\(\int \frac {(a+b x+c x^2)^2}{(d+e x)^7} \, dx\) [2129]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 156 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {\left (c d^2-b d e+a e^2\right )^2}{6 e^5 (d+e x)^6}+\frac {2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )}{5 e^5 (d+e x)^5}-\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{4 e^5 (d+e x)^4}+\frac {2 c (2 c d-b e)}{3 e^5 (d+e x)^3}-\frac {c^2}{2 e^5 (d+e x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 156, 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 = {712} \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2}{4 e^5 (d+e x)^4}+\frac {2 (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{5 e^5 (d+e x)^5}-\frac {\left (a e^2-b d e+c d^2\right )^2}{6 e^5 (d+e x)^6}+\frac {2 c (2 c d-b e)}{3 e^5 (d+e x)^3}-\frac {c^2}{2 e^5 (d+e x)^2} \]

[In]

Int[(a + b*x + c*x^2)^2/(d + e*x)^7,x]

[Out]

-1/6*(c*d^2 - b*d*e + a*e^2)^2/(e^5*(d + e*x)^6) + (2*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2))/(5*e^5*(d + e*x)^
5) - (6*c^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))/(4*e^5*(d + e*x)^4) + (2*c*(2*c*d - b*e))/(3*e^5*(d + e*x)^3)
 - c^2/(2*e^5*(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 {\left (c d^2-b d e+a e^2\right )^2}{e^4 (d+e x)^7}+\frac {2 (-2 c d+b e) \left (c d^2-b d e+a e^2\right )}{e^4 (d+e x)^6}+\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{e^4 (d+e x)^5}-\frac {2 c (2 c d-b e)}{e^4 (d+e x)^4}+\frac {c^2}{e^4 (d+e x)^3}\right ) \, dx \\ & = -\frac {\left (c d^2-b d e+a e^2\right )^2}{6 e^5 (d+e x)^6}+\frac {2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )}{5 e^5 (d+e x)^5}-\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{4 e^5 (d+e x)^4}+\frac {2 c (2 c d-b e)}{3 e^5 (d+e x)^3}-\frac {c^2}{2 e^5 (d+e x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {2 c^2 \left (d^4+6 d^3 e x+15 d^2 e^2 x^2+20 d e^3 x^3+15 e^4 x^4\right )+e^2 \left (10 a^2 e^2+4 a b e (d+6 e x)+b^2 \left (d^2+6 d e x+15 e^2 x^2\right )\right )+2 c e \left (a e \left (d^2+6 d e x+15 e^2 x^2\right )+b \left (d^3+6 d^2 e x+15 d e^2 x^2+20 e^3 x^3\right )\right )}{60 e^5 (d+e x)^6} \]

[In]

Integrate[(a + b*x + c*x^2)^2/(d + e*x)^7,x]

[Out]

-1/60*(2*c^2*(d^4 + 6*d^3*e*x + 15*d^2*e^2*x^2 + 20*d*e^3*x^3 + 15*e^4*x^4) + e^2*(10*a^2*e^2 + 4*a*b*e*(d + 6
*e*x) + b^2*(d^2 + 6*d*e*x + 15*e^2*x^2)) + 2*c*e*(a*e*(d^2 + 6*d*e*x + 15*e^2*x^2) + b*(d^3 + 6*d^2*e*x + 15*
d*e^2*x^2 + 20*e^3*x^3)))/(e^5*(d + e*x)^6)

Maple [A] (verified)

