\(\int \frac {(d+e x)^7}{(c d^2+2 c d e x+c e^2 x^2)^3} \, dx\) [1019]

Optimal result

Integrand size = 30, antiderivative size = 17 \[ \int \frac {(d+e x)^7}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {(d+e x)^2}{2 c^3 e} \]

[Out]

1/2*(e*x+d)^2/c^3/e

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {27, 9} \[ \int \frac {(d+e x)^7}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {(d+e x)^2}{2 c^3 e} \]

[In]

Int[(d + e*x)^7/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^3,x]

[Out]

(d + e*x)^2/(2*c^3*e)

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[a*((b + c*x)^2/(2*c)), x] /; FreeQ[{a, b, c}, x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {d+e x}{c^3} \, dx \\ & = \frac {(d+e x)^2}{2 c^3 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x)^7}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {d x+\frac {e x^2}{2}}{c^3} \]

[In]

Integrate[(d + e*x)^7/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^3,x]

[Out]

(d*x + (e*x^2)/2)/c^3

Maple [A] (verified)

Time = 2.91 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {x \left (e x +2 d \right )}{2 c^{3}}\)	\(14\)
default	\(\frac {\frac {1}{2} e \,x^{2}+d x}{c^{3}}\)	\(15\)
parallelrisch	\(\frac {e \,x^{2}+2 d x}{2 c^{3}}\)	\(16\)
risch	\(\frac {e \,x^{2}}{2 c^{3}}+\frac {d x}{c^{3}}\)	\(17\)
norman	\(\frac {-\frac {10 d^{7}}{e c}-\frac {49 d^{6} x}{c}+\frac {e^{6} x^{7}}{2 c}+\frac {7 d \,e^{5} x^{6}}{2 c}-\frac {35 e^{3} d^{3} x^{4}}{c}-\frac {175 d^{4} e^{2} x^{3}}{2 c}-\frac {189 d^{5} e \,x^{2}}{2 c}}{c^{2} \left (e x +d \right )^{5}}\)	\(96\)

[In]

int((e*x+d)^7/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x,method=_RETURNVERBOSE)

[Out]

1/2*x*(e*x+2*d)/c^3

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^7}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {e x^{2} + 2 \, d x}{2 \, c^{3}} \]

[In]

integrate((e*x+d)^7/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x, algorithm="fricas")

[Out]

1/2*(e*x^2 + 2*d*x)/c^3

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^7}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {d x}{c^{3}} + \frac {e x^{2}}{2 c^{3}} \]

[In]

integrate((e*x+d)**7/(c*e**2*x**2+2*c*d*e*x+c*d**2)**3,x)

[Out]

d*x/c**3 + e*x**2/(2*c**3)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^7}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {e x^{2} + 2 \, d x}{2 \, c^{3}} \]

[In]

integrate((e*x+d)^7/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x, algorithm="maxima")

[Out]

1/2*(e*x^2 + 2*d*x)/c^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^7}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {e x^{2} + 2 \, d x}{2 \, c^{3}} \]

[In]

integrate((e*x+d)^7/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x, algorithm="giac")

[Out]

1/2*(e*x^2 + 2*d*x)/c^3

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {(d+e x)^7}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {x\,\left (2\,d+e\,x\right )}{2\,c^3} \]

[In]

int((d + e*x)^7/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^3,x)

[Out]

(x*(2*d + e*x))/(2*c^3)