\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^2}{(d+e x)^8} \, dx\) [1849]

Optimal result

Integrand size = 35, antiderivative size = 77 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^8} \, dx=-\frac {\left (c d^2-a e^2\right )^2}{5 e^3 (d+e x)^5}+\frac {c d \left (c d^2-a e^2\right )}{2 e^3 (d+e x)^4}-\frac {c^2 d^2}{3 e^3 (d+e x)^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {640, 45} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^8} \, dx=\frac {c d \left (c d^2-a e^2\right )}{2 e^3 (d+e x)^4}-\frac {\left (c d^2-a e^2\right )^2}{5 e^3 (d+e x)^5}-\frac {c^2 d^2}{3 e^3 (d+e x)^3} \]

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^8,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(a e+c d x)^2}{(d+e x)^6} \, dx \\ & = \int \left (\frac {\left (-c d^2+a e^2\right )^2}{e^2 (d+e x)^6}-\frac {2 c d \left (c d^2-a e^2\right )}{e^2 (d+e x)^5}+\frac {c^2 d^2}{e^2 (d+e x)^4}\right ) \, dx \\ & = -\frac {\left (c d^2-a e^2\right )^2}{5 e^3 (d+e x)^5}+\frac {c d \left (c d^2-a e^2\right )}{2 e^3 (d+e x)^4}-\frac {c^2 d^2}{3 e^3 (d+e x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^8} \, dx=-\frac {6 a^2 e^4+3 a c d e^2 (d+5 e x)+c^2 d^2 \left (d^2+5 d e x+10 e^2 x^2\right )}{30 e^3 (d+e x)^5} \]

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^8,x]

[Out]

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

Maple [A] (verified)

Time = 2.97 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94

method	result	size
gosper	\(-\frac {10 x^{2} c^{2} d^{2} e^{2}+15 x a c d \,e^{3}+5 x \,c^{2} d^{3} e +6 a^{2} e^{4}+3 a c \,d^{2} e^{2}+c^{2} d^{4}}{30 e^{3} \left (e x +d \right )^{5}}\)	\(72\)
risch	\(\frac {-\frac {d^{2} c^{2} x^{2}}{3 e}-\frac {d c \left (3 e^{2} a +c \,d^{2}\right ) x}{6 e^{2}}-\frac {6 a^{2} e^{4}+3 a c \,d^{2} e^{2}+c^{2} d^{4}}{30 e^{3}}}{\left (e x +d \right )^{5}}\)	\(75\)
parallelrisch	\(\frac {-10 c^{2} d^{2} x^{2} e^{4}-15 a c d \,e^{5} x -5 c^{2} d^{3} e^{3} x -6 a^{2} e^{6}-3 a c \,d^{2} e^{4}-c^{2} d^{4} e^{2}}{30 e^{5} \left (e x +d \right )^{5}}\)	\(78\)
default	\(-\frac {a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}{5 e^{3} \left (e x +d \right )^{5}}-\frac {c^{2} d^{2}}{3 e^{3} \left (e x +d \right )^{3}}-\frac {c d \left (e^{2} a -c \,d^{2}\right )}{2 e^{3} \left (e x +d \right )^{4}}\)	\(83\)
norman	\(\frac {-\frac {d^{2} \left (6 a^{2} e^{8}+3 a c \,d^{2} e^{6}+c^{2} d^{4} e^{4}\right )}{30 e^{7}}-\frac {\left (2 a^{2} e^{8}+11 a c \,d^{2} e^{6}+7 c^{2} d^{4} e^{4}\right ) x^{2}}{10 e^{5}}-\frac {d \left (3 a c \,e^{6}+5 c^{2} d^{2} e^{4}\right ) x^{3}}{6 e^{4}}-\frac {d \left (12 a^{2} e^{8}+21 a c \,d^{2} e^{6}+7 c^{2} d^{4} e^{4}\right ) x}{30 e^{6}}-\frac {e \,c^{2} d^{2} x^{4}}{3}}{\left (e x +d \right )^{7}}\)	\(162\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^8,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.55 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^8} \, dx=-\frac {10 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 3 \, a c d^{2} e^{2} + 6 \, a^{2} e^{4} + 5 \, {\left (c^{2} d^{3} e + 3 \, a c d e^{3}\right )} x}{30 \, {\left (e^{8} x^{5} + 5 \, d e^{7} x^{4} + 10 \, d^{2} e^{6} x^{3} + 10 \, d^{3} e^{5} x^{2} + 5 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^8,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 3.59 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.64 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^8} \, dx=\frac {- 6 a^{2} e^{4} - 3 a c d^{2} e^{2} - c^{2} d^{4} - 10 c^{2} d^{2} e^{2} x^{2} + x \left (- 15 a c d e^{3} - 5 c^{2} d^{3} e\right )}{30 d^{5} e^{3} + 150 d^{4} e^{4} x + 300 d^{3} e^{5} x^{2} + 300 d^{2} e^{6} x^{3} + 150 d e^{7} x^{4} + 30 e^{8} x^{5}} \]

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2/(e*x+d)**8,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.55 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^8} \, dx=-\frac {10 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 3 \, a c d^{2} e^{2} + 6 \, a^{2} e^{4} + 5 \, {\left (c^{2} d^{3} e + 3 \, a c d e^{3}\right )} x}{30 \, {\left (e^{8} x^{5} + 5 \, d e^{7} x^{4} + 10 \, d^{2} e^{6} x^{3} + 10 \, d^{3} e^{5} x^{2} + 5 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^8,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^8} \, dx=-\frac {10 \, c^{2} d^{2} e^{2} x^{2} + 5 \, c^{2} d^{3} e x + 15 \, a c d e^{3} x + c^{2} d^{4} + 3 \, a c d^{2} e^{2} + 6 \, a^{2} e^{4}}{30 \, {\left (e x + d\right )}^{5} e^{3}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^8,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.55 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^8} \, dx=-\frac {\frac {6\,a^2\,e^4+3\,a\,c\,d^2\,e^2+c^2\,d^4}{30\,e^3}+\frac {c^2\,d^2\,x^2}{3\,e}+\frac {c\,d\,x\,\left (c\,d^2+3\,a\,e^2\right )}{6\,e^2}}{d^5+5\,d^4\,e\,x+10\,d^3\,e^2\,x^2+10\,d^2\,e^3\,x^3+5\,d\,e^4\,x^4+e^5\,x^5} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2/(d + e*x)^8,x)

[Out]

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