\(\int \frac {(b x+c x^2)^2}{(d+e x)^4} \, dx\) [239]

Optimal result

Integrand size = 19, antiderivative size = 120 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {c^2 x}{e^4}-\frac {d^2 (c d-b e)^2}{3 e^5 (d+e x)^3}+\frac {d (c d-b e) (2 c d-b e)}{e^5 (d+e x)^2}-\frac {6 c^2 d^2-6 b c d e+b^2 e^2}{e^5 (d+e x)}-\frac {2 c (2 c d-b e) \log (d+e x)}{e^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 120, 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} \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=-\frac {b^2 e^2-6 b c d e+6 c^2 d^2}{e^5 (d+e x)}-\frac {d^2 (c d-b e)^2}{3 e^5 (d+e x)^3}+\frac {d (c d-b e) (2 c d-b e)}{e^5 (d+e x)^2}-\frac {2 c (2 c d-b e) \log (d+e x)}{e^5}+\frac {c^2 x}{e^4} \]

[In]

Int[(b*x + c*x^2)^2/(d + e*x)^4,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.12 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {-b^2 e^2 \left (d^2+3 d e x+3 e^2 x^2\right )+b c d e \left (11 d^2+27 d e x+18 e^2 x^2\right )+c^2 \left (-13 d^4-27 d^3 e x-9 d^2 e^2 x^2+9 d e^3 x^3+3 e^4 x^4\right )-6 c (2 c d-b e) (d+e x)^3 \log (d+e x)}{3 e^5 (d+e x)^3} \]

[In]

Integrate[(b*x + c*x^2)^2/(d + e*x)^4,x]

[Out]

(-(b^2*e^2*(d^2 + 3*d*e*x + 3*e^2*x^2)) + b*c*d*e*(11*d^2 + 27*d*e*x + 18*e^2*x^2) + c^2*(-13*d^4 - 27*d^3*e*x
- 9*d^2*e^2*x^2 + 9*d*e^3*x^3 + 3*e^4*x^4) - 6*c*(2*c*d - b*e)*(d + e*x)^3*Log[d + e*x])/(3*e^5*(d + e*x)^3)

Maple [A] (verified)

Time = 2.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.08

method	result	size
norman	\(\frac {\frac {c^{2} x^{4}}{e}-\frac {d^{2} \left (b^{2} e^{2}-11 b c d e +22 c^{2} d^{2}\right )}{3 e^{5}}-\frac {\left (b^{2} e^{2}-6 b c d e +12 c^{2} d^{2}\right ) x^{2}}{e^{3}}-\frac {d \left (b^{2} e^{2}-9 b c d e +18 c^{2} d^{2}\right ) x}{e^{4}}}{\left (e x +d \right )^{3}}+\frac {2 c \left (b e -2 c d \right ) \ln \left (e x +d \right )}{e^{5}}\)	\(130\)
default	\(\frac {c^{2} x}{e^{4}}-\frac {b^{2} e^{2}-6 b c d e +6 c^{2} d^{2}}{e^{5} \left (e x +d \right )}-\frac {d^{2} \left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right )}{3 e^{5} \left (e x +d \right )^{3}}+\frac {d \left (b^{2} e^{2}-3 b c d e +2 c^{2} d^{2}\right )}{e^{5} \left (e x +d \right )^{2}}+\frac {2 c \left (b e -2 c d \right ) \ln \left (e x +d \right )}{e^{5}}\)	\(134\)
risch	\(\frac {c^{2} x}{e^{4}}+\frac {\left (-e^{3} b^{2}+6 b c d \,e^{2}-6 d^{2} e \,c^{2}\right ) x^{2}-d \left (b^{2} e^{2}-9 b c d e +10 c^{2} d^{2}\right ) x -\frac {d^{2} \left (b^{2} e^{2}-11 b c d e +13 c^{2} d^{2}\right )}{3 e}}{e^{4} \left (e x +d \right )^{3}}+\frac {2 c \ln \left (e x +d \right ) b}{e^{4}}-\frac {4 c^{2} d \ln \left (e x +d \right )}{e^{5}}\)	\(136\)
parallelrisch	\(\frac {6 \ln \left (e x +d \right ) x^{3} b c \,e^{4}-12 \ln \left (e x +d \right ) x^{3} c^{2} d \,e^{3}+3 c^{2} x^{4} e^{4}+18 \ln \left (e x +d \right ) x^{2} b c d \,e^{3}-36 \ln \left (e x +d \right ) x^{2} c^{2} d^{2} e^{2}+18 \ln \left (e x +d \right ) x b c \,d^{2} e^{2}-36 \ln \left (e x +d \right ) x \,c^{2} d^{3} e -3 x^{2} b^{2} e^{4}+18 x^{2} b c d \,e^{3}-36 x^{2} c^{2} d^{2} e^{2}+6 \ln \left (e x +d \right ) b c \,d^{3} e -12 \ln \left (e x +d \right ) c^{2} d^{4}-3 x \,b^{2} d \,e^{3}+27 x b c \,d^{2} e^{2}-54 x \,c^{2} d^{3} e -b^{2} d^{2} e^{2}+11 d^{3} e b c -22 c^{2} d^{4}}{3 e^{5} \left (e x +d \right )^{3}}\)	\(251\)

