\(\int \frac {(c d^2+2 c d e x+c e^2 x^2)^2}{(d+e x)^5} \, dx\) [994]

Optimal result

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

[Out]

c^2*ln(e*x+d)/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} \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^2 \log (d+e x)}{e} \]

[In]

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

[Out]

(c^2*Log[d + e*x])/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 {c^2}{d+e x} \, dx \\ & = c^2 \int \frac {1}{d+e x} \, dx \\ & = \frac {c^2 \log (d+e x)}{e} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

(c^2*Log[d + e*x])/e

Maple [A] (verified)

Time = 2.79 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {c^{2} \ln \left (e x +d \right )}{e}\)	\(14\)
norman	\(\frac {c^{2} \ln \left (e x +d \right )}{e}\)	\(14\)
risch	\(\frac {c^{2} \ln \left (e x +d \right )}{e}\)	\(14\)
parallelrisch	\(\frac {c^{2} \ln \left (e x +d \right )}{e}\)	\(14\)

[In]

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

[Out]

c^2*ln(e*x+d)/e

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^{2} \log \left (e x + d\right )}{e} \]

[In]

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

[Out]

c^2*log(e*x + d)/e

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^{2} \log {\left (d + e x \right )}}{e} \]

[In]

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

[Out]

c**2*log(d + e*x)/e

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^{2} \log \left (e x + d\right )}{e} \]

[In]

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

[Out]

c^2*log(e*x + d)/e

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (13) = 26\).

Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 2.08 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^5} \, dx=-\frac {c^{2} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e} \]

[In]

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

[Out]

-c^2*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^2\,\ln \left (d+e\,x\right )}{e} \]

[In]

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

[Out]

(c^2*log(d + e*x))/e