\(\int \frac {b x+c x^2}{(d+e x)^2} \, dx\) [227]

Optimal result

Integrand size = 17, antiderivative size = 48 \[ \int \frac {b x+c x^2}{(d+e x)^2} \, dx=\frac {c x}{e^2}-\frac {d (c d-b e)}{e^3 (d+e x)}-\frac {(2 c d-b e) \log (d+e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, 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 = {712} \[ \int \frac {b x+c x^2}{(d+e x)^2} \, dx=-\frac {d (c d-b e)}{e^3 (d+e x)}-\frac {(2 c d-b e) \log (d+e x)}{e^3}+\frac {c x}{e^2} \]

[In]

Int[(b*x + c*x^2)/(d + e*x)^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.85 \[ \int \frac {b x+c x^2}{(d+e x)^2} \, dx=\frac {c e x+\frac {d (-c d+b e)}{d+e x}+(-2 c d+b e) \log (d+e x)}{e^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.97 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {c x}{e^{2}}+\frac {d \left (b e -c d \right )}{e^{3} \left (e x +d \right )}+\frac {\left (b e -2 c d \right ) \ln \left (e x +d \right )}{e^{3}}\)	\(46\)
norman	\(\frac {\frac {c \,x^{2}}{e}+\frac {d \left (b e -2 c d \right )}{e^{3}}}{e x +d}+\frac {\left (b e -2 c d \right ) \ln \left (e x +d \right )}{e^{3}}\)	\(50\)
risch	\(\frac {c x}{e^{2}}+\frac {d b}{e^{2} \left (e x +d \right )}-\frac {d^{2} c}{e^{3} \left (e x +d \right )}+\frac {\ln \left (e x +d \right ) b}{e^{2}}-\frac {2 c d \ln \left (e x +d \right )}{e^{3}}\)	\(61\)
parallelrisch	\(\frac {\ln \left (e x +d \right ) x b \,e^{2}-2 \ln \left (e x +d \right ) x c d e +c \,x^{2} e^{2}+\ln \left (e x +d \right ) b d e -2 \ln \left (e x +d \right ) c \,d^{2}+b d e -2 c \,d^{2}}{e^{3} \left (e x +d \right )}\)	\(77\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.50 \[ \int \frac {b x+c x^2}{(d+e x)^2} \, dx=\frac {c e^{2} x^{2} + c d e x - c d^{2} + b d e - {\left (2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} x\right )} \log \left (e x + d\right )}{e^{4} x + d e^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.92 \[ \int \frac {b x+c x^2}{(d+e x)^2} \, dx=\frac {c x}{e^{2}} + \frac {b d e - c d^{2}}{d e^{3} + e^{4} x} + \frac {\left (b e - 2 c d\right ) \log {\left (d + e x \right )}}{e^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {b x+c x^2}{(d+e x)^2} \, dx=-\frac {c d^{2} - b d e}{e^{4} x + d e^{3}} + \frac {c x}{e^{2}} - \frac {{\left (2 \, c d - b e\right )} \log \left (e x + d\right )}{e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.00 \[ \int \frac {b x+c x^2}{(d+e x)^2} \, dx=c {\left (\frac {2 \, d \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{3}} + \frac {e x + d}{e^{3}} - \frac {d^{2}}{{\left (e x + d\right )} e^{3}}\right )} - \frac {b {\left (\frac {\log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e} - \frac {d}{{\left (e x + d\right )} e}\right )}}{e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.82 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12 \[ \int \frac {b x+c x^2}{(d+e x)^2} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (b\,e-2\,c\,d\right )}{e^3}-\frac {c\,d^2-b\,d\,e}{e\,\left (x\,e^3+d\,e^2\right )}+\frac {c\,x}{e^2} \]

[In]

int((b*x + c*x^2)/(d + e*x)^2,x)

[Out]

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