\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^3}{(d+e x)^4} \, dx\) [1857]

Optimal result

Integrand size = 35, antiderivative size = 89 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^4} \, dx=\frac {c d \left (c d^2-a e^2\right )^2 x}{e^3}+\frac {1}{2} \left (a-\frac {c d^2}{e^2}\right ) (a e+c d x)^2+\frac {(a e+c d x)^3}{3 e}-\frac {\left (c d^2-a e^2\right )^3 \log (d+e x)}{e^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 89, 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} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^4} \, dx=\frac {1}{2} \left (a-\frac {c d^2}{e^2}\right ) (a e+c d x)^2-\frac {\left (c d^2-a e^2\right )^3 \log (d+e x)}{e^4}+\frac {c d x \left (c d^2-a e^2\right )^2}{e^3}+\frac {(a e+c d x)^3}{3 e} \]

[In]

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

[Out]

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

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)^3}{d+e x} \, dx \\ & = \int \left (\frac {c d \left (c d^2-a e^2\right )^2}{e^3}-\frac {c d \left (c d^2-a e^2\right ) (a e+c d x)}{e^2}+\frac {c d (a e+c d x)^2}{e}+\frac {\left (-c d^2+a e^2\right )^3}{e^3 (d+e x)}\right ) \, dx \\ & = \frac {c d \left (c d^2-a e^2\right )^2 x}{e^3}+\frac {1}{2} \left (a-\frac {c d^2}{e^2}\right ) (a e+c d x)^2+\frac {(a e+c d x)^3}{3 e}-\frac {\left (c d^2-a e^2\right )^3 \log (d+e x)}{e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^4} \, dx=\frac {c d e x \left (18 a^2 e^4+9 a c d e^2 (-2 d+e x)+c^2 d^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )-6 \left (c d^2-a e^2\right )^3 \log (d+e x)}{6 e^4} \]

[In]

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

[Out]

(c*d*e*x*(18*a^2*e^4 + 9*a*c*d*e^2*(-2*d + e*x) + c^2*d^2*(6*d^2 - 3*d*e*x + 2*e^2*x^2)) - 6*(c*d^2 - a*e^2)^3
*Log[d + e*x])/(6*e^4)

Maple [A] (verified)

Time = 2.84 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.39

method	result	size
default	\(\frac {c d \left (\frac {1}{3} x^{3} c^{2} d^{2} e^{2}+\frac {3}{2} x^{2} a c d \,e^{3}-\frac {1}{2} x^{2} c^{2} d^{3} e +3 a^{2} e^{4} x -3 a c \,d^{2} e^{2} x +c^{2} d^{4} x \right )}{e^{3}}+\frac {\left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}\right ) \ln \left (e x +d \right )}{e^{4}}\)	\(124\)
risch	\(\frac {c^{3} d^{3} x^{3}}{3 e}+\frac {3 a \,c^{2} d^{2} x^{2}}{2}-\frac {c^{3} d^{4} x^{2}}{2 e^{2}}+3 a^{2} x c d e -\frac {3 c^{2} d^{3} a x}{e}+\frac {c^{3} d^{5} x}{e^{3}}+e^{2} \ln \left (e x +d \right ) a^{3}-3 \ln \left (e x +d \right ) d^{2} a^{2} c +\frac {3 \ln \left (e x +d \right ) d^{4} c^{2} a}{e^{2}}-\frac {\ln \left (e x +d \right ) c^{3} d^{6}}{e^{4}}\)	\(138\)
parallelrisch	\(\frac {2 x^{3} c^{3} d^{3} e^{3}+9 x^{2} a \,c^{2} d^{2} e^{4}-3 x^{2} c^{3} d^{4} e^{2}+6 \ln \left (e x +d \right ) a^{3} e^{6}-18 \ln \left (e x +d \right ) a^{2} c \,d^{2} e^{4}+18 \ln \left (e x +d \right ) a \,c^{2} d^{4} e^{2}-6 \ln \left (e x +d \right ) c^{3} d^{6}+18 x \,a^{2} c d \,e^{5}-18 x a \,c^{2} d^{3} e^{3}+6 x \,c^{3} d^{5} e}{6 e^{4}}\)	\(148\)
norman	\(\frac {\left (\frac {3}{2} e^{3} a \,c^{2} d^{2}+\frac {1}{2} d^{4} e \,c^{3}\right ) x^{5}+\left (3 d \,e^{4} a^{2} c +\frac {3}{2} d^{3} e^{2} c^{2} a +\frac {1}{2} d^{5} c^{3}\right ) x^{4}-\frac {d^{3} \left (54 d^{2} e^{4} a^{2} c -27 d^{4} e^{2} c^{2} a +11 c^{3} d^{6}\right )}{6 e^{4}}-\frac {3 d \left (6 d^{2} e^{4} a^{2} c -2 d^{4} e^{2} c^{2} a +c^{3} d^{6}\right ) x^{2}}{e^{2}}-\frac {3 d^{2} \left (16 d^{2} e^{4} a^{2} c -7 d^{4} e^{2} c^{2} a +3 c^{3} d^{6}\right ) x}{2 e^{3}}+\frac {e^{2} c^{3} d^{3} x^{6}}{3}}{\left (e x +d \right )^{3}}+\frac {\left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}\right ) \ln \left (e x +d \right )}{e^{4}}\)	\(260\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

