3.1.65 \(\int \frac {x^5 (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

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

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Rubi [A]  time = 0.05, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1584, 444, 43} \begin {gather*} \frac {b B-A c}{2 c^2 \left (b+c x^2\right )}+\frac {B \log \left (b+c x^2\right )}{2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

(b*B - A*c)/(2*c^2*(b + c*x^2)) + (B*Log[b + c*x^2])/(2*c^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]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{(b+c x)^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {-b B+A c}{c (b+c x)^2}+\frac {B}{c (b+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {b B-A c}{2 c^2 \left (b+c x^2\right )}+\frac {B \log \left (b+c x^2\right )}{2 c^2}\\ \end {align*}

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

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

(b*B - A*c)/(2*c^2*(b + c*x^2)) + (B*Log[b + c*x^2])/(2*c^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^5 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^5*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

IntegrateAlgebraic[(x^5*(A + B*x^2))/(b*x^2 + c*x^4)^2, x]

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fricas [A]  time = 0.39, size = 44, normalized size = 1.07 \begin {gather*} \frac {B b - A c + {\left (B c x^{2} + B b\right )} \log \left (c x^{2} + b\right )}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

1/2*(B*b - A*c + (B*c*x^2 + B*b)*log(c*x^2 + b))/(c^3*x^2 + b*c^2)

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giac [A]  time = 0.19, size = 37, normalized size = 0.90 \begin {gather*} \frac {B \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{2}} - \frac {B x^{2} + A}{2 \, {\left (c x^{2} + b\right )} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 47, normalized size = 1.15 \begin {gather*} -\frac {A}{2 \left (c \,x^{2}+b \right ) c}+\frac {B b}{2 \left (c \,x^{2}+b \right ) c^{2}}+\frac {B \ln \left (c \,x^{2}+b \right )}{2 c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

1/2*B*ln(c*x^2+b)/c^2-1/2/c/(c*x^2+b)*A+1/2/c^2/(c*x^2+b)*b*B

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maxima [A]  time = 1.31, size = 40, normalized size = 0.98 \begin {gather*} \frac {B b - A c}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}} + \frac {B \log \left (c x^{2} + b\right )}{2 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

1/2*(B*b - A*c)/(c^3*x^2 + b*c^2) + 1/2*B*log(c*x^2 + b)/c^2

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mupad [B]  time = 0.09, size = 37, normalized size = 0.90 \begin {gather*} \frac {B\,\ln \left (c\,x^2+b\right )}{2\,c^2}-\frac {A\,c-B\,b}{2\,c^2\,\left (c\,x^2+b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)

[Out]

(B*log(b + c*x^2))/(2*c^2) - (A*c - B*b)/(2*c^2*(b + c*x^2))

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sympy [A]  time = 0.37, size = 36, normalized size = 0.88 \begin {gather*} \frac {B \log {\left (b + c x^{2} \right )}}{2 c^{2}} + \frac {- A c + B b}{2 b c^{2} + 2 c^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

B*log(b + c*x**2)/(2*c**2) + (-A*c + B*b)/(2*b*c**2 + 2*c**3*x**2)

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