) }}-{\frac{d\ln \left ( bx+a \right ) }{ \left ( ad-bc \right ) ^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.0461, size = 124, normalized size = 2.18 \begin{align*} -\frac{d \log \left (b x + a\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} + \frac{d \log \left (d x + c\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} - \frac{1}{a b c - a^{2} d +{\left (b^{2} c - a b d\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d*log(b*x + a)/(b^2*c^2 - 2*a*b*c*d + a^2*d^2) + d*log(d*x + c)/(b^2*c^2 - 2*a*b*c*d + a^2*d^2) - 1/(a*b*c -
a^2*d + (b^2*c - a*b*d)*x)

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Fricas [A]  time = 1.67954, size = 200, normalized size = 3.51 \begin{align*} -\frac{b c - a d +{\left (b d x + a d\right )} \log \left (b x + a\right ) -{\left (b d x + a d\right )} \log \left (d x + c\right )}{a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} +{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.17334, size = 233, normalized size = 4.09 \begin{align*} \frac{d \log{\left (x + \frac{- \frac{a^{3} d^{4}}{\left (a d - b c\right )^{2}} + \frac{3 a^{2} b c d^{3}}{\left (a d - b c\right )^{2}} - \frac{3 a b^{2} c^{2} d^{2}}{\left (a d - b c\right )^{2}} + a d^{2} + \frac{b^{3} c^{3} d}{\left (a d - b c\right )^{2}} + b c d}{2 b d^{2}} \right )}}{\left (a d - b c\right )^{2}} - \frac{d \log{\left (x + \frac{\frac{a^{3} d^{4}}{\left (a d - b c\right )^{2}} - \frac{3 a^{2} b c d^{3}}{\left (a d - b c\right )^{2}} + \frac{3 a b^{2} c^{2} d^{2}}{\left (a d - b c\right )^{2}} + a d^{2} - \frac{b^{3} c^{3} d}{\left (a d - b c\right )^{2}} + b c d}{2 b d^{2}} \right )}}{\left (a d - b c\right )^{2}} + \frac{1}{a^{2} d - a b c + x \left (a b d - b^{2} c\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*log(x + (-a**3*d**4/(a*d - b*c)**2 + 3*a**2*b*c*d**3/(a*d - b*c)**2 - 3*a*b**2*c**2*d**2/(a*d - b*c)**2 + a*
d**2 + b**3*c**3*d/(a*d - b*c)**2 + b*c*d)/(2*b*d**2))/(a*d - b*c)**2 - d*log(x + (a**3*d**4/(a*d - b*c)**2 -
3*a**2*b*c*d**3/(a*d - b*c)**2 + 3*a*b**2*c**2*d**2/(a*d - b*c)**2 + a*d**2 - b**3*c**3*d/(a*d - b*c)**2 + b*c
*d)/(2*b*d**2))/(a*d - b*c)**2 + 1/(a**2*d - a*b*c + x*(a*b*d - b**2*c))

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Giac [A]  time = 1.21447, size = 127, normalized size = 2.23 \begin{align*} -\frac{b d \log \left ({\left | b x + a \right |}\right )}{b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}} + \frac{d^{2} \log \left ({\left | d x + c \right |}\right )}{b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}} - \frac{1}{{\left (b c - a d\right )}{\left (b x + a\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*d*log(abs(b*x + a))/(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2) + d^2*log(abs(d*x + c))/(b^2*c^2*d - 2*a*b*c*d^2 +
a^2*d^3) - 1/((b*c - a*d)*(b*x + a))