3.1815 \(\int \frac{a+b x}{(a c+(b c+a d) x+b d x^2)^2} \, dx\)

Optimal. Leaf size=56 \[ \frac{1}{(c+d x) (b c-a d)}+\frac{b \log (a+b x)}{(b c-a d)^2}-\frac{b \log (c+d x)}{(b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.0316406, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {626, 44} \[ \frac{1}{(c+d x) (b c-a d)}+\frac{b \log (a+b x)}{(b c-a d)^2}-\frac{b \log (c+d x)}{(b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0237145, size = 53, normalized size = 0.95 \[ \frac{b (c+d x) \log (a+b x)-a d-b (c+d x) \log (c+d x)+b c}{(c+d x) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.055, size = 58, normalized size = 1. \begin{align*} -{\frac{1}{ \left ( ad-bc \right ) \left ( dx+c \right ) }}-{\frac{b\ln \left ( dx+c \right ) }{ \left ( ad-bc \right ) ^{2}}}+{\frac{b\ln \left ( bx+a \right ) }{ \left ( ad-bc \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.14036, size = 122, normalized size = 2.18 \begin{align*} \frac{b \log \left (b x + a\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} - \frac{b \log \left (d x + c\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} + \frac{1}{b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.65618, size = 198, normalized size = 3.54 \begin{align*} \frac{b c - a d +{\left (b d x + b c\right )} \log \left (b x + a\right ) -{\left (b d x + b c\right )} \log \left (d x + c\right )}{b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} +{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.19887, size = 233, normalized size = 4.16 \begin{align*} - \frac{b \log{\left (x + \frac{- \frac{a^{3} b d^{3}}{\left (a d - b c\right )^{2}} + \frac{3 a^{2} b^{2} c d^{2}}{\left (a d - b c\right )^{2}} - \frac{3 a b^{3} c^{2} d}{\left (a d - b c\right )^{2}} + a b d + \frac{b^{4} c^{3}}{\left (a d - b c\right )^{2}} + b^{2} c}{2 b^{2} d} \right )}}{\left (a d - b c\right )^{2}} + \frac{b \log{\left (x + \frac{\frac{a^{3} b d^{3}}{\left (a d - b c\right )^{2}} - \frac{3 a^{2} b^{2} c d^{2}}{\left (a d - b c\right )^{2}} + \frac{3 a b^{3} c^{2} d}{\left (a d - b c\right )^{2}} + a b d - \frac{b^{4} c^{3}}{\left (a d - b c\right )^{2}} + b^{2} c}{2 b^{2} d} \right )}}{\left (a d - b c\right )^{2}} - \frac{1}{a c d - b c^{2} + x \left (a d^{2} - b c d\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18335, size = 126, normalized size = 2.25 \begin{align*} \frac{b^{2} \log \left ({\left | b x + a \right |}\right )}{b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}} - \frac{b d \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 (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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