3.2.29 \(\int \frac {x (a c+b c x^2)}{(a+b x^2)^2} \, dx\) [129]

Optimal. Leaf size=16 \[ \frac {c \log \left (a+b x^2\right )}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 16, 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, 266} \begin {gather*} \frac {c \log \left (a+b x^2\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Log[a + b*x^2])/(2*b)

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 266

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

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 16, normalized size = 1.00 \begin {gather*} \frac {c \log \left (a+b x^2\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Log[a + b*x^2])/(2*b)

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Maple [A]
time = 0.07, size = 15, normalized size = 0.94

method	result	size
default	\(\frac {c \ln \left (b \,x^{2}+a \right )}{2 b}\)	\(15\)
norman	\(\frac {c \ln \left (b \,x^{2}+a \right )}{2 b}\)	\(15\)
risch	\(\frac {c \ln \left (b \,x^{2}+a \right )}{2 b}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 14, normalized size = 0.88 \begin {gather*} \frac {c \log \left (b x^{2} + a\right )}{2 \, 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)^2,x, algorithm="maxima")

[Out]

1/2*c*log(b*x^2 + a)/b

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Fricas [A]
time = 0.96, size = 14, normalized size = 0.88 \begin {gather*} \frac {c \log \left (b x^{2} + a\right )}{2 \, 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)^2,x, algorithm="fricas")

[Out]

1/2*c*log(b*x^2 + a)/b

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Sympy [A]
time = 0.04, size = 12, normalized size = 0.75 \begin {gather*} \frac {c \log {\left (a + b x^{2} \right )}}{2 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)**2,x)

[Out]

c*log(a + b*x**2)/(2*b)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (14) = 28\).
time = 1.14, size = 63, normalized size = 3.94 \begin {gather*} -\frac {1}{2} \, c {\left (\frac {\log \left (\frac {{\left | b x^{2} + a \right |}}{{\left (b x^{2} + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b x^{2} + a\right )} b}\right )} - \frac {a 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)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.02, size = 14, normalized size = 0.88 \begin {gather*} \frac {c\,\ln \left (b\,x^2+a\right )}{2\,b} \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)^2,x)

[Out]

(c*log(a + b*x^2))/(2*b)

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