3.1113 \(\int \frac{a+b x+c x^2}{b d+2 c d x} \, dx\)

Optimal. Leaf size=48 \[ -\frac{\left (b^2-4 a c\right ) \log (b+2 c x)}{8 c^2 d}+\frac{b x}{4 c d}+\frac{x^2}{4 d} \]

[Out]

(b*x)/(4*c*d) + x^2/(4*d) - ((b^2 - 4*a*c)*Log[b + 2*c*x])/(8*c^2*d)

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Rubi [A]  time = 0.0350725, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {683} \[ -\frac{\left (b^2-4 a c\right ) \log (b+2 c x)}{8 c^2 d}+\frac{b x}{4 c d}+\frac{x^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x)/(4*c*d) + x^2/(4*d) - ((b^2 - 4*a*c)*Log[b + 2*c*x])/(8*c^2*d)

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps

\begin{align*} \int \frac{a+b x+c x^2}{b d+2 c d x} \, dx &=\int \left (\frac{b}{4 c d}+\frac{x}{2 d}+\frac{-b^2+4 a c}{4 c d (b+2 c x)}\right ) \, dx\\ &=\frac{b x}{4 c d}+\frac{x^2}{4 d}-\frac{\left (b^2-4 a c\right ) \log (b+2 c x)}{8 c^2 d}\\ \end{align*}

Mathematica [A]  time = 0.0133888, size = 37, normalized size = 0.77 \[ \frac{2 c x (b+c x)-\left (b^2-4 a c\right ) \log (b+2 c x)}{8 c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*x*(b + c*x) - (b^2 - 4*a*c)*Log[b + 2*c*x])/(8*c^2*d)

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Maple [A]  time = 0.041, size = 54, normalized size = 1.1 \begin{align*}{\frac{{x}^{2}}{4\,d}}+{\frac{bx}{4\,cd}}+{\frac{\ln \left ( 2\,cx+b \right ) a}{2\,cd}}-{\frac{\ln \left ( 2\,cx+b \right ){b}^{2}}{8\,{c}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^2/d+1/4*b*x/c/d+1/2/d/c*ln(2*c*x+b)*a-1/8/d/c^2*ln(2*c*x+b)*b^2

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Maxima [A]  time = 1.1732, size = 55, normalized size = 1.15 \begin{align*} \frac{c x^{2} + b x}{4 \, c d} - \frac{{\left (b^{2} - 4 \, a c\right )} \log \left (2 \, c x + b\right )}{8 \, c^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c*x^2 + b*x)/(c*d) - 1/8*(b^2 - 4*a*c)*log(2*c*x + b)/(c^2*d)

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Fricas [A]  time = 2.22093, size = 89, normalized size = 1.85 \begin{align*} \frac{2 \, c^{2} x^{2} + 2 \, b c x -{\left (b^{2} - 4 \, a c\right )} \log \left (2 \, c x + b\right )}{8 \, c^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*c^2*x^2 + 2*b*c*x - (b^2 - 4*a*c)*log(2*c*x + b))/(c^2*d)

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Sympy [A]  time = 0.434418, size = 37, normalized size = 0.77 \begin{align*} \frac{b x}{4 c d} + \frac{x^{2}}{4 d} + \frac{\left (4 a c - b^{2}\right ) \log{\left (b + 2 c x \right )}}{8 c^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x/(4*c*d) + x**2/(4*d) + (4*a*c - b**2)*log(b + 2*c*x)/(8*c**2*d)

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Giac [A]  time = 1.17559, size = 63, normalized size = 1.31 \begin{align*} -\frac{{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, c x + b \right |}\right )}{8 \, c^{2} d} + \frac{c^{2} d x^{2} + b c d x}{4 \, c^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b^2 - 4*a*c)*log(abs(2*c*x + b))/(c^2*d) + 1/4*(c^2*d*x^2 + b*c*d*x)/(c^2*d^2)