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

Optimal. Leaf size=43 \[ -\frac{\log (x) (c d-b e)}{b^2}+\frac{(c d-b e) \log (b+c x)}{b^2}-\frac{d}{b x} \]

[Out]

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

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Rubi [A]  time = 0.037177, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {765} \[ -\frac{\log (x) (c d-b e)}{b^2}+\frac{(c d-b e) \log (b+c x)}{b^2}-\frac{d}{b x} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(x*(b*x + c*x^2)),x]

[Out]

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

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

Mathematica [A]  time = 0.0178734, size = 42, normalized size = 0.98 \[ \frac{\log (x) (b e-c d)}{b^2}+\frac{(c d-b e) \log (b+c x)}{b^2}-\frac{d}{b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.015, size = 51, normalized size = 1.2 \begin{align*} -{\frac{d}{bx}}+{\frac{e\ln \left ( x \right ) }{b}}-{\frac{\ln \left ( x \right ) cd}{{b}^{2}}}-{\frac{\ln \left ( cx+b \right ) e}{b}}+{\frac{\ln \left ( cx+b \right ) cd}{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00179, size = 58, normalized size = 1.35 \begin{align*} \frac{{\left (c d - b e\right )} \log \left (c x + b\right )}{b^{2}} - \frac{{\left (c d - b e\right )} \log \left (x\right )}{b^{2}} - \frac{d}{b x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.82349, size = 90, normalized size = 2.09 \begin{align*} \frac{{\left (c d - b e\right )} x \log \left (c x + b\right ) -{\left (c d - b e\right )} x \log \left (x\right ) - b d}{b^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.834891, size = 95, normalized size = 2.21 \begin{align*} - \frac{d}{b x} + \frac{\left (b e - c d\right ) \log{\left (x + \frac{b^{2} e - b c d - b \left (b e - c d\right )}{2 b c e - 2 c^{2} d} \right )}}{b^{2}} - \frac{\left (b e - c d\right ) \log{\left (x + \frac{b^{2} e - b c d + b \left (b e - c d\right )}{2 b c e - 2 c^{2} d} \right )}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20286, size = 72, normalized size = 1.67 \begin{align*} -\frac{{\left (c d - b e\right )} \log \left ({\left | x \right |}\right )}{b^{2}} - \frac{d}{b x} + \frac{{\left (c^{2} d - b c e\right )} \log \left ({\left | c x + b \right |}\right )}{b^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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