\(\int \frac {d+e x}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\) [1881]

Optimal result

Integrand size = 33, antiderivative size = 75 \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x)}-\frac {e \log (a e+c d x)}{\left (c d^2-a e^2\right )^2}+\frac {e \log (d+e x)}{\left (c d^2-a e^2\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 75, 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 = {640, 46} \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x)}-\frac {e \log (a e+c d x)}{\left (c d^2-a e^2\right )^2}+\frac {e \log (d+e x)}{\left (c d^2-a e^2\right )^2} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {1}{\left (-c d^2+a e^2\right ) (a e+c d x)}-\frac {e \log (a e+c d x)}{\left (-c d^2+a e^2\right )^2}+\frac {e \log (d+e x)}{\left (-c d^2+a e^2\right )^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {1}{\left (e^{2} a -c \,d^{2}\right ) \left (c d x +a e \right )}-\frac {e \ln \left (c d x +a e \right )}{\left (e^{2} a -c \,d^{2}\right )^{2}}+\frac {e \ln \left (e x +d \right )}{\left (e^{2} a -c \,d^{2}\right )^{2}}\)	\(75\)
risch	\(\frac {1}{\left (e^{2} a -c \,d^{2}\right ) \left (c d x +a e \right )}-\frac {e \ln \left (c d x +a e \right )}{a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}+\frac {e \ln \left (-e x -d \right )}{a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}\)	\(104\)
parallelrisch	\(\frac {\ln \left (e x +d \right ) x \,c^{2} d^{2} e -\ln \left (c d x +a e \right ) x \,c^{2} d^{2} e +\ln \left (e x +d \right ) a c d \,e^{2}-\ln \left (c d x +a e \right ) a c d \,e^{2}+d \,e^{2} a c -c^{2} d^{3}}{\left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \left (c d x +a e \right ) c d}\)	\(125\)
norman	\(\frac {\frac {d}{e^{2} a -c \,d^{2}}+\frac {e x}{e^{2} a -c \,d^{2}}}{\left (c d x +a e \right ) \left (e x +d \right )}+\frac {e \ln \left (e x +d \right )}{a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}-\frac {e \ln \left (c d x +a e \right )}{a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}\)	\(128\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.55 \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {c d^{2} - a e^{2} + {\left (c d e x + a e^{2}\right )} \log \left (c d x + a e\right ) - {\left (c d e x + a e^{2}\right )} \log \left (e x + d\right )}{a c^{2} d^{4} e - 2 \, a^{2} c d^{2} e^{3} + a^{3} e^{5} + {\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} x} \]

[In]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (63) = 126\).

Time = 0.40 (sec) , antiderivative size = 287, normalized size of antiderivative = 3.83 \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {e \log {\left (x + \frac {- \frac {a^{3} e^{7}}{\left (a e^{2} - c d^{2}\right )^{2}} + \frac {3 a^{2} c d^{2} e^{5}}{\left (a e^{2} - c d^{2}\right )^{2}} - \frac {3 a c^{2} d^{4} e^{3}}{\left (a e^{2} - c d^{2}\right )^{2}} + a e^{3} + \frac {c^{3} d^{6} e}{\left (a e^{2} - c d^{2}\right )^{2}} + c d^{2} e}{2 c d e^{2}} \right )}}{\left (a e^{2} - c d^{2}\right )^{2}} - \frac {e \log {\left (x + \frac {\frac {a^{3} e^{7}}{\left (a e^{2} - c d^{2}\right )^{2}} - \frac {3 a^{2} c d^{2} e^{5}}{\left (a e^{2} - c d^{2}\right )^{2}} + \frac {3 a c^{2} d^{4} e^{3}}{\left (a e^{2} - c d^{2}\right )^{2}} + a e^{3} - \frac {c^{3} d^{6} e}{\left (a e^{2} - c d^{2}\right )^{2}} + c d^{2} e}{2 c d e^{2}} \right )}}{\left (a e^{2} - c d^{2}\right )^{2}} + \frac {1}{a^{2} e^{3} - a c d^{2} e + x \left (a c d e^{2} - c^{2} d^{3}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.51 \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {e \log \left (c d x + a e\right )}{c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}} + \frac {e \log \left (e x + d\right )}{c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}} - \frac {1}{a c d^{2} e - a^{2} e^{3} + {\left (c^{2} d^{3} - a c d e^{2}\right )} x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.49 \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {c d e \log \left ({\left | c d x + a e \right |}\right )}{c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}} + \frac {e^{2} \log \left ({\left | e x + d \right |}\right )}{c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}} - \frac {1}{{\left (c d^{2} - a e^{2}\right )} {\left (c d x + a e\right )}} \]

[In]

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

[Out]

-c*d*e*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) + e^2*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)*(c*d*x + a*e))

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.28 \[ \int \frac {d+e x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {1}{\left (a\,e+c\,d\,x\right )\,\left (a\,e^2-c\,d^2\right )}-\frac {2\,e\,\mathrm {atanh}\left (\frac {a^2\,e^4-c^2\,d^4}{{\left (a\,e^2-c\,d^2\right )}^2}+\frac {2\,c\,d\,e\,x}{a\,e^2-c\,d^2}\right )}{{\left (a\,e^2-c\,d^2\right )}^2} \]

[In]

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

[Out]

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