\(\int \frac {1}{(d+e x)^2 (a d e+(c d^2+a e^2) x+c d e x^2)} \, dx\) [1872]

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

Optimal result

Integrand size = 35, antiderivative size = 108 \[ \int \frac {1}{(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {1}{2 \left (c d^2-a e^2\right ) (d+e x)^2}+\frac {c d}{\left (c d^2-a e^2\right )^2 (d+e x)}+\frac {c^2 d^2 \log (a e+c d x)}{\left (c d^2-a e^2\right )^3}-\frac {c^2 d^2 \log (d+e x)}{\left (c d^2-a e^2\right )^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 108, 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, 46} \[ \int \frac {1}{(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {c^2 d^2 \log (a e+c d x)}{\left (c d^2-a e^2\right )^3}-\frac {c^2 d^2 \log (d+e x)}{\left (c d^2-a e^2\right )^3}+\frac {c d}{(d+e x) \left (c d^2-a e^2\right )^2}+\frac {1}{2 (d+e x)^2 \left (c d^2-a e^2\right )} \]

[In]

Int[1/((d + e*x)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)),x]

[Out]

1/(2*(c*d^2 - a*e^2)*(d + e*x)^2) + (c*d)/((c*d^2 - a*e^2)^2*(d + e*x)) + (c^2*d^2*Log[a*e + c*d*x])/(c*d^2 -
a*e^2)^3 - (c^2*d^2*Log[d + e*x])/(c*d^2 - a*e^2)^3

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[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 {1}{(a e+c d x) (d+e x)^3} \, dx \\ & = \int \left (\frac {c^3 d^3}{\left (c d^2-a e^2\right )^3 (a e+c d x)}-\frac {e}{\left (c d^2-a e^2\right ) (d+e x)^3}-\frac {c d e}{\left (c d^2-a e^2\right )^2 (d+e x)^2}-\frac {c^2 d^2 e}{\left (c d^2-a e^2\right )^3 (d+e x)}\right ) \, dx \\ & = \frac {1}{2 \left (c d^2-a e^2\right ) (d+e x)^2}+\frac {c d}{\left (c d^2-a e^2\right )^2 (d+e x)}+\frac {c^2 d^2 \log (a e+c d x)}{\left (c d^2-a e^2\right )^3}-\frac {c^2 d^2 \log (d+e x)}{\left (c d^2-a e^2\right )^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {\left (c d^2-a e^2\right ) \left (-a e^2+c d (3 d+2 e x)\right )+2 c^2 d^2 (d+e x)^2 \log (a e+c d x)-2 c^2 d^2 (d+e x)^2 \log (d+e x)}{2 \left (c d^2-a e^2\right )^3 (d+e x)^2} \]

[In]

Integrate[1/((d + e*x)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)),x]

[Out]

((c*d^2 - a*e^2)*(-(a*e^2) + c*d*(3*d + 2*e*x)) + 2*c^2*d^2*(d + e*x)^2*Log[a*e + c*d*x] - 2*c^2*d^2*(d + e*x)
^2*Log[d + e*x])/(2*(c*d^2 - a*e^2)^3*(d + e*x)^2)

Maple [A] (verified)

Time = 2.59 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.99

method result size
default \(-\frac {c^{2} d^{2} \ln \left (c d x +a e \right )}{\left (e^{2} a -c \,d^{2}\right )^{3}}-\frac {1}{2 \left (e^{2} a -c \,d^{2}\right ) \left (e x +d \right )^{2}}+\frac {c^{2} d^{2} \ln \left (e x +d \right )}{\left (e^{2} a -c \,d^{2}\right )^{3}}+\frac {c d}{\left (e^{2} a -c \,d^{2}\right )^{2} \left (e x +d \right )}\) \(107\)
risch \(\frac {\frac {c d e x}{a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}-\frac {e^{2} a -3 c \,d^{2}}{2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}}{\left (e x +d \right )^{2}}+\frac {c^{2} d^{2} \ln \left (-e x -d \right )}{e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}}-\frac {c^{2} d^{2} \ln \left (c d x +a e \right )}{e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}}\) \(201\)
norman \(\frac {\frac {-a \,e^{3}+2 d^{2} e c}{2 e \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}-\frac {c \,e^{2} x^{2}}{2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}}{\left (e x +d \right )^{2}}+\frac {c^{2} d^{2} \ln \left (e x +d \right )}{e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}}-\frac {c^{2} d^{2} \ln \left (c d x +a e \right )}{e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}}\) \(207\)
parallelrisch \(\frac {2 \ln \left (e x +d \right ) x^{2} c^{2} d^{2} e^{4}-2 \ln \left (c d x +a e \right ) x^{2} c^{2} d^{2} e^{4}+4 \ln \left (e x +d \right ) x \,c^{2} d^{3} e^{3}-4 \ln \left (c d x +a e \right ) x \,c^{2} d^{3} e^{3}+2 \ln \left (e x +d \right ) c^{2} d^{4} e^{2}-2 \ln \left (c d x +a e \right ) c^{2} d^{4} e^{2}+2 x a c d \,e^{5}-2 x \,c^{2} d^{3} e^{3}-a^{2} e^{6}+4 a c \,d^{2} e^{4}-3 c^{2} d^{4} e^{2}}{2 \left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}\right ) \left (e x +d \right )^{2} e^{2}}\) \(225\)

[In]

int(1/(e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (106) = 212\).

