3.1151 \(\int \frac{A+B x}{(d+e x) (b x+c x^2)^2} \, dx\)

Optimal. Leaf size=147 \[ \frac{c \log (b+c x) \left (-b c (3 A e+B d)+2 A c^2 d+2 b^2 B e\right )}{b^3 (c d-b e)^2}+\frac{\log (x) (-A b e-2 A c d+b B d)}{b^3 d^2}+\frac{c (b B-A c)}{b^2 (b+c x) (c d-b e)}-\frac{A}{b^2 d x}-\frac{e^2 (B d-A e) \log (d+e x)}{d^2 (c d-b e)^2} \]

[Out]

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

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Rubi [A]  time = 0.198891, antiderivative size = 147, 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} \[ \frac{c \log (b+c x) \left (-b c (3 A e+B d)+2 A c^2 d+2 b^2 B e\right )}{b^3 (c d-b e)^2}+\frac{\log (x) (-A b e-2 A c d+b B d)}{b^3 d^2}+\frac{c (b B-A c)}{b^2 (b+c x) (c d-b e)}-\frac{A}{b^2 d x}-\frac{e^2 (B d-A e) \log (d+e x)}{d^2 (c d-b e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.180847, size = 146, normalized size = 0.99 \[ \frac{c \log (b+c x) \left (-b c (3 A e+B d)+2 A c^2 d+2 b^2 B e\right )}{b^3 (c d-b e)^2}+\frac{\log (x) (-A b e-2 A c d+b B d)}{b^3 d^2}+\frac{c (A c-b B)}{b^2 (b+c x) (b e-c d)}-\frac{A}{b^2 d x}+\frac{e^2 (A e-B d) \log (d+e x)}{d^2 (c d-b e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 248, normalized size = 1.7 \begin{align*} -{\frac{A}{{b}^{2}dx}}-{\frac{\ln \left ( x \right ) Ae}{{d}^{2}{b}^{2}}}-2\,{\frac{Ac\ln \left ( x \right ) }{{b}^{3}d}}+{\frac{\ln \left ( x \right ) B}{{b}^{2}d}}+{\frac{{e}^{3}\ln \left ( ex+d \right ) A}{{d}^{2} \left ( be-cd \right ) ^{2}}}-{\frac{{e}^{2}\ln \left ( ex+d \right ) B}{d \left ( be-cd \right ) ^{2}}}-3\,{\frac{{c}^{2}\ln \left ( cx+b \right ) Ae}{{b}^{2} \left ( be-cd \right ) ^{2}}}+2\,{\frac{{c}^{3}\ln \left ( cx+b \right ) Ad}{{b}^{3} \left ( be-cd \right ) ^{2}}}+2\,{\frac{c\ln \left ( cx+b \right ) Be}{b \left ( be-cd \right ) ^{2}}}-{\frac{{c}^{2}\ln \left ( cx+b \right ) Bd}{{b}^{2} \left ( be-cd \right ) ^{2}}}+{\frac{A{c}^{2}}{{b}^{2} \left ( be-cd \right ) \left ( cx+b \right ) }}-{\frac{cB}{b \left ( be-cd \right ) \left ( cx+b \right ) }} \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)^2,x)

[Out]

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

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Maxima [A]  time = 1.17625, size = 306, normalized size = 2.08 \begin{align*} -\frac{{\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d -{\left (2 \, B b^{2} c - 3 \, A b c^{2}\right )} e\right )} \log \left (c x + b\right )}{b^{3} c^{2} d^{2} - 2 \, b^{4} c d e + b^{5} e^{2}} - \frac{{\left (B d e^{2} - A e^{3}\right )} \log \left (e x + d\right )}{c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}} - \frac{A b c d - A b^{2} e -{\left (A b c e +{\left (B b c - 2 \, A c^{2}\right )} d\right )} x}{{\left (b^{2} c^{2} d^{2} - b^{3} c d e\right )} x^{2} +{\left (b^{3} c d^{2} - b^{4} d e\right )} x} - \frac{{\left (A b e -{\left (B b - 2 \, A c\right )} d\right )} \log \left (x\right )}{b^{3} d^{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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 137.443, size = 895, normalized size = 6.09 \begin{align*} -\frac{A b^{2} c^{2} d^{3} - 2 \, A b^{3} c d^{2} e + A b^{4} d e^{2} +{\left (A b^{3} c d e^{2} -{\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d^{3} +{\left (B b^{3} c - 3 \, A b^{2} c^{2}\right )} d^{2} e\right )} x +{\left ({\left ({\left (B b c^{3} - 2 \, A c^{4}\right )} d^{3} -{\left (2 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} d^{2} e\right )} x^{2} +{\left ({\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d^{3} -{\left (2 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} d^{2} e\right )} x\right )} \log \left (c x + b\right ) +{\left ({\left (B b^{3} c d e^{2} - A b^{3} c e^{3}\right )} x^{2} +{\left (B b^{4} d e^{2} - A b^{4} e^{3}\right )} x\right )} \log \left (e x + d\right ) -{\left ({\left (B b^{3} c d e^{2} - A b^{3} c e^{3} +{\left (B b c^{3} - 2 \, A c^{4}\right )} d^{3} -{\left (2 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} d^{2} e\right )} x^{2} +{\left (B b^{4} d e^{2} - A b^{4} e^{3} +{\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} d^{3} -{\left (2 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} d^{2} e\right )} x\right )} \log \left (x\right )}{{\left (b^{3} c^{3} d^{4} - 2 \, b^{4} c^{2} d^{3} e + b^{5} c d^{2} e^{2}\right )} x^{2} +{\left (b^{4} c^{2} d^{4} - 2 \, b^{5} c d^{3} e + b^{6} d^{2} e^{2}\right )} x} \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)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.23696, size = 362, normalized size = 2.46 \begin{align*} -\frac{{\left (B b c^{3} d - 2 \, A c^{4} d - 2 \, B b^{2} c^{2} e + 3 \, A b c^{3} e\right )} \log \left ({\left | c x + b \right |}\right )}{b^{3} c^{3} d^{2} - 2 \, b^{4} c^{2} d e + b^{5} c e^{2}} - \frac{{\left (B d e^{3} - A e^{4}\right )} \log \left ({\left | x e + d \right |}\right )}{c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3}} + \frac{{\left (B b d - 2 \, A c d - A b e\right )} \log \left ({\left | x \right |}\right )}{b^{3} d^{2}} - \frac{A b c^{2} d^{3} - 2 \, A b^{2} c d^{2} e + A b^{3} d e^{2} -{\left (B b c^{2} d^{3} - 2 \, A c^{3} d^{3} - B b^{2} c d^{2} e + 3 \, A b c^{2} d^{2} e - A b^{2} c d e^{2}\right )} x}{{\left (c d - b e\right )}^{2}{\left (c x + b\right )} b^{2} d^{2} x} \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)^2,x, algorithm="giac")

[Out]

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