3.1.88 \(\int \frac {x^{11} (A+B x^2)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=150 \[ \frac {a^5 (A b-a B)}{4 b^7 \left (a+b x^2\right )^2}-\frac {a^4 (5 A b-6 a B)}{2 b^7 \left (a+b x^2\right )}-\frac {5 a^3 (2 A b-3 a B) \log \left (a+b x^2\right )}{2 b^7}+\frac {a^2 x^2 (3 A b-5 a B)}{b^6}-\frac {3 a x^4 (A b-2 a B)}{4 b^5}+\frac {x^6 (A b-3 a B)}{6 b^4}+\frac {B x^8}{8 b^3} \]

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Rubi [A]  time = 0.23, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {446, 77} \begin {gather*} \frac {a^2 x^2 (3 A b-5 a B)}{b^6}-\frac {a^4 (5 A b-6 a B)}{2 b^7 \left (a+b x^2\right )}+\frac {a^5 (A b-a B)}{4 b^7 \left (a+b x^2\right )^2}-\frac {5 a^3 (2 A b-3 a B) \log \left (a+b x^2\right )}{2 b^7}+\frac {x^6 (A b-3 a B)}{6 b^4}-\frac {3 a x^4 (A b-2 a B)}{4 b^5}+\frac {B x^8}{8 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(3*A*b - 5*a*B)*x^2)/b^6 - (3*a*(A*b - 2*a*B)*x^4)/(4*b^5) + ((A*b - 3*a*B)*x^6)/(6*b^4) + (B*x^8)/(8*b^3
) + (a^5*(A*b - a*B))/(4*b^7*(a + b*x^2)^2) - (a^4*(5*A*b - 6*a*B))/(2*b^7*(a + b*x^2)) - (5*a^3*(2*A*b - 3*a*
B)*Log[a + b*x^2])/(2*b^7)

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]]

Rubi steps

\begin {align*} \int \frac {x^{11} \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^5 (A+B x)}{(a+b x)^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {2 a^2 (-3 A b+5 a B)}{b^6}+\frac {3 a (-A b+2 a B) x}{b^5}+\frac {(A b-3 a B) x^2}{b^4}+\frac {B x^3}{b^3}+\frac {a^5 (-A b+a B)}{b^6 (a+b x)^3}-\frac {a^4 (-5 A b+6 a B)}{b^6 (a+b x)^2}+\frac {5 a^3 (-2 A b+3 a B)}{b^6 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=\frac {a^2 (3 A b-5 a B) x^2}{b^6}-\frac {3 a (A b-2 a B) x^4}{4 b^5}+\frac {(A b-3 a B) x^6}{6 b^4}+\frac {B x^8}{8 b^3}+\frac {a^5 (A b-a B)}{4 b^7 \left (a+b x^2\right )^2}-\frac {a^4 (5 A b-6 a B)}{2 b^7 \left (a+b x^2\right )}-\frac {5 a^3 (2 A b-3 a B) \log \left (a+b x^2\right )}{2 b^7}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 136, normalized size = 0.91 \begin {gather*} \frac {\frac {6 a^5 (A b-a B)}{\left (a+b x^2\right )^2}+\frac {12 a^4 (6 a B-5 A b)}{a+b x^2}+60 a^3 (3 a B-2 A b) \log \left (a+b x^2\right )-24 a^2 b x^2 (5 a B-3 A b)+4 b^3 x^6 (A b-3 a B)+18 a b^2 x^4 (2 a B-A b)+3 b^4 B x^8}{24 b^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*a^2*b*(-3*A*b + 5*a*B)*x^2 + 18*a*b^2*(-(A*b) + 2*a*B)*x^4 + 4*b^3*(A*b - 3*a*B)*x^6 + 3*b^4*B*x^8 + (6*a
^5*(A*b - a*B))/(a + b*x^2)^2 + (12*a^4*(-5*A*b + 6*a*B))/(a + b*x^2) + 60*a^3*(-2*A*b + 3*a*B)*Log[a + b*x^2]
)/(24*b^7)

