\(\int \frac {d+e x}{x (b x+c x^2)^2} \, dx\) [57]

Optimal result

Integrand size = 20, antiderivative size = 85 \[ \int \frac {d+e x}{x \left (b x+c x^2\right )^2} \, dx=-\frac {d}{2 b^2 x^2}+\frac {2 c d-b e}{b^3 x}+\frac {c (c d-b e)}{b^3 (b+c x)}+\frac {c (3 c d-2 b e) \log (x)}{b^4}-\frac {c (3 c d-2 b e) \log (b+c x)}{b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {779} \[ \int \frac {d+e x}{x \left (b x+c x^2\right )^2} \, dx=\frac {c \log (x) (3 c d-2 b e)}{b^4}-\frac {c (3 c d-2 b e) \log (b+c x)}{b^4}+\frac {2 c d-b e}{b^3 x}+\frac {c (c d-b e)}{b^3 (b+c x)}-\frac {d}{2 b^2 x^2} \]

[In]

Int[(d + e*x)/(x*(b*x + c*x^2)^2),x]

[Out]

-1/2*d/(b^2*x^2) + (2*c*d - b*e)/(b^3*x) + (c*(c*d - b*e))/(b^3*(b + c*x)) + (c*(3*c*d - 2*b*e)*Log[x])/b^4 -
(c*(3*c*d - 2*b*e)*Log[b + c*x])/b^4

Rule 779

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00 \[ \int \frac {d+e x}{x \left (b x+c x^2\right )^2} \, dx=\frac {-\frac {b \left (-6 c^2 d x^2+b^2 (d+2 e x)+b c x (-3 d+4 e x)\right )}{x^2 (b+c x)}+2 c (3 c d-2 b e) \log (x)+2 c (-3 c d+2 b e) \log (b+c x)}{2 b^4} \]

[In]

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

[Out]

(-((b*(-6*c^2*d*x^2 + b^2*(d + 2*e*x) + b*c*x*(-3*d + 4*e*x)))/(x^2*(b + c*x))) + 2*c*(3*c*d - 2*b*e)*Log[x] +
2*c*(-3*c*d + 2*b*e)*Log[b + c*x])/(2*b^4)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {d}{2 b^{2} x^{2}}-\frac {b e -2 c d}{b^{3} x}-\frac {c \left (2 b e -3 c d \right ) \ln \left (x \right )}{b^{4}}+\frac {c \left (2 b e -3 c d \right ) \ln \left (c x +b \right )}{b^{4}}-\frac {\left (b e -c d \right ) c}{b^{3} \left (c x +b \right )}\)	\(85\)
norman	\(\frac {\frac {c \left (2 c e b -3 c^{2} d \right ) x^{3}}{b^{4}}-\frac {d}{2 b}-\frac {\left (2 b e -3 c d \right ) x}{2 b^{2}}}{x^{2} \left (c x +b \right )}+\frac {c \left (2 b e -3 c d \right ) \ln \left (c x +b \right )}{b^{4}}-\frac {c \left (2 b e -3 c d \right ) \ln \left (x \right )}{b^{4}}\)	\(92\)
risch	\(\frac {-\frac {c \left (2 b e -3 c d \right ) x^{2}}{b^{3}}-\frac {\left (2 b e -3 c d \right ) x}{2 b^{2}}-\frac {d}{2 b}}{x^{2} \left (c x +b \right )}+\frac {2 c \ln \left (-c x -b \right ) e}{b^{3}}-\frac {3 c^{2} \ln \left (-c x -b \right ) d}{b^{4}}-\frac {2 c \ln \left (x \right ) e}{b^{3}}+\frac {3 c^{2} \ln \left (x \right ) d}{b^{4}}\)	\(107\)
parallelrisch	\(-\frac {4 \ln \left (x \right ) x^{3} b \,c^{2} e -6 \ln \left (x \right ) x^{3} c^{3} d -4 \ln \left (c x +b \right ) x^{3} b \,c^{2} e +6 \ln \left (c x +b \right ) x^{3} c^{3} d +4 \ln \left (x \right ) x^{2} b^{2} c e -6 \ln \left (x \right ) x^{2} b \,c^{2} d -4 \ln \left (c x +b \right ) x^{2} b^{2} c e +6 \ln \left (c x +b \right ) x^{2} b \,c^{2} d -4 b \,c^{2} e \,x^{3}+6 c^{3} d \,x^{3}+2 b^{3} e x -3 b^{2} c d x +d \,b^{3}}{2 b^{4} x^{2} \left (c x +b \right )}\)	\(166\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.78 \[ \int \frac {d+e x}{x \left (b x+c x^2\right )^2} \, dx=-\frac {b^{3} d - 2 \, {\left (3 \, b c^{2} d - 2 \, b^{2} c e\right )} x^{2} - {\left (3 \, b^{2} c d - 2 \, b^{3} e\right )} x + 2 \, {\left ({\left (3 \, c^{3} d - 2 \, b c^{2} e\right )} x^{3} + {\left (3 \, b c^{2} d - 2 \, b^{2} c e\right )} x^{2}\right )} \log \left (c x + b\right ) - 2 \, {\left ({\left (3 \, c^{3} d - 2 \, b c^{2} e\right )} x^{3} + {\left (3 \, b c^{2} d - 2 \, b^{2} c e\right )} x^{2}\right )} \log \left (x\right )}{2 \, {\left (b^{4} c x^{3} + b^{5} x^{2}\right )}} \]

