\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^3}{(d+e x)^{10}} \, dx\) [1863]

Optimal result

Integrand size = 35, antiderivative size = 111 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{10}} \, dx=\frac {\left (c d^2-a e^2\right )^3}{6 e^4 (d+e x)^6}-\frac {3 c d \left (c d^2-a e^2\right )^2}{5 e^4 (d+e x)^5}+\frac {3 c^2 d^2 \left (c d^2-a e^2\right )}{4 e^4 (d+e x)^4}-\frac {c^3 d^3}{3 e^4 (d+e x)^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 111, 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} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{10}} \, dx=\frac {3 c^2 d^2 \left (c d^2-a e^2\right )}{4 e^4 (d+e x)^4}-\frac {3 c d \left (c d^2-a e^2\right )^2}{5 e^4 (d+e x)^5}+\frac {\left (c d^2-a e^2\right )^3}{6 e^4 (d+e x)^6}-\frac {c^3 d^3}{3 e^4 (d+e x)^3} \]

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^10,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{10}} \, dx=-\frac {10 a^3 e^6+6 a^2 c d e^4 (d+6 e x)+3 a c^2 d^2 e^2 \left (d^2+6 d e x+15 e^2 x^2\right )+c^3 d^3 \left (d^3+6 d^2 e x+15 d e^2 x^2+20 e^3 x^3\right )}{60 e^4 (d+e x)^6} \]

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^10,x]

[Out]

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

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.16

method	result	size
risch	\(\frac {-\frac {c^{3} d^{3} x^{3}}{3 e}-\frac {d^{2} c^{2} \left (3 e^{2} a +c \,d^{2}\right ) x^{2}}{4 e^{2}}-\frac {d c \left (6 a^{2} e^{4}+3 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) x}{10 e^{3}}-\frac {10 e^{6} a^{3}+6 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a +c^{3} d^{6}}{60 e^{4}}}{\left (e x +d \right )^{6}}\)	\(129\)
gosper	\(-\frac {20 x^{3} c^{3} d^{3} e^{3}+45 x^{2} a \,c^{2} d^{2} e^{4}+15 x^{2} c^{3} d^{4} e^{2}+36 x \,a^{2} c d \,e^{5}+18 x a \,c^{2} d^{3} e^{3}+6 x \,c^{3} d^{5} e +10 e^{6} a^{3}+6 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a +c^{3} d^{6}}{60 e^{4} \left (e x +d \right )^{6}}\)	\(130\)
parallelrisch	\(\frac {-20 c^{3} d^{3} x^{3} e^{5}-45 a \,c^{2} d^{2} e^{6} x^{2}-15 c^{3} d^{4} e^{4} x^{2}-36 a^{2} c d \,e^{7} x -18 a \,c^{2} d^{3} e^{5} x -6 c^{3} d^{5} e^{3} x -10 a^{3} e^{8}-6 a^{2} c \,d^{2} e^{6}-3 a \,c^{2} d^{4} e^{4}-c^{3} d^{6} e^{2}}{60 e^{6} \left (e x +d \right )^{6}}\)	\(136\)
default	\(-\frac {3 c d \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{5 e^{4} \left (e x +d \right )^{5}}-\frac {c^{3} d^{3}}{3 e^{4} \left (e x +d \right )^{3}}-\frac {3 c^{2} d^{2} \left (e^{2} a -c \,d^{2}\right )}{4 e^{4} \left (e x +d \right )^{4}}-\frac {e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}}{6 e^{4} \left (e x +d \right )^{6}}\)	\(141\)
norman	\(\frac {-\frac {d^{3} \left (10 a^{3} e^{11}+6 a^{2} c \,d^{2} e^{9}+3 d^{4} c^{2} a \,e^{7}+c^{3} d^{6} e^{5}\right )}{60 e^{9}}-\frac {\left (5 a^{3} e^{11}+57 a^{2} c \,d^{2} e^{9}+96 d^{4} c^{2} a \,e^{7}+42 c^{3} d^{6} e^{5}\right ) x^{3}}{30 e^{6}}-\frac {d \left (12 a^{2} c \,e^{9}+51 a \,c^{2} d^{2} e^{7}+37 d^{4} c^{3} e^{5}\right ) x^{4}}{20 e^{5}}-\frac {d \left (5 a^{3} e^{11}+21 a^{2} c \,d^{2} e^{9}+18 d^{4} c^{2} a \,e^{7}+6 c^{3} d^{6} e^{5}\right ) x^{2}}{10 e^{7}}-\frac {d^{2} \left (3 e^{7} c^{2} a +5 c^{3} d^{2} e^{5}\right ) x^{5}}{4 e^{4}}-\frac {d^{2} \left (10 a^{3} e^{11}+18 a^{2} c \,d^{2} e^{9}+9 d^{4} c^{2} a \,e^{7}+3 c^{3} d^{6} e^{5}\right ) x}{20 e^{8}}-\frac {e^{2} c^{3} d^{3} x^{6}}{3}}{\left (e x +d \right )^{9}}\)	\(307\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^10,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.68 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{10}} \, dx=-\frac {20 \, c^{3} d^{3} e^{3} x^{3} + c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 6 \, a^{2} c d^{2} e^{4} + 10 \, a^{3} e^{6} + 15 \, {\left (c^{3} d^{4} e^{2} + 3 \, a c^{2} d^{2} e^{4}\right )} x^{2} + 6 \, {\left (c^{3} d^{5} e + 3 \, a c^{2} d^{3} e^{3} + 6 \, a^{2} c d e^{5}\right )} x}{60 \, {\left (e^{10} x^{6} + 6 \, d e^{9} x^{5} + 15 \, d^{2} e^{8} x^{4} + 20 \, d^{3} e^{7} x^{3} + 15 \, d^{4} e^{6} x^{2} + 6 \, d^{5} e^{5} x + d^{6} e^{4}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^10,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{10}} \, dx=\text {Timed out} \]

