3.1.76 \(\int \frac {x^{11} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=89 \[ \frac {b^2 (b B-A c)}{4 c^4 \left (b+c x^2\right )^2}-\frac {b (3 b B-2 A c)}{2 c^4 \left (b+c x^2\right )}-\frac {(3 b B-A c) \log \left (b+c x^2\right )}{2 c^4}+\frac {B x^2}{2 c^3} \]

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Rubi [A]  time = 0.10, antiderivative size = 89, 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 {b^2 (b B-A c)}{4 c^4 \left (b+c x^2\right )^2}-\frac {b (3 b B-2 A c)}{2 c^4 \left (b+c x^2\right )}-\frac {(3 b B-A c) \log \left (b+c x^2\right )}{2 c^4}+\frac {B x^2}{2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^11*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

(B*x^2)/(2*c^3) + (b^2*(b*B - A*c))/(4*c^4*(b + c*x^2)^2) - (b*(3*b*B - 2*A*c))/(2*c^4*(b + c*x^2)) - ((3*b*B
- A*c)*Log[b + c*x^2])/(2*c^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 {x^{11} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {x^5 \left (A+B x^2\right )}{\left (b+c x^2\right )^3} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2 (A+B x)}{(b+c x)^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {B}{c^3}-\frac {b^2 (b B-A c)}{c^3 (b+c x)^3}+\frac {b (3 b B-2 A c)}{c^3 (b+c x)^2}+\frac {-3 b B+A c}{c^3 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {B x^2}{2 c^3}+\frac {b^2 (b B-A c)}{4 c^4 \left (b+c x^2\right )^2}-\frac {b (3 b B-2 A c)}{2 c^4 \left (b+c x^2\right )}-\frac {(3 b B-A c) \log \left (b+c x^2\right )}{2 c^4}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 92, normalized size = 1.03 \begin {gather*} \frac {2 A b c-3 b^2 B}{2 c^4 \left (b+c x^2\right )}+\frac {b^3 B-A b^2 c}{4 c^4 \left (b+c x^2\right )^2}+\frac {(A c-3 b B) \log \left (b+c x^2\right )}{2 c^4}+\frac {B x^2}{2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^11*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{11} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^11*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]

[Out]

IntegrateAlgebraic[(x^11*(A + B*x^2))/(b*x^2 + c*x^4)^3, x]

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fricas [A]  time = 0.40, size = 142, normalized size = 1.60 \begin {gather*} \frac {2 \, B c^{3} x^{6} + 4 \, B b c^{2} x^{4} - 5 \, B b^{3} + 3 \, A b^{2} c - 4 \, {\left (B b^{2} c - A b c^{2}\right )} x^{2} - 2 \, {\left ({\left (3 \, B b c^{2} - A c^{3}\right )} x^{4} + 3 \, B b^{3} - A b^{2} c + 2 \, {\left (3 \, B b^{2} c - A b c^{2}\right )} x^{2}\right )} \log \left (c x^{2} + b\right )}{4 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

1/4*(2*B*c^3*x^6 + 4*B*b*c^2*x^4 - 5*B*b^3 + 3*A*b^2*c - 4*(B*b^2*c - A*b*c^2)*x^2 - 2*((3*B*b*c^2 - A*c^3)*x^
4 + 3*B*b^3 - A*b^2*c + 2*(3*B*b^2*c - A*b*c^2)*x^2)*log(c*x^2 + b))/(c^6*x^4 + 2*b*c^5*x^2 + b^2*c^4)

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giac [A]  time = 0.16, size = 93, normalized size = 1.04 \begin {gather*} \frac {B x^{2}}{2 \, c^{3}} - \frac {{\left (3 \, B b - A c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{4}} + \frac {9 \, B b c^{2} x^{4} - 3 \, A c^{3} x^{4} + 12 \, B b^{2} c x^{2} - 2 \, A b c^{2} x^{2} + 4 \, B b^{3}}{4 \, {\left (c x^{2} + b\right )}^{2} c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