[In]

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

[Out]

-1/2*(b^3*d - 2*(3*b*c^2*d - 2*b^2*c*e)*x^2 - (3*b^2*c*d - 2*b^3*e)*x + 2*((3*c^3*d - 2*b*c^2*e)*x^3 + (3*b*c^
2*d - 2*b^2*c*e)*x^2)*log(c*x + b) - 2*((3*c^3*d - 2*b*c^2*e)*x^3 + (3*b*c^2*d - 2*b^2*c*e)*x^2)*log(x))/(b^4*
c*x^3 + b^5*x^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (80) = 160\).

Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.16 \[ \int \frac {d+e x}{x \left (b x+c x^2\right )^2} \, dx=\frac {- b^{2} d + x^{2} \left (- 4 b c e + 6 c^{2} d\right ) + x \left (- 2 b^{2} e + 3 b c d\right )}{2 b^{4} x^{2} + 2 b^{3} c x^{3}} - \frac {c \left (2 b e - 3 c d\right ) \log {\left (x + \frac {2 b^{2} c e - 3 b c^{2} d - b c \left (2 b e - 3 c d\right )}{4 b c^{2} e - 6 c^{3} d} \right )}}{b^{4}} + \frac {c \left (2 b e - 3 c d\right ) \log {\left (x + \frac {2 b^{2} c e - 3 b c^{2} d + b c \left (2 b e - 3 c d\right )}{4 b c^{2} e - 6 c^{3} d} \right )}}{b^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.18 \[ \int \frac {d+e x}{x \left (b x+c x^2\right )^2} \, dx=-\frac {b^{2} d - 2 \, {\left (3 \, c^{2} d - 2 \, b c e\right )} x^{2} - {\left (3 \, b c d - 2 \, b^{2} e\right )} x}{2 \, {\left (b^{3} c x^{3} + b^{4} x^{2}\right )}} - \frac {{\left (3 \, c^{2} d - 2 \, b c e\right )} \log \left (c x + b\right )}{b^{4}} + \frac {{\left (3 \, c^{2} d - 2 \, b c e\right )} \log \left (x\right )}{b^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.26 \[ \int \frac {d+e x}{x \left (b x+c x^2\right )^2} \, dx=\frac {{\left (3 \, c^{2} d - 2 \, b c e\right )} \log \left ({\left | x \right |}\right )}{b^{4}} - \frac {{\left (3 \, c^{3} d - 2 \, b c^{2} e\right )} \log \left ({\left | c x + b \right |}\right )}{b^{4} c} - \frac {b^{3} d - 2 \, {\left (3 \, b c^{2} d - 2 \, b^{2} c e\right )} x^{2} - {\left (3 \, b^{2} c d - 2 \, b^{3} e\right )} x}{2 \, {\left (c x + b\right )} b^{4} x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.90 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.24 \[ \int \frac {d+e x}{x \left (b x+c x^2\right )^2} \, dx=-\frac {\frac {d}{2\,b}+\frac {x\,\left (2\,b\,e-3\,c\,d\right )}{2\,b^2}+\frac {c\,x^2\,\left (2\,b\,e-3\,c\,d\right )}{b^3}}{c\,x^3+b\,x^2}-\frac {2\,c\,\mathrm {atanh}\left (\frac {c\,\left (2\,b\,e-3\,c\,d\right )\,\left (b+2\,c\,x\right )}{b\,\left (3\,c^2\,d-2\,b\,c\,e\right )}\right )\,\left (2\,b\,e-3\,c\,d\right )}{b^4} \]

[In]

int((d + e*x)/(x*(b*x + c*x^2)^2),x)

[Out]

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