\(\int \frac {b x+c x^2}{(d+e x)^3} \, dx\) [228]

Optimal result

Integrand size = 17, antiderivative size = 55 \[ \int \frac {b x+c x^2}{(d+e x)^3} \, dx=-\frac {d (c d-b e)}{2 e^3 (d+e x)^2}+\frac {2 c d-b e}{e^3 (d+e x)}+\frac {c \log (d+e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 55, 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} \[ \int \frac {b x+c x^2}{(d+e x)^3} \, dx=-\frac {d (c d-b e)}{2 e^3 (d+e x)^2}+\frac {2 c d-b e}{e^3 (d+e x)}+\frac {c \log (d+e x)}{e^3} \]

[In]

Int[(b*x + c*x^2)/(d + e*x)^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int \frac {b x+c x^2}{(d+e x)^3} \, dx=\frac {-b e (d+2 e x)+c d (3 d+4 e x)+2 c (d+e x)^2 \log (d+e x)}{2 e^3 (d+e x)^2} \]

[In]

Integrate[(b*x + c*x^2)/(d + e*x)^3,x]

[Out]

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

Maple [A] (verified)

Time = 1.94 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91

method	result	size
norman	\(\frac {-\frac {d \left (b e -3 c d \right )}{2 e^{3}}-\frac {\left (b e -2 c d \right ) x}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {c \ln \left (e x +d \right )}{e^{3}}\)	\(50\)
risch	\(\frac {-\frac {d \left (b e -3 c d \right )}{2 e^{3}}-\frac {\left (b e -2 c d \right ) x}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {c \ln \left (e x +d \right )}{e^{3}}\)	\(50\)
default	\(-\frac {b e -2 c d}{e^{3} \left (e x +d \right )}+\frac {d \left (b e -c d \right )}{2 e^{3} \left (e x +d \right )^{2}}+\frac {c \ln \left (e x +d \right )}{e^{3}}\)	\(54\)
parallelrisch	\(\frac {2 \ln \left (e x +d \right ) x^{2} c \,e^{2}+4 \ln \left (e x +d \right ) x c d e +2 \ln \left (e x +d \right ) c \,d^{2}-2 x b \,e^{2}+4 x c d e -b d e +3 c \,d^{2}}{2 e^{3} \left (e x +d \right )^{2}}\)	\(77\)

[In]

int((c*x^2+b*x)/(e*x+d)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.47 \[ \int \frac {b x+c x^2}{(d+e x)^3} \, dx=\frac {3 \, c d^{2} - b d e + 2 \, {\left (2 \, c d e - b e^{2}\right )} x + 2 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]

[In]

integrate((c*x^2+b*x)/(e*x+d)^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int \frac {b x+c x^2}{(d+e x)^3} \, dx=\frac {c \log {\left (d + e x \right )}}{e^{3}} + \frac {- b d e + 3 c d^{2} + x \left (- 2 b e^{2} + 4 c d e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \]

[In]

integrate((c*x**2+b*x)/(e*x+d)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.18 \[ \int \frac {b x+c x^2}{(d+e x)^3} \, dx=\frac {3 \, c d^{2} - b d e + 2 \, {\left (2 \, c d e - b e^{2}\right )} x}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} + \frac {c \log \left (e x + d\right )}{e^{3}} \]

[In]

integrate((c*x^2+b*x)/(e*x+d)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int \frac {b x+c x^2}{(d+e x)^3} \, dx=\frac {c \log \left ({\left | e x + d \right |}\right )}{e^{3}} + \frac {2 \, {\left (2 \, c d - b e\right )} x + \frac {3 \, c d^{2} - b d e}{e}}{2 \, {\left (e x + d\right )}^{2} e^{2}} \]

[In]

integrate((c*x^2+b*x)/(e*x+d)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.94 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int \frac {b x+c x^2}{(d+e x)^3} \, dx=\frac {\frac {3\,c\,d^2-b\,d\,e}{2\,e^3}-\frac {x\,\left (b\,e-2\,c\,d\right )}{e^2}}{d^2+2\,d\,e\,x+e^2\,x^2}+\frac {c\,\ln \left (d+e\,x\right )}{e^3} \]

[In]

int((b*x + c*x^2)/(d + e*x)^3,x)

[Out]

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