\(\int \frac {(b x+c x^2)^2}{(d+e x)^6} \, dx\) [241]

Optimal result

Integrand size = 19, antiderivative size = 132 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {d^2 (c d-b e)^2}{5 e^5 (d+e x)^5}+\frac {d (c d-b e) (2 c d-b e)}{2 e^5 (d+e x)^4}-\frac {6 c^2 d^2-6 b c d e+b^2 e^2}{3 e^5 (d+e x)^3}+\frac {c (2 c d-b e)}{e^5 (d+e x)^2}-\frac {c^2}{e^5 (d+e x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {712} \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {b^2 e^2-6 b c d e+6 c^2 d^2}{3 e^5 (d+e x)^3}-\frac {d^2 (c d-b e)^2}{5 e^5 (d+e x)^5}+\frac {c (2 c d-b e)}{e^5 (d+e x)^2}+\frac {d (c d-b e) (2 c d-b e)}{2 e^5 (d+e x)^4}-\frac {c^2}{e^5 (d+e x)} \]

[In]

Int[(b*x + c*x^2)^2/(d + e*x)^6,x]

[Out]

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

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^2 (c d-b e)^2}{e^4 (d+e x)^6}+\frac {2 d (c d-b e) (-2 c d+b e)}{e^4 (d+e x)^5}+\frac {6 c^2 d^2-6 b c d e+b^2 e^2}{e^4 (d+e x)^4}-\frac {2 c (2 c d-b e)}{e^4 (d+e x)^3}+\frac {c^2}{e^4 (d+e x)^2}\right ) \, dx \\ & = -\frac {d^2 (c d-b e)^2}{5 e^5 (d+e x)^5}+\frac {d (c d-b e) (2 c d-b e)}{2 e^5 (d+e x)^4}-\frac {6 c^2 d^2-6 b c d e+b^2 e^2}{3 e^5 (d+e x)^3}+\frac {c (2 c d-b e)}{e^5 (d+e x)^2}-\frac {c^2}{e^5 (d+e x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.88 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {b^2 e^2 \left (d^2+5 d e x+10 e^2 x^2\right )+3 b c e \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )+6 c^2 \left (d^4+5 d^3 e x+10 d^2 e^2 x^2+10 d e^3 x^3+5 e^4 x^4\right )}{30 e^5 (d+e x)^5} \]

[In]

Integrate[(b*x + c*x^2)^2/(d + e*x)^6,x]

[Out]

-1/30*(b^2*e^2*(d^2 + 5*d*e*x + 10*e^2*x^2) + 3*b*c*e*(d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3) + 6*c^2*(d
^4 + 5*d^3*e*x + 10*d^2*e^2*x^2 + 10*d*e^3*x^3 + 5*e^4*x^4))/(e^5*(d + e*x)^5)

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {-\frac {c^{2} x^{4}}{e}-\frac {c \left (b e +2 c d \right ) x^{3}}{e^{2}}-\frac {\left (b^{2} e^{2}+3 b c d e +6 c^{2} d^{2}\right ) x^{2}}{3 e^{3}}-\frac {d \left (b^{2} e^{2}+3 b c d e +6 c^{2} d^{2}\right ) x}{6 e^{4}}-\frac {d^{2} \left (b^{2} e^{2}+3 b c d e +6 c^{2} d^{2}\right )}{30 e^{5}}}{\left (e x +d \right )^{5}}\)	\(127\)
norman	\(\frac {-\frac {c^{2} x^{4}}{e}-\frac {\left (b c e +2 c^{2} d \right ) x^{3}}{e^{2}}-\frac {\left (b^{2} e^{2}+3 b c d e +6 c^{2} d^{2}\right ) x^{2}}{3 e^{3}}-\frac {d \left (b^{2} e^{2}+3 b c d e +6 c^{2} d^{2}\right ) x}{6 e^{4}}-\frac {d^{2} \left (b^{2} e^{2}+3 b c d e +6 c^{2} d^{2}\right )}{30 e^{5}}}{\left (e x +d \right )^{5}}\)	\(129\)
gosper	\(-\frac {30 c^{2} x^{4} e^{4}+30 x^{3} b c \,e^{4}+60 x^{3} c^{2} d \,e^{3}+10 x^{2} b^{2} e^{4}+30 x^{2} b c d \,e^{3}+60 x^{2} c^{2} d^{2} e^{2}+5 x \,b^{2} d \,e^{3}+15 x b c \,d^{2} e^{2}+30 x \,c^{2} d^{3} e +b^{2} d^{2} e^{2}+3 d^{3} e b c +6 c^{2} d^{4}}{30 e^{5} \left (e x +d \right )^{5}}\)	\(140\)
parallelrisch	\(\frac {-30 c^{2} x^{4} e^{4}-30 x^{3} b c \,e^{4}-60 x^{3} c^{2} d \,e^{3}-10 x^{2} b^{2} e^{4}-30 x^{2} b c d \,e^{3}-60 x^{2} c^{2} d^{2} e^{2}-5 x \,b^{2} d \,e^{3}-15 x b c \,d^{2} e^{2}-30 x \,c^{2} d^{3} e -b^{2} d^{2} e^{2}-3 d^{3} e b c -6 c^{2} d^{4}}{30 e^{5} \left (e x +d \right )^{5}}\)	\(141\)
default	\(-\frac {c^{2}}{e^{5} \left (e x +d \right )}-\frac {b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}}{3 e^{5} \left (e x +d \right )^{3}}+\frac {d \left (b^{2} e^{2}-3 b c d e +2 c^{2} d^{2}\right )}{2 e^{5} \left (e x +d \right )^{4}}-\frac {d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}{5 e^{5} \left (e x +d \right )^{5}}-\frac {c \left (b e -2 c d \right )}{e^{5} \left (e x +d \right )^{2}}\)	\(143\)

