\(\int \frac {a d e+(c d^2+a e^2) x+c d e x^2}{(d+e x)^2} \, dx\) [1834]

Optimal result

Integrand size = 33, antiderivative size = 26 \[ \int \frac {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}{(d+e x)^2} \, dx=\frac {c d x}{e}+\left (a-\frac {c d^2}{e^2}\right ) \log (d+e x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {24, 45} \[ \int \frac {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}{(d+e x)^2} \, dx=\left (a-\frac {c d^2}{e^2}\right ) \log (d+e x)+\frac {c d x}{e} \]

[In]

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

[Out]

(c*d*x)/e + (a - (c*d^2)/e^2)*Log[d + e*x]

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

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])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}{(d+e x)^2} \, dx=\frac {c d x}{e}+\frac {\left (-c d^2+a e^2\right ) \log (d+e x)}{e^2} \]

[In]

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

[Out]

(c*d*x)/e + ((-(c*d^2) + a*e^2)*Log[d + e*x])/e^2

Maple [A] (verified)

Time = 2.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {c d x}{e}+\frac {\left (e^{2} a -c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{2}}\)	\(31\)
risch	\(\frac {c d x}{e}+\ln \left (e x +d \right ) a -\frac {\ln \left (e x +d \right ) c \,d^{2}}{e^{2}}\)	\(32\)
parallelrisch	\(\frac {\ln \left (e x +d \right ) a \,e^{2}-\ln \left (e x +d \right ) c \,d^{2}+x c d e}{e^{2}}\)	\(34\)
norman	\(\frac {c d \,x^{2}-\frac {d^{3} c}{e^{2}}}{e x +d}+\frac {\left (e^{2} a -c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{2}}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}{(d+e x)^2} \, dx=\frac {c d e x - {\left (c d^{2} - a e^{2}\right )} \log \left (e x + d\right )}{e^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}{(d+e x)^2} \, dx=\frac {c d x}{e} + \frac {\left (a e^{2} - c d^{2}\right ) \log {\left (d + e x \right )}}{e^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}{(d+e x)^2} \, dx=\frac {c d x}{e} - \frac {{\left (c d^{2} - a e^{2}\right )} \log \left (e x + d\right )}{e^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (26) = 52\).

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 4.58 \[ \int \frac {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}{(d+e x)^2} \, dx=c d e {\left (\frac {2 \, d \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{3}} + \frac {e x + d}{e^{3}} - \frac {d^{2}}{{\left (e x + d\right )} e^{3}}\right )} - \frac {a d}{e x + d} - \frac {{\left (c d^{2} + a e^{2}\right )} {\left (\frac {\log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e} - \frac {d}{{\left (e x + d\right )} e}\right )}}{e} \]

[In]

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

[Out]

c*d*e*(2*d*log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e^3 + (e*x + d)/e^3 - d^2/((e*x + d)*e^3)) - a*d/(e*x + d) -
(c*d^2 + a*e^2)*(log(abs(e*x + d)/((e*x + d)^2*abs(e)))/e - d/((e*x + d)*e))/e

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}{(d+e x)^2} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (a\,e^2-c\,d^2\right )}{e^2}+\frac {c\,d\,x}{e} \]

[In]

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

[Out]

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