3.3.27 \(\int \frac {b x+c x^2}{(d+e x)^2} \, dx\) [227]

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

[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 = 0.03, 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 = {712} \begin {gather*} -\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} \end {gather*}

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 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*} \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 \begin {gather*} \frac {c e x+\frac {d (-c d+b e)}{d+e x}+(-2 c d+b e) \log (d+e x)}{e^3} \end {gather*}

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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Maple [A]
time = 0.40, size = 46, normalized size = 0.96

method	result	size
default	\(\frac {c x}{e^{2}}+\frac {d \left (b e -c d \right )}{e^{3} \left (e x +d \right )}+\frac {\left (b e -2 c d \right ) \ln \left (e x +d \right )}{e^{3}}\)	\(46\)
norman	\(\frac {\frac {c \,x^{2}}{e}+\frac {d \left (b e -2 c d \right )}{e^{3}}}{e x +d}+\frac {\left (b e -2 c d \right ) \ln \left (e x +d \right )}{e^{3}}\)	\(50\)
risch	\(\frac {c x}{e^{2}}+\frac {d b}{e^{2} \left (e x +d \right )}-\frac {d^{2} c}{e^{3} \left (e x +d \right )}+\frac {\ln \left (e x +d \right ) b}{e^{2}}-\frac {2 c d \ln \left (e x +d \right )}{e^{3}}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 52, normalized size = 1.08 \begin {gather*} c x e^{\left (-2\right )} - {\left (2 \, c d - b e\right )} e^{\left (-3\right )} \log \left (x e + d\right ) - \frac {c d^{2} - b d e}{x e^{4} + d e^{3}} \end {gather*}

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

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Fricas [A]
time = 1.16, size = 72, normalized size = 1.50 \begin {gather*} \frac {c x^{2} e^{2} - c d^{2} + {\left (c d x + b d\right )} e - {\left (2 \, c d^{2} - b x e^{2} + {\left (2 \, c d x - b d\right )} e\right )} \log \left (x e + d\right )}{x e^{4} + d e^{3}} \end {gather*}

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

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Sympy [A]
time = 0.13, size = 44, normalized size = 0.92 \begin {gather*} \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}} \end {gather*}

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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Giac [A]
time = 1.18, size = 93, normalized size = 1.94 \begin {gather*} -{\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 \end {gather*}

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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Mupad [B]
time = 0.18, size = 54, normalized size = 1.12 \begin {gather*} \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} \end {gather*}

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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