\(\int \frac {(d+e x)^2}{b x+c x^2} \, dx\) [261]

Optimal result

Integrand size = 19, antiderivative size = 42 \[ \int \frac {(d+e x)^2}{b x+c x^2} \, dx=\frac {e^2 x}{c}+\frac {d^2 \log (x)}{b}-\frac {(c d-b e)^2 \log (b+c x)}{b c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, 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} \[ \int \frac {(d+e x)^2}{b x+c x^2} \, dx=-\frac {(c d-b e)^2 \log (b+c x)}{b c^2}+\frac {d^2 \log (x)}{b}+\frac {e^2 x}{c} \]

[In]

Int[(d + e*x)^2/(b*x + c*x^2),x]

[Out]

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

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}{c}+\frac {d^2}{b x}-\frac {(-c d+b e)^2}{b c (b+c x)}\right ) \, dx \\ & = \frac {e^2 x}{c}+\frac {d^2 \log (x)}{b}-\frac {(c d-b e)^2 \log (b+c x)}{b c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x)^2}{b x+c x^2} \, dx=\frac {b c e^2 x+c^2 d^2 \log (x)-(c d-b e)^2 \log (b+c x)}{b c^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.89 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.29

method	result	size
norman	\(\frac {e^{2} x}{c}+\frac {d^{2} \ln \left (x \right )}{b}-\frac {\left (b^{2} e^{2}-2 b c d e +c^{2} d^{2}\right ) \ln \left (c x +b \right )}{b \,c^{2}}\)	\(54\)
default	\(\frac {e^{2} x}{c}+\frac {d^{2} \ln \left (x \right )}{b}+\frac {\left (-b^{2} e^{2}+2 b c d e -c^{2} d^{2}\right ) \ln \left (c x +b \right )}{b \,c^{2}}\)	\(55\)
risch	\(\frac {e^{2} x}{c}+\frac {d^{2} \ln \left (-x \right )}{b}-\frac {b \ln \left (c x +b \right ) e^{2}}{c^{2}}+\frac {2 \ln \left (c x +b \right ) d e}{c}-\frac {\ln \left (c x +b \right ) d^{2}}{b}\)	\(63\)
parallelrisch	\(\frac {d^{2} \ln \left (x \right ) c^{2}-\ln \left (c x +b \right ) b^{2} e^{2}+2 \ln \left (c x +b \right ) b c d e -\ln \left (c x +b \right ) c^{2} d^{2}+b c \,e^{2} x}{b \,c^{2}}\)	\(65\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.26 \[ \int \frac {(d+e x)^2}{b x+c x^2} \, dx=\frac {b c e^{2} x + c^{2} d^{2} \log \left (x\right ) - {\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \log \left (c x + b\right )}{b c^{2}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (34) = 68\).

Time = 0.45 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.74 \[ \int \frac {(d+e x)^2}{b x+c x^2} \, dx=\frac {e^{2} x}{c} + \frac {d^{2} \log {\left (x \right )}}{b} - \frac {\left (b e - c d\right )^{2} \log {\left (x + \frac {b c d^{2} + \frac {b \left (b e - c d\right )^{2}}{c}}{b^{2} e^{2} - 2 b c d e + 2 c^{2} d^{2}} \right )}}{b c^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.26 \[ \int \frac {(d+e x)^2}{b x+c x^2} \, dx=\frac {e^{2} x}{c} + \frac {d^{2} \log \left (x\right )}{b} - \frac {{\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \log \left (c x + b\right )}{b c^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.31 \[ \int \frac {(d+e x)^2}{b x+c x^2} \, dx=\frac {e^{2} x}{c} + \frac {d^{2} \log \left ({\left | x \right |}\right )}{b} - \frac {{\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \log \left ({\left | c x + b \right |}\right )}{b c^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.61 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.17 \[ \int \frac {(d+e x)^2}{b x+c x^2} \, dx=\frac {e^2\,x}{c}-\ln \left (b+c\,x\right )\,\left (\frac {d^2}{b}+\frac {b\,e^2}{c^2}-\frac {2\,d\,e}{c}\right )+\frac {d^2\,\ln \left (x\right )}{b} \]

[In]

int((d + e*x)^2/(b*x + c*x^2),x)

[Out]

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