\(\int \frac {(b x+c x^2)^2}{(d+e x)^5} \, dx\) [240]

Optimal result

Integrand size = 19, antiderivative size = 131 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=-\frac {d^2 (c d-b e)^2}{4 e^5 (d+e x)^4}+\frac {2 d (c d-b e) (2 c d-b e)}{3 e^5 (d+e x)^3}-\frac {6 c^2 d^2-6 b c d e+b^2 e^2}{2 e^5 (d+e x)^2}+\frac {2 c (2 c d-b e)}{e^5 (d+e x)}+\frac {c^2 \log (d+e x)}{e^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 131, 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} \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=-\frac {b^2 e^2-6 b c d e+6 c^2 d^2}{2 e^5 (d+e x)^2}-\frac {d^2 (c d-b e)^2}{4 e^5 (d+e x)^4}+\frac {2 c (2 c d-b e)}{e^5 (d+e x)}+\frac {2 d (c d-b e) (2 c d-b e)}{3 e^5 (d+e x)^3}+\frac {c^2 \log (d+e x)}{e^5} \]

[In]

Int[(b*x + c*x^2)^2/(d + e*x)^5,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.96 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {-b^2 e^2 \left (d^2+4 d e x+6 e^2 x^2\right )-6 b c e \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )+c^2 d \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )+12 c^2 (d+e x)^4 \log (d+e x)}{12 e^5 (d+e x)^4} \]

[In]

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

[Out]

(-(b^2*e^2*(d^2 + 4*d*e*x + 6*e^2*x^2)) - 6*b*c*e*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3) + c^2*d*(25*d^3
+ 88*d^2*e*x + 108*d*e^2*x^2 + 48*e^3*x^3) + 12*c^2*(d + e*x)^4*Log[d + e*x])/(12*e^5*(d + e*x)^4)

Maple [A] (verified)

Time = 2.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.99

method	result	size
risch	\(\frac {-\frac {2 c \left (b e -2 c d \right ) x^{3}}{e^{2}}-\frac {\left (b^{2} e^{2}+6 b c d e -18 c^{2} d^{2}\right ) x^{2}}{2 e^{3}}-\frac {d \left (b^{2} e^{2}+6 b c d e -22 c^{2} d^{2}\right ) x}{3 e^{4}}-\frac {d^{2} \left (b^{2} e^{2}+6 b c d e -25 c^{2} d^{2}\right )}{12 e^{5}}}{\left (e x +d \right )^{4}}+\frac {c^{2} \ln \left (e x +d \right )}{e^{5}}\)	\(130\)
norman	\(\frac {-\frac {d^{2} \left (b^{2} e^{2}+6 b c d e -25 c^{2} d^{2}\right )}{12 e^{5}}-\frac {2 \left (b c e -2 c^{2} d \right ) x^{3}}{e^{2}}-\frac {\left (b^{2} e^{2}+6 b c d e -18 c^{2} d^{2}\right ) x^{2}}{2 e^{3}}-\frac {d \left (b^{2} e^{2}+6 b c d e -22 c^{2} d^{2}\right ) x}{3 e^{4}}}{\left (e x +d \right )^{4}}+\frac {c^{2} \ln \left (e x +d \right )}{e^{5}}\)	\(132\)
default	\(-\frac {2 c \left (b e -2 c d \right )}{e^{5} \left (e x +d \right )}+\frac {2 d \left (b^{2} e^{2}-3 b c d e +2 c^{2} d^{2}\right )}{3 e^{5} \left (e x +d \right )^{3}}-\frac {d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}{4 e^{5} \left (e x +d \right )^{4}}-\frac {b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}}{2 e^{5} \left (e x +d \right )^{2}}+\frac {c^{2} \ln \left (e x +d \right )}{e^{5}}\)	\(141\)
parallelrisch	\(\frac {12 \ln \left (e x +d \right ) x^{4} c^{2} e^{4}+48 \ln \left (e x +d \right ) x^{3} c^{2} d \,e^{3}+72 \ln \left (e x +d \right ) x^{2} c^{2} d^{2} e^{2}-24 x^{3} b c \,e^{4}+48 x^{3} c^{2} d \,e^{3}+48 \ln \left (e x +d \right ) x \,c^{2} d^{3} e -6 x^{2} b^{2} e^{4}-36 x^{2} b c d \,e^{3}+108 x^{2} c^{2} d^{2} e^{2}+12 \ln \left (e x +d \right ) c^{2} d^{4}-4 x \,b^{2} d \,e^{3}-24 x b c \,d^{2} e^{2}+88 x \,c^{2} d^{3} e -b^{2} d^{2} e^{2}-6 d^{3} e b c +25 c^{2} d^{4}}{12 e^{5} \left (e x +d \right )^{4}}\)	\(215\)

