3.1152 \(\int \frac{A+B x}{(d+e x)^2 (b x+c x^2)^2} \, dx\)

Optimal. Leaf size=201 \[ \frac{c^2 (b B-A c)}{b^2 (b+c x) (c d-b e)^2}+\frac{c^2 \log (b+c x) \left (-b c (4 A e+B d)+2 A c^2 d+3 b^2 B e\right )}{b^3 (c d-b e)^3}+\frac{\log (x) (-2 A b e-2 A c d+b B d)}{b^3 d^3}-\frac{A}{b^2 d^2 x}+\frac{e^2 (B d-A e)}{d^2 (d+e x) (c d-b e)^2}+\frac{e^2 \log (d+e x) (2 A e (2 c d-b e)-B d (3 c d-b e))}{d^3 (c d-b e)^3} \]

[Out]

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

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Rubi [A]  time = 0.330545, antiderivative size = 201, 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} \[ \frac{c^2 (b B-A c)}{b^2 (b+c x) (c d-b e)^2}+\frac{c^2 \log (b+c x) \left (-b c (4 A e+B d)+2 A c^2 d+3 b^2 B e\right )}{b^3 (c d-b e)^3}+\frac{\log (x) (-2 A b e-2 A c d+b B d)}{b^3 d^3}-\frac{A}{b^2 d^2 x}+\frac{e^2 (B d-A e)}{d^2 (d+e x) (c d-b e)^2}+\frac{e^2 \log (d+e x) (2 A e (2 c d-b e)-B d (3 c d-b e))}{d^3 (c d-b e)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 )^2} \, dx &=\int \left (\frac{A}{b^2 d^2 x^2}+\frac{b B d-2 A c d-2 A b e}{b^3 d^3 x}-\frac{c^3 (b B-A c)}{b^2 (-c d+b e)^2 (b+c x)^2}+\frac{c^3 \left (2 A c^2 d+3 b^2 B e-b c (B d+4 A e)\right )}{b^3 (c d-b e)^3 (b+c x)}-\frac{e^3 (B d-A e)}{d^2 (c d-b e)^2 (d+e x)^2}+\frac{e^3 (2 A e (2 c d-b e)-B d (3 c d-b e))}{d^3 (c d-b e)^3 (d+e x)}\right ) \, dx\\ &=-\frac{A}{b^2 d^2 x}+\frac{c^2 (b B-A c)}{b^2 (c d-b e)^2 (b+c x)}+\frac{e^2 (B d-A e)}{d^2 (c d-b e)^2 (d+e x)}+\frac{(b B d-2 A c d-2 A b e) \log (x)}{b^3 d^3}+\frac{c^2 \left (2 A c^2 d+3 b^2 B e-b c (B d+4 A e)\right ) \log (b+c x)}{b^3 (c d-b e)^3}+\frac{e^2 (2 A e (2 c d-b e)-B d (3 c d-b e)) \log (d+e x)}{d^3 (c d-b e)^3}\\ \end{align*}

