\(\int \frac {d+e x}{(b x+c x^2)^2} \, dx\) [271]

Optimal result

Integrand size = 17, antiderivative size = 65 \[ \int \frac {d+e x}{\left (b x+c x^2\right )^2} \, dx=-\frac {d}{b^2 x}-\frac {c d-b e}{b^2 (b+c x)}-\frac {(2 c d-b e) \log (x)}{b^3}+\frac {(2 c d-b e) \log (b+c x)}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {645} \[ \int \frac {d+e x}{\left (b x+c x^2\right )^2} \, dx=-\frac {\log (x) (2 c d-b e)}{b^3}+\frac {(2 c d-b e) \log (b+c x)}{b^3}-\frac {c d-b e}{b^2 (b+c x)}-\frac {d}{b^2 x} \]

[In]

Int[(d + e*x)/(b*x + c*x^2)^2,x]

[Out]

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

Rule 645

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \frac {d+e x}{\left (b x+c x^2\right )^2} \, dx=\frac {-\frac {b d}{x}+\frac {b (-c d+b e)}{b+c x}+(-2 c d+b e) \log (x)+(2 c d-b e) \log (b+c x)}{b^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {d}{b^{2} x}+\frac {\left (b e -2 c d \right ) \ln \left (x \right )}{b^{3}}-\frac {\left (b e -2 c d \right ) \ln \left (c x +b \right )}{b^{3}}+\frac {b e -c d}{b^{2} \left (c x +b \right )}\)	\(63\)
norman	\(\frac {\frac {c \left (-b e +2 c d \right ) x^{2}}{b^{3}}-\frac {d}{b}}{x \left (c x +b \right )}+\frac {\left (b e -2 c d \right ) \ln \left (x \right )}{b^{3}}-\frac {\left (b e -2 c d \right ) \ln \left (c x +b \right )}{b^{3}}\)	\(70\)
risch	\(\frac {\frac {\left (b e -2 c d \right ) x}{b^{2}}-\frac {d}{b}}{x \left (c x +b \right )}-\frac {\ln \left (c x +b \right ) e}{b^{2}}+\frac {2 \ln \left (c x +b \right ) c d}{b^{3}}+\frac {\ln \left (-x \right ) e}{b^{2}}-\frac {2 \ln \left (-x \right ) c d}{b^{3}}\)	\(78\)
parallelrisch	\(\frac {\ln \left (x \right ) x^{2} b \,c^{2} e -2 \ln \left (x \right ) x^{2} c^{3} d -\ln \left (c x +b \right ) x^{2} b \,c^{2} e +2 \ln \left (c x +b \right ) x^{2} c^{3} d +\ln \left (x \right ) x \,b^{2} c e -2 \ln \left (x \right ) x b \,c^{2} d -\ln \left (c x +b \right ) x \,b^{2} c e +2 \ln \left (c x +b \right ) x b \,c^{2} d +b^{2} c e x -2 b \,c^{2} d x -b^{2} c d}{b^{3} c x \left (c x +b \right )}\)	\(141\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.71 \[ \int \frac {d+e x}{\left (b x+c x^2\right )^2} \, dx=-\frac {b^{2} d + {\left (2 \, b c d - b^{2} e\right )} x - {\left ({\left (2 \, c^{2} d - b c e\right )} x^{2} + {\left (2 \, b c d - b^{2} e\right )} x\right )} \log \left (c x + b\right ) + {\left ({\left (2 \, c^{2} d - b c e\right )} x^{2} + {\left (2 \, b c d - b^{2} e\right )} x\right )} \log \left (x\right )}{b^{3} c x^{2} + b^{4} x} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (54) = 108\).

Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.97 \[ \int \frac {d+e x}{\left (b x+c x^2\right )^2} \, dx=\frac {- b d + x \left (b e - 2 c d\right )}{b^{3} x + b^{2} c x^{2}} + \frac {\left (b e - 2 c d\right ) \log {\left (x + \frac {b^{2} e - 2 b c d - b \left (b e - 2 c d\right )}{2 b c e - 4 c^{2} d} \right )}}{b^{3}} - \frac {\left (b e - 2 c d\right ) \log {\left (x + \frac {b^{2} e - 2 b c d + b \left (b e - 2 c d\right )}{2 b c e - 4 c^{2} d} \right )}}{b^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06 \[ \int \frac {d+e x}{\left (b x+c x^2\right )^2} \, dx=-\frac {b d + {\left (2 \, c d - b e\right )} x}{b^{2} c x^{2} + b^{3} x} + \frac {{\left (2 \, c d - b e\right )} \log \left (c x + b\right )}{b^{3}} - \frac {{\left (2 \, c d - b e\right )} \log \left (x\right )}{b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.14 \[ \int \frac {d+e x}{\left (b x+c x^2\right )^2} \, dx=-\frac {{\left (2 \, c d - b e\right )} \log \left ({\left | x \right |}\right )}{b^{3}} - \frac {2 \, c d x - b e x + b d}{{\left (c x^{2} + b x\right )} b^{2}} + \frac {{\left (2 \, c^{2} d - b c e\right )} \log \left ({\left | c x + b \right |}\right )}{b^{3} c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.56 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \frac {d+e x}{\left (b x+c x^2\right )^2} \, dx=-\frac {\frac {d}{b}-\frac {x\,\left (b\,e-2\,c\,d\right )}{b^2}}{c\,x^2+b\,x}-\frac {2\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )\,\left (b\,e-2\,c\,d\right )}{b^3} \]

[In]

int((d + e*x)/(b*x + c*x^2)^2,x)

[Out]

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