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

Optimal. Leaf size=41 \[ \frac {a d-b c}{2 b^2 \left (a+b x^2\right )}+\frac {d \log \left (a+b x^2\right )}{2 b^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {444, 43} \begin {gather*} \frac {d \log \left (a+b x^2\right )}{2 b^2}-\frac {b c-a d}{2 b^2 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(c + d*x^2))/(a + b*x^2)^2,x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rubi steps

\begin {align*} \int \frac {x \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {c+d x}{(a+b x)^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {b c-a d}{b (a+b x)^2}+\frac {d}{b (a+b x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {b c-a d}{2 b^2 \left (a+b x^2\right )}+\frac {d \log \left (a+b x^2\right )}{2 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 1.00 \begin {gather*} \frac {a d-b c}{2 b^2 \left (a+b x^2\right )}+\frac {d \log \left (a+b x^2\right )}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.85, size = 45, normalized size = 1.10 \begin {gather*} -\frac {b c - a d - {\left (b d x^{2} + a d\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*c - a*d - (b*d*x^2 + a*d)*log(b*x^2 + a))/(b^3*x^2 + a*b^2)

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giac [A]  time = 0.34, size = 65, normalized size = 1.59 \begin {gather*} -\frac {d {\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 )}}{2 \, b} - \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*(d*x^2+c)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 47, normalized size = 1.15 \begin {gather*} \frac {a d}{2 \left (b \,x^{2}+a \right ) b^{2}}-\frac {c}{2 \left (b \,x^{2}+a \right ) b}+\frac {d \ln \left (b \,x^{2}+a \right )}{2 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.07, size = 40, normalized size = 0.98 \begin {gather*} -\frac {b c - a d}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} + \frac {d \log \left (b x^{2} + a\right )}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*c - a*d)/(b^3*x^2 + a*b^2) + 1/2*d*log(b*x^2 + a)/b^2

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mupad [B]  time = 0.15, size = 37, normalized size = 0.90 \begin {gather*} \frac {d\,\ln \left (b\,x^2+a\right )}{2\,b^2}+\frac {a\,d-b\,c}{2\,b^2\,\left (b\,x^2+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(c + d*x^2))/(a + b*x^2)^2,x)

[Out]

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

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sympy [A]  time = 0.36, size = 36, normalized size = 0.88 \begin {gather*} \frac {a d - b c}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {d \log {\left (a + b x^{2} \right )}}{2 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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