Time = 0.37 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.46 \[ \int \frac {1}{(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {3 \, c^{2} d^{4} - 4 \, a c d^{2} e^{2} + a^{2} e^{4} + 2 \, {\left (c^{2} d^{3} e - a c d e^{3}\right )} x + 2 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, c^{2} d^{3} e x + c^{2} d^{4}\right )} \log \left (c d x + a e\right ) - 2 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, c^{2} d^{3} e x + c^{2} d^{4}\right )} \log \left (e x + d\right )}{2 \, {\left (c^{3} d^{8} - 3 \, a c^{2} d^{6} e^{2} + 3 \, a^{2} c d^{4} e^{4} - a^{3} d^{2} e^{6} + {\left (c^{3} d^{6} e^{2} - 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} - a^{3} e^{8}\right )} x^{2} + 2 \, {\left (c^{3} d^{7} e - 3 \, a c^{2} d^{5} e^{3} + 3 \, a^{2} c d^{3} e^{5} - a^{3} d e^{7}\right )} x\right )}} \]

[In]

integrate(1/(e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="fricas")

[Out]

1/2*(3*c^2*d^4 - 4*a*c*d^2*e^2 + a^2*e^4 + 2*(c^2*d^3*e - a*c*d*e^3)*x + 2*(c^2*d^2*e^2*x^2 + 2*c^2*d^3*e*x +
c^2*d^4)*log(c*d*x + a*e) - 2*(c^2*d^2*e^2*x^2 + 2*c^2*d^3*e*x + c^2*d^4)*log(e*x + d))/(c^3*d^8 - 3*a*c^2*d^6
*e^2 + 3*a^2*c*d^4*e^4 - a^3*d^2*e^6 + (c^3*d^6*e^2 - 3*a*c^2*d^4*e^4 + 3*a^2*c*d^2*e^6 - a^3*e^8)*x^2 + 2*(c^
3*d^7*e - 3*a*c^2*d^5*e^3 + 3*a^2*c*d^3*e^5 - a^3*d*e^7)*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 471 vs. \(2 (94) = 188\).

Time = 0.62 (sec) , antiderivative size = 471, normalized size of antiderivative = 4.36 \[ \int \frac {1}{(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {c^{2} d^{2} \log {\left (x + \frac {- \frac {a^{4} c^{2} d^{2} e^{8}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac {4 a^{3} c^{3} d^{4} e^{6}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac {6 a^{2} c^{4} d^{6} e^{4}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac {4 a c^{5} d^{8} e^{2}}{\left (a e^{2} - c d^{2}\right )^{3}} + a c^{2} d^{2} e^{2} - \frac {c^{6} d^{10}}{\left (a e^{2} - c d^{2}\right )^{3}} + c^{3} d^{4}}{2 c^{3} d^{3} e} \right )}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac {c^{2} d^{2} \log {\left (x + \frac {\frac {a^{4} c^{2} d^{2} e^{8}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac {4 a^{3} c^{3} d^{4} e^{6}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac {6 a^{2} c^{4} d^{6} e^{4}}{\left (a e^{2} - c d^{2}\right )^{3}} - \frac {4 a c^{5} d^{8} e^{2}}{\left (a e^{2} - c d^{2}\right )^{3}} + a c^{2} d^{2} e^{2} + \frac {c^{6} d^{10}}{\left (a e^{2} - c d^{2}\right )^{3}} + c^{3} d^{4}}{2 c^{3} d^{3} e} \right )}}{\left (a e^{2} - c d^{2}\right )^{3}} + \frac {- a e^{2} + 3 c d^{2} + 2 c d e x}{2 a^{2} d^{2} e^{4} - 4 a c d^{4} e^{2} + 2 c^{2} d^{6} + x^{2} \cdot \left (2 a^{2} e^{6} - 4 a c d^{2} e^{4} + 2 c^{2} d^{4} e^{2}\right ) + x \left (4 a^{2} d e^{5} - 8 a c d^{3} e^{3} + 4 c^{2} d^{5} e\right )} \]

[In]

integrate(1/(e*x+d)**2/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2),x)

