\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^2}{(d+e x)^3} \, dx\) [1844]

Optimal result

Integrand size = 35, antiderivative size = 62 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^3} \, dx=-\frac {c d \left (c d^2-a e^2\right ) x}{e^2}+\frac {(a e+c d x)^2}{2 e}+\frac {\left (c d^2-a e^2\right )^2 \log (d+e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, 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} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^3} \, dx=-\frac {c d x \left (c d^2-a e^2\right )}{e^2}+\frac {\left (c d^2-a e^2\right )^2 \log (d+e x)}{e^3}+\frac {(a e+c d x)^2}{2 e} \]

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^3} \, dx=\frac {c d e x \left (4 a e^2+c d (-2 d+e x)\right )+2 \left (c d^2-a e^2\right )^2 \log (d+e x)}{2 e^3} \]

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^3,x]

[Out]

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

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.06

method	result	size
default	\(\frac {c d \left (\frac {1}{2} c d e \,x^{2}+2 a \,e^{2} x -c \,d^{2} x \right )}{e^{2}}+\frac {\left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \ln \left (e x +d \right )}{e^{3}}\)	\(66\)
risch	\(\frac {c^{2} d^{2} x^{2}}{2 e}+2 a d x c -\frac {c^{2} d^{3} x}{e^{2}}+e \ln \left (e x +d \right ) a^{2}-\frac {2 \ln \left (e x +d \right ) a c \,d^{2}}{e}+\frac {\ln \left (e x +d \right ) c^{2} d^{4}}{e^{3}}\)	\(77\)
parallelrisch	\(\frac {x^{2} c^{2} d^{2} e^{2}+2 \ln \left (e x +d \right ) a^{2} e^{4}-4 \ln \left (e x +d \right ) a c \,d^{2} e^{2}+2 \ln \left (e x +d \right ) c^{2} d^{4}+4 x a c d \,e^{3}-2 x \,c^{2} d^{3} e}{2 e^{3}}\)	\(83\)
norman	\(\frac {-\frac {d^{2} \left (8 a c \,d^{2} e^{2}-3 c^{2} d^{4}\right )}{2 e^{3}}-\frac {2 d \left (3 a c \,d^{2} e^{2}-c^{2} d^{4}\right ) x}{e^{2}}+\frac {e \,c^{2} d^{2} x^{4}}{2}+2 x^{3} d \,e^{2} a c}{\left (e x +d \right )^{2}}+\frac {\left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \ln \left (e x +d \right )}{e^{3}}\)	\(122\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^3} \, dx=\frac {c^{2} d^{2} e^{2} x^{2} - 2 \, {\left (c^{2} d^{3} e - 2 \, a c d e^{3}\right )} x + 2 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \log \left (e x + d\right )}{2 \, e^{3}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^3} \, dx=\frac {c^{2} d^{2} x^{2}}{2 e} + x \left (2 a c d - \frac {c^{2} d^{3}}{e^{2}}\right ) + \frac {\left (a e^{2} - c d^{2}\right )^{2} \log {\left (d + e x \right )}}{e^{3}} \]

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2/(e*x+d)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^3} \, dx=\frac {c^{2} d^{2} e x^{2} - 2 \, {\left (c^{2} d^{3} - 2 \, a c d e^{2}\right )} x}{2 \, e^{2}} + \frac {{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \log \left (e x + d\right )}{e^{3}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^3} \, dx=\frac {c^{2} d^{2} e x^{2} - 2 \, c^{2} d^{3} x + 4 \, a c d e^{2} x}{2 \, e^{2}} + \frac {{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{3}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2/(e*x+d)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^3} \, dx=x\,\left (2\,a\,c\,d-\frac {c^2\,d^3}{e^2}\right )+\frac {\ln \left (d+e\,x\right )\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{e^3}+\frac {c^2\,d^2\,x^2}{2\,e} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2/(d + e*x)^3,x)

[Out]

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