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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 (a+b x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 231, normalized size = 1.54 \begin {gather*} \frac {3 \, B b^{6} x^{12} - 2 \, {\left (3 \, B a b^{5} - 2 \, A b^{6}\right )} x^{10} + 5 \, {\left (3 \, B a^{2} b^{4} - 2 \, A a b^{5}\right )} x^{8} + 66 \, B a^{6} - 54 \, A a^{5} b - 20 \, {\left (3 \, B a^{3} b^{3} - 2 \, A a^{2} b^{4}\right )} x^{6} - 6 \, {\left (34 \, B a^{4} b^{2} - 21 \, A a^{3} b^{3}\right )} x^{4} - 12 \, {\left (4 \, B a^{5} b - A a^{4} b^{2}\right )} x^{2} + 60 \, {\left (3 \, B a^{6} - 2 \, A a^{5} b + {\left (3 \, B a^{4} b^{2} - 2 \, A a^{3} b^{3}\right )} x^{4} + 2 \, {\left (3 \, B a^{5} b - 2 \, A a^{4} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{24 \, {\left (b^{9} x^{4} + 2 \, a b^{8} x^{2} + a^{2} b^{7}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*B*b^6*x^12 - 2*(3*B*a*b^5 - 2*A*b^6)*x^10 + 5*(3*B*a^2*b^4 - 2*A*a*b^5)*x^8 + 66*B*a^6 - 54*A*a^5*b -
20*(3*B*a^3*b^3 - 2*A*a^2*b^4)*x^6 - 6*(34*B*a^4*b^2 - 21*A*a^3*b^3)*x^4 - 12*(4*B*a^5*b - A*a^4*b^2)*x^2 + 60
*(3*B*a^6 - 2*A*a^5*b + (3*B*a^4*b^2 - 2*A*a^3*b^3)*x^4 + 2*(3*B*a^5*b - 2*A*a^4*b^2)*x^2)*log(b*x^2 + a))/(b^
9*x^4 + 2*a*b^8*x^2 + a^2*b^7)

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giac [A]  time = 0.40, size = 183, normalized size = 1.22 \begin {gather*} \frac {5 \, {\left (3 \, B a^{4} - 2 \, A a^{3} b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{7}} - \frac {45 \, B a^{4} b^{2} x^{4} - 30 \, A a^{3} b^{3} x^{4} + 78 \, B a^{5} b x^{2} - 50 \, A a^{4} b^{2} x^{2} + 34 \, B a^{6} - 21 \, A a^{5} b}{4 \, {\left (b x^{2} + a\right )}^{2} b^{7}} + \frac {3 \, B b^{9} x^{8} - 12 \, B a b^{8} x^{6} + 4 \, A b^{9} x^{6} + 36 \, B a^{2} b^{7} x^{4} - 18 \, A a b^{8} x^{4} - 120 \, B a^{3} b^{6} x^{2} + 72 \, A a^{2} b^{7} x^{2}}{24 \, b^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*(3*B*a^4 - 2*A*a^3*b)*log(abs(b*x^2 + a))/b^7 - 1/4*(45*B*a^4*b^2*x^4 - 30*A*a^3*b^3*x^4 + 78*B*a^5*b*x^2
- 50*A*a^4*b^2*x^2 + 34*B*a^6 - 21*A*a^5*b)/((b*x^2 + a)^2*b^7) + 1/24*(3*B*b^9*x^8 - 12*B*a*b^8*x^6 + 4*A*b^9
*x^6 + 36*B*a^2*b^7*x^4 - 18*A*a*b^8*x^4 - 120*B*a^3*b^6*x^2 + 72*A*a^2*b^7*x^2)/b^12

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maple [A]  time = 0.02, size = 182, normalized size = 1.21 \begin {gather*} \frac {B \,x^{8}}{8 b^{3}}+\frac {A \,x^{6}}{6 b^{3}}-\frac {B a \,x^{6}}{2 b^{4}}-\frac {3 A a \,x^{4}}{4 b^{4}}+\frac {3 B \,a^{2} x^{4}}{2 b^{5}}+\frac {A \,a^{5}}{4 \left (b \,x^{2}+a \right )^{2} b^{6}}+\frac {3 A \,a^{2} x^{2}}{b^{5}}-\frac {B \,a^{6}}{4 \left (b \,x^{2}+a \right )^{2} b^{7}}-\frac {5 B \,a^{3} x^{2}}{b^{6}}-\frac {5 A \,a^{4}}{2 \left (b \,x^{2}+a \right ) b^{6}}-\frac {5 A \,a^{3} \ln \left (b \,x^{2}+a \right )}{b^{6}}+\frac {3 B \,a^{5}}{\left (b \,x^{2}+a \right ) b^{7}}+\frac {15 B \,a^{4} \ln \left (b \,x^{2}+a \right )}{2 b^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*x^8/b^3+1/6/b^3*x^6*A-1/2/b^4*x^6*B*a-3/4/b^4*x^4*a*A+3/2/b^5*x^4*a^2*B+3/b^5*A*x^2*a^2-5/b^6*B*x^2*a^3-
5/2*a^4/b^6/(b*x^2+a)*A+3*a^5/b^7/(b*x^2+a)*B+1/4*a^5/b^6/(b*x^2+a)^2*A-1/4*a^6/b^7/(b*x^2+a)^2*B-5*a^3/b^6*ln
(b*x^2+a)*A+15/2*a^4/b^7*ln(b*x^2+a)*B

