3.1140 \(\int \frac{(A+B x) (d+e x)}{b x+c x^2} \, dx\)

Optimal. Leaf size=45 \[ \frac{(b B-A c) (c d-b e) \log (b+c x)}{b c^2}+\frac{A d \log (x)}{b}+\frac{B e x}{c} \]

[Out]

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

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Rubi [A]  time = 0.0447597, antiderivative size = 45, 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 = {771} \[ \frac{(b B-A c) (c d-b e) \log (b+c x)}{b c^2}+\frac{A d \log (x)}{b}+\frac{B e x}{c} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x))/(b*x + c*x^2),x]

[Out]

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

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

Mathematica [A]  time = 0.0239048, size = 46, normalized size = 1.02 \[ \frac{-(b B-A c) (b e-c d) \log (b+c x)+A c^2 d \log (x)+b B c e x}{b c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 68, normalized size = 1.5 \begin{align*}{\frac{Bex}{c}}+{\frac{Ad\ln \left ( x \right ) }{b}}+{\frac{\ln \left ( cx+b \right ) Ae}{c}}-{\frac{\ln \left ( cx+b \right ) Ad}{b}}-{\frac{b\ln \left ( cx+b \right ) Be}{{c}^{2}}}+{\frac{\ln \left ( cx+b \right ) Bd}{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.07457, size = 77, normalized size = 1.71 \begin{align*} \frac{B e x}{c} + \frac{A d \log \left (x\right )}{b} + \frac{{\left ({\left (B b c - A c^{2}\right )} d -{\left (B b^{2} - A b c\right )} e\right )} \log \left (c x + b\right )}{b c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.46243, size = 126, normalized size = 2.8 \begin{align*} \frac{B b c e x + A c^{2} d \log \left (x\right ) +{\left ({\left (B b c - A c^{2}\right )} d -{\left (B b^{2} - A b c\right )} e\right )} \log \left (c x + b\right )}{b c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.85819, size = 88, normalized size = 1.96 \begin{align*} \frac{A d \log{\left (x \right )}}{b} + \frac{B e x}{c} - \frac{\left (- A c + B b\right ) \left (b e - c d\right ) \log{\left (x + \frac{A b c d + \frac{b \left (- A c + B b\right ) \left (b e - c d\right )}{c}}{- A b c e + 2 A c^{2} d + B b^{2} e - B b c d} \right )}}{b c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20963, size = 80, normalized size = 1.78 \begin{align*} \frac{B x e}{c} + \frac{A d \log \left ({\left | x \right |}\right )}{b} + \frac{{\left (B b c d - A c^{2} d - B b^{2} e + A b c e\right )} \log \left ({\left | c x + b \right |}\right )}{b c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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