Time = 2.93 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.14

method result size
risch \(\frac {-\frac {c^{2} x^{4}}{2 e}-\frac {2 c \left (b e +c d \right ) x^{3}}{3 e^{2}}-\frac {\left (2 a c \,e^{2}+b^{2} e^{2}+2 b c d e +2 c^{2} d^{2}\right ) x^{2}}{4 e^{3}}-\frac {\left (4 a b \,e^{3}+2 d \,e^{2} a c +b^{2} d \,e^{2}+2 b c e \,d^{2}+2 c^{2} d^{3}\right ) x}{10 e^{4}}-\frac {10 a^{2} e^{4}+4 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}+2 d^{3} e b c +2 c^{2} d^{4}}{60 e^{5}}}{\left (e x +d \right )^{6}}\) \(178\)
norman \(\frac {-\frac {c^{2} x^{4}}{2 e}-\frac {2 \left (b c \,e^{2}+d e \,c^{2}\right ) x^{3}}{3 e^{3}}-\frac {\left (2 a c \,e^{3}+b^{2} e^{3}+2 b c d \,e^{2}+2 d^{2} e \,c^{2}\right ) x^{2}}{4 e^{4}}-\frac {\left (4 a b \,e^{4}+2 a c d \,e^{3}+b^{2} d \,e^{3}+2 b c \,d^{2} e^{2}+2 d^{3} e \,c^{2}\right ) x}{10 e^{5}}-\frac {10 a^{2} e^{5}+4 a b d \,e^{4}+2 a \,d^{2} e^{3} c +b^{2} d^{2} e^{3}+2 b c \,d^{3} e^{2}+2 d^{4} e \,c^{2}}{60 e^{6}}}{\left (e x +d \right )^{6}}\) \(192\)
gosper \(-\frac {30 c^{2} x^{4} e^{4}+40 x^{3} b c \,e^{4}+40 x^{3} c^{2} d \,e^{3}+30 x^{2} a c \,e^{4}+15 x^{2} b^{2} e^{4}+30 x^{2} b c d \,e^{3}+30 x^{2} c^{2} d^{2} e^{2}+24 x a b \,e^{4}+12 x a c d \,e^{3}+6 x \,b^{2} d \,e^{3}+12 x b c \,d^{2} e^{2}+12 x \,c^{2} d^{3} e +10 a^{2} e^{4}+4 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}+2 d^{3} e b c +2 c^{2} d^{4}}{60 e^{5} \left (e x +d \right )^{6}}\) \(193\)
default \(-\frac {2 a b \,e^{3}-4 d \,e^{2} a c -2 b^{2} d \,e^{2}+6 b c e \,d^{2}-4 c^{2} d^{3}}{5 e^{5} \left (e x +d \right )^{5}}-\frac {2 c \left (b e -2 c d \right )}{3 e^{5} \left (e x +d \right )^{3}}-\frac {2 a c \,e^{2}+b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}}{4 e^{5} \left (e x +d \right )^{4}}-\frac {c^{2}}{2 e^{5} \left (e x +d \right )^{2}}-\frac {a^{2} e^{4}-2 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}-2 d^{3} e b c +c^{2} d^{4}}{6 e^{5} \left (e x +d \right )^{6}}\) \(195\)
parallelrisch \(\frac {-30 c^{2} e^{5} x^{4}-40 b c \,e^{5} x^{3}-40 c^{2} d \,e^{4} x^{3}-30 a c \,e^{5} x^{2}-15 b^{2} e^{5} x^{2}-30 b c d \,e^{4} x^{2}-30 c^{2} d^{2} e^{3} x^{2}-24 a b \,e^{5} x -12 a c d \,e^{4} x -6 b^{2} d \,e^{4} x -12 b c \,d^{2} e^{3} x -12 c^{2} d^{3} e^{2} x -10 a^{2} e^{5}-4 a b d \,e^{4}-2 a \,d^{2} e^{3} c -b^{2} d^{2} e^{3}-2 b c \,d^{3} e^{2}-2 d^{4} e \,c^{2}}{60 e^{6} \left (e x +d \right )^{6}}\) \(199\)

[In]

int((c*x^2+b*x+a)^2/(e*x+d)^7,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.64 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {30 \, c^{2} e^{4} x^{4} + 2 \, c^{2} d^{4} + 2 \, b c d^{3} e + 4 \, a b d e^{3} + 10 \, a^{2} e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + 40 \, {\left (c^{2} d e^{3} + b c e^{4}\right )} x^{3} + 15 \, {\left (2 \, c^{2} d^{2} e^{2} + 2 \, b c d e^{3} + {\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} + 6 \, {\left (2 \, c^{2} d^{3} e + 2 \, b c d^{2} e^{2} + 4 \, a b e^{4} + {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x}{60 \, {\left (e^{11} x^{6} + 6 \, d e^{10} x^{5} + 15 \, d^{2} e^{9} x^{4} + 20 \, d^{3} e^{8} x^{3} + 15 \, d^{4} e^{7} x^{2} + 6 \, d^{5} e^{6} x + d^{6} e^{5}\right )}} \]

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^7,x, algorithm="fricas")

[Out]

-1/60*(30*c^2*e^4*x^4 + 2*c^2*d^4 + 2*b*c*d^3*e + 4*a*b*d*e^3 + 10*a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2 + 40*(c^2*d
*e^3 + b*c*e^4)*x^3 + 15*(2*c^2*d^2*e^2 + 2*b*c*d*e^3 + (b^2 + 2*a*c)*e^4)*x^2 + 6*(2*c^2*d^3*e + 2*b*c*d^2*e^
2 + 4*a*b*e^4 + (b^2 + 2*a*c)*d*e^3)*x)/(e^11*x^6 + 6*d*e^10*x^5 + 15*d^2*e^9*x^4 + 20*d^3*e^8*x^3 + 15*d^4*e^
7*x^2 + 6*d^5*e^6*x + d^6*e^5)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=\text {Timed out} \]

[In]

integrate((c*x**2+b*x+a)**2/(e*x+d)**7,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {30 \, c^{2} e^{4} x^{4} + 2 \, c^{2} d^{4} + 2 \, b c d^{3} e + 4 \, a b d e^{3} + 10 \, a^{2} e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} + 40 \, {\left (c^{2} d e^{3} + b c e^{4}\right )} x^{3} + 15 \, {\left (2 \, c^{2} d^{2} e^{2} + 2 \, b c d e^{3} + {\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} + 6 \, {\left (2 \, c^{2} d^{3} e + 2 \, b c d^{2} e^{2} + 4 \, a b e^{4} + {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x}{60 \, {\left (e^{11} x^{6} + 6 \, d e^{10} x^{5} + 15 \, d^{2} e^{9} x^{4} + 20 \, d^{3} e^{8} x^{3} + 15 \, d^{4} e^{7} x^{2} + 6 \, d^{5} e^{6} x + d^{6} e^{5}\right )}} \]