Mathematica [A]  time = 0.346054, size = 201, normalized size = 1. \[ \frac{c^2 (b B-A c)}{b^2 (b+c x) (c d-b e)^2}-\frac{c^2 \log (b+c x) \left (-b c (4 A e+B d)+2 A c^2 d+3 b^2 B e\right )}{b^3 (b e-c d)^3}+\frac{\log (x) (-2 A b e-2 A c d+b B d)}{b^3 d^3}-\frac{A}{b^2 d^2 x}+\frac{e^2 (B d-A e)}{d^2 (d+e x) (c d-b e)^2}-\frac{e^2 \log (d+e x) (2 A e (b e-2 c d)+B d (3 c d-b e))}{d^3 (c d-b e)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 357, normalized size = 1.8 \begin{align*} -{\frac{A}{{d}^{2}{b}^{2}x}}-2\,{\frac{\ln \left ( x \right ) Ae}{{d}^{3}{b}^{2}}}-2\,{\frac{Ac\ln \left ( x \right ) }{{d}^{2}{b}^{3}}}+{\frac{\ln \left ( x \right ) B}{{d}^{2}{b}^{2}}}+2\,{\frac{{e}^{4}\ln \left ( ex+d \right ) Ab}{{d}^{3} \left ( be-cd \right ) ^{3}}}-4\,{\frac{{e}^{3}\ln \left ( ex+d \right ) Ac}{{d}^{2} \left ( be-cd \right ) ^{3}}}-{\frac{{e}^{3}\ln \left ( ex+d \right ) Bb}{{d}^{2} \left ( be-cd \right ) ^{3}}}+3\,{\frac{{e}^{2}\ln \left ( ex+d \right ) Bc}{d \left ( be-cd \right ) ^{3}}}-{\frac{A{e}^{3}}{{d}^{2} \left ( be-cd \right ) ^{2} \left ( ex+d \right ) }}+{\frac{B{e}^{2}}{d \left ( be-cd \right ) ^{2} \left ( ex+d \right ) }}+4\,{\frac{{c}^{3}\ln \left ( cx+b \right ) Ae}{{b}^{2} \left ( be-cd \right ) ^{3}}}-2\,{\frac{{c}^{4}\ln \left ( cx+b \right ) Ad}{{b}^{3} \left ( be-cd \right ) ^{3}}}-3\,{\frac{{c}^{2}\ln \left ( cx+b \right ) Be}{b \left ( be-cd \right ) ^{3}}}+{\frac{{c}^{3}\ln \left ( cx+b \right ) Bd}{{b}^{2} \left ( be-cd \right ) ^{3}}}-{\frac{A{c}^{3}}{{b}^{2} \left ( be-cd \right ) ^{2} \left ( cx+b \right ) }}+{\frac{B{c}^{2}}{b \left ( be-cd \right ) ^{2} \left ( cx+b \right ) }} \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)^2,x)

[Out]

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

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Maxima [B]  time = 1.177, size = 630, normalized size = 3.13 \begin{align*} -\frac{{\left ({\left (B b c^{3} - 2 \, A c^{4}\right )} d -{\left (3 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} e\right )} \log \left (c x + b\right )}{b^{3} c^{3} d^{3} - 3 \, b^{4} c^{2} d^{2} e + 3 \, b^{5} c d e^{2} - b^{6} e^{3}} - \frac{{\left (3 \, B c d^{2} e^{2} + 2 \, A b e^{4} -{\left (B b + 4 \, A c\right )} d e^{3}\right )} \log \left (e x + d\right )}{c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}} - \frac{A b c^{2} d^{3} - 2 \, A b^{2} c d^{2} e + A b^{3} d e^{2} +{\left (2 \, A b^{2} c e^{3} -{\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} e -{\left (B b^{2} c + 2 \, A b c^{2}\right )} d e^{2}\right )} x^{2} -{\left (A b c^{2} d^{2} e - 2 \, A b^{3} e^{3} +{\left (B b c^{2} - 2 \, A c^{3}\right )} d^{3} +{\left (B b^{3} + A b^{2} c\right )} d e^{2}\right )} x}{{\left (b^{2} c^{3} d^{4} e - 2 \, b^{3} c^{2} d^{3} e^{2} + b^{4} c d^{2} e^{3}\right )} x^{3} +{\left (b^{2} c^{3} d^{5} - b^{3} c^{2} d^{4} e - b^{4} c d^{3} e^{2} + b^{5} d^{2} e^{3}\right )} x^{2} +{\left (b^{3} c^{2} d^{5} - 2 \, b^{4} c d^{4} e + b^{5} d^{3} e^{2}\right )} x} - \frac{{\left (2 \, A b e -{\left (B b - 2 \, A c\right )} d\right )} \log \left (x\right )}{b^{3} d^{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)^2,x, algorithm="maxima")

[Out]

-((B*b*c^3 - 2*A*c^4)*d - (3*B*b^2*c^2 - 4*A*b*c^3)*e)*log(c*x + b)/(b^3*c^3*d^3 - 3*b^4*c^2*d^2*e + 3*b^5*c*d
*e^2 - b^6*e^3) - (3*B*c*d^2*e^2 + 2*A*b*e^4 - (B*b + 4*A*c)*d*e^3)*log(e*x + d)/(c^3*d^6 - 3*b*c^2*d^5*e + 3*
b^2*c*d^4*e^2 - b^3*d^3*e^3) - (A*b*c^2*d^3 - 2*A*b^2*c*d^2*e + A*b^3*d*e^2 + (2*A*b^2*c*e^3 - (B*b*c^2 - 2*A*
c^3)*d^2*e - (B*b^2*c + 2*A*b*c^2)*d*e^2)*x^2 - (A*b*c^2*d^2*e - 2*A*b^3*e^3 + (B*b*c^2 - 2*A*c^3)*d^3 + (B*b^
3 + A*b^2*c)*d*e^2)*x)/((b^2*c^3*d^4*e - 2*b^3*c^2*d^3*e^2 + b^4*c*d^2*e^3)*x^3 + (b^2*c^3*d^5 - b^3*c^2*d^4*e
- b^4*c*d^3*e^2 + b^5*d^2*e^3)*x^2 + (b^3*c^2*d^5 - 2*b^4*c*d^4*e + b^5*d^3*e^2)*x) - (2*A*b*e - (B*b - 2*A*c
)*d)*log(x)/(b^3*d^3)

