\(\int \frac {x^9 (A+B x^2)}{(a+b x^2)^3} \, dx\) [89]

Optimal result

Integrand size = 20, antiderivative size = 128 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {3 a (A b-2 a B) x^2}{2 b^5}+\frac {(A b-3 a B) x^4}{4 b^4}+\frac {B x^6}{6 b^3}-\frac {a^4 (A b-a B)}{4 b^6 \left (a+b x^2\right )^2}+\frac {a^3 (4 A b-5 a B)}{2 b^6 \left (a+b x^2\right )}+\frac {a^2 (3 A b-5 a B) \log \left (a+b x^2\right )}{b^6} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {457, 78} \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {a^4 (A b-a B)}{4 b^6 \left (a+b x^2\right )^2}+\frac {a^3 (4 A b-5 a B)}{2 b^6 \left (a+b x^2\right )}+\frac {a^2 (3 A b-5 a B) \log \left (a+b x^2\right )}{b^6}-\frac {3 a x^2 (A b-2 a B)}{2 b^5}+\frac {x^4 (A b-3 a B)}{4 b^4}+\frac {B x^6}{6 b^3} \]

[In]

Int[(x^9*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

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

Rule 78

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 457

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.91 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {18 a b (-A b+2 a B) x^2+3 b^2 (A b-3 a B) x^4+2 b^3 B x^6+\frac {3 a^4 (-A b+a B)}{\left (a+b x^2\right )^2}+\frac {6 a^3 (4 A b-5 a B)}{a+b x^2}+12 a^2 (3 A b-5 a B) \log \left (a+b x^2\right )}{12 b^6} \]

[In]

Integrate[(x^9*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

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

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.98

method	result	size
norman	\(\frac {\frac {B \,x^{10}}{6 b}+\frac {a^{2} \left (9 a^{2} b A -15 a^{3} B \right )}{2 b^{6}}+\frac {\left (3 A b -5 B a \right ) x^{8}}{12 b^{2}}-\frac {a \left (3 A b -5 B a \right ) x^{6}}{3 b^{3}}+\frac {2 a \left (3 a^{2} b A -5 a^{3} B \right ) x^{2}}{b^{5}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {a^{2} \left (3 A b -5 B a \right ) \ln \left (b \,x^{2}+a \right )}{b^{6}}\)	\(126\)
default	\(-\frac {-\frac {b^{2} B \,x^{6}}{6}+\frac {\left (-b^{2} A +3 a b B \right ) x^{4}}{4}+\frac {\left (3 a b A -6 a^{2} B \right ) x^{2}}{2}}{b^{5}}+\frac {a^{2} \left (\frac {\left (6 A b -10 B a \right ) \ln \left (b \,x^{2}+a \right )}{b}-\frac {a^{2} \left (A b -B a \right )}{2 b \left (b \,x^{2}+a \right )^{2}}+\frac {a \left (4 A b -5 B a \right )}{b \left (b \,x^{2}+a \right )}\right )}{2 b^{5}}\)	\(129\)
risch	\(\frac {B \,x^{6}}{6 b^{3}}+\frac {A \,x^{4}}{4 b^{3}}-\frac {3 B a \,x^{4}}{4 b^{4}}-\frac {3 a A \,x^{2}}{2 b^{4}}+\frac {3 a^{2} B \,x^{2}}{b^{5}}+\frac {\left (2 A \,a^{3} b -\frac {5}{2} B \,a^{4}\right ) x^{2}+\frac {a^{4} \left (7 A b -9 B a \right )}{4 b}}{b^{5} \left (b \,x^{2}+a \right )^{2}}+\frac {3 a^{2} \ln \left (b \,x^{2}+a \right ) A}{b^{5}}-\frac {5 a^{3} \ln \left (b \,x^{2}+a \right ) B}{b^{6}}\)	\(135\)
parallelrisch	\(\frac {2 b^{5} B \,x^{10}+3 A \,b^{5} x^{8}-5 B a \,b^{4} x^{8}-12 A a \,b^{4} x^{6}+20 B \,a^{2} b^{3} x^{6}+36 A \ln \left (b \,x^{2}+a \right ) x^{4} a^{2} b^{3}-60 B \ln \left (b \,x^{2}+a \right ) x^{4} a^{3} b^{2}+72 A \ln \left (b \,x^{2}+a \right ) x^{2} a^{3} b^{2}-120 B \ln \left (b \,x^{2}+a \right ) x^{2} a^{4} b +72 a^{3} A \,b^{2} x^{2}-120 B \,a^{4} b \,x^{2}+36 A \ln \left (b \,x^{2}+a \right ) a^{4} b -60 B \ln \left (b \,x^{2}+a \right ) a^{5}+54 a^{4} b A -90 a^{5} B}{12 b^{6} \left (b \,x^{2}+a \right )^{2}}\)	\(208\)

[In]

int(x^9*(B*x^2+A)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

(1/6*B/b*x^10+1/2*a^2*(9*A*a^2*b-15*B*a^3)/b^6+1/12*(3*A*b-5*B*a)/b^2*x^8-1/3*a*(3*A*b-5*B*a)/b^3*x^6+2*a*(3*A
*a^2*b-5*B*a^3)/b^5*x^2)/(b*x^2+a)^2+a^2*(3*A*b-5*B*a)*ln(b*x^2+a)/b^6