[In]

int((c*x^2+b*x)^2/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

(c^2*x^4/e-1/3*d^2*(b^2*e^2-11*b*c*d*e+22*c^2*d^2)/e^5-(b^2*e^2-6*b*c*d*e+12*c^2*d^2)/e^3*x^2-d*(b^2*e^2-9*b*c
*d*e+18*c^2*d^2)/e^4*x)/(e*x+d)^3+2/e^5*c*(b*e-2*c*d)*ln(e*x+d)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (118) = 236\).

Time = 0.26 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.04 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {3 \, c^{2} e^{4} x^{4} + 9 \, c^{2} d e^{3} x^{3} - 13 \, c^{2} d^{4} + 11 \, b c d^{3} e - b^{2} d^{2} e^{2} - 3 \, {\left (3 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} + b^{2} e^{4}\right )} x^{2} - 3 \, {\left (9 \, c^{2} d^{3} e - 9 \, b c d^{2} e^{2} + b^{2} d e^{3}\right )} x - 6 \, {\left (2 \, c^{2} d^{4} - b c d^{3} e + {\left (2 \, c^{2} d e^{3} - b c e^{4}\right )} x^{3} + 3 \, {\left (2 \, c^{2} d^{2} e^{2} - b c d e^{3}\right )} x^{2} + 3 \, {\left (2 \, c^{2} d^{3} e - b c d^{2} e^{2}\right )} x\right )} \log \left (e x + d\right )}{3 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} \]

[In]

integrate((c*x^2+b*x)^2/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/3*(3*c^2*e^4*x^4 + 9*c^2*d*e^3*x^3 - 13*c^2*d^4 + 11*b*c*d^3*e - b^2*d^2*e^2 - 3*(3*c^2*d^2*e^2 - 6*b*c*d*e^
3 + b^2*e^4)*x^2 - 3*(9*c^2*d^3*e - 9*b*c*d^2*e^2 + b^2*d*e^3)*x - 6*(2*c^2*d^4 - b*c*d^3*e + (2*c^2*d*e^3 - b
*c*e^4)*x^3 + 3*(2*c^2*d^2*e^2 - b*c*d*e^3)*x^2 + 3*(2*c^2*d^3*e - b*c*d^2*e^2)*x)*log(e*x + d))/(e^8*x^3 + 3*
d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5)

Sympy [A] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.36 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {c^{2} x}{e^{4}} + \frac {2 c \left (b e - 2 c d\right ) \log {\left (d + e x \right )}}{e^{5}} + \frac {- b^{2} d^{2} e^{2} + 11 b c d^{3} e - 13 c^{2} d^{4} + x^{2} \left (- 3 b^{2} e^{4} + 18 b c d e^{3} - 18 c^{2} d^{2} e^{2}\right ) + x \left (- 3 b^{2} d e^{3} + 27 b c d^{2} e^{2} - 30 c^{2} d^{3} e\right )}{3 d^{3} e^{5} + 9 d^{2} e^{6} x + 9 d e^{7} x^{2} + 3 e^{8} x^{3}} \]

[In]

integrate((c*x**2+b*x)**2/(e*x+d)**4,x)