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Fricas [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)^2,x, algorithm="fricas")

[Out]

Timed out

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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)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.45793, size = 905, normalized size = 4.5 \begin{align*} \frac{{\left (2 \, B b c^{3} d^{4} e^{2} - 4 \, A c^{4} d^{4} e^{2} - 6 \, B b^{2} c^{2} d^{3} e^{3} + 8 \, A b c^{3} d^{3} e^{3} + 3 \, B b^{3} c d^{2} e^{4} - B b^{4} d e^{5} - 4 \, A b^{3} c d e^{5} + 2 \, A b^{4} e^{6}\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 (b^{2} c^{3} d^{6} - 3 \, b^{3} c^{2} d^{5} e + 3 \, b^{4} c d^{4} e^{2} - b^{5} d^{3} e^{3}\right )}{\left | b \right |}} + \frac{{\left (3 \, B c d^{2} e^{2} - B b d e^{3} - 4 \, A c d e^{3} + 2 \, A b e^{4}\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^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}\right )}} + \frac{\frac{B d e^{6}}{x e + d} - \frac{A e^{7}}{x e + d}}{c^{2} d^{4} e^{4} - 2 \, b c d^{3} e^{5} + b^{2} d^{2} e^{6}} + \frac{\frac{B b c^{3} d^{3} e - 2 \, A c^{4} d^{3} e + 3 \, A b c^{3} d^{2} e^{2} - 3 \, A b^{2} c^{2} d e^{3} + A b^{3} c e^{4}}{c d^{2} - b d e} - \frac{{\left (B b c^{3} d^{4} e^{2} - 2 \, A c^{4} d^{4} e^{2} + 4 \, A b c^{3} d^{3} e^{3} - 6 \, A b^{2} c^{2} d^{2} e^{4} + 4 \, A b^{3} c d e^{5} - A b^{4} e^{6}\right )} e^{\left (-1\right )}}{{\left (c d^{2} - b d e\right )}{\left (x e + d\right )}}}{{\left (c d - b e\right )}^{2} b^{2}{\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 )} 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)^2,x, algorithm="giac")

[Out]

1/2*(2*B*b*c^3*d^4*e^2 - 4*A*c^4*d^4*e^2 - 6*B*b^2*c^2*d^3*e^3 + 8*A*b*c^3*d^3*e^3 + 3*B*b^3*c*d^2*e^4 - B*b^4
*d*e^5 - 4*A*b^3*c*d*e^5 + 2*A*b^4*e^6)*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))/((b^2*c^3*d^
6 - 3*b^3*c^2*d^5*e + 3*b^4*c*d^4*e^2 - b^5*d^3*e^3)*abs(b)) + 1/2*(3*B*c*d^2*e^2 - B*b*d*e^3 - 4*A*c*d*e^3 +
2*A*b*e^4)*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^3*d^6 - 3*
b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3) + (B*d*e^6/(x*e + d) - A*e^7/(x*e + d))/(c^2*d^4*e^4 - 2*b*c*d^3*
e^5 + b^2*d^2*e^6) + ((B*b*c^3*d^3*e - 2*A*c^4*d^3*e + 3*A*b*c^3*d^2*e^2 - 3*A*b^2*c^2*d*e^3 + A*b^3*c*e^4)/(c
*d^2 - b*d*e) - (B*b*c^3*d^4*e^2 - 2*A*c^4*d^4*e^2 + 4*A*b*c^3*d^3*e^3 - 6*A*b^2*c^2*d^2*e^4 + 4*A*b^3*c*d*e^5
- A*b^4*e^6)*e^(-1)/((c*d^2 - b*d*e)*(x*e + d)))/((c*d - b*e)^2*b^2*(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)*d^2)