[Out]

c**2*d**2*log(x + (-a**4*c**2*d**2*e**8/(a*e**2 - c*d**2)**3 + 4*a**3*c**3*d**4*e**6/(a*e**2 - c*d**2)**3 - 6*
a**2*c**4*d**6*e**4/(a*e**2 - c*d**2)**3 + 4*a*c**5*d**8*e**2/(a*e**2 - c*d**2)**3 + a*c**2*d**2*e**2 - c**6*d
**10/(a*e**2 - c*d**2)**3 + c**3*d**4)/(2*c**3*d**3*e))/(a*e**2 - c*d**2)**3 - c**2*d**2*log(x + (a**4*c**2*d*
*2*e**8/(a*e**2 - c*d**2)**3 - 4*a**3*c**3*d**4*e**6/(a*e**2 - c*d**2)**3 + 6*a**2*c**4*d**6*e**4/(a*e**2 - c*
d**2)**3 - 4*a*c**5*d**8*e**2/(a*e**2 - c*d**2)**3 + a*c**2*d**2*e**2 + c**6*d**10/(a*e**2 - c*d**2)**3 + c**3
*d**4)/(2*c**3*d**3*e))/(a*e**2 - c*d**2)**3 + (-a*e**2 + 3*c*d**2 + 2*c*d*e*x)/(2*a**2*d**2*e**4 - 4*a*c*d**4
*e**2 + 2*c**2*d**6 + x**2*(2*a**2*e**6 - 4*a*c*d**2*e**4 + 2*c**2*d**4*e**2) + x*(4*a**2*d*e**5 - 8*a*c*d**3*
e**3 + 4*c**2*d**5*e))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (106) = 212\).

Time = 0.21 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.11 \[ \int \frac {1}{(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {c^{2} d^{2} \log \left (c d x + a e\right )}{c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}} - \frac {c^{2} d^{2} \log \left (e x + d\right )}{c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}} + \frac {2 \, c d e x + 3 \, c d^{2} - a e^{2}}{2 \, {\left (c^{2} d^{6} - 2 \, a c d^{4} e^{2} + a^{2} d^{2} e^{4} + {\left (c^{2} d^{4} e^{2} - 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )} x^{2} + 2 \, {\left (c^{2} d^{5} e - 2 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )} x\right )}} \]

[In]

integrate(1/(e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="maxima")

[Out]

c^2*d^2*log(c*d*x + a*e)/(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6) - c^2*d^2*log(e*x + d)/(c^3*d
^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6) + 1/2*(2*c*d*e*x + 3*c*d^2 - a*e^2)/(c^2*d^6 - 2*a*c*d^4*e^2
 + a^2*d^2*e^4 + (c^2*d^4*e^2 - 2*a*c*d^2*e^4 + a^2*e^6)*x^2 + 2*(c^2*d^5*e - 2*a*c*d^3*e^3 + a^2*d*e^5)*x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.46 \[ \int \frac {1}{(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {c^{2} d^{2} e \log \left ({\left | c d - \frac {c d^{2}}{e x + d} + \frac {a e^{2}}{e x + d} \right |}\right )}{c^{3} d^{6} e - 3 \, a c^{2} d^{4} e^{3} + 3 \, a^{2} c d^{2} e^{5} - a^{3} e^{7}} + \frac {\frac {2 \, c d e^{2}}{e x + d} + \frac {c d^{2} e^{2}}{{\left (e x + d\right )}^{2}} - \frac {a e^{4}}{{\left (e x + d\right )}^{2}}}{2 \, {\left (c^{2} d^{4} e^{2} - 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )}} \]

[In]

integrate(1/(e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="giac")

[Out]

c^2*d^2*e*log(abs(c*d - c*d^2/(e*x + d) + a*e^2/(e*x + d)))/(c^3*d^6*e - 3*a*c^2*d^4*e^3 + 3*a^2*c*d^2*e^5 - a
^3*e^7) + 1/2*(2*c*d*e^2/(e*x + d) + c*d^2*e^2/(e*x + d)^2 - a*e^4/(e*x + d)^2)/(c^2*d^4*e^2 - 2*a*c*d^2*e^4 +
 a^2*e^6)

Mupad [B] (verification not implemented)

Time = 9.76 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.04 \[ \int \frac {1}{(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=-\frac {\frac {a\,e^2-3\,c\,d^2}{2\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}-\frac {c\,d\,e\,x}{a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4}}{d^2+2\,d\,e\,x+e^2\,x^2}-\frac {2\,c^2\,d^2\,\mathrm {atanh}\left (\frac {a^3\,e^6-a^2\,c\,d^2\,e^4-a\,c^2\,d^4\,e^2+c^3\,d^6}{{\left (a\,e^2-c\,d^2\right )}^3}+\frac {2\,c\,d\,e\,x\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{{\left (a\,e^2-c\,d^2\right )}^3}\right )}{{\left (a\,e^2-c\,d^2\right )}^3} \]

[In]

int(1/((d + e*x)^2*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)),x)

[Out]

- ((a*e^2 - 3*c*d^2)/(2*(a^2*e^4 + c^2*d^4 - 2*a*c*d^2*e^2)) - (c*d*e*x)/(a^2*e^4 + c^2*d^4 - 2*a*c*d^2*e^2))/
(d^2 + e^2*x^2 + 2*d*e*x) - (2*c^2*d^2*atanh((a^3*e^6 + c^3*d^6 - a*c^2*d^4*e^2 - a^2*c*d^2*e^4)/(a*e^2 - c*d^
2)^3 + (2*c*d*e*x*(a^2*e^4 + c^2*d^4 - 2*a*c*d^2*e^2))/(a*e^2 - c*d^2)^3))/(a*e^2 - c*d^2)^3