3.273 \(\int \frac{1}{(d+e x) (b x+c x^2)^2} \, dx\)

Optimal. Leaf size=110 \[ -\frac{c^2}{b^2 (b+c x) (c d-b e)}+\frac{c^2 (2 c d-3 b e) \log (b+c x)}{b^3 (c d-b e)^2}-\frac{\log (x) (b e+2 c d)}{b^3 d^2}-\frac{1}{b^2 d x}+\frac{e^3 \log (d+e x)}{d^2 (c d-b e)^2} \]

[Out]

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

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Rubi [A]  time = 0.118041, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {698} \[ -\frac{c^2}{b^2 (b+c x) (c d-b e)}+\frac{c^2 (2 c d-3 b e) \log (b+c x)}{b^3 (c d-b e)^2}-\frac{\log (x) (b e+2 c d)}{b^3 d^2}-\frac{1}{b^2 d x}+\frac{e^3 \log (d+e x)}{d^2 (c d-b e)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(b*x + c*x^2)^2),x]

[Out]

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

Rule 698

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

Rubi steps

\begin{align*} \int \frac{1}{(d+e x) \left (b x+c x^2\right )^2} \, dx &=\int \left (\frac{1}{b^2 d x^2}+\frac{-2 c d-b e}{b^3 d^2 x}-\frac{c^3}{b^2 (-c d+b e) (b+c x)^2}-\frac{c^3 (-2 c d+3 b e)}{b^3 (-c d+b e)^2 (b+c x)}+\frac{e^4}{d^2 (c d-b e)^2 (d+e x)}\right ) \, dx\\ &=-\frac{1}{b^2 d x}-\frac{c^2}{b^2 (c d-b e) (b+c x)}-\frac{(2 c d+b e) \log (x)}{b^3 d^2}+\frac{c^2 (2 c d-3 b e) \log (b+c x)}{b^3 (c d-b e)^2}+\frac{e^3 \log (d+e x)}{d^2 (c d-b e)^2}\\ \end{align*}

Mathematica [A]  time = 0.0960025, size = 111, normalized size = 1.01 \[ \frac{c^2}{b^2 (b+c x) (b e-c d)}+\frac{\left (2 c^3 d-3 b c^2 e\right ) \log (b+c x)}{b^3 (b e-c d)^2}+\frac{\log (x) (-b e-2 c d)}{b^3 d^2}-\frac{1}{b^2 d x}+\frac{e^3 \log (d+e x)}{d^2 (c d-b e)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)*(b*x + c*x^2)^2),x]

[Out]

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

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Maple [A]  time = 0.105, size = 132, normalized size = 1.2 \begin{align*} -{\frac{1}{{b}^{2}dx}}-{\frac{\ln \left ( x \right ) e}{{b}^{2}{d}^{2}}}-2\,{\frac{\ln \left ( x \right ) c}{d{b}^{3}}}+{\frac{{c}^{2}}{ \left ( be-cd \right ){b}^{2} \left ( cx+b \right ) }}-3\,{\frac{{c}^{2}\ln \left ( cx+b \right ) e}{ \left ( be-cd \right ) ^{2}{b}^{2}}}+2\,{\frac{{c}^{3}\ln \left ( cx+b \right ) d}{ \left ( be-cd \right ) ^{2}{b}^{3}}}+{\frac{{e}^{3}\ln \left ( ex+d \right ) }{{d}^{2} \left ( be-cd \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)/(c*x^2+b*x)^2,x)

[Out]

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

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Maxima [A]  time = 1.17669, size = 239, normalized size = 2.17 \begin{align*} \frac{e^{3} \log \left (e x + d\right )}{c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}} + \frac{{\left (2 \, c^{3} d - 3 \, b c^{2} 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{b c d - b^{2} e +{\left (2 \, c^{2} d - b c e\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 (2 \, c d + b e\right )} \log \left (x\right )}{b^{3} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 24.7447, size = 574, normalized size = 5.22 \begin{align*} -\frac{b^{2} c^{2} d^{3} - 2 \, b^{3} c d^{2} e + b^{4} d e^{2} +{\left (2 \, b c^{3} d^{3} - 3 \, b^{2} c^{2} d^{2} e + b^{3} c d e^{2}\right )} x -{\left ({\left (2 \, c^{4} d^{3} - 3 \, b c^{3} d^{2} e\right )} x^{2} +{\left (2 \, b c^{3} d^{3} - 3 \, b^{2} c^{2} d^{2} e\right )} x\right )} \log \left (c x + b\right ) -{\left (b^{3} c e^{3} x^{2} + b^{4} e^{3} x\right )} \log \left (e x + d\right ) +{\left ({\left (2 \, c^{4} d^{3} - 3 \, b c^{3} d^{2} e + b^{3} c e^{3}\right )} x^{2} +{\left (2 \, b c^{3} d^{3} - 3 \, b^{2} c^{2} d^{2} e + b^{4} e^{3}\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(1/(e*x+d)/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.29849, size = 273, normalized size = 2.48 \begin{align*} \frac{{\left (2 \, c^{4} d - 3 \, 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{e^{4} \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 (2 \, c d + b e\right )} \log \left ({\left | x \right |}\right )}{b^{3} d^{2}} - \frac{b c^{2} d^{3} - 2 \, b^{2} c d^{2} e + b^{3} d e^{2} +{\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 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(1/(e*x+d)/(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

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