[In]

int((c*x^2+b*x)^2/(e*x+d)^6,x,method=_RETURNVERBOSE)

[Out]

(-c^2*x^4/e-c*(b*e+2*c*d)/e^2*x^3-1/3*(b^2*e^2+3*b*c*d*e+6*c^2*d^2)/e^3*x^2-1/6*d*(b^2*e^2+3*b*c*d*e+6*c^2*d^2
)/e^4*x-1/30*d^2*(b^2*e^2+3*b*c*d*e+6*c^2*d^2)/e^5)/(e*x+d)^5

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.37 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {30 \, c^{2} e^{4} x^{4} + 6 \, c^{2} d^{4} + 3 \, b c d^{3} e + b^{2} d^{2} e^{2} + 30 \, {\left (2 \, c^{2} d e^{3} + b c e^{4}\right )} x^{3} + 10 \, {\left (6 \, c^{2} d^{2} e^{2} + 3 \, b c d e^{3} + b^{2} e^{4}\right )} x^{2} + 5 \, {\left (6 \, c^{2} d^{3} e + 3 \, b c d^{2} e^{2} + b^{2} d e^{3}\right )} x}{30 \, {\left (e^{10} x^{5} + 5 \, d e^{9} x^{4} + 10 \, d^{2} e^{8} x^{3} + 10 \, d^{3} e^{7} x^{2} + 5 \, d^{4} e^{6} x + d^{5} e^{5}\right )}} \]

[In]

integrate((c*x^2+b*x)^2/(e*x+d)^6,x, algorithm="fricas")

[Out]

-1/30*(30*c^2*e^4*x^4 + 6*c^2*d^4 + 3*b*c*d^3*e + b^2*d^2*e^2 + 30*(2*c^2*d*e^3 + b*c*e^4)*x^3 + 10*(6*c^2*d^2
*e^2 + 3*b*c*d*e^3 + b^2*e^4)*x^2 + 5*(6*c^2*d^3*e + 3*b*c*d^2*e^2 + b^2*d*e^3)*x)/(e^10*x^5 + 5*d*e^9*x^4 + 1
0*d^2*e^8*x^3 + 10*d^3*e^7*x^2 + 5*d^4*e^6*x + d^5*e^5)

Sympy [A] (verification not implemented)

Time = 4.04 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.48 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=\frac {- b^{2} d^{2} e^{2} - 3 b c d^{3} e - 6 c^{2} d^{4} - 30 c^{2} e^{4} x^{4} + x^{3} \left (- 30 b c e^{4} - 60 c^{2} d e^{3}\right ) + x^{2} \left (- 10 b^{2} e^{4} - 30 b c d e^{3} - 60 c^{2} d^{2} e^{2}\right ) + x \left (- 5 b^{2} d e^{3} - 15 b c d^{2} e^{2} - 30 c^{2} d^{3} e\right )}{30 d^{5} e^{5} + 150 d^{4} e^{6} x + 300 d^{3} e^{7} x^{2} + 300 d^{2} e^{8} x^{3} + 150 d e^{9} x^{4} + 30 e^{10} x^{5}} \]

[In]

integrate((c*x**2+b*x)**2/(e*x+d)**6,x)

[Out]

