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

Optimal. Leaf size=48 \[ -\frac {d (c d-b e)}{e^3 (d+e x)}-\frac {(2 c d-b e) \log (d+e x)}{e^3}+\frac {c x}{e^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {698} \[ -\frac {d (c d-b e)}{e^3 (d+e x)}-\frac {(2 c d-b e) \log (d+e x)}{e^3}+\frac {c x}{e^2} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)/(d + e*x)^2,x]

[Out]

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

Rule 698

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

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Mathematica [A]  time = 0.02, size = 41, normalized size = 0.85 \[ \frac {\frac {d (b e-c d)}{d+e x}+(b e-2 c d) \log (d+e x)+c e x}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.07, size = 72, normalized size = 1.50 \[ \frac {c e^{2} x^{2} + c d e x - c d^{2} + b d e - {\left (2 \, c d^{2} - b d e + {\left (2 \, c d e - b e^{2}\right )} x\right )} \log \left (e x + d\right )}{e^{4} x + d e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 93, normalized size = 1.94 \[ -{\left (e^{\left (-1\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) - \frac {d e^{\left (-1\right )}}{x e + d}\right )} b e^{\left (-1\right )} + {\left (2 \, d e^{\left (-3\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) + {\left (x e + d\right )} e^{\left (-3\right )} - \frac {d^{2} e^{\left (-3\right )}}{x e + d}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 61, normalized size = 1.27 \[ \frac {b d}{\left (e x +d \right ) e^{2}}+\frac {b \ln \left (e x +d \right )}{e^{2}}-\frac {c \,d^{2}}{\left (e x +d \right ) e^{3}}-\frac {2 c d \ln \left (e x +d \right )}{e^{3}}+\frac {c x}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.30, size = 53, normalized size = 1.10 \[ -\frac {c d^{2} - b d e}{e^{4} x + d e^{3}} + \frac {c x}{e^{2}} - \frac {{\left (2 \, c d - b e\right )} \log \left (e x + d\right )}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.18, size = 54, normalized size = 1.12 \[ \frac {\ln \left (d+e\,x\right )\,\left (b\,e-2\,c\,d\right )}{e^3}-\frac {c\,d^2-b\,d\,e}{e\,\left (x\,e^3+d\,e^2\right )}+\frac {c\,x}{e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)/(d + e*x)^2,x)

[Out]

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

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sympy [A]  time = 0.28, size = 44, normalized size = 0.92 \[ \frac {c x}{e^{2}} + \frac {b d e - c d^{2}}{d e^{3} + e^{4} x} + \frac {\left (b e - 2 c d\right ) \log {\left (d + e x \right )}}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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