[In]

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

[Out]

(-2*c*(b*e-2*c*d)/e^2*x^3-1/2*(b^2*e^2+6*b*c*d*e-18*c^2*d^2)/e^3*x^2-1/3*d*(b^2*e^2+6*b*c*d*e-22*c^2*d^2)/e^4*
x-1/12*d^2*(b^2*e^2+6*b*c*d*e-25*c^2*d^2)/e^5)/(e*x+d)^4+c^2*ln(e*x+d)/e^5

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.72 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {25 \, c^{2} d^{4} - 6 \, b c d^{3} e - b^{2} d^{2} e^{2} + 24 \, {\left (2 \, c^{2} d e^{3} - b c e^{4}\right )} x^{3} + 6 \, {\left (18 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} - b^{2} e^{4}\right )} x^{2} + 4 \, {\left (22 \, c^{2} d^{3} e - 6 \, b c d^{2} e^{2} - b^{2} d e^{3}\right )} x + 12 \, {\left (c^{2} e^{4} x^{4} + 4 \, c^{2} d e^{3} x^{3} + 6 \, c^{2} d^{2} e^{2} x^{2} + 4 \, c^{2} d^{3} e x + c^{2} d^{4}\right )} \log \left (e x + d\right )}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} \]

[In]

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

[Out]

1/12*(25*c^2*d^4 - 6*b*c*d^3*e - b^2*d^2*e^2 + 24*(2*c^2*d*e^3 - b*c*e^4)*x^3 + 6*(18*c^2*d^2*e^2 - 6*b*c*d*e^
3 - b^2*e^4)*x^2 + 4*(22*c^2*d^3*e - 6*b*c*d^2*e^2 - b^2*d*e^3)*x + 12*(c^2*e^4*x^4 + 4*c^2*d*e^3*x^3 + 6*c^2*
d^2*e^2*x^2 + 4*c^2*d^3*e*x + c^2*d^4)*log(e*x + d))/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*d^3*e^6*x + d^
4*e^5)

Sympy [A] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.37 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^{2} \log {\left (d + e x \right )}}{e^{5}} + \frac {- b^{2} d^{2} e^{2} - 6 b c d^{3} e + 25 c^{2} d^{4} + x^{3} \left (- 24 b c e^{4} + 48 c^{2} d e^{3}\right ) + x^{2} \left (- 6 b^{2} e^{4} - 36 b c d e^{3} + 108 c^{2} d^{2} e^{2}\right ) + x \left (- 4 b^{2} d e^{3} - 24 b c d^{2} e^{2} + 88 c^{2} d^{3} e\right )}{12 d^{4} e^{5} + 48 d^{3} e^{6} x + 72 d^{2} e^{7} x^{2} + 48 d e^{8} x^{3} + 12 e^{9} x^{4}} \]

[In]

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

[Out]

c**2*log(d + e*x)/e**5 + (-b**2*d**2*e**2 - 6*b*c*d**3*e + 25*c**2*d**4 + x**3*(-24*b*c*e**4 + 48*c**2*d*e**3)
+ x**2*(-6*b**2*e**4 - 36*b*c*d*e**3 + 108*c**2*d**2*e**2) + x*(-4*b**2*d*e**3 - 24*b*c*d**2*e**2 + 88*c**2*d
**3*e))/(12*d**4*e**5 + 48*d**3*e**6*x + 72*d**2*e**7*x**2 + 48*d*e**8*x**3 + 12*e**9*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.35 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {25 \, c^{2} d^{4} - 6 \, b c d^{3} e - b^{2} d^{2} e^{2} + 24 \, {\left (2 \, c^{2} d e^{3} - b c e^{4}\right )} x^{3} + 6 \, {\left (18 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} - b^{2} e^{4}\right )} x^{2} + 4 \, {\left (22 \, c^{2} d^{3} e - 6 \, b c d^{2} e^{2} - b^{2} d e^{3}\right )} x}{12 \, {\left (e^{9} x^{4} + 4 \, d e^{8} x^{3} + 6 \, d^{2} e^{7} x^{2} + 4 \, d^{3} e^{6} x + d^{4} e^{5}\right )}} + \frac {c^{2} \log \left (e x + d\right )}{e^{5}} \]