(-b**2*d**2*e**2 - 3*b*c*d**3*e - 6*c**2*d**4 - 30*c**2*e**4*x**4 + x**3*(-30*b*c*e**4 - 60*c**2*d*e**3) + x**
2*(-10*b**2*e**4 - 30*b*c*d*e**3 - 60*c**2*d**2*e**2) + x*(-5*b**2*d*e**3 - 15*b*c*d**2*e**2 - 30*c**2*d**3*e)
)/(30*d**5*e**5 + 150*d**4*e**6*x + 300*d**3*e**7*x**2 + 300*d**2*e**8*x**3 + 150*d*e**9*x**4 + 30*e**10*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.37 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {30 \, c^{2} e^{4} x^{4} + 6 \, c^{2} d^{4} + 3 \, b c d^{3} e + b^{2} d^{2} e^{2} + 30 \, {\left (2 \, c^{2} d e^{3} + b c e^{4}\right )} x^{3} + 10 \, {\left (6 \, c^{2} d^{2} e^{2} + 3 \, b c d e^{3} + b^{2} e^{4}\right )} x^{2} + 5 \, {\left (6 \, c^{2} d^{3} e + 3 \, b c d^{2} e^{2} + b^{2} d e^{3}\right )} x}{30 \, {\left (e^{10} x^{5} + 5 \, d e^{9} x^{4} + 10 \, d^{2} e^{8} x^{3} + 10 \, d^{3} e^{7} x^{2} + 5 \, d^{4} e^{6} x + d^{5} e^{5}\right )}} \]

[In]

integrate((c*x^2+b*x)^2/(e*x+d)^6,x, algorithm="maxima")

[Out]

-1/30*(30*c^2*e^4*x^4 + 6*c^2*d^4 + 3*b*c*d^3*e + b^2*d^2*e^2 + 30*(2*c^2*d*e^3 + b*c*e^4)*x^3 + 10*(6*c^2*d^2
*e^2 + 3*b*c*d*e^3 + b^2*e^4)*x^2 + 5*(6*c^2*d^3*e + 3*b*c*d^2*e^2 + b^2*d*e^3)*x)/(e^10*x^5 + 5*d*e^9*x^4 + 1
0*d^2*e^8*x^3 + 10*d^3*e^7*x^2 + 5*d^4*e^6*x + d^5*e^5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {30 \, c^{2} e^{4} x^{4} + 60 \, c^{2} d e^{3} x^{3} + 30 \, b c e^{4} x^{3} + 60 \, c^{2} d^{2} e^{2} x^{2} + 30 \, b c d e^{3} x^{2} + 10 \, b^{2} e^{4} x^{2} + 30 \, c^{2} d^{3} e x + 15 \, b c d^{2} e^{2} x + 5 \, b^{2} d e^{3} x + 6 \, c^{2} d^{4} + 3 \, b c d^{3} e + b^{2} d^{2} e^{2}}{30 \, {\left (e x + d\right )}^{5} e^{5}} \]

[In]

integrate((c*x^2+b*x)^2/(e*x+d)^6,x, algorithm="giac")

[Out]

-1/30*(30*c^2*e^4*x^4 + 60*c^2*d*e^3*x^3 + 30*b*c*e^4*x^3 + 60*c^2*d^2*e^2*x^2 + 30*b*c*d*e^3*x^2 + 10*b^2*e^4
*x^2 + 30*c^2*d^3*e*x + 15*b*c*d^2*e^2*x + 5*b^2*d*e^3*x + 6*c^2*d^4 + 3*b*c*d^3*e + b^2*d^2*e^2)/((e*x + d)^5
*e^5)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.28 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^6} \, dx=-\frac {\frac {x^2\,\left (b^2\,e^2+3\,b\,c\,d\,e+6\,c^2\,d^2\right )}{3\,e^3}+\frac {c^2\,x^4}{e}+\frac {d^2\,\left (b^2\,e^2+3\,b\,c\,d\,e+6\,c^2\,d^2\right )}{30\,e^5}+\frac {c\,x^3\,\left (b\,e+2\,c\,d\right )}{e^2}+\frac {d\,x\,\left (b^2\,e^2+3\,b\,c\,d\,e+6\,c^2\,d^2\right )}{6\,e^4}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \]

[In]

int((b*x + c*x^2)^2/(d + e*x)^6,x)

[Out]

-((x^2*(b^2*e^2 + 6*c^2*d^2 + 3*b*c*d*e))/(3*e^3) + (c^2*x^4)/e + (d^2*(b^2*e^2 + 6*c^2*d^2 + 3*b*c*d*e))/(30*
e^5) + (c*x^3*(b*e + 2*c*d))/e^2 + (d*x*(b^2*e^2 + 6*c^2*d^2 + 3*b*c*d*e))/(6*e^4))/(d^5 + e^5*x^5 + 5*d*e^4*x
^4 + 10*d^3*e^2*x^2 + 10*d^2*e^3*x^3 + 5*d^4*e*x)