\(\int \frac {b x+c x^2}{(d+e x)^5} \, dx\) [230]

Optimal result

Integrand size = 17, antiderivative size = 62 \[ \int \frac {b x+c x^2}{(d+e x)^5} \, dx=-\frac {d (c d-b e)}{4 e^3 (d+e x)^4}+\frac {2 c d-b e}{3 e^3 (d+e x)^3}-\frac {c}{2 e^3 (d+e x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 62, 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} \[ \int \frac {b x+c x^2}{(d+e x)^5} \, dx=\frac {2 c d-b e}{3 e^3 (d+e x)^3}-\frac {d (c d-b e)}{4 e^3 (d+e x)^4}-\frac {c}{2 e^3 (d+e x)^2} \]

[In]

Int[(b*x + c*x^2)/(d + e*x)^5,x]

[Out]

-1/4*(d*(c*d - b*e))/(e^3*(d + e*x)^4) + (2*c*d - b*e)/(3*e^3*(d + e*x)^3) - c/(2*e^3*(d + e*x)^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 {d (c d-b e)}{e^2 (d+e x)^5}+\frac {-2 c d+b e}{e^2 (d+e x)^4}+\frac {c}{e^2 (d+e x)^3}\right ) \, dx \\ & = -\frac {d (c d-b e)}{4 e^3 (d+e x)^4}+\frac {2 c d-b e}{3 e^3 (d+e x)^3}-\frac {c}{2 e^3 (d+e x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.69 \[ \int \frac {b x+c x^2}{(d+e x)^5} \, dx=-\frac {b e (d+4 e x)+c \left (d^2+4 d e x+6 e^2 x^2\right )}{12 e^3 (d+e x)^4} \]

[In]

Integrate[(b*x + c*x^2)/(d + e*x)^5,x]

[Out]

-1/12*(b*e*(d + 4*e*x) + c*(d^2 + 4*d*e*x + 6*e^2*x^2))/(e^3*(d + e*x)^4)

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.73

method	result	size
gosper	\(-\frac {6 c \,x^{2} e^{2}+4 x b \,e^{2}+4 x c d e +b d e +c \,d^{2}}{12 e^{3} \left (e x +d \right )^{4}}\)	\(45\)
risch	\(\frac {-\frac {c \,x^{2}}{2 e}-\frac {\left (b e +c d \right ) x}{3 e^{2}}-\frac {d \left (b e +c d \right )}{12 e^{3}}}{\left (e x +d \right )^{4}}\)	\(45\)
norman	\(\frac {-\frac {c \,x^{2}}{2 e}-\frac {\left (e^{2} b +c d e \right ) x}{3 e^{3}}-\frac {d \left (e^{2} b +c d e \right )}{12 e^{4}}}{\left (e x +d \right )^{4}}\)	\(51\)
parallelrisch	\(\frac {-6 c \,x^{2} e^{3}-4 b \,e^{3} x -4 c d \,e^{2} x -b d \,e^{2}-d^{2} e c}{12 e^{4} \left (e x +d \right )^{4}}\)	\(52\)
default	\(-\frac {b e -2 c d}{3 e^{3} \left (e x +d \right )^{3}}+\frac {d \left (b e -c d \right )}{4 e^{3} \left (e x +d \right )^{4}}-\frac {c}{2 e^{3} \left (e x +d \right )^{2}}\)	\(56\)

[In]

int((c*x^2+b*x)/(e*x+d)^5,x,method=_RETURNVERBOSE)

[Out]

-1/12/e^3*(6*c*e^2*x^2+4*b*e^2*x+4*c*d*e*x+b*d*e+c*d^2)/(e*x+d)^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.29 \[ \int \frac {b x+c x^2}{(d+e x)^5} \, dx=-\frac {6 \, c e^{2} x^{2} + c d^{2} + b d e + 4 \, {\left (c d e + b e^{2}\right )} x}{12 \, {\left (e^{7} x^{4} + 4 \, d e^{6} x^{3} + 6 \, d^{2} e^{5} x^{2} + 4 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]

[In]

integrate((c*x^2+b*x)/(e*x+d)^5,x, algorithm="fricas")

[Out]

-1/12*(6*c*e^2*x^2 + c*d^2 + b*d*e + 4*(c*d*e + b*e^2)*x)/(e^7*x^4 + 4*d*e^6*x^3 + 6*d^2*e^5*x^2 + 4*d^3*e^4*x
+ d^4*e^3)

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.37 \[ \int \frac {b x+c x^2}{(d+e x)^5} \, dx=\frac {- b d e - c d^{2} - 6 c e^{2} x^{2} + x \left (- 4 b e^{2} - 4 c d e\right )}{12 d^{4} e^{3} + 48 d^{3} e^{4} x + 72 d^{2} e^{5} x^{2} + 48 d e^{6} x^{3} + 12 e^{7} x^{4}} \]

[In]

integrate((c*x**2+b*x)/(e*x+d)**5,x)

[Out]

(-b*d*e - c*d**2 - 6*c*e**2*x**2 + x*(-4*b*e**2 - 4*c*d*e))/(12*d**4*e**3 + 48*d**3*e**4*x + 72*d**2*e**5*x**2
+ 48*d*e**6*x**3 + 12*e**7*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.29 \[ \int \frac {b x+c x^2}{(d+e x)^5} \, dx=-\frac {6 \, c e^{2} x^{2} + c d^{2} + b d e + 4 \, {\left (c d e + b e^{2}\right )} x}{12 \, {\left (e^{7} x^{4} + 4 \, d e^{6} x^{3} + 6 \, d^{2} e^{5} x^{2} + 4 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} \]

[In]

integrate((c*x^2+b*x)/(e*x+d)^5,x, algorithm="maxima")

[Out]

-1/12*(6*c*e^2*x^2 + c*d^2 + b*d*e + 4*(c*d*e + b*e^2)*x)/(e^7*x^4 + 4*d*e^6*x^3 + 6*d^2*e^5*x^2 + 4*d^3*e^4*x
+ d^4*e^3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.23 \[ \int \frac {b x+c x^2}{(d+e x)^5} \, dx=-\frac {\frac {6 \, c}{{\left (e x + d\right )}^{2} e^{2}} - \frac {8 \, c d}{{\left (e x + d\right )}^{3} e^{2}} + \frac {3 \, c d^{2}}{{\left (e x + d\right )}^{4} e^{2}} + \frac {4 \, b}{{\left (e x + d\right )}^{3} e} - \frac {3 \, b d}{{\left (e x + d\right )}^{4} e}}{12 \, e} \]

[In]

integrate((c*x^2+b*x)/(e*x+d)^5,x, algorithm="giac")

[Out]

-1/12*(6*c/((e*x + d)^2*e^2) - 8*c*d/((e*x + d)^3*e^2) + 3*c*d^2/((e*x + d)^4*e^2) + 4*b/((e*x + d)^3*e) - 3*b
*d/((e*x + d)^4*e))/e

Mupad [B] (verification not implemented)

Time = 9.96 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.26 \[ \int \frac {b x+c x^2}{(d+e x)^5} \, dx=-\frac {\frac {d\,\left (b\,e+c\,d\right )}{12\,e^3}+\frac {x\,\left (b\,e+c\,d\right )}{3\,e^2}+\frac {c\,x^2}{2\,e}}{d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4} \]

[In]

int((b*x + c*x^2)/(d + e*x)^5,x)

[Out]

-((d*(b*e + c*d))/(12*e^3) + (x*(b*e + c*d))/(3*e^2) + (c*x^2)/(2*e))/(d^4 + e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2
*x^2 + 4*d^3*e*x)