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

Optimal. Leaf size=92 \[ -\frac {c^2 (b B-A c) \log \left (b+c x^2\right )}{2 b^4}+\frac {c^2 \log (x) (b B-A c)}{b^4}+\frac {c (b B-A c)}{2 b^3 x^2}-\frac {b B-A c}{4 b^2 x^4}-\frac {A}{6 b x^6} \]

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Rubi [A]  time = 0.09, antiderivative size = 92, 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, 446, 77} \begin {gather*} -\frac {c^2 (b B-A c) \log \left (b+c x^2\right )}{2 b^4}+\frac {c^2 \log (x) (b B-A c)}{b^4}+\frac {c (b B-A c)}{2 b^3 x^2}-\frac {b B-A c}{4 b^2 x^4}-\frac {A}{6 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(6*b*x^6) - (b*B - A*c)/(4*b^2*x^4) + (c*(b*B - A*c))/(2*b^3*x^2) + (c^2*(b*B - A*c)*Log[x])/b^4 - (c^2*(b*
B - A*c)*Log[b + c*x^2])/(2*b^4)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

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 {A+B x^2}{x^5 \left (b x^2+c x^4\right )} \, dx &=\int \frac {A+B x^2}{x^7 \left (b+c x^2\right )} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {A+B x}{x^4 (b+c x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {A}{b x^4}+\frac {b B-A c}{b^2 x^3}-\frac {c (b B-A c)}{b^3 x^2}+\frac {c^2 (b B-A c)}{b^4 x}-\frac {c^3 (b B-A c)}{b^4 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {A}{6 b x^6}-\frac {b B-A c}{4 b^2 x^4}+\frac {c (b B-A c)}{2 b^3 x^2}+\frac {c^2 (b B-A c) \log (x)}{b^4}-\frac {c^2 (b B-A c) \log \left (b+c x^2\right )}{2 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 96, normalized size = 1.04 \begin {gather*} \frac {\left (A c^3-b B c^2\right ) \log \left (b+c x^2\right )}{2 b^4}+\frac {\log (x) \left (b B c^2-A c^3\right )}{b^4}+\frac {c (b B-A c)}{2 b^3 x^2}+\frac {A c-b B}{4 b^2 x^4}-\frac {A}{6 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*A/(b*x^6) + (-(b*B) + A*c)/(4*b^2*x^4) + (c*(b*B - A*c))/(2*b^3*x^2) + ((b*B*c^2 - A*c^3)*Log[x])/b^4 + (
(-(b*B*c^2) + A*c^3)*Log[b + c*x^2])/(2*b^4)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 98, normalized size = 1.07 \begin {gather*} -\frac {6 \, {\left (B b c^{2} - A c^{3}\right )} x^{6} \log \left (c x^{2} + b\right ) - 12 \, {\left (B b c^{2} - A c^{3}\right )} x^{6} \log \relax (x) - 6 \, {\left (B b^{2} c - A b c^{2}\right )} x^{4} + 2 \, A b^{3} + 3 \, {\left (B b^{3} - A b^{2} c\right )} x^{2}}{12 \, b^{4} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*(B*b*c^2 - A*c^3)*x^6*log(c*x^2 + b) - 12*(B*b*c^2 - A*c^3)*x^6*log(x) - 6*(B*b^2*c - A*b*c^2)*x^4 +
2*A*b^3 + 3*(B*b^3 - A*b^2*c)*x^2)/(b^4*x^6)

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giac [A]  time = 0.16, size = 126, normalized size = 1.37 \begin {gather*} \frac {{\left (B b c^{2} - A c^{3}\right )} \log \left (x^{2}\right )}{2 \, b^{4}} - \frac {{\left (B b c^{3} - A c^{4}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{4} c} - \frac {11 \, B b c^{2} x^{6} - 11 \, A c^{3} x^{6} - 6 \, B b^{2} c x^{4} + 6 \, A b c^{2} x^{4} + 3 \, B b^{3} x^{2} - 3 \, A b^{2} c x^{2} + 2 \, A b^{3}}{12 \, b^{4} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*b*c^2 - A*c^3)*log(x^2)/b^4 - 1/2*(B*b*c^3 - A*c^4)*log(abs(c*x^2 + b))/(b^4*c) - 1/12*(11*B*b*c^2*x^6
- 11*A*c^3*x^6 - 6*B*b^2*c*x^4 + 6*A*b*c^2*x^4 + 3*B*b^3*x^2 - 3*A*b^2*c*x^2 + 2*A*b^3)/(b^4*x^6)

