\(\int \frac {(d+e x)^3}{b x+c x^2} \, dx\) [260]

Optimal result

Integrand size = 19, antiderivative size = 64 \[ \int \frac {(d+e x)^3}{b x+c x^2} \, dx=\frac {e^2 (3 c d-b e) x}{c^2}+\frac {e^3 x^2}{2 c}+\frac {d^3 \log (x)}{b}-\frac {(c d-b e)^3 \log (b+c x)}{b c^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {712} \[ \int \frac {(d+e x)^3}{b x+c x^2} \, dx=-\frac {(c d-b e)^3 \log (b+c x)}{b c^3}+\frac {e^2 x (3 c d-b e)}{c^2}+\frac {d^3 \log (x)}{b}+\frac {e^3 x^2}{2 c} \]

[In]

Int[(d + e*x)^3/(b*x + c*x^2),x]

[Out]

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

Rule 712

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^2 (3 c d-b e)}{c^2}+\frac {d^3}{b x}+\frac {e^3 x}{c}+\frac {(-c d+b e)^3}{b c^2 (b+c x)}\right ) \, dx \\ & = \frac {e^2 (3 c d-b e) x}{c^2}+\frac {e^3 x^2}{2 c}+\frac {d^3 \log (x)}{b}-\frac {(c d-b e)^3 \log (b+c x)}{b c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.92 \[ \int \frac {(d+e x)^3}{b x+c x^2} \, dx=\frac {b c e^2 x (6 c d-2 b e+c e x)+2 c^3 d^3 \log (x)-2 (c d-b e)^3 \log (b+c x)}{2 b c^3} \]

[In]

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

[Out]

(b*c*e^2*x*(6*c*d - 2*b*e + c*e*x) + 2*c^3*d^3*Log[x] - 2*(c*d - b*e)^3*Log[b + c*x])/(2*b*c^3)

Maple [A] (verified)

Time = 1.88 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.33

method	result	size
default	\(-\frac {e^{2} \left (-\frac {1}{2} c e \,x^{2}+b e x -3 c d x \right )}{c^{2}}+\frac {d^{3} \ln \left (x \right )}{b}+\frac {\left (b^{3} e^{3}-3 b^{2} d \,e^{2} c +3 b \,c^{2} d^{2} e -c^{3} d^{3}\right ) \ln \left (c x +b \right )}{c^{3} b}\)	\(85\)
norman	\(\frac {e^{3} x^{2}}{2 c}-\frac {e^{2} \left (b e -3 c d \right ) x}{c^{2}}+\frac {d^{3} \ln \left (x \right )}{b}+\frac {\left (b^{3} e^{3}-3 b^{2} d \,e^{2} c +3 b \,c^{2} d^{2} e -c^{3} d^{3}\right ) \ln \left (c x +b \right )}{c^{3} b}\)	\(88\)
parallelrisch	\(\frac {e^{3} x^{2} c^{2} b +2 d^{3} \ln \left (x \right ) c^{3}+2 \ln \left (c x +b \right ) b^{3} e^{3}-6 \ln \left (c x +b \right ) b^{2} c d \,e^{2}+6 \ln \left (c x +b \right ) b \,c^{2} d^{2} e -2 \ln \left (c x +b \right ) c^{3} d^{3}-2 x \,b^{2} c \,e^{3}+6 x b \,c^{2} d \,e^{2}}{2 c^{3} b}\)	\(112\)
risch	\(\frac {e^{3} x^{2}}{2 c}-\frac {e^{3} b x}{c^{2}}+\frac {3 d \,e^{2} x}{c}+\frac {d^{3} \ln \left (x \right )}{b}+\frac {b^{2} \ln \left (-c x -b \right ) e^{3}}{c^{3}}-\frac {3 b \ln \left (-c x -b \right ) d \,e^{2}}{c^{2}}+\frac {3 \ln \left (-c x -b \right ) d^{2} e}{c}-\frac {\ln \left (-c x -b \right ) d^{3}}{b}\)	\(115\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.52 \[ \int \frac {(d+e x)^3}{b x+c x^2} \, dx=\frac {b c^{2} e^{3} x^{2} + 2 \, c^{3} d^{3} \log \left (x\right ) + 2 \, {\left (3 \, b c^{2} d e^{2} - b^{2} c e^{3}\right )} x - 2 \, {\left (c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - b^{3} e^{3}\right )} \log \left (c x + b\right )}{2 \, b c^{3}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (54) = 108\).

Time = 0.70 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.75 \[ \int \frac {(d+e x)^3}{b x+c x^2} \, dx=x \left (- \frac {b e^{3}}{c^{2}} + \frac {3 d e^{2}}{c}\right ) + \frac {e^{3} x^{2}}{2 c} + \frac {d^{3} \log {\left (x \right )}}{b} + \frac {\left (b e - c d\right )^{3} \log {\left (x + \frac {- b c^{2} d^{3} + \frac {b \left (b e - c d\right )^{3}}{c}}{b^{3} e^{3} - 3 b^{2} c d e^{2} + 3 b c^{2} d^{2} e - 2 c^{3} d^{3}} \right )}}{b c^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.42 \[ \int \frac {(d+e x)^3}{b x+c x^2} \, dx=\frac {d^{3} \log \left (x\right )}{b} + \frac {c e^{3} x^{2} + 2 \, {\left (3 \, c d e^{2} - b e^{3}\right )} x}{2 \, c^{2}} - \frac {{\left (c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - b^{3} e^{3}\right )} \log \left (c x + b\right )}{b c^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.42 \[ \int \frac {(d+e x)^3}{b x+c x^2} \, dx=\frac {d^{3} \log \left ({\left | x \right |}\right )}{b} + \frac {c e^{3} x^{2} + 6 \, c d e^{2} x - 2 \, b e^{3} x}{2 \, c^{2}} - \frac {{\left (c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - b^{3} e^{3}\right )} \log \left ({\left | c x + b \right |}\right )}{b c^{3}} \]

[In]

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

[Out]

d^3*log(abs(x))/b + 1/2*(c*e^3*x^2 + 6*c*d*e^2*x - 2*b*e^3*x)/c^2 - (c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 -
b^3*e^3)*log(abs(c*x + b))/(b*c^3)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.02 \[ \int \frac {(d+e x)^3}{b x+c x^2} \, dx=\frac {e^3\,x^2}{2\,c}-x\,\left (\frac {b\,e^3}{c^2}-\frac {3\,d\,e^2}{c}\right )+\frac {d^3\,\ln \left (x\right )}{b}+\frac {\ln \left (b+c\,x\right )\,{\left (b\,e-c\,d\right )}^3}{b\,c^3} \]

[In]

int((d + e*x)^3/(b*x + c*x^2),x)

[Out]

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