3.1143 \(\int \frac{A+B x}{(d+e x)^2 (b x+c x^2)} \, dx\)

Optimal. Leaf size=110 \[ -\frac{\log (d+e x) \left (B c d^2-A e (2 c d-b e)\right )}{d^2 (c d-b e)^2}+\frac{B d-A e}{d (d+e x) (c d-b e)}+\frac{c (b B-A c) \log (b+c x)}{b (c d-b e)^2}+\frac{A \log (x)}{b d^2} \]

[Out]

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

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Rubi [A]  time = 0.134021, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.042, Rules used = {771} \[ -\frac{\log (d+e x) \left (B c d^2-A e (2 c d-b e)\right )}{d^2 (c d-b e)^2}+\frac{B d-A e}{d (d+e x) (c d-b e)}+\frac{c (b B-A c) \log (b+c x)}{b (c d-b e)^2}+\frac{A \log (x)}{b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.153422, size = 106, normalized size = 0.96 \[ \frac{\frac{c d^2 (d+e x) (b B-A c) \log (b+c x)-b (d+e x) \log (d+e x) \left (A e (b e-2 c d)+B c d^2\right )+b d (B d-A e) (c d-b e)}{(d+e x) (c d-b e)^2}+A \log (x)}{b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 169, normalized size = 1.5 \begin{align*}{\frac{A\ln \left ( x \right ) }{{d}^{2}b}}-{\frac{\ln \left ( ex+d \right ) Ab{e}^{2}}{{d}^{2} \left ( be-cd \right ) ^{2}}}+2\,{\frac{\ln \left ( ex+d \right ) Ace}{d \left ( be-cd \right ) ^{2}}}-{\frac{\ln \left ( ex+d \right ) Bc}{ \left ( be-cd \right ) ^{2}}}+{\frac{Ae}{d \left ( be-cd \right ) \left ( ex+d \right ) }}-{\frac{B}{ \left ( be-cd \right ) \left ( ex+d \right ) }}-{\frac{{c}^{2}\ln \left ( cx+b \right ) A}{b \left ( be-cd \right ) ^{2}}}+{\frac{c\ln \left ( cx+b \right ) B}{ \left ( be-cd \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.104, size = 203, normalized size = 1.85 \begin{align*} \frac{{\left (B b c - A c^{2}\right )} \log \left (c x + b\right )}{b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}} - \frac{{\left (B c d^{2} - 2 \, A c d e + A b e^{2}\right )} \log \left (e x + d\right )}{c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}} + \frac{B d - A e}{c d^{3} - b d^{2} e +{\left (c d^{2} e - b d e^{2}\right )} x} + \frac{A \log \left (x\right )}{b d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*b*c - A*c^2)*log(c*x + b)/(b*c^2*d^2 - 2*b^2*c*d*e + b^3*e^2) - (B*c*d^2 - 2*A*c*d*e + A*b*e^2)*log(e*x + d
)/(c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2) + (B*d - A*e)/(c*d^3 - b*d^2*e + (c*d^2*e - b*d*e^2)*x) + A*log(x)/(b*
d^2)

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Fricas [B]  time = 56.9818, size = 548, normalized size = 4.98 \begin{align*} \frac{B b c d^{3} + A b^{2} d e^{2} -{\left (B b^{2} + A b c\right )} d^{2} e +{\left ({\left (B b c - A c^{2}\right )} d^{2} e x +{\left (B b c - A c^{2}\right )} d^{3}\right )} \log \left (c x + b\right ) -{\left (B b c d^{3} - 2 \, A b c d^{2} e + A b^{2} d e^{2} +{\left (B b c d^{2} e - 2 \, A b c d e^{2} + A b^{2} e^{3}\right )} x\right )} \log \left (e x + d\right ) +{\left (A c^{2} d^{3} - 2 \, A b c d^{2} e + A b^{2} d e^{2} +{\left (A c^{2} d^{2} e - 2 \, A b c d e^{2} + A b^{2} e^{3}\right )} x\right )} \log \left (x\right )}{b c^{2} d^{5} - 2 \, b^{2} c d^{4} e + b^{3} d^{3} e^{2} +{\left (b c^{2} d^{4} e - 2 \, b^{2} c d^{3} e^{2} + b^{3} d^{2} e^{3}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*b*c*d^3 + A*b^2*d*e^2 - (B*b^2 + A*b*c)*d^2*e + ((B*b*c - A*c^2)*d^2*e*x + (B*b*c - A*c^2)*d^3)*log(c*x + b
) - (B*b*c*d^3 - 2*A*b*c*d^2*e + A*b^2*d*e^2 + (B*b*c*d^2*e - 2*A*b*c*d*e^2 + A*b^2*e^3)*x)*log(e*x + d) + (A*
c^2*d^3 - 2*A*b*c*d^2*e + A*b^2*d*e^2 + (A*c^2*d^2*e - 2*A*b*c*d*e^2 + A*b^2*e^3)*x)*log(x))/(b*c^2*d^5 - 2*b^
2*c*d^4*e + b^3*d^3*e^2 + (b*c^2*d^4*e - 2*b^2*c*d^3*e^2 + b^3*d^2*e^3)*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.38805, size = 436, normalized size = 3.96 \begin{align*} -\frac{{\left (B b c d^{2} e^{2} - 2 \, A c^{2} d^{2} e^{2} + 2 \, A b c d e^{3} - A b^{2} e^{4}\right )} e^{\left (-2\right )} \log \left (\frac{{\left | 2 \, c d e - \frac{2 \, c d^{2} e}{x e + d} - b e^{2} + \frac{2 \, b d e^{2}}{x e + d} -{\left | b \right |} e^{2} \right |}}{{\left | 2 \, c d e - \frac{2 \, c d^{2} e}{x e + d} - b e^{2} + \frac{2 \, b d e^{2}}{x e + d} +{\left | b \right |} e^{2} \right |}}\right )}{2 \,{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )}{\left | b \right |}} + \frac{{\left (B c d^{2} - 2 \, A c d e + A b e^{2}\right )} \log \left ({\left | c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{b e}{x e + d} - \frac{b d e}{{\left (x e + d\right )}^{2}} \right |}\right )}{2 \,{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )}} + \frac{\frac{B d e^{2}}{x e + d} - \frac{A e^{3}}{x e + d}}{c d^{2} e^{2} - b d e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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