\(\int \frac {b x+c x^2}{d+e x} \, dx\) [226]

Optimal result

Integrand size = 17, antiderivative size = 45 \[ \int \frac {b x+c x^2}{d+e x} \, dx=-\frac {(c d-b e) x}{e^2}+\frac {c x^2}{2 e}+\frac {d (c d-b e) \log (d+e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {712} \[ \int \frac {b x+c x^2}{d+e x} \, dx=\frac {d (c d-b e) \log (d+e x)}{e^3}-\frac {x (c d-b e)}{e^2}+\frac {c x^2}{2 e} \]

[In]

Int[(b*x + c*x^2)/(d + e*x),x]

[Out]

-(((c*d - b*e)*x)/e^2) + (c*x^2)/(2*e) + (d*(c*d - b*e)*Log[d + e*x])/e^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 {-c d+b e}{e^2}+\frac {c x}{e}+\frac {d (c d-b e)}{e^2 (d+e x)}\right ) \, dx \\ & = -\frac {(c d-b e) x}{e^2}+\frac {c x^2}{2 e}+\frac {d (c d-b e) \log (d+e x)}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.91 \[ \int \frac {b x+c x^2}{d+e x} \, dx=\frac {e x (-2 c d+2 b e+c e x)+2 d (c d-b e) \log (d+e x)}{2 e^3} \]

[In]

Integrate[(b*x + c*x^2)/(d + e*x),x]

[Out]

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

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {\frac {1}{2} c e \,x^{2}+b e x -c d x}{e^{2}}-\frac {d \left (b e -c d \right ) \ln \left (e x +d \right )}{e^{3}}\)	\(43\)
norman	\(\frac {\left (b e -c d \right ) x}{e^{2}}+\frac {c \,x^{2}}{2 e}-\frac {d \left (b e -c d \right ) \ln \left (e x +d \right )}{e^{3}}\)	\(44\)
risch	\(\frac {c \,x^{2}}{2 e}+\frac {b x}{e}-\frac {c d x}{e^{2}}-\frac {d \ln \left (e x +d \right ) b}{e^{2}}+\frac {d^{2} \ln \left (e x +d \right ) c}{e^{3}}\)	\(52\)
parallelrisch	\(-\frac {-c \,x^{2} e^{2}+2 \ln \left (e x +d \right ) b d e -2 \ln \left (e x +d \right ) c \,d^{2}-2 x b \,e^{2}+2 x c d e}{2 e^{3}}\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.04 \[ \int \frac {b x+c x^2}{d+e x} \, dx=\frac {c e^{2} x^{2} - 2 \, {\left (c d e - b e^{2}\right )} x + 2 \, {\left (c d^{2} - b d e\right )} \log \left (e x + d\right )}{2 \, e^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {b x+c x^2}{d+e x} \, dx=\frac {c x^{2}}{2 e} - \frac {d \left (b e - c d\right ) \log {\left (d + e x \right )}}{e^{3}} + x \left (\frac {b}{e} - \frac {c d}{e^{2}}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {b x+c x^2}{d+e x} \, dx=\frac {c e x^{2} - 2 \, {\left (c d - b e\right )} x}{2 \, e^{2}} + \frac {{\left (c d^{2} - b d e\right )} \log \left (e x + d\right )}{e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {b x+c x^2}{d+e x} \, dx=\frac {c e x^{2} - 2 \, c d x + 2 \, b e x}{2 \, e^{2}} + \frac {{\left (c d^{2} - b d e\right )} \log \left ({\left | e x + d \right |}\right )}{e^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.02 \[ \int \frac {b x+c x^2}{d+e x} \, dx=x\,\left (\frac {b}{e}-\frac {c\,d}{e^2}\right )+\frac {c\,x^2}{2\,e}+\frac {\ln \left (d+e\,x\right )\,\left (c\,d^2-b\,d\,e\right )}{e^3} \]

[In]

int((b*x + c*x^2)/(d + e*x),x)

[Out]

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