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.60 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {2 \, B b^{5} x^{10} - {\left (5 \, B a b^{4} - 3 \, A b^{5}\right )} x^{8} + 4 \, {\left (5 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{6} - 27 \, B a^{5} + 21 \, A a^{4} b + 3 \, {\left (21 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{4} + 6 \, {\left (B a^{4} b + A a^{3} b^{2}\right )} x^{2} - 12 \, {\left (5 \, B a^{5} - 3 \, A a^{4} b + {\left (5 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{4} + 2 \, {\left (5 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{8} x^{4} + 2 \, a b^{7} x^{2} + a^{2} b^{6}\right )}} \]

[In]

integrate(x^9*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

1/12*(2*B*b^5*x^10 - (5*B*a*b^4 - 3*A*b^5)*x^8 + 4*(5*B*a^2*b^3 - 3*A*a*b^4)*x^6 - 27*B*a^5 + 21*A*a^4*b + 3*(
21*B*a^3*b^2 - 11*A*a^2*b^3)*x^4 + 6*(B*a^4*b + A*a^3*b^2)*x^2 - 12*(5*B*a^5 - 3*A*a^4*b + (5*B*a^3*b^2 - 3*A*
a^2*b^3)*x^4 + 2*(5*B*a^4*b - 3*A*a^3*b^2)*x^2)*log(b*x^2 + a))/(b^8*x^4 + 2*a*b^7*x^2 + a^2*b^6)

Sympy [A] (verification not implemented)

Time = 0.86 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.12 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B x^{6}}{6 b^{3}} - \frac {a^{2} \left (- 3 A b + 5 B a\right ) \log {\left (a + b x^{2} \right )}}{b^{6}} + x^{4} \left (\frac {A}{4 b^{3}} - \frac {3 B a}{4 b^{4}}\right ) + x^{2} \left (- \frac {3 A a}{2 b^{4}} + \frac {3 B a^{2}}{b^{5}}\right ) + \frac {7 A a^{4} b - 9 B a^{5} + x^{2} \cdot \left (8 A a^{3} b^{2} - 10 B a^{4} b\right )}{4 a^{2} b^{6} + 8 a b^{7} x^{2} + 4 b^{8} x^{4}} \]

[In]

integrate(x**9*(B*x**2+A)/(b*x**2+a)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.10 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {9 \, B a^{5} - 7 \, A a^{4} b + 2 \, {\left (5 \, B a^{4} b - 4 \, A a^{3} b^{2}\right )} x^{2}}{4 \, {\left (b^{8} x^{4} + 2 \, a b^{7} x^{2} + a^{2} b^{6}\right )}} + \frac {2 \, B b^{2} x^{6} - 3 \, {\left (3 \, B a b - A b^{2}\right )} x^{4} + 18 \, {\left (2 \, B a^{2} - A a b\right )} x^{2}}{12 \, b^{5}} - \frac {{\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (b x^{2} + a\right )}{b^{6}} \]

[In]

integrate(x^9*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

-1/4*(9*B*a^5 - 7*A*a^4*b + 2*(5*B*a^4*b - 4*A*a^3*b^2)*x^2)/(b^8*x^4 + 2*a*b^7*x^2 + a^2*b^6) + 1/12*(2*B*b^2
*x^6 - 3*(3*B*a*b - A*b^2)*x^4 + 18*(2*B*a^2 - A*a*b)*x^2)/b^5 - (5*B*a^3 - 3*A*a^2*b)*log(b*x^2 + a)/b^6

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.24 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {{\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{b^{6}} + \frac {30 \, B a^{3} b^{2} x^{4} - 18 \, A a^{2} b^{3} x^{4} + 50 \, B a^{4} b x^{2} - 28 \, A a^{3} b^{2} x^{2} + 21 \, B a^{5} - 11 \, A a^{4} b}{4 \, {\left (b x^{2} + a\right )}^{2} b^{6}} + \frac {2 \, B b^{6} x^{6} - 9 \, B a b^{5} x^{4} + 3 \, A b^{6} x^{4} + 36 \, B a^{2} b^{4} x^{2} - 18 \, A a b^{5} x^{2}}{12 \, b^{9}} \]

[In]

integrate(x^9*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

-(5*B*a^3 - 3*A*a^2*b)*log(abs(b*x^2 + a))/b^6 + 1/4*(30*B*a^3*b^2*x^4 - 18*A*a^2*b^3*x^4 + 50*B*a^4*b*x^2 - 2
8*A*a^3*b^2*x^2 + 21*B*a^5 - 11*A*a^4*b)/((b*x^2 + a)^2*b^6) + 1/12*(2*B*b^6*x^6 - 9*B*a*b^5*x^4 + 3*A*b^6*x^4
+ 36*B*a^2*b^4*x^2 - 18*A*a*b^5*x^2)/b^9

Mupad [B] (verification not implemented)

Time = 5.15 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.21 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=x^4\,\left (\frac {A}{4\,b^3}-\frac {3\,B\,a}{4\,b^4}\right )-\frac {\frac {9\,B\,a^5-7\,A\,a^4\,b}{4\,b}+x^2\,\left (\frac {5\,B\,a^4}{2}-2\,A\,a^3\,b\right )}{a^2\,b^5+2\,a\,b^6\,x^2+b^7\,x^4}-x^2\,\left (\frac {3\,a\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )}{2\,b}+\frac {3\,B\,a^2}{2\,b^5}\right )+\frac {B\,x^6}{6\,b^3}-\frac {\ln \left (b\,x^2+a\right )\,\left (5\,B\,a^3-3\,A\,a^2\,b\right )}{b^6} \]

[In]

int((x^9*(A + B*x^2))/(a + b*x^2)^3,x)

[Out]

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