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

Optimal. Leaf size=82 \[ -\frac{a^2 (A b-a B)}{2 b^4 \left (a+b x^2\right )}+\frac{x^2 (A b-2 a B)}{2 b^3}-\frac{a (2 A b-3 a B) \log \left (a+b x^2\right )}{2 b^4}+\frac{B x^4}{4 b^2} \]

[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)

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Rubi [A]  time = 0.0873322, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {446, 77} \[ -\frac{a^2 (A b-a B)}{2 b^4 \left (a+b x^2\right )}+\frac{x^2 (A b-2 a B)}{2 b^3}-\frac{a (2 A b-3 a B) \log \left (a+b x^2\right )}{2 b^4}+\frac{B x^4}{4 b^2} \]

Antiderivative was successfully verified.

[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 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 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]))))

Rubi steps

\begin{align*} \int \frac{x^5 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2 (A+B x)}{(a+b x)^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{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]  time = 0.0645405, size = 72, normalized size = 0.88 \[ \frac{\frac{2 a^2 (a B-A b)}{a+b x^2}+2 b x^2 (A b-2 a B)+2 a (3 a B-2 A b) \log \left (a+b x^2\right )+b^2 B x^4}{4 b^4} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.01, size = 98, normalized size = 1.2 \begin{align*}{\frac{B{x}^{4}}{4\,{b}^{2}}}+{\frac{A{x}^{2}}{2\,{b}^{2}}}-{\frac{B{x}^{2}a}{{b}^{3}}}-{\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}}}-{\frac{{a}^{2}A}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{B{a}^{3}}{2\,{b}^{4} \left ( b{x}^{2}+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01722, size = 111, normalized size = 1.35 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.17782, size = 251, normalized size = 3.06 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.798151, size = 78, normalized size = 0.95 \begin{align*} \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}} + \frac{- A a^{2} b + B a^{3}}{2 a b^{4} + 2 b^{5} x^{2}} - \frac{x^{2} \left (- A b + 2 B a\right )}{2 b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Giac [A]  time = 1.16731, size = 143, normalized size = 1.74 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)