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

Optimal. Leaf size=32 \[ \frac{e \log (b+c x)}{c^2}-\frac{c d-b e}{c^2 (b+c x)} \]

[Out]

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

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Rubi [A]  time = 0.029932, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {765} \[ \frac{e \log (b+c x)}{c^2}-\frac{c d-b e}{c^2 (b+c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

Mathematica [A]  time = 0.010746, size = 31, normalized size = 0.97 \[ \frac{b e-c d}{c^2 (b+c x)}+\frac{e \log (b+c x)}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 39, normalized size = 1.2 \begin{align*}{\frac{be}{{c}^{2} \left ( cx+b \right ) }}-{\frac{d}{c \left ( cx+b \right ) }}+{\frac{e\ln \left ( cx+b \right ) }{{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.12139, size = 47, normalized size = 1.47 \begin{align*} -\frac{c d - b e}{c^{3} x + b c^{2}} + \frac{e \log \left (c x + b\right )}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.74865, size = 80, normalized size = 2.5 \begin{align*} -\frac{c d - b e -{\left (c e x + b e\right )} \log \left (c x + b\right )}{c^{3} x + b c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.384825, size = 27, normalized size = 0.84 \begin{align*} \frac{b e - c d}{b c^{2} + c^{3} x} + \frac{e \log{\left (b + c x \right )}}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10128, size = 47, normalized size = 1.47 \begin{align*} \frac{e \log \left ({\left | c x + b \right |}\right )}{c^{2}} - \frac{c d - b e}{{\left (c x + b\right )} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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