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3/(e*x+d)**10,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.68 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{10}} \, dx=-\frac {20 \, c^{3} d^{3} e^{3} x^{3} + c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 6 \, a^{2} c d^{2} e^{4} + 10 \, a^{3} e^{6} + 15 \, {\left (c^{3} d^{4} e^{2} + 3 \, a c^{2} d^{2} e^{4}\right )} x^{2} + 6 \, {\left (c^{3} d^{5} e + 3 \, a c^{2} d^{3} e^{3} + 6 \, a^{2} c d e^{5}\right )} x}{60 \, {\left (e^{10} x^{6} + 6 \, d e^{9} x^{5} + 15 \, d^{2} e^{8} x^{4} + 20 \, d^{3} e^{7} x^{3} + 15 \, d^{4} e^{6} x^{2} + 6 \, d^{5} e^{5} x + d^{6} e^{4}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^10,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{10}} \, dx=-\frac {20 \, c^{3} d^{3} e^{3} x^{3} + 15 \, c^{3} d^{4} e^{2} x^{2} + 45 \, a c^{2} d^{2} e^{4} x^{2} + 6 \, c^{3} d^{5} e x + 18 \, a c^{2} d^{3} e^{3} x + 36 \, a^{2} c d e^{5} x + c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 6 \, a^{2} c d^{2} e^{4} + 10 \, a^{3} e^{6}}{60 \, {\left (e x + d\right )}^{6} e^{4}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^10,x, algorithm="giac")

[Out]

-1/60*(20*c^3*d^3*e^3*x^3 + 15*c^3*d^4*e^2*x^2 + 45*a*c^2*d^2*e^4*x^2 + 6*c^3*d^5*e*x + 18*a*c^2*d^3*e^3*x + 3
6*a^2*c*d*e^5*x + c^3*d^6 + 3*a*c^2*d^4*e^2 + 6*a^2*c*d^2*e^4 + 10*a^3*e^6)/((e*x + d)^6*e^4)

Mupad [B] (verification not implemented)

Time = 9.90 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.66 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^{10}} \, dx=-\frac {\frac {10\,a^3\,e^6+6\,a^2\,c\,d^2\,e^4+3\,a\,c^2\,d^4\,e^2+c^3\,d^6}{60\,e^4}+\frac {c^3\,d^3\,x^3}{3\,e}+\frac {c\,d\,x\,\left (6\,a^2\,e^4+3\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{10\,e^3}+\frac {c^2\,d^2\,x^2\,\left (c\,d^2+3\,a\,e^2\right )}{4\,e^2}}{d^6+6\,d^5\,e\,x+15\,d^4\,e^2\,x^2+20\,d^3\,e^3\,x^3+15\,d^2\,e^4\,x^4+6\,d\,e^5\,x^5+e^6\,x^6} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3/(d + e*x)^10,x)

[Out]

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