\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^2}{(d+e x)^5} \, dx\) [1846]

Optimal result

Integrand size = 35, antiderivative size = 71 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^5} \, dx=-\frac {\left (c d^2-a e^2\right )^2}{2 e^3 (d+e x)^2}+\frac {2 c d \left (c d^2-a e^2\right )}{e^3 (d+e x)}+\frac {c^2 d^2 \log (d+e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 71, 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} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^5} \, dx=\frac {2 c d \left (c d^2-a e^2\right )}{e^3 (d+e x)}-\frac {\left (c d^2-a e^2\right )^2}{2 e^3 (d+e x)^2}+\frac {c^2 d^2 \log (d+e x)}{e^3} \]

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^5,x]

[Out]

-1/2*(c*d^2 - a*e^2)^2/(e^3*(d + e*x)^2) + (2*c*d*(c*d^2 - a*e^2))/(e^3*(d + e*x)) + (c^2*d^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)^3} \, dx \\ & = \int \left (\frac {\left (-c d^2+a e^2\right )^2}{e^2 (d+e x)^3}-\frac {2 c d \left (c d^2-a e^2\right )}{e^2 (d+e x)^2}+\frac {c^2 d^2}{e^2 (d+e x)}\right ) \, dx \\ & = -\frac {\left (c d^2-a e^2\right )^2}{2 e^3 (d+e x)^2}+\frac {2 c d \left (c d^2-a e^2\right )}{e^3 (d+e x)}+\frac {c^2 d^2 \log (d+e x)}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^5} \, dx=\frac {\frac {\left (c d^2-a e^2\right ) \left (a e^2+c d (3 d+4 e x)\right )}{(d+e x)^2}+2 c^2 d^2 \log (d+e x)}{2 e^3} \]

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^5,x]

[Out]

(((c*d^2 - a*e^2)*(a*e^2 + c*d*(3*d + 4*e*x)))/(d + e*x)^2 + 2*c^2*d^2*Log[d + e*x])/(2*e^3)

Maple [A] (verified)

Time = 2.87 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.10

method	result	size
risch	\(\frac {-\frac {2 c d \left (e^{2} a -c \,d^{2}\right ) x}{e^{2}}-\frac {a^{2} e^{4}+2 a c \,d^{2} e^{2}-3 c^{2} d^{4}}{2 e^{3}}}{\left (e x +d \right )^{2}}+\frac {c^{2} d^{2} \ln \left (e x +d \right )}{e^{3}}\)	\(78\)
default	\(-\frac {2 c d \left (e^{2} a -c \,d^{2}\right )}{e^{3} \left (e x +d \right )}-\frac {a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}}{2 e^{3} \left (e x +d \right )^{2}}+\frac {c^{2} d^{2} \ln \left (e x +d \right )}{e^{3}}\)	\(81\)
parallelrisch	\(\frac {2 \ln \left (e x +d \right ) x^{2} c^{2} d^{2} e^{2}+4 \ln \left (e x +d \right ) x \,c^{2} d^{3} e +2 \ln \left (e x +d \right ) c^{2} d^{4}-4 x a c d \,e^{3}+4 x \,c^{2} d^{3} e -a^{2} e^{4}-2 a c \,d^{2} e^{2}+3 c^{2} d^{4}}{2 e^{3} \left (e x +d \right )^{2}}\)	\(109\)
norman	\(\frac {-\frac {d^{2} \left (a^{2} e^{5}+2 a \,d^{2} e^{3} c -3 d^{4} e \,c^{2}\right )}{2 e^{4}}-\frac {\left (a^{2} e^{5}+10 a \,d^{2} e^{3} c -11 d^{4} e \,c^{2}\right ) x^{2}}{2 e^{2}}-\frac {2 d \left (a c \,e^{3}-e \,c^{2} d^{2}\right ) x^{3}}{e}-\frac {d \left (a^{2} e^{5}+4 a \,d^{2} e^{3} c -5 d^{4} e \,c^{2}\right ) x}{e^{3}}}{\left (e x +d \right )^{4}}+\frac {c^{2} d^{2} \ln \left (e x +d \right )}{e^{3}}\)	\(156\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.58 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^5} \, dx=\frac {3 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - a^{2} e^{4} + 4 \, {\left (c^{2} d^{3} e - a c d e^{3}\right )} x + 2 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, c^{2} d^{3} e x + c^{2} d^{4}\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^5,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.27 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^{2} d^{2} \log {\left (d + e x \right )}}{e^{3}} + \frac {- a^{2} e^{4} - 2 a c d^{2} e^{2} + 3 c^{2} d^{4} + x \left (- 4 a c d e^{3} + 4 c^{2} d^{3} e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \]

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2/(e*x+d)**5,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.27 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^{2} d^{2} \log \left (e x + d\right )}{e^{3}} + \frac {3 \, c^{2} d^{4} - 2 \, a c d^{2} e^{2} - a^{2} e^{4} + 4 \, {\left (c^{2} d^{3} e - a c d e^{3}\right )} x}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^5,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.69 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^5} \, dx=-\frac {c^{2} d^{2} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{3}} + \frac {\frac {4 \, c^{2} d^{3} e^{3}}{e x + d} - \frac {c^{2} d^{4} e^{3}}{{\left (e x + d\right )}^{2}} - \frac {4 \, a c d e^{5}}{e x + d} + \frac {2 \, a c d^{2} e^{5}}{{\left (e x + d\right )}^{2}} - \frac {a^{2} e^{7}}{{\left (e x + d\right )}^{2}}}{2 \, e^{6}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^5,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^5} \, dx=\frac {c^2\,d^2\,\ln \left (d+e\,x\right )}{e^3}-\frac {\frac {a^2\,e^4+2\,a\,c\,d^2\,e^2-3\,c^2\,d^4}{2\,e^3}+\frac {2\,c\,d\,x\,\left (a\,e^2-c\,d^2\right )}{e^2}}{d^2+2\,d\,e\,x+e^2\,x^2} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2/(d + e*x)^5,x)

[Out]

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