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maple [A]  time = 0.05, size = 107, normalized size = 1.16 \begin {gather*} -\frac {A \,c^{3} \ln \relax (x )}{b^{4}}+\frac {A \,c^{3} \ln \left (c \,x^{2}+b \right )}{2 b^{4}}+\frac {B \,c^{2} \ln \relax (x )}{b^{3}}-\frac {B \,c^{2} \ln \left (c \,x^{2}+b \right )}{2 b^{3}}-\frac {A \,c^{2}}{2 b^{3} x^{2}}+\frac {B c}{2 b^{2} x^{2}}+\frac {A c}{4 b^{2} x^{4}}-\frac {B}{4 b \,x^{4}}-\frac {A}{6 b \,x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c^3/b^4*ln(c*x^2+b)*A-1/2*c^2/b^3*ln(c*x^2+b)*B-1/6*A/b/x^6+1/4/b^2/x^4*A*c-1/4/b/x^4*B-1/2*c^2/b^3/x^2*A+
1/2*c/b^2/x^2*B-c^3/b^4*ln(x)*A+c^2/b^3*ln(x)*B

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maxima [A]  time = 1.28, size = 96, normalized size = 1.04 \begin {gather*} -\frac {{\left (B b c^{2} - A c^{3}\right )} \log \left (c x^{2} + b\right )}{2 \, b^{4}} + \frac {{\left (B b c^{2} - A c^{3}\right )} \log \left (x^{2}\right )}{2 \, b^{4}} + \frac {6 \, {\left (B b c - A c^{2}\right )} x^{4} - 2 \, A b^{2} - 3 \, {\left (B b^{2} - A b c\right )} x^{2}}{12 \, b^{3} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*b*c^2 - A*c^3)*log(c*x^2 + b)/b^4 + 1/2*(B*b*c^2 - A*c^3)*log(x^2)/b^4 + 1/12*(6*(B*b*c - A*c^2)*x^4 -
2*A*b^2 - 3*(B*b^2 - A*b*c)*x^2)/(b^3*x^6)

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mupad [B]  time = 0.15, size = 92, normalized size = 1.00 \begin {gather*} \frac {\ln \left (c\,x^2+b\right )\,\left (A\,c^3-B\,b\,c^2\right )}{2\,b^4}-\frac {\frac {A}{6\,b}-\frac {x^2\,\left (A\,c-B\,b\right )}{4\,b^2}+\frac {c\,x^4\,\left (A\,c-B\,b\right )}{2\,b^3}}{x^6}-\frac {\ln \relax (x)\,\left (A\,c^3-B\,b\,c^2\right )}{b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(b + c*x^2)*(A*c^3 - B*b*c^2))/(2*b^4) - (A/(6*b) - (x^2*(A*c - B*b))/(4*b^2) + (c*x^4*(A*c - B*b))/(2*b^3
))/x^6 - (log(x)*(A*c^3 - B*b*c^2))/b^4

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sympy [A]  time = 1.41, size = 88, normalized size = 0.96 \begin {gather*} \frac {- 2 A b^{2} + x^{4} \left (- 6 A c^{2} + 6 B b c\right ) + x^{2} \left (3 A b c - 3 B b^{2}\right )}{12 b^{3} x^{6}} + \frac {c^{2} \left (- A c + B b\right ) \log {\relax (x )}}{b^{4}} - \frac {c^{2} \left (- A c + B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*A*b**2 + x**4*(-6*A*c**2 + 6*B*b*c) + x**2*(3*A*b*c - 3*B*b**2))/(12*b**3*x**6) + c**2*(-A*c + B*b)*log(x)
/b**4 - c**2*(-A*c + B*b)*log(b/c + x**2)/(2*b**4)

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