\(\int \frac {x^5 (A+B x^2)}{(a+b x^2)^2} \, dx\) [75]

Optimal result

Integrand size = 20, antiderivative size = 82 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {(A b-2 a B) x^2}{2 b^3}+\frac {B x^4}{4 b^2}-\frac {a^2 (A b-a B)}{2 b^4 \left (a+b x^2\right )}-\frac {a (2 A b-3 a B) \log \left (a+b x^2\right )}{2 b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 82, 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} \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {a^2 (A b-a B)}{2 b^4 \left (a+b x^2\right )}-\frac {a (2 A b-3 a B) \log \left (a+b x^2\right )}{2 b^4}+\frac {x^2 (A b-2 a B)}{2 b^3}+\frac {B x^4}{4 b^2} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.88 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {2 b (A b-2 a B) x^2+b^2 B x^4+\frac {2 a^2 (-A b+a B)}{a+b x^2}+2 a (-2 A b+3 a B) \log \left (a+b x^2\right )}{4 b^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\left (b B \,x^{2}+A b -2 B a \right )^{2}}{4 b^{4} B}-\frac {a \left (\frac {\left (2 A b -3 B a \right ) \ln \left (b \,x^{2}+a \right )}{b}+\frac {a \left (A b -B a \right )}{b \left (b \,x^{2}+a \right )}\right )}{2 b^{3}}\)	\(76\)
norman	\(\frac {\frac {B \,x^{6}}{4 b}-\frac {a \left (2 a b A -3 a^{2} B \right )}{2 b^{4}}+\frac {\left (2 A b -3 B a \right ) x^{4}}{4 b^{2}}}{b \,x^{2}+a}-\frac {a \left (2 A b -3 B a \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\)	\(80\)
parallelrisch	\(-\frac {-b^{3} B \,x^{6}-2 A \,x^{4} b^{3}+3 B \,x^{4} a \,b^{2}+4 A \ln \left (b \,x^{2}+a \right ) x^{2} a \,b^{2}-6 B \ln \left (b \,x^{2}+a \right ) x^{2} a^{2} b +4 A \ln \left (b \,x^{2}+a \right ) a^{2} b -6 B \ln \left (b \,x^{2}+a \right ) a^{3}+4 a^{2} b A -6 a^{3} B}{4 b^{4} \left (b \,x^{2}+a \right )}\)	\(122\)
risch	\(\frac {B \,x^{4}}{4 b^{2}}+\frac {A \,x^{2}}{2 b^{2}}-\frac {B a \,x^{2}}{b^{3}}+\frac {A^{2}}{4 b^{2} B}-\frac {A a}{b^{3}}+\frac {B \,a^{2}}{b^{4}}-\frac {a^{2} A}{2 b^{3} \left (b \,x^{2}+a \right )}+\frac {a^{3} B}{2 b^{4} \left (b \,x^{2}+a \right )}-\frac {a \ln \left (b \,x^{2}+a \right ) A}{b^{3}}+\frac {3 a^{2} \ln \left (b \,x^{2}+a \right ) B}{2 b^{4}}\)	\(124\)

[In]

int(x^5*(B*x^2+A)/(b*x^2+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.48 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B b^{3} x^{6} - {\left (3 \, B a b^{2} - 2 \, A b^{3}\right )} x^{4} + 2 \, B a^{3} - 2 \, A a^{2} b - 2 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} x^{2} + 2 \, {\left (3 \, B a^{3} - 2 \, A a^{2} b + {\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (b^{5} x^{2} + a b^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.95 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B x^{4}}{4 b^{2}} + \frac {a \left (- 2 A b + 3 B a\right ) \log {\left (a + b x^{2} \right )}}{2 b^{4}} + x^{2} \left (\frac {A}{2 b^{2}} - \frac {B a}{b^{3}}\right ) + \frac {- A a^{2} b + B a^{3}}{2 a b^{4} + 2 b^{5} x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B a^{3} - A a^{2} b}{2 \, {\left (b^{5} x^{2} + a b^{4}\right )}} + \frac {B b x^{4} - 2 \, {\left (2 \, B a - A b\right )} x^{2}}{4 \, b^{3}} + \frac {{\left (3 \, B a^{2} - 2 \, A a b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.29 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (3 \, B a^{2} - 2 \, A a b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{4}} + \frac {B b^{2} x^{4} - 4 \, B a b x^{2} + 2 \, A b^{2} x^{2}}{4 \, b^{4}} - \frac {3 \, B a^{2} b x^{2} - 2 \, A a b^{2} x^{2} + 2 \, B a^{3} - A a^{2} b}{2 \, {\left (b x^{2} + a\right )} b^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=x^2\,\left (\frac {A}{2\,b^2}-\frac {B\,a}{b^3}\right )+\frac {\ln \left (b\,x^2+a\right )\,\left (3\,B\,a^2-2\,A\,a\,b\right )}{2\,b^4}+\frac {B\,x^4}{4\,b^2}+\frac {B\,a^3-A\,a^2\,b}{2\,b\,\left (b^4\,x^2+a\,b^3\right )} \]

[In]

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

[Out]

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