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

Optimal. Leaf size=105 \[ -\frac {b^3 (b B-A c)}{2 c^5 \left (b+c x^2\right )}-\frac {b^2 (4 b B-3 A c) \log \left (b+c x^2\right )}{2 c^5}+\frac {b x^2 (3 b B-2 A c)}{2 c^4}-\frac {x^4 (2 b B-A c)}{4 c^3}+\frac {B x^6}{6 c^2} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 105, 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} \[ -\frac {b^3 (b B-A c)}{2 c^5 \left (b+c x^2\right )}-\frac {b^2 (4 b B-3 A c) \log \left (b+c x^2\right )}{2 c^5}-\frac {x^4 (2 b B-A c)}{4 c^3}+\frac {b x^2 (3 b B-2 A c)}{2 c^4}+\frac {B x^6}{6 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(3*b*B - 2*A*c)*x^2)/(2*c^4) - ((2*b*B - A*c)*x^4)/(4*c^3) + (B*x^6)/(6*c^2) - (b^3*(b*B - A*c))/(2*c^5*(b
+ c*x^2)) - (b^2*(4*b*B - 3*A*c)*Log[b + c*x^2])/(2*c^5)

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

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Mathematica [A]  time = 0.08, size = 93, normalized size = 0.89 \[ \frac {\frac {6 b^3 (A c-b B)}{b+c x^2}+6 b^2 (3 A c-4 b B) \log \left (b+c x^2\right )+3 c^2 x^4 (A c-2 b B)+6 b c x^2 (3 b B-2 A c)+2 B c^3 x^6}{12 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b*c*(3*b*B - 2*A*c)*x^2 + 3*c^2*(-2*b*B + A*c)*x^4 + 2*B*c^3*x^6 + (6*b^3*(-(b*B) + A*c))/(b + c*x^2) + 6*b
^2*(-4*b*B + 3*A*c)*Log[b + c*x^2])/(12*c^5)

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fricas [A]  time = 0.95, size = 148, normalized size = 1.41 \[ \frac {2 \, B c^{4} x^{8} - {\left (4 \, B b c^{3} - 3 \, A c^{4}\right )} x^{6} - 6 \, B b^{4} + 6 \, A b^{3} c + 3 \, {\left (4 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{4} + 6 \, {\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{2} - 6 \, {\left (4 \, B b^{4} - 3 \, A b^{3} c + {\left (4 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} x^{2}\right )} \log \left (c x^{2} + b\right )}{12 \, {\left (c^{6} x^{2} + b c^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*B*c^4*x^8 - (4*B*b*c^3 - 3*A*c^4)*x^6 - 6*B*b^4 + 6*A*b^3*c + 3*(4*B*b^2*c^2 - 3*A*b*c^3)*x^4 + 6*(3*B
*b^3*c - 2*A*b^2*c^2)*x^2 - 6*(4*B*b^4 - 3*A*b^3*c + (4*B*b^3*c - 3*A*b^2*c^2)*x^2)*log(c*x^2 + b))/(c^6*x^2 +
b*c^5)

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giac [A]  time = 0.15, size = 135, normalized size = 1.29 \[ -\frac {{\left (4 \, B b^{3} - 3 \, A b^{2} c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{5}} + \frac {2 \, B c^{4} x^{6} - 6 \, B b c^{3} x^{4} + 3 \, A c^{4} x^{4} + 18 \, B b^{2} c^{2} x^{2} - 12 \, A b c^{3} x^{2}}{12 \, c^{6}} + \frac {4 \, B b^{3} c x^{2} - 3 \, A b^{2} c^{2} x^{2} + 3 \, B b^{4} - 2 \, A b^{3} c}{2 \, {\left (c x^{2} + b\right )} c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*B*b^3 - 3*A*b^2*c)*log(abs(c*x^2 + b))/c^5 + 1/12*(2*B*c^4*x^6 - 6*B*b*c^3*x^4 + 3*A*c^4*x^4 + 18*B*b^
2*c^2*x^2 - 12*A*b*c^3*x^2)/c^6 + 1/2*(4*B*b^3*c*x^2 - 3*A*b^2*c^2*x^2 + 3*B*b^4 - 2*A*b^3*c)/((c*x^2 + b)*c^5
)

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maple [A]  time = 0.05, size = 122, normalized size = 1.16 \[ \frac {B \,x^{6}}{6 c^{2}}+\frac {A \,x^{4}}{4 c^{2}}-\frac {B b \,x^{4}}{2 c^{3}}-\frac {A b \,x^{2}}{c^{3}}+\frac {3 B \,b^{2} x^{2}}{2 c^{4}}+\frac {A \,b^{3}}{2 \left (c \,x^{2}+b \right ) c^{4}}+\frac {3 A \,b^{2} \ln \left (c \,x^{2}+b \right )}{2 c^{4}}-\frac {B \,b^{4}}{2 \left (c \,x^{2}+b \right ) c^{5}}-\frac {2 B \,b^{3} \ln \left (c \,x^{2}+b \right )}{c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.38, size = 107, normalized size = 1.02 \[ -\frac {B b^{4} - A b^{3} c}{2 \, {\left (c^{6} x^{2} + b c^{5}\right )}} + \frac {2 \, B c^{2} x^{6} - 3 \, {\left (2 \, B b c - A c^{2}\right )} x^{4} + 6 \, {\left (3 \, B b^{2} - 2 \, A b c\right )} x^{2}}{12 \, c^{4}} - \frac {{\left (4 \, B b^{3} - 3 \, A b^{2} c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*b^4 - A*b^3*c)/(c^6*x^2 + b*c^5) + 1/12*(2*B*c^2*x^6 - 3*(2*B*b*c - A*c^2)*x^4 + 6*(3*B*b^2 - 2*A*b*c)
*x^2)/c^4 - 1/2*(4*B*b^3 - 3*A*b^2*c)*log(c*x^2 + b)/c^5

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mupad [B]  time = 0.10, size = 121, normalized size = 1.15 \[ x^4\,\left (\frac {A}{4\,c^2}-\frac {B\,b}{2\,c^3}\right )-x^2\,\left (\frac {b\,\left (\frac {A}{c^2}-\frac {2\,B\,b}{c^3}\right )}{c}+\frac {B\,b^2}{2\,c^4}\right )+\frac {B\,x^6}{6\,c^2}-\frac {\ln \left (c\,x^2+b\right )\,\left (4\,B\,b^3-3\,A\,b^2\,c\right )}{2\,c^5}-\frac {B\,b^4-A\,b^3\,c}{2\,c\,\left (c^5\,x^2+b\,c^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.93, size = 104, normalized size = 0.99 \[ \frac {B x^{6}}{6 c^{2}} - \frac {b^{2} \left (- 3 A c + 4 B b\right ) \log {\left (b + c x^{2} \right )}}{2 c^{5}} + x^{4} \left (\frac {A}{4 c^{2}} - \frac {B b}{2 c^{3}}\right ) + x^{2} \left (- \frac {A b}{c^{3}} + \frac {3 B b^{2}}{2 c^{4}}\right ) + \frac {A b^{3} c - B b^{4}}{2 b c^{5} + 2 c^{6} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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