\(\int \frac {(d+e x)^4}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\) [1878]

Optimal result

Integrand size = 35, antiderivative size = 74 \[ \int \frac {(d+e x)^4}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {e^2 x}{c^2 d^2}-\frac {\left (c d^2-a e^2\right )^2}{c^3 d^3 (a e+c d x)}+\frac {2 e \left (c d^2-a e^2\right ) \log (a e+c d x)}{c^3 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 74, 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} \[ \int \frac {(d+e x)^4}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {\left (c d^2-a e^2\right )^2}{c^3 d^3 (a e+c d x)}+\frac {2 e \left (c d^2-a e^2\right ) \log (a e+c d x)}{c^3 d^3}+\frac {e^2 x}{c^2 d^2} \]

[In]

Int[(d + e*x)^4/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^4}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {c d e^2 x-\frac {\left (c d^2-a e^2\right )^2}{a e+c d x}+2 \left (c d^2 e-a e^3\right ) \log (a e+c d x)}{c^3 d^3} \]

[In]

Integrate[(d + e*x)^4/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

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

Maple [A] (verified)

Time = 2.41 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16

method	result	size
default	\(\frac {e^{2} x}{c^{2} d^{2}}-\frac {2 e \left (e^{2} a -c \,d^{2}\right ) \ln \left (c d x +a e \right )}{c^{3} d^{3}}-\frac {a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}{c^{3} d^{3} \left (c d x +a e \right )}\)	\(86\)
risch	\(\frac {e^{2} x}{c^{2} d^{2}}-\frac {2 e^{3} \ln \left (c d x +a e \right ) a}{c^{3} d^{3}}+\frac {2 e \ln \left (c d x +a e \right )}{c^{2} d}-\frac {a^{2} e^{4}}{c^{3} d^{3} \left (c d x +a e \right )}+\frac {2 a \,e^{2}}{c^{2} d \left (c d x +a e \right )}-\frac {d}{c \left (c d x +a e \right )}\)	\(114\)
parallelrisch	\(-\frac {2 \ln \left (c d x +a e \right ) x a c d \,e^{3}-2 \ln \left (c d x +a e \right ) x \,c^{2} d^{3} e -x^{2} c^{2} d^{2} e^{2}+2 \ln \left (c d x +a e \right ) a^{2} e^{4}-2 \ln \left (c d x +a e \right ) a c \,d^{2} e^{2}+2 a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}{c^{3} d^{3} \left (c d x +a e \right )}\)	\(132\)
norman	\(\frac {\frac {e^{3} x^{3}}{c d}-\frac {2 a^{2} e^{4}-a c \,d^{2} e^{2}+c^{2} d^{4}}{d^{2} c^{3}}-\frac {\left (2 a^{2} e^{6}-a c \,d^{2} e^{4}+2 c^{2} d^{4} e^{2}\right ) x}{c^{3} d^{3} e}}{\left (c d x +a e \right ) \left (e x +d \right )}-\frac {2 e \left (e^{2} a -c \,d^{2}\right ) \ln \left (c d x +a e \right )}{c^{3} d^{3}}\)	\(140\)

[In]

int((e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.58 \[ \int \frac {(d+e x)^4}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {c^{2} d^{2} e^{2} x^{2} + a c d e^{3} x - c^{2} d^{4} + 2 \, a c d^{2} e^{2} - a^{2} e^{4} + 2 \, {\left (a c d^{2} e^{2} - a^{2} e^{4} + {\left (c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \log \left (c d x + a e\right )}{c^{4} d^{4} x + a c^{3} d^{3} e} \]

[In]

integrate((e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x)^4}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {- a^{2} e^{4} + 2 a c d^{2} e^{2} - c^{2} d^{4}}{a c^{3} d^{3} e + c^{4} d^{4} x} + \frac {e^{2} x}{c^{2} d^{2}} - \frac {2 e \left (a e^{2} - c d^{2}\right ) \log {\left (a e + c d x \right )}}{c^{3} d^{3}} \]

[In]

integrate((e*x+d)**4/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.20 \[ \int \frac {(d+e x)^4}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}}{c^{4} d^{4} x + a c^{3} d^{3} e} + \frac {e^{2} x}{c^{2} d^{2}} + \frac {2 \, {\left (c d^{2} e - a e^{3}\right )} \log \left (c d x + a e\right )}{c^{3} d^{3}} \]

[In]

integrate((e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16 \[ \int \frac {(d+e x)^4}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {e^{2} x}{c^{2} d^{2}} + \frac {2 \, {\left (c d^{2} e - a e^{3}\right )} \log \left ({\left | c d x + a e \right |}\right )}{c^{3} d^{3}} - \frac {c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}}{{\left (c d x + a e\right )} c^{3} d^{3}} \]

[In]

integrate((e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.30 \[ \int \frac {(d+e x)^4}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {e^2\,x}{c^2\,d^2}-\frac {\ln \left (a\,e+c\,d\,x\right )\,\left (2\,a\,e^3-2\,c\,d^2\,e\right )}{c^3\,d^3}-\frac {a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4}{c\,d\,\left (x\,c^3\,d^3+a\,e\,c^2\,d^2\right )} \]

[In]

int((d + e*x)^4/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2,x)

[Out]

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