[In]

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

[Out]

1/12*(25*c^2*d^4 - 6*b*c*d^3*e - b^2*d^2*e^2 + 24*(2*c^2*d*e^3 - b*c*e^4)*x^3 + 6*(18*c^2*d^2*e^2 - 6*b*c*d*e^
3 - b^2*e^4)*x^2 + 4*(22*c^2*d^3*e - 6*b*c*d^2*e^2 - b^2*d*e^3)*x)/(e^9*x^4 + 4*d*e^8*x^3 + 6*d^2*e^7*x^2 + 4*
d^3*e^6*x + d^4*e^5) + c^2*log(e*x + d)/e^5

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.65 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=-\frac {c^{2} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{5}} + \frac {\frac {48 \, c^{2} d e^{15}}{e x + d} - \frac {36 \, c^{2} d^{2} e^{15}}{{\left (e x + d\right )}^{2}} + \frac {16 \, c^{2} d^{3} e^{15}}{{\left (e x + d\right )}^{3}} - \frac {3 \, c^{2} d^{4} e^{15}}{{\left (e x + d\right )}^{4}} - \frac {24 \, b c e^{16}}{e x + d} + \frac {36 \, b c d e^{16}}{{\left (e x + d\right )}^{2}} - \frac {24 \, b c d^{2} e^{16}}{{\left (e x + d\right )}^{3}} + \frac {6 \, b c d^{3} e^{16}}{{\left (e x + d\right )}^{4}} - \frac {6 \, b^{2} e^{17}}{{\left (e x + d\right )}^{2}} + \frac {8 \, b^{2} d e^{17}}{{\left (e x + d\right )}^{3}} - \frac {3 \, b^{2} d^{2} e^{17}}{{\left (e x + d\right )}^{4}}}{12 \, e^{20}} \]

[In]

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

[Out]

-c^2*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^5 + 1/12*(48*c^2*d*e^15/(e*x + d) - 36*c^2*d^2*e^15/(e*x + d)^2
+ 16*c^2*d^3*e^15/(e*x + d)^3 - 3*c^2*d^4*e^15/(e*x + d)^4 - 24*b*c*e^16/(e*x + d) + 36*b*c*d*e^16/(e*x + d)^2
- 24*b*c*d^2*e^16/(e*x + d)^3 + 6*b*c*d^3*e^16/(e*x + d)^4 - 6*b^2*e^17/(e*x + d)^2 + 8*b^2*d*e^17/(e*x + d)^
3 - 3*b^2*d^2*e^17/(e*x + d)^4)/e^20

Mupad [B] (verification not implemented)

Time = 9.59 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.27 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^2\,\ln \left (d+e\,x\right )}{e^5}-\frac {\frac {b^2\,d^2\,e^2+6\,b\,c\,d^3\,e-25\,c^2\,d^4}{12\,e^5}+\frac {x^2\,\left (b^2\,e^2+6\,b\,c\,d\,e-18\,c^2\,d^2\right )}{2\,e^3}+\frac {x\,\left (b^2\,d\,e^2+6\,b\,c\,d^2\,e-22\,c^2\,d^3\right )}{3\,e^4}+\frac {2\,c\,x^3\,\left (b\,e-2\,c\,d\right )}{e^2}}{d^4+4\,d^3\,e\,x+6\,d^2\,e^2\,x^2+4\,d\,e^3\,x^3+e^4\,x^4} \]

[In]

int((b*x + c*x^2)^2/(d + e*x)^5,x)

[Out]

(c^2*log(d + e*x))/e^5 - ((b^2*d^2*e^2 - 25*c^2*d^4 + 6*b*c*d^3*e)/(12*e^5) + (x^2*(b^2*e^2 - 18*c^2*d^2 + 6*b
*c*d*e))/(2*e^3) + (x*(b^2*d*e^2 - 22*c^2*d^3 + 6*b*c*d^2*e))/(3*e^4) + (2*c*x^3*(b*e - 2*c*d))/e^2)/(d^4 + e^
4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x)