[Out]

c**2*x/e**4 + 2*c*(b*e - 2*c*d)*log(d + e*x)/e**5 + (-b**2*d**2*e**2 + 11*b*c*d**3*e - 13*c**2*d**4 + x**2*(-3
*b**2*e**4 + 18*b*c*d*e**3 - 18*c**2*d**2*e**2) + x*(-3*b**2*d*e**3 + 27*b*c*d**2*e**2 - 30*c**2*d**3*e))/(3*d
**3*e**5 + 9*d**2*e**6*x + 9*d*e**7*x**2 + 3*e**8*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.32 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=-\frac {13 \, c^{2} d^{4} - 11 \, b c d^{3} e + b^{2} d^{2} e^{2} + 3 \, {\left (6 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} + b^{2} e^{4}\right )} x^{2} + 3 \, {\left (10 \, c^{2} d^{3} e - 9 \, b c d^{2} e^{2} + b^{2} d e^{3}\right )} x}{3 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} + \frac {c^{2} x}{e^{4}} - \frac {2 \, {\left (2 \, c^{2} d - b c e\right )} \log \left (e x + d\right )}{e^{5}} \]

[In]

integrate((c*x^2+b*x)^2/(e*x+d)^4,x, algorithm="maxima")

[Out]

-1/3*(13*c^2*d^4 - 11*b*c*d^3*e + b^2*d^2*e^2 + 3*(6*c^2*d^2*e^2 - 6*b*c*d*e^3 + b^2*e^4)*x^2 + 3*(10*c^2*d^3*
e - 9*b*c*d^2*e^2 + b^2*d*e^3)*x)/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5) + c^2*x/e^4 - 2*(2*c^2*d - b
*c*e)*log(e*x + d)/e^5

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.12 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {c^{2} x}{e^{4}} - \frac {2 \, {\left (2 \, c^{2} d - b c e\right )} \log \left ({\left | e x + d \right |}\right )}{e^{5}} - \frac {13 \, c^{2} d^{4} - 11 \, b c d^{3} e + b^{2} d^{2} e^{2} + 3 \, {\left (6 \, c^{2} d^{2} e^{2} - 6 \, b c d e^{3} + b^{2} e^{4}\right )} x^{2} + 3 \, {\left (10 \, c^{2} d^{3} e - 9 \, b c d^{2} e^{2} + b^{2} d e^{3}\right )} x}{3 \, {\left (e x + d\right )}^{3} e^{5}} \]

[In]

integrate((c*x^2+b*x)^2/(e*x+d)^4,x, algorithm="giac")

[Out]

c^2*x/e^4 - 2*(2*c^2*d - b*c*e)*log(abs(e*x + d))/e^5 - 1/3*(13*c^2*d^4 - 11*b*c*d^3*e + b^2*d^2*e^2 + 3*(6*c^
2*d^2*e^2 - 6*b*c*d*e^3 + b^2*e^4)*x^2 + 3*(10*c^2*d^3*e - 9*b*c*d^2*e^2 + b^2*d*e^3)*x)/((e*x + d)^3*e^5)

Mupad [B] (verification not implemented)

Time = 9.59 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.32 \[ \int \frac {\left (b x+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {c^2\,x}{e^4}-\frac {x^2\,\left (b^2\,e^3-6\,b\,c\,d\,e^2+6\,c^2\,d^2\,e\right )+\frac {b^2\,d^2\,e^2-11\,b\,c\,d^3\,e+13\,c^2\,d^4}{3\,e}+x\,\left (b^2\,d\,e^2-9\,b\,c\,d^2\,e+10\,c^2\,d^3\right )}{d^3\,e^4+3\,d^2\,e^5\,x+3\,d\,e^6\,x^2+e^7\,x^3}-\frac {\ln \left (d+e\,x\right )\,\left (4\,c^2\,d-2\,b\,c\,e\right )}{e^5} \]

[In]

int((b*x + c*x^2)^2/(d + e*x)^4,x)

[Out]

(c^2*x)/e^4 - (x^2*(b^2*e^3 + 6*c^2*d^2*e - 6*b*c*d*e^2) + (13*c^2*d^4 + b^2*d^2*e^2 - 11*b*c*d^3*e)/(3*e) + x
*(10*c^2*d^3 + b^2*d*e^2 - 9*b*c*d^2*e))/(d^3*e^4 + e^7*x^3 + 3*d^2*e^5*x + 3*d*e^6*x^2) - (log(d + e*x)*(4*c^
2*d - 2*b*c*e))/e^5