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^7,x, algorithm="maxima")

[Out]

-1/60*(30*c^2*e^4*x^4 + 2*c^2*d^4 + 2*b*c*d^3*e + 4*a*b*d*e^3 + 10*a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2 + 40*(c^2*d
*e^3 + b*c*e^4)*x^3 + 15*(2*c^2*d^2*e^2 + 2*b*c*d*e^3 + (b^2 + 2*a*c)*e^4)*x^2 + 6*(2*c^2*d^3*e + 2*b*c*d^2*e^
2 + 4*a*b*e^4 + (b^2 + 2*a*c)*d*e^3)*x)/(e^11*x^6 + 6*d*e^10*x^5 + 15*d^2*e^9*x^4 + 20*d^3*e^8*x^3 + 15*d^4*e^
7*x^2 + 6*d^5*e^6*x + d^6*e^5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.23 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {30 \, c^{2} e^{4} x^{4} + 40 \, c^{2} d e^{3} x^{3} + 40 \, b c e^{4} x^{3} + 30 \, c^{2} d^{2} e^{2} x^{2} + 30 \, b c d e^{3} x^{2} + 15 \, b^{2} e^{4} x^{2} + 30 \, a c e^{4} x^{2} + 12 \, c^{2} d^{3} e x + 12 \, b c d^{2} e^{2} x + 6 \, b^{2} d e^{3} x + 12 \, a c d e^{3} x + 24 \, a b e^{4} x + 2 \, c^{2} d^{4} + 2 \, b c d^{3} e + b^{2} d^{2} e^{2} + 2 \, a c d^{2} e^{2} + 4 \, a b d e^{3} + 10 \, a^{2} e^{4}}{60 \, {\left (e x + d\right )}^{6} e^{5}} \]

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^7,x, algorithm="giac")

[Out]

-1/60*(30*c^2*e^4*x^4 + 40*c^2*d*e^3*x^3 + 40*b*c*e^4*x^3 + 30*c^2*d^2*e^2*x^2 + 30*b*c*d*e^3*x^2 + 15*b^2*e^4
*x^2 + 30*a*c*e^4*x^2 + 12*c^2*d^3*e*x + 12*b*c*d^2*e^2*x + 6*b^2*d*e^3*x + 12*a*c*d*e^3*x + 24*a*b*e^4*x + 2*
c^2*d^4 + 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 + 4*a*b*d*e^3 + 10*a^2*e^4)/((e*x + d)^6*e^5)

Mupad [B] (verification not implemented)

Time = 9.90 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^7} \, dx=-\frac {\frac {10\,a^2\,e^4+4\,a\,b\,d\,e^3+2\,a\,c\,d^2\,e^2+b^2\,d^2\,e^2+2\,b\,c\,d^3\,e+2\,c^2\,d^4}{60\,e^5}+\frac {x\,\left (b^2\,d\,e^2+2\,b\,c\,d^2\,e+4\,a\,b\,e^3+2\,c^2\,d^3+2\,a\,c\,d\,e^2\right )}{10\,e^4}+\frac {c^2\,x^4}{2\,e}+\frac {x^2\,\left (b^2\,e^2+2\,b\,c\,d\,e+2\,c^2\,d^2+2\,a\,c\,e^2\right )}{4\,e^3}+\frac {2\,c\,x^3\,\left (b\,e+c\,d\right )}{3\,e^2}}{d^6+6\,d^5\,e\,x+15\,d^4\,e^2\,x^2+20\,d^3\,e^3\,x^3+15\,d^2\,e^4\,x^4+6\,d\,e^5\,x^5+e^6\,x^6} \]

[In]

int((a + b*x + c*x^2)^2/(d + e*x)^7,x)

[Out]

-((10*a^2*e^4 + 2*c^2*d^4 + b^2*d^2*e^2 + 4*a*b*d*e^3 + 2*b*c*d^3*e + 2*a*c*d^2*e^2)/(60*e^5) + (x*(2*c^2*d^3
+ b^2*d*e^2 + 4*a*b*e^3 + 2*a*c*d*e^2 + 2*b*c*d^2*e))/(10*e^4) + (c^2*x^4)/(2*e) + (x^2*(b^2*e^2 + 2*c^2*d^2 +
 2*a*c*e^2 + 2*b*c*d*e))/(4*e^3) + (2*c*x^3*(b*e + c*d))/(3*e^2))/(d^6 + e^6*x^6 + 6*d*e^5*x^5 + 15*d^4*e^2*x^
2 + 20*d^3*e^3*x^3 + 15*d^2*e^4*x^4 + 6*d^5*e*x)