\(\int \frac {1}{(d+e x) (a d e+(c d^2+a e^2) x+c d e x^2)} \, dx\) [1871]

Optimal result

Integrand size = 35, antiderivative size = 73 \[ \int \frac {1}{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {1}{\left (c d^2-a e^2\right ) (d+e x)}+\frac {c d \log (a e+c d x)}{\left (c d^2-a e^2\right )^2}-\frac {c d \log (d+e x)}{\left (c d^2-a e^2\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 73, 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} \[ \int \frac {1}{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {1}{(d+e x) \left (c d^2-a e^2\right )}+\frac {c d \log (a e+c d x)}{\left (c d^2-a e^2\right )^2}-\frac {c d \log (d+e x)}{\left (c d^2-a e^2\right )^2} \]

[In]

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

[Out]

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

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)^2} \, dx \\ & = \int \left (\frac {c^2 d^2}{\left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac {e}{\left (c d^2-a e^2\right ) (d+e x)^2}-\frac {c d e}{\left (c d^2-a e^2\right )^2 (d+e x)}\right ) \, dx \\ & = \frac {1}{\left (c d^2-a e^2\right ) (d+e x)}+\frac {c d \log (a e+c d x)}{\left (c d^2-a e^2\right )^2}-\frac {c d \log (d+e x)}{\left (c d^2-a e^2\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {c d^2-a e^2+c d (d+e x) \log (a e+c d x)-c d (d+e x) \log (d+e x)}{\left (c d^2-a e^2\right )^2 (d+e x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.43 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {c d \ln \left (c d x +a e \right )}{\left (e^{2} a -c \,d^{2}\right )^{2}}-\frac {1}{\left (e^{2} a -c \,d^{2}\right ) \left (e x +d \right )}-\frac {c d \ln \left (e x +d \right )}{\left (e^{2} a -c \,d^{2}\right )^{2}}\)	\(75\)
risch	\(-\frac {1}{\left (e^{2} a -c \,d^{2}\right ) \left (e x +d \right )}+\frac {c d \ln \left (-c d x -a e \right )}{a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}-\frac {c d \ln \left (e x +d \right )}{a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}\)	\(103\)
norman	\(\frac {e x}{d \left (e^{2} a -c \,d^{2}\right ) \left (e x +d \right )}+\frac {c d \ln \left (c d x +a e \right )}{a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}-\frac {c d \ln \left (e x +d \right )}{a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}\)	\(105\)
parallelrisch	\(-\frac {\ln \left (e x +d \right ) x c d \,e^{2}-\ln \left (c d x +a e \right ) x c d \,e^{2}+\ln \left (e x +d \right ) c \,d^{2} e -\ln \left (c d x +a e \right ) c \,d^{2} e +a \,e^{3}-d^{2} e c}{\left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \left (e x +d \right ) e}\)	\(111\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.49 \[ \int \frac {1}{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {c d^{2} - a e^{2} + {\left (c d e x + c d^{2}\right )} \log \left (c d x + a e\right ) - {\left (c d e x + c d^{2}\right )} \log \left (e x + d\right )}{c^{2} d^{5} - 2 \, a c d^{3} e^{2} + a^{2} d e^{4} + {\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (63) = 126\).

Time = 0.38 (sec) , antiderivative size = 301, normalized size of antiderivative = 4.12 \[ \int \frac {1}{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=- \frac {c d \log {\left (x + \frac {- \frac {a^{3} c d e^{6}}{\left (a e^{2} - c d^{2}\right )^{2}} + \frac {3 a^{2} c^{2} d^{3} e^{4}}{\left (a e^{2} - c d^{2}\right )^{2}} - \frac {3 a c^{3} d^{5} e^{2}}{\left (a e^{2} - c d^{2}\right )^{2}} + a c d e^{2} + \frac {c^{4} d^{7}}{\left (a e^{2} - c d^{2}\right )^{2}} + c^{2} d^{3}}{2 c^{2} d^{2} e} \right )}}{\left (a e^{2} - c d^{2}\right )^{2}} + \frac {c d \log {\left (x + \frac {\frac {a^{3} c d e^{6}}{\left (a e^{2} - c d^{2}\right )^{2}} - \frac {3 a^{2} c^{2} d^{3} e^{4}}{\left (a e^{2} - c d^{2}\right )^{2}} + \frac {3 a c^{3} d^{5} e^{2}}{\left (a e^{2} - c d^{2}\right )^{2}} + a c d e^{2} - \frac {c^{4} d^{7}}{\left (a e^{2} - c d^{2}\right )^{2}} + c^{2} d^{3}}{2 c^{2} d^{2} e} \right )}}{\left (a e^{2} - c d^{2}\right )^{2}} - \frac {1}{a d e^{2} - c d^{3} + x \left (a e^{3} - c d^{2} e\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.47 \[ \int \frac {1}{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {c d \log \left (c d x + a e\right )}{c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}} - \frac {c d \log \left (e x + d\right )}{c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {1}{c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.52 \[ \int \frac {1}{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {c^{2} d^{2} \log \left ({\left | c d x + a e \right |}\right )}{c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}} - \frac {c d e \log \left ({\left | e x + d \right |}\right )}{c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}} + \frac {1}{{\left (c d^{2} - a e^{2}\right )} {\left (e x + d\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {2\,c\,d\,\mathrm {atanh}\left (\frac {a^2\,e^4-c^2\,d^4}{{\left (a\,e^2-c\,d^2\right )}^2}+\frac {2\,c\,d\,e\,x}{a\,e^2-c\,d^2}\right )}{{\left (a\,e^2-c\,d^2\right )}^2}-\frac {1}{\left (a\,e^2-c\,d^2\right )\,\left (d+e\,x\right )} \]

[In]

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

[Out]

(2*c*d*atanh((a^2*e^4 - c^2*d^4)/(a*e^2 - c*d^2)^2 + (2*c*d*e*x)/(a*e^2 - c*d^2)))/(a*e^2 - c*d^2)^2 - 1/((a*e
^2 - c*d^2)*(d + e*x))