### 3.1456 $$\int \frac{a^2+2 a b x+b^2 x^2}{d+e x} \, dx$$

Optimal. Leaf size=50 $-\frac{b x (b d-a e)}{e^2}+\frac{(b d-a e)^2 \log (d+e x)}{e^3}+\frac{(a+b x)^2}{2 e}$

[Out]

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

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Rubi [A]  time = 0.021483, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {27, 43} $-\frac{b x (b d-a e)}{e^2}+\frac{(b d-a e)^2 \log (d+e x)}{e^3}+\frac{(a+b x)^2}{2 e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 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^2+2 a b x+b^2 x^2}{d+e x} \, dx &=\int \frac{(a+b x)^2}{d+e x} \, dx\\ &=\int \left (-\frac{b (b d-a e)}{e^2}+\frac{b (a+b x)}{e}+\frac{(-b d+a e)^2}{e^2 (d+e x)}\right ) \, dx\\ &=-\frac{b (b d-a e) x}{e^2}+\frac{(a+b x)^2}{2 e}+\frac{(b d-a e)^2 \log (d+e x)}{e^3}\\ \end{align*}

Mathematica [A]  time = 0.0182524, size = 43, normalized size = 0.86 $\frac{b e x (4 a e-2 b d+b e x)+2 (b d-a e)^2 \log (d+e x)}{2 e^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(b*e*x*(-2*b*d + 4*a*e + b*e*x) + 2*(b*d - a*e)^2*Log[d + e*x])/(2*e^3)

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Maple [A]  time = 0.04, size = 74, normalized size = 1.5 \begin{align*}{\frac{{b}^{2}{x}^{2}}{2\,e}}+2\,{\frac{abx}{e}}-{\frac{{b}^{2}dx}{{e}^{2}}}+{\frac{\ln \left ( ex+d \right ){a}^{2}}{e}}-2\,{\frac{\ln \left ( ex+d \right ) abd}{{e}^{2}}}+{\frac{{d}^{2}\ln \left ( ex+d \right ){b}^{2}}{{e}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.19115, size = 81, normalized size = 1.62 \begin{align*} \frac{b^{2} e x^{2} - 2 \,{\left (b^{2} d - 2 \, a b e\right )} x}{2 \, e^{2}} + \frac{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (e x + d\right )}{e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.64159, size = 135, normalized size = 2.7 \begin{align*} \frac{b^{2} e^{2} x^{2} - 2 \,{\left (b^{2} d e - 2 \, a b e^{2}\right )} x + 2 \,{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (e x + d\right )}{2 \, e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.401215, size = 44, normalized size = 0.88 \begin{align*} \frac{b^{2} x^{2}}{2 e} + \frac{x \left (2 a b e - b^{2} d\right )}{e^{2}} + \frac{\left (a e - b d\right )^{2} \log{\left (d + e x \right )}}{e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14471, size = 82, normalized size = 1.64 \begin{align*}{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{2} \,{\left (b^{2} x^{2} e - 2 \, b^{2} d x + 4 \, a b x e\right )} e^{\left (-2\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*d^2 - 2*a*b*d*e + a^2*e^2)*e^(-3)*log(abs(x*e + d)) + 1/2*(b^2*x^2*e - 2*b^2*d*x + 4*a*b*x*e)*e^(-2)