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maxima [A]  time = 1.12, size = 165, normalized size = 1.10 \begin {gather*} \frac {11 \, B a^{6} - 9 \, A a^{5} b + 2 \, {\left (6 \, B a^{5} b - 5 \, A a^{4} b^{2}\right )} x^{2}}{4 \, {\left (b^{9} x^{4} + 2 \, a b^{8} x^{2} + a^{2} b^{7}\right )}} + \frac {3 \, B b^{3} x^{8} - 4 \, {\left (3 \, B a b^{2} - A b^{3}\right )} x^{6} + 18 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x^{4} - 24 \, {\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} x^{2}}{24 \, b^{6}} + \frac {5 \, {\left (3 \, B a^{4} - 2 \, A a^{3} b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(11*B*a^6 - 9*A*a^5*b + 2*(6*B*a^5*b - 5*A*a^4*b^2)*x^2)/(b^9*x^4 + 2*a*b^8*x^2 + a^2*b^7) + 1/24*(3*B*b^3
*x^8 - 4*(3*B*a*b^2 - A*b^3)*x^6 + 18*(2*B*a^2*b - A*a*b^2)*x^4 - 24*(5*B*a^3 - 3*A*a^2*b)*x^2)/b^6 + 5/2*(3*B
*a^4 - 2*A*a^3*b)*log(b*x^2 + a)/b^7

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mupad [B]  time = 0.12, size = 225, normalized size = 1.50 \begin {gather*} \frac {\frac {11\,B\,a^6-9\,A\,a^5\,b}{4\,b}+x^2\,\left (3\,B\,a^5-\frac {5\,A\,a^4\,b}{2}\right )}{a^2\,b^6+2\,a\,b^7\,x^2+b^8\,x^4}-x^2\,\left (\frac {B\,a^3}{2\,b^6}-\frac {3\,a\,\left (\frac {3\,a\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )}{b}+\frac {3\,B\,a^2}{b^5}\right )}{2\,b}+\frac {3\,a^2\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )}{2\,b^2}\right )+x^6\,\left (\frac {A}{6\,b^3}-\frac {B\,a}{2\,b^4}\right )-x^4\,\left (\frac {3\,a\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )}{4\,b}+\frac {3\,B\,a^2}{4\,b^5}\right )+\frac {B\,x^8}{8\,b^3}+\frac {\ln \left (b\,x^2+a\right )\,\left (15\,B\,a^4-10\,A\,a^3\,b\right )}{2\,b^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((11*B*a^6 - 9*A*a^5*b)/(4*b) + x^2*(3*B*a^5 - (5*A*a^4*b)/2))/(a^2*b^6 + b^8*x^4 + 2*a*b^7*x^2) - x^2*((B*a^3
)/(2*b^6) - (3*a*((3*a*(A/b^3 - (3*B*a)/b^4))/b + (3*B*a^2)/b^5))/(2*b) + (3*a^2*(A/b^3 - (3*B*a)/b^4))/(2*b^2
)) + x^6*(A/(6*b^3) - (B*a)/(2*b^4)) - x^4*((3*a*(A/b^3 - (3*B*a)/b^4))/(4*b) + (3*B*a^2)/(4*b^5)) + (B*x^8)/(
8*b^3) + (log(a + b*x^2)*(15*B*a^4 - 10*A*a^3*b))/(2*b^7)

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sympy [A]  time = 1.61, size = 170, normalized size = 1.13 \begin {gather*} \frac {B x^{8}}{8 b^{3}} + \frac {5 a^{3} \left (- 2 A b + 3 B a\right ) \log {\left (a + b x^{2} \right )}}{2 b^{7}} + x^{6} \left (\frac {A}{6 b^{3}} - \frac {B a}{2 b^{4}}\right ) + x^{4} \left (- \frac {3 A a}{4 b^{4}} + \frac {3 B a^{2}}{2 b^{5}}\right ) + x^{2} \left (\frac {3 A a^{2}}{b^{5}} - \frac {5 B a^{3}}{b^{6}}\right ) + \frac {- 9 A a^{5} b + 11 B a^{6} + x^{2} \left (- 10 A a^{4} b^{2} + 12 B a^{5} b\right )}{4 a^{2} b^{7} + 8 a b^{8} x^{2} + 4 b^{9} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x**8/(8*b**3) + 5*a**3*(-2*A*b + 3*B*a)*log(a + b*x**2)/(2*b**7) + x**6*(A/(6*b**3) - B*a/(2*b**4)) + x**4*(
-3*A*a/(4*b**4) + 3*B*a**2/(2*b**5)) + x**2*(3*A*a**2/b**5 - 5*B*a**3/b**6) + (-9*A*a**5*b + 11*B*a**6 + x**2*
(-10*A*a**4*b**2 + 12*B*a**5*b))/(4*a**2*b**7 + 8*a*b**8*x**2 + 4*b**9*x**4)

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