\(\int \frac {(d+e x)^5}{(c d^2+2 c d e x+c e^2 x^2)^3} \, dx\) [1021]

Optimal result

Integrand size = 30, antiderivative size = 13 \[ \int \frac {(d+e x)^5}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {\log (d+e x)}{c^3 e} \]

[Out]

ln(e*x+d)/c^3/e

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {27, 12, 31} \[ \int \frac {(d+e x)^5}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {\log (d+e x)}{c^3 e} \]

[In]

Int[(d + e*x)^5/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^3,x]

[Out]

Log[d + e*x]/(c^3*e)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{c^3 (d+e x)} \, dx \\ & = \frac {\int \frac {1}{d+e x} \, dx}{c^3} \\ & = \frac {\log (d+e x)}{c^3 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x)^5}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {\log (d+e x)}{c^3 e} \]

[In]

Integrate[(d + e*x)^5/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^3,x]

[Out]

Log[d + e*x]/(c^3*e)

Maple [A] (verified)

Time = 2.39 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {\ln \left (e x +d \right )}{c^{3} e}\)	\(14\)
norman	\(\frac {\ln \left (e x +d \right )}{c^{3} e}\)	\(14\)
risch	\(\frac {\ln \left (e x +d \right )}{c^{3} e}\)	\(14\)
parallelrisch	\(\frac {\ln \left (e x +d \right )}{c^{3} e}\)	\(14\)

[In]

int((e*x+d)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x,method=_RETURNVERBOSE)

[Out]

ln(e*x+d)/c^3/e

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x)^5}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {\log \left (e x + d\right )}{c^{3} e} \]

[In]

integrate((e*x+d)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x, algorithm="fricas")

[Out]

log(e*x + d)/(c^3*e)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {(d+e x)^5}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {\log {\left (c^{3} d + c^{3} e x \right )}}{c^{3} e} \]

[In]

integrate((e*x+d)**5/(c*e**2*x**2+2*c*d*e*x+c*d**2)**3,x)

[Out]

log(c**3*d + c**3*e*x)/(c**3*e)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x)^5}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {\log \left (e x + d\right )}{c^{3} e} \]

[In]

integrate((e*x+d)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x, algorithm="maxima")

[Out]

log(e*x + d)/(c^3*e)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08 \[ \int \frac {(d+e x)^5}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {\log \left ({\left | e x + d \right |}\right )}{c^{3} e} \]

[In]

integrate((e*x+d)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2)^3,x, algorithm="giac")

[Out]

log(abs(e*x + d))/(c^3*e)

Mupad [B] (verification not implemented)

Time = 9.56 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x)^5}{\left (c d^2+2 c d e x+c e^2 x^2\right )^3} \, dx=\frac {\ln \left (d+e\,x\right )}{c^3\,e} \]

[In]

int((d + e*x)^5/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^3,x)

[Out]

log(d + e*x)/(c^3*e)