### 3.1776 $$\int \frac{(a c+(b c+a d) x+b d x^2)^2}{(a+b x)^4} \, dx$$

Optimal. Leaf size=51 $-\frac{(b c-a d)^2}{b^3 (a+b x)}+\frac{2 d (b c-a d) \log (a+b x)}{b^3}+\frac{d^2 x}{b^2}$

[Out]

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

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Rubi [A]  time = 0.0442666, antiderivative size = 51, 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} $-\frac{(b c-a d)^2}{b^3 (a+b x)}+\frac{2 d (b c-a d) \log (a+b x)}{b^3}+\frac{d^2 x}{b^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*c + (b*c + a*d)*x + b*d*x^2)^2/(a + b*x)^4,x]

[Out]

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

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{\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^4} \, dx &=\int \frac{(c+d x)^2}{(a+b x)^2} \, dx\\ &=\int \left (\frac{d^2}{b^2}+\frac{(b c-a d)^2}{b^2 (a+b x)^2}+\frac{2 d (b c-a d)}{b^2 (a+b x)}\right ) \, dx\\ &=\frac{d^2 x}{b^2}-\frac{(b c-a d)^2}{b^3 (a+b x)}+\frac{2 d (b c-a d) \log (a+b x)}{b^3}\\ \end{align*}

Mathematica [A]  time = 0.0364032, size = 47, normalized size = 0.92 $\frac{-\frac{(b c-a d)^2}{a+b x}+2 d (b c-a d) \log (a+b x)+b d^2 x}{b^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*c + (b*c + a*d)*x + b*d*x^2)^2/(a + b*x)^4,x]

[Out]

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

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Maple [A]  time = 0.045, size = 86, normalized size = 1.7 \begin{align*}{\frac{{d}^{2}x}{{b}^{2}}}-2\,{\frac{{d}^{2}\ln \left ( bx+a \right ) a}{{b}^{3}}}+2\,{\frac{d\ln \left ( bx+a \right ) c}{{b}^{2}}}-{\frac{{a}^{2}{d}^{2}}{{b}^{3} \left ( bx+a \right ) }}+2\,{\frac{acd}{{b}^{2} \left ( bx+a \right ) }}-{\frac{{c}^{2}}{b \left ( bx+a \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a*c+(a*d+b*c)*x+b*d*x^2)^2/(b*x+a)^4,x)

[Out]

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

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Maxima [A]  time = 1.0366, size = 90, normalized size = 1.76 \begin{align*} \frac{d^{2} x}{b^{2}} - \frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{b^{4} x + a b^{3}} + \frac{2 \,{\left (b c d - a d^{2}\right )} \log \left (b x + a\right )}{b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^2/(b*x+a)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.52461, size = 184, normalized size = 3.61 \begin{align*} \frac{b^{2} d^{2} x^{2} + a b d^{2} x - b^{2} c^{2} + 2 \, a b c d - a^{2} d^{2} + 2 \,{\left (a b c d - a^{2} d^{2} +{\left (b^{2} c d - a b d^{2}\right )} x\right )} \log \left (b x + a\right )}{b^{4} x + a b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^2/(b*x+a)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.765751, size = 60, normalized size = 1.18 \begin{align*} - \frac{a^{2} d^{2} - 2 a b c d + b^{2} c^{2}}{a b^{3} + b^{4} x} + \frac{d^{2} x}{b^{2}} - \frac{2 d \left (a d - b c\right ) \log{\left (a + b x \right )}}{b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x**2)**2/(b*x+a)**4,x)

[Out]

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

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Giac [A]  time = 1.22063, size = 88, normalized size = 1.73 \begin{align*} \frac{d^{2} x}{b^{2}} + \frac{2 \,{\left (b c d - a d^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} - \frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{{\left (b x + a\right )} b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^2/(b*x+a)^4,x, algorithm="giac")

[Out]

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