\(\int \frac {1}{(d+e x)^2 (b x+c x^2)} \, dx\) [265]

Optimal result

Integrand size = 19, antiderivative size = 87 \[ \int \frac {1}{(d+e x)^2 \left (b x+c x^2\right )} \, dx=-\frac {e}{d (c d-b e) (d+e x)}+\frac {\log (x)}{b d^2}-\frac {c^2 \log (b+c x)}{b (c d-b e)^2}+\frac {e (2 c d-b e) \log (d+e x)}{d^2 (c d-b e)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {712} \[ \int \frac {1}{(d+e x)^2 \left (b x+c x^2\right )} \, dx=-\frac {c^2 \log (b+c x)}{b (c d-b e)^2}+\frac {e (2 c d-b e) \log (d+e x)}{d^2 (c d-b e)^2}-\frac {e}{d (d+e x) (c d-b e)}+\frac {\log (x)}{b d^2} \]

[In]

Int[1/((d + e*x)^2*(b*x + c*x^2)),x]

[Out]

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

Rule 712

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(d+e x)^2 \left (b x+c x^2\right )} \, dx=\frac {\log (x)+\frac {-c^2 d^2 (d+e x) \log (b+c x)+b e (d (-c d+b e)+(2 c d-b e) (d+e x) \log (d+e x))}{(c d-b e)^2 (d+e x)}}{b d^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.93 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {\ln \left (x \right )}{b \,d^{2}}-\frac {c^{2} \ln \left (c x +b \right )}{\left (b e -c d \right )^{2} b}+\frac {e}{d \left (b e -c d \right ) \left (e x +d \right )}-\frac {e \left (b e -2 c d \right ) \ln \left (e x +d \right )}{d^{2} \left (b e -c d \right )^{2}}\)	\(87\)
norman	\(-\frac {e^{2} x}{d^{2} \left (b e -c d \right ) \left (e x +d \right )}+\frac {\ln \left (x \right )}{b \,d^{2}}-\frac {c^{2} \ln \left (c x +b \right )}{b \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}-\frac {e \left (b e -2 c d \right ) \ln \left (e x +d \right )}{d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}\)	\(117\)
risch	\(\frac {e}{d \left (b e -c d \right ) \left (e x +d \right )}+\frac {\ln \left (-x \right )}{d^{2} b}-\frac {e^{2} \ln \left (-e x -d \right ) b}{d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}+\frac {2 e \ln \left (-e x -d \right ) c}{d \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}-\frac {c^{2} \ln \left (c x +b \right )}{b \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}\)	\(152\)
parallelrisch	\(\frac {\ln \left (x \right ) x \,b^{2} e^{4}-2 \ln \left (x \right ) x b c d \,e^{3}+\ln \left (x \right ) x \,c^{2} d^{2} e^{2}-\ln \left (c x +b \right ) x \,c^{2} d^{2} e^{2}-\ln \left (e x +d \right ) x \,b^{2} e^{4}+2 \ln \left (e x +d \right ) x b c d \,e^{3}+\ln \left (x \right ) b^{2} d \,e^{3}-2 \ln \left (x \right ) b c \,d^{2} e^{2}+\ln \left (x \right ) c^{2} d^{3} e -\ln \left (c x +b \right ) c^{2} d^{3} e -\ln \left (e x +d \right ) b^{2} d \,e^{3}+2 \ln \left (e x +d \right ) b c \,d^{2} e^{2}+d \,e^{3} b^{2}-b c \,d^{2} e^{2}}{\left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right ) b \left (e x +d \right ) d^{2} e}\)	\(220\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (87) = 174\).

Time = 0.69 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.38 \[ \int \frac {1}{(d+e x)^2 \left (b x+c x^2\right )} \, dx=-\frac {b c d^{2} e - b^{2} d e^{2} + {\left (c^{2} d^{2} e x + c^{2} d^{3}\right )} \log \left (c x + b\right ) - {\left (2 \, b c d^{2} e - b^{2} d e^{2} + {\left (2 \, b c d e^{2} - b^{2} e^{3}\right )} x\right )} \log \left (e x + d\right ) - {\left (c^{2} d^{3} - 2 \, b c d^{2} e + b^{2} d e^{2} + {\left (c^{2} d^{2} e - 2 \, b c d e^{2} + b^{2} e^{3}\right )} x\right )} \log \left (x\right )}{b c^{2} d^{5} - 2 \, b^{2} c d^{4} e + b^{3} d^{3} e^{2} + {\left (b c^{2} d^{4} e - 2 \, b^{2} c d^{3} e^{2} + b^{3} d^{2} e^{3}\right )} x} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^2 \left (b x+c x^2\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(d+e x)^2 \left (b x+c x^2\right )} \, dx=-\frac {c^{2} \log \left (c x + b\right )}{b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}} + \frac {{\left (2 \, c d e - b e^{2}\right )} \log \left (e x + d\right )}{c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}} - \frac {e}{c d^{3} - b d^{2} e + {\left (c d^{2} e - b d e^{2}\right )} x} + \frac {\log \left (x\right )}{b d^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (87) = 174\).

Time = 0.27 (sec) , antiderivative size = 286, normalized size of antiderivative = 3.29 \[ \int \frac {1}{(d+e x)^2 \left (b x+c x^2\right )} \, dx=-\frac {e^{3}}{{\left (c d^{2} e^{2} - b d e^{3}\right )} {\left (e x + d\right )}} - \frac {{\left (2 \, c d e - b e^{2}\right )} \log \left ({\left | -c + \frac {2 \, c d}{e x + d} - \frac {c d^{2}}{{\left (e x + d\right )}^{2}} - \frac {b e}{e x + d} + \frac {b d e}{{\left (e x + d\right )}^{2}} \right |}\right )}{2 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )}} - \frac {{\left (2 \, c^{2} d^{2} e^{2} - 2 \, b c d e^{3} + b^{2} e^{4}\right )} \log \left (\frac {{\left | -2 \, c d e + \frac {2 \, c d^{2} e}{e x + d} + b e^{2} - \frac {2 \, b d e^{2}}{e x + d} - e^{2} {\left | b \right |} \right |}}{{\left | -2 \, c d e + \frac {2 \, c d^{2} e}{e x + d} + b e^{2} - \frac {2 \, b d e^{2}}{e x + d} + e^{2} {\left | b \right |} \right |}}\right )}{2 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )} e^{2} {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.94 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.33 \[ \int \frac {1}{(d+e x)^2 \left (b x+c x^2\right )} \, dx=\frac {\ln \left (x\right )}{b\,d^2}-\frac {c^2\,\ln \left (b+c\,x\right )}{b^3\,e^2-2\,b^2\,c\,d\,e+b\,c^2\,d^2}-\frac {\ln \left (d+e\,x\right )\,\left (b\,e^2-2\,c\,d\,e\right )}{b^2\,d^2\,e^2-2\,b\,c\,d^3\,e+c^2\,d^4}+\frac {e}{d\,\left (b\,e-c\,d\right )\,\left (d+e\,x\right )} \]

[In]

int(1/((b*x + c*x^2)*(d + e*x)^2),x)

[Out]

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