3.2.36 \(\int \frac {x (a c+b c x^2)}{(a+b x^2)^3} \, dx\) [136]

Optimal. Leaf size=17 \[ -\frac {c}{2 b \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {21, 267} \begin {gather*} -\frac {c}{2 b \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*c/(b*(a + b*x^2))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 17, normalized size = 1.00 \begin {gather*} -\frac {c}{2 b \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*c/(b*(a + b*x^2))

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Maple [A]
time = 0.06, size = 16, normalized size = 0.94

method	result	size
gosper	\(-\frac {c}{2 b \left (b \,x^{2}+a \right )}\)	\(16\)
default	\(-\frac {c}{2 b \left (b \,x^{2}+a \right )}\)	\(16\)
risch	\(-\frac {c}{2 b \left (b \,x^{2}+a \right )}\)	\(16\)
norman	\(\frac {-\frac {a c}{2 b}-\frac {c \,x^{2}}{2}}{\left (b \,x^{2}+a \right )^{2}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 16, normalized size = 0.94 \begin {gather*} -\frac {c}{2 \, {\left (b^{2} x^{2} + a b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c/(b^2*x^2 + a*b)

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Fricas [A]
time = 0.83, size = 16, normalized size = 0.94 \begin {gather*} -\frac {c}{2 \, {\left (b^{2} x^{2} + a b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c/(b^2*x^2 + a*b)

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Sympy [A]
time = 0.09, size = 15, normalized size = 0.88 \begin {gather*} - \frac {c}{2 a b + 2 b^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c/(2*a*b + 2*b**2*x**2)

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Giac [A]
time = 0.90, size = 15, normalized size = 0.88 \begin {gather*} -\frac {c}{2 \, {\left (b x^{2} + a\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c/((b*x^2 + a)*b)

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Mupad [B]
time = 0.02, size = 15, normalized size = 0.88 \begin {gather*} -\frac {c}{2\,b\,\left (b\,x^2+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c/(2*b*(a + b*x^2))

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