### 3.1813 $$\int \frac{(a+b x)^3}{(a c+(b c+a d) x+b d x^2)^2} \, dx$$

Optimal. Leaf size=31 $\frac{b c-a d}{d^2 (c+d x)}+\frac{b \log (c+d x)}{d^2}$

[Out]

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

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Rubi [A]  time = 0.0254543, antiderivative size = 31, 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, 43} $\frac{b c-a d}{d^2 (c+d x)}+\frac{b \log (c+d x)}{d^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x)^3/(a*c + (b*c + a*d)*x + b*d*x^2)^2,x]

[Out]

(b*c - a*d)/(d^2*(c + d*x)) + (b*Log[c + d*x])/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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0101975, size = 31, normalized size = 1. $\frac{b c-a d}{d^2 (c+d x)}+\frac{b \log (c+d x)}{d^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x)^3/(a*c + (b*c + a*d)*x + b*d*x^2)^2,x]

[Out]

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

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Maple [A]  time = 0.044, size = 39, normalized size = 1.3 \begin{align*}{\frac{b\ln \left ( dx+c \right ) }{{d}^{2}}}-{\frac{a}{d \left ( dx+c \right ) }}+{\frac{bc}{{d}^{2} \left ( dx+c \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x)

[Out]

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

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Maxima [A]  time = 1.04856, size = 46, normalized size = 1.48 \begin{align*} \frac{b c - a d}{d^{3} x + c d^{2}} + \frac{b \log \left (d x + c\right )}{d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.58518, size = 78, normalized size = 2.52 \begin{align*} \frac{b c - a d +{\left (b d x + b c\right )} \log \left (d x + c\right )}{d^{3} x + c d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]  time = 0.44771, size = 27, normalized size = 0.87 \begin{align*} \frac{b \log{\left (c + d x \right )}}{d^{2}} - \frac{a d - b c}{c d^{2} + d^{3} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3/(a*c+(a*d+b*c)*x+b*d*x**2)**2,x)

[Out]

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

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Giac [A]  time = 1.19807, size = 43, normalized size = 1.39 \begin{align*} \frac{b \log \left ({\left | d x + c \right |}\right )}{d^{2}} + \frac{b c - a d}{{\left (d x + c\right )} d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(a*c+(a*d+b*c)*x+b*d*x^2)^2,x, algorithm="giac")

[Out]

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