c*d/e^3*(1/3*x^3*c^2*d^2*e^2+3/2*x^2*a*c*d*e^3-1/2*x^2*c^2*d^3*e+3*a^2*e^4*x-3*a*c*d^2*e^2*x+c^2*d^4*x)+(a^3*e
^6-3*a^2*c*d^2*e^4+3*a*c^2*d^4*e^2-c^3*d^6)/e^4*ln(e*x+d)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^4} \, dx=\frac {2 \, c^{3} d^{3} e^{3} x^{3} - 3 \, {\left (c^{3} d^{4} e^{2} - 3 \, a c^{2} d^{2} e^{4}\right )} x^{2} + 6 \, {\left (c^{3} d^{5} e - 3 \, a c^{2} d^{3} e^{3} + 3 \, a^{2} c d e^{5}\right )} x - 6 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \log \left (e x + d\right )}{6 \, e^{4}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^4} \, dx=\frac {c^{3} d^{3} x^{3}}{3 e} + x^{2} \cdot \left (\frac {3 a c^{2} d^{2}}{2} - \frac {c^{3} d^{4}}{2 e^{2}}\right ) + x \left (3 a^{2} c d e - \frac {3 a c^{2} d^{3}}{e} + \frac {c^{3} d^{5}}{e^{3}}\right ) + \frac {\left (a e^{2} - c d^{2}\right )^{3} \log {\left (d + e x \right )}}{e^{4}} \]

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3/(e*x+d)**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^4} \, dx=\frac {2 \, c^{3} d^{3} e^{2} x^{3} - 3 \, {\left (c^{3} d^{4} e - 3 \, a c^{2} d^{2} e^{3}\right )} x^{2} + 6 \, {\left (c^{3} d^{5} - 3 \, a c^{2} d^{3} e^{2} + 3 \, a^{2} c d e^{4}\right )} x}{6 \, e^{3}} - \frac {{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \log \left (e x + d\right )}{e^{4}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^4} \, dx=\frac {2 \, c^{3} d^{3} e^{2} x^{3} - 3 \, c^{3} d^{4} e x^{2} + 9 \, a c^{2} d^{2} e^{3} x^{2} + 6 \, c^{3} d^{5} x - 18 \, a c^{2} d^{3} e^{2} x + 18 \, a^{2} c d e^{4} x}{6 \, e^{3}} - \frac {{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^4,x, algorithm="giac")

[Out]

1/6*(2*c^3*d^3*e^2*x^3 - 3*c^3*d^4*e*x^2 + 9*a*c^2*d^2*e^3*x^2 + 6*c^3*d^5*x - 18*a*c^2*d^3*e^2*x + 18*a^2*c*d
*e^4*x)/e^3 - (c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*log(abs(e*x + d))/e^4

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^4} \, dx=x^2\,\left (\frac {3\,a\,c^2\,d^2}{2}-\frac {c^3\,d^4}{2\,e^2}\right )-x\,\left (\frac {d\,\left (3\,a\,c^2\,d^2-\frac {c^3\,d^4}{e^2}\right )}{e}-3\,a^2\,c\,d\,e\right )+\frac {\ln \left (d+e\,x\right )\,\left (a^3\,e^6-3\,a^2\,c\,d^2\,e^4+3\,a\,c^2\,d^4\,e^2-c^3\,d^6\right )}{e^4}+\frac {c^3\,d^3\,x^3}{3\,e} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3/(d + e*x)^4,x)

[Out]

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