\(\int \frac {(d+e x)^3}{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\) [1867]

Optimal result

Integrand size = 35, antiderivative size = 69 \[ \int \frac {(d+e x)^3}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {e \left (c d^2-a e^2\right ) x}{c^2 d^2}+\frac {(d+e x)^2}{2 c d}+\frac {\left (c d^2-a e^2\right )^2 \log (a e+c d x)}{c^3 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 69, 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} \[ \int \frac {(d+e x)^3}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {\left (c d^2-a e^2\right )^2 \log (a e+c d x)}{c^3 d^3}+\frac {e x \left (c d^2-a e^2\right )}{c^2 d^2}+\frac {(d+e x)^2}{2 c d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {(d+e x)^3}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {c d e x \left (-2 a e^2+c d (4 d+e x)\right )+2 \left (c d^2-a e^2\right )^2 \log (a e+c d x)}{2 c^3 d^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07

method	result	size
default	\(-\frac {e \left (-\frac {1}{2} c d e \,x^{2}+a \,e^{2} x -2 c \,d^{2} x \right )}{c^{2} d^{2}}+\frac {\left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \ln \left (c d x +a e \right )}{c^{3} d^{3}}\)	\(74\)
norman	\(\frac {e^{2} x^{2}}{2 d c}-\frac {e \left (e^{2} a -2 c \,d^{2}\right ) x}{c^{2} d^{2}}+\frac {\left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \ln \left (c d x +a e \right )}{c^{3} d^{3}}\)	\(79\)
risch	\(\frac {e^{2} x^{2}}{2 d c}-\frac {e^{3} a x}{c^{2} d^{2}}+\frac {2 e x}{c}+\frac {\ln \left (c d x +a e \right ) a^{2} e^{4}}{c^{3} d^{3}}-\frac {2 \ln \left (c d x +a e \right ) a \,e^{2}}{c^{2} d}+\frac {d \ln \left (c d x +a e \right )}{c}\)	\(93\)
parallelrisch	\(\frac {x^{2} c^{2} d^{2} e^{2}+2 \ln \left (c d x +a e \right ) a^{2} e^{4}-4 \ln \left (c d x +a e \right ) a c \,d^{2} e^{2}+2 \ln \left (c d x +a e \right ) c^{2} d^{4}-2 x a c d \,e^{3}+4 x \,c^{2} d^{3} e}{2 c^{3} d^{3}}\)	\(95\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14 \[ \int \frac {(d+e x)^3}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {c^{2} d^{2} e^{2} x^{2} + 2 \, {\left (2 \, c^{2} d^{3} e - a c d e^{3}\right )} x + 2 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \log \left (c d x + a e\right )}{2 \, c^{3} d^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {(d+e x)^3}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=x \left (- \frac {a e^{3}}{c^{2} d^{2}} + \frac {2 e}{c}\right ) + \frac {e^{2} x^{2}}{2 c d} + \frac {\left (a e^{2} - c d^{2}\right )^{2} \log {\left (a e + c d x \right )}}{c^{3} d^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.12 \[ \int \frac {(d+e x)^3}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {c d e^{2} x^{2} + 2 \, {\left (2 \, c d^{2} e - a e^{3}\right )} x}{2 \, c^{2} d^{2}} + \frac {{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \log \left (c d x + a e\right )}{c^{3} d^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \frac {(d+e x)^3}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {c d e^{2} x^{2} + 4 \, c d^{2} e x - 2 \, a e^{3} x}{2 \, c^{2} d^{2}} + \frac {{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \log \left ({\left | c d x + a e \right |}\right )}{c^{3} d^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.12 \[ \int \frac {(d+e x)^3}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=x\,\left (\frac {2\,e}{c}-\frac {a\,e^3}{c^2\,d^2}\right )+\frac {\ln \left (a\,e+c\,d\,x\right )\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{c^3\,d^3}+\frac {e^2\,x^2}{2\,c\,d} \]

[In]

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

[Out]

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