\(\int \frac {(d+e x)^3}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\) [1879]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 35, antiderivative size = 48 \[ \int \frac {(d+e x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {c d^2-a e^2}{c^2 d^2 (a e+c d x)}+\frac {e \log (a e+c d x)}{c^2 d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, 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} \[ \int \frac {(d+e x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {e \log (a e+c d x)}{c^2 d^2}-\frac {c d^2-a e^2}{c^2 d^2 (a e+c d x)} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {-c d^2+a e^2}{c^2 d^2 (a e+c d x)}+\frac {e \log (a e+c d x)}{c^2 d^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.41 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.02

method result size
default \(\frac {e \ln \left (c d x +a e \right )}{c^{2} d^{2}}-\frac {-e^{2} a +c \,d^{2}}{c^{2} d^{2} \left (c d x +a e \right )}\) \(49\)
risch \(\frac {e^{2} a}{c^{2} d^{2} \left (c d x +a e \right )}-\frac {1}{c \left (c d x +a e \right )}+\frac {e \ln \left (c d x +a e \right )}{c^{2} d^{2}}\) \(55\)
parallelrisch \(\frac {\ln \left (c d x +a e \right ) x c d e +\ln \left (c d x +a e \right ) a \,e^{2}+e^{2} a -c \,d^{2}}{c^{2} d^{2} \left (c d x +a e \right )}\) \(58\)
norman \(\frac {\frac {e^{2} a -c \,d^{2}}{c^{2} d}+\frac {\left (e^{4} a -d^{2} e^{2} c \right ) x}{c^{2} d^{2} e}}{\left (c d x +a e \right ) \left (e x +d \right )}+\frac {e \ln \left (c d x +a e \right )}{c^{2} d^{2}}\) \(83\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.17 \[ \int \frac {(d+e x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {c d^{2} - a e^{2} - {\left (c d e x + a e^{2}\right )} \log \left (c d x + a e\right )}{c^{3} d^{3} x + a c^{2} d^{2} e} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96 \[ \int \frac {(d+e x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {a e^{2} - c d^{2}}{a c^{2} d^{2} e + c^{3} d^{3} x} + \frac {e \log {\left (a e + c d x \right )}}{c^{2} d^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08 \[ \int \frac {(d+e x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {c d^{2} - a e^{2}}{c^{3} d^{3} x + a c^{2} d^{2} e} + \frac {e \log \left (c d x + a e\right )}{c^{2} d^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.02 \[ \int \frac {(d+e x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {e \log \left ({\left | c d x + a e \right |}\right )}{c^{2} d^{2}} - \frac {c d^{2} - a e^{2}}{{\left (c d x + a e\right )} c^{2} d^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {a\,e^2-c\,d^2}{c^2\,d^2\,\left (a\,e+c\,d\,x\right )}+\frac {e\,\ln \left (a\,e+c\,d\,x\right )}{c^2\,d^2} \]

[In]

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

[Out]

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