3.1.62 \(\int \frac {x^3 (d+e x)}{(b x+c x^2)^3} \, dx\) [62]

Optimal. Leaf size=36 \[ -\frac {c d-b e}{2 c^2 (b+c x)^2}-\frac {e}{c^2 (b+c x)} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {779} \begin {gather*} -\frac {c d-b e}{2 c^2 (b+c x)^2}-\frac {e}{c^2 (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(d + e*x))/(b*x + c*x^2)^3,x]

[Out]

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

Rule 779

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

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Mathematica [A]
time = 0.01, size = 26, normalized size = 0.72 \begin {gather*} -\frac {b e+c (d+2 e x)}{2 c^2 (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d + e*x))/(b*x + c*x^2)^3,x]

[Out]

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

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Maple [A]
time = 0.54, size = 35, normalized size = 0.97

method	result	size
gosper	\(-\frac {2 c e x +b e +c d}{2 \left (c x +b \right )^{2} c^{2}}\)	\(25\)
risch	\(\frac {-\frac {e x}{c}-\frac {b e +c d}{2 c^{2}}}{\left (c x +b \right )^{2}}\)	\(29\)
default	\(-\frac {e}{c^{2} \left (c x +b \right )}-\frac {-b e +c d}{2 c^{2} \left (c x +b \right )^{2}}\)	\(35\)
norman	\(\frac {-\frac {e \,x^{3}}{c}+\frac {\left (-b e -c d \right ) x^{2}}{2 c^{2}}}{x^{2} \left (c x +b \right )^{2}}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x+d)/(c*x^2+b*x)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 40, normalized size = 1.11 \begin {gather*} -\frac {2 \, c x e + c d + b e}{2 \, {\left (c^{4} x^{2} + 2 \, b c^{3} x + b^{2} c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

-1/2*(2*c*x*e + c*d + b*e)/(c^4*x^2 + 2*b*c^3*x + b^2*c^2)

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Fricas [A]
time = 3.12, size = 39, normalized size = 1.08 \begin {gather*} -\frac {c d + {\left (2 \, c x + b\right )} e}{2 \, {\left (c^{4} x^{2} + 2 \, b c^{3} x + b^{2} c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

-1/2*(c*d + (2*c*x + b)*e)/(c^4*x^2 + 2*b*c^3*x + b^2*c^2)

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Sympy [A]
time = 0.15, size = 39, normalized size = 1.08 \begin {gather*} \frac {- b e - c d - 2 c e x}{2 b^{2} c^{2} + 4 b c^{3} x + 2 c^{4} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x+d)/(c*x**2+b*x)**3,x)

[Out]

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

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Giac [A]
time = 1.43, size = 26, normalized size = 0.72 \begin {gather*} -\frac {2 \, c x e + c d + b e}{2 \, {\left (c x + b\right )}^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)/(c*x^2+b*x)^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.03, size = 39, normalized size = 1.08 \begin {gather*} -\frac {\frac {b\,e+c\,d}{2\,c^2}+\frac {e\,x}{c}}{b^2+2\,b\,c\,x+c^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(d + e*x))/(b*x + c*x^2)^3,x)

[Out]

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

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