1/2*B*x^2/c^3 - 1/2*(3*B*b - A*c)*log(abs(c*x^2 + b))/c^4 + 1/4*(9*B*b*c^2*x^4 - 3*A*c^3*x^4 + 12*B*b^2*c*x^2
- 2*A*b*c^2*x^2 + 4*B*b^3)/((c*x^2 + b)^2*c^4)

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maple [A]  time = 0.06, size = 109, normalized size = 1.22 \begin {gather*} -\frac {A \,b^{2}}{4 \left (c \,x^{2}+b \right )^{2} c^{3}}+\frac {B \,b^{3}}{4 \left (c \,x^{2}+b \right )^{2} c^{4}}+\frac {B \,x^{2}}{2 c^{3}}+\frac {A b}{\left (c \,x^{2}+b \right ) c^{3}}+\frac {A \ln \left (c \,x^{2}+b \right )}{2 c^{3}}-\frac {3 B \,b^{2}}{2 \left (c \,x^{2}+b \right ) c^{4}}-\frac {3 B b \ln \left (c \,x^{2}+b \right )}{2 c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(B*x^2+A)/(c*x^4+b*x^2)^3,x)

[Out]

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

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maxima [A]  time = 1.36, size = 94, normalized size = 1.06 \begin {gather*} -\frac {5 \, B b^{3} - 3 \, A b^{2} c + 2 \, {\left (3 \, B b^{2} c - 2 \, A b c^{2}\right )} x^{2}}{4 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}} + \frac {B x^{2}}{2 \, c^{3}} - \frac {{\left (3 \, B b - A c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

-1/4*(5*B*b^3 - 3*A*b^2*c + 2*(3*B*b^2*c - 2*A*b*c^2)*x^2)/(c^6*x^4 + 2*b*c^5*x^2 + b^2*c^4) + 1/2*B*x^2/c^3 -
1/2*(3*B*b - A*c)*log(c*x^2 + b)/c^4

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mupad [B]  time = 0.09, size = 95, normalized size = 1.07 \begin {gather*} \frac {B\,x^2}{2\,c^3}-\frac {x^2\,\left (\frac {3\,B\,b^2}{2}-A\,b\,c\right )+\frac {5\,B\,b^3-3\,A\,b^2\,c}{4\,c}}{b^2\,c^3+2\,b\,c^4\,x^2+c^5\,x^4}+\frac {\ln \left (c\,x^2+b\right )\,\left (A\,c-3\,B\,b\right )}{2\,c^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^11*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)

[Out]

(B*x^2)/(2*c^3) - (x^2*((3*B*b^2)/2 - A*b*c) + (5*B*b^3 - 3*A*b^2*c)/(4*c))/(b^2*c^3 + c^5*x^4 + 2*b*c^4*x^2)
+ (log(b + c*x^2)*(A*c - 3*B*b))/(2*c^4)

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sympy [A]  time = 1.29, size = 94, normalized size = 1.06 \begin {gather*} \frac {B x^{2}}{2 c^{3}} + \frac {3 A b^{2} c - 5 B b^{3} + x^{2} \left (4 A b c^{2} - 6 B b^{2} c\right )}{4 b^{2} c^{4} + 8 b c^{5} x^{2} + 4 c^{6} x^{4}} - \frac {\left (- A c + 3 B b\right ) \log {\left (b + c x^{2} \right )}}{2 c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11*(B*x**2+A)/(c*x**4+b*x**2)**3,x)

[Out]

B*x**2/(2*c**3) + (3*A*b**2*c - 5*B*b**3 + x**2*(4*A*b*c**2 - 6*B*b**2*c))/(4*b**2*c**4 + 8*b*c**5*x**2 + 4*c*
*6*x**4) - (-A*c + 3*B*b)*log(b + c*x**2)/(2*c**4)

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