\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^2}{(d+e x)^4} \, dx\) [1845]

Optimal result

Integrand size = 35, antiderivative size = 63 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^4} \, dx=\frac {c^2 d^2 x}{e^2}-\frac {\left (c d^2-a e^2\right )^2}{e^3 (d+e x)}-\frac {2 c d \left (c d^2-a e^2\right ) \log (d+e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, 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} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^4} \, dx=-\frac {\left (c d^2-a e^2\right )^2}{e^3 (d+e x)}-\frac {2 c d \left (c d^2-a e^2\right ) \log (d+e x)}{e^3}+\frac {c^2 d^2 x}{e^2} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^4} \, dx=\frac {c^2 d^2 e x-\frac {\left (c d^2-a e^2\right )^2}{d+e x}+2 c d \left (-c d^2+a e^2\right ) \log (d+e x)}{e^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {c^{2} d^{2} x}{e^{2}}-\frac {a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}{e^{3} \left (e x +d \right )}+\frac {2 c d \left (e^{2} a -c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{3}}\)	\(75\)
risch	\(\frac {c^{2} d^{2} x}{e^{2}}-\frac {e \,a^{2}}{e x +d}+\frac {2 a c \,d^{2}}{e \left (e x +d \right )}-\frac {c^{2} d^{4}}{e^{3} \left (e x +d \right )}+\frac {2 c d \ln \left (e x +d \right ) a}{e}-\frac {2 c^{2} d^{3} \ln \left (e x +d \right )}{e^{3}}\)	\(92\)
parallelrisch	\(\frac {2 \ln \left (e x +d \right ) x a c d \,e^{3}-2 \ln \left (e x +d \right ) x \,c^{2} d^{3} e +x^{2} c^{2} d^{2} e^{2}+2 \ln \left (e x +d \right ) a c \,d^{2} e^{2}-2 \ln \left (e x +d \right ) c^{2} d^{4}-a^{2} e^{4}+2 a c \,d^{2} e^{2}-2 c^{2} d^{4}}{e^{3} \left (e x +d \right )}\)	\(113\)
norman	\(\frac {e \,c^{2} d^{2} x^{4}-\frac {d^{2} \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+4 c^{2} d^{4}\right )}{e^{3}}-\frac {\left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+7 c^{2} d^{4}\right ) x^{2}}{e}-\frac {d \left (2 a^{2} e^{4}-4 a c \,d^{2} e^{2}+10 c^{2} d^{4}\right ) x}{e^{2}}}{\left (e x +d \right )^{3}}+\frac {2 c d \left (e^{2} a -c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{3}}\)	\(149\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.71 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^4} \, dx=\frac {c^{2} d^{2} e^{2} x^{2} + c^{2} d^{3} e x - c^{2} d^{4} + 2 \, a c d^{2} e^{2} - a^{2} e^{4} - 2 \, {\left (c^{2} d^{4} - a c d^{2} e^{2} + {\left (c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \log \left (e x + d\right )}{e^{4} x + d e^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^4} \, dx=\frac {c^{2} d^{2} x}{e^{2}} + \frac {2 c d \left (a e^{2} - c d^{2}\right ) \log {\left (d + e x \right )}}{e^{3}} + \frac {- a^{2} e^{4} + 2 a c d^{2} e^{2} - c^{2} d^{4}}{d e^{3} + e^{4} x} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^4} \, dx=\frac {c^{2} d^{2} x}{e^{2}} - \frac {c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}}{e^{4} x + d e^{3}} - \frac {2 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} \log \left (e x + d\right )}{e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^4} \, dx=\frac {c^{2} d^{2} x}{e^{2}} - \frac {2 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{3}} - \frac {c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}}{{\left (e x + d\right )} e^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^4} \, dx=\frac {c^2\,d^2\,x}{e^2}-\frac {a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4}{e\,\left (x\,e^3+d\,e^2\right )}-\frac {\ln \left (d+e\,x\right )\,\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )}{e^3} \]

[In]

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

[Out]

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