3.91 \(\int \frac{x^5 (A+B x^2)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=88 \[ -\frac{a^2 (A b-a B)}{4 b^4 \left (a+b x^2\right )^2}+\frac{a (2 A b-3 a B)}{2 b^4 \left (a+b x^2\right )}+\frac{(A b-3 a B) \log \left (a+b x^2\right )}{2 b^4}+\frac{B x^2}{2 b^3} \]

[Out]

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

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Rubi [A]  time = 0.09024, antiderivative size = 88, 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} \[ -\frac{a^2 (A b-a B)}{4 b^4 \left (a+b x^2\right )^2}+\frac{a (2 A b-3 a B)}{2 b^4 \left (a+b x^2\right )}+\frac{(A b-3 a B) \log \left (a+b x^2\right )}{2 b^4}+\frac{B x^2}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

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

Mathematica [A]  time = 0.0368467, size = 92, normalized size = 1.05 \[ \frac{2 a A b-3 a^2 B}{2 b^4 \left (a+b x^2\right )}+\frac{a^3 B-a^2 A b}{4 b^4 \left (a+b x^2\right )^2}+\frac{(A b-3 a B) \log \left (a+b x^2\right )}{2 b^4}+\frac{B x^2}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

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

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Maple [A]  time = 0.009, size = 109, normalized size = 1.2 \begin{align*}{\frac{B{x}^{2}}{2\,{b}^{3}}}-{\frac{{a}^{2}A}{4\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{B{a}^{3}}{4\,{b}^{4} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{\ln \left ( b{x}^{2}+a \right ) A}{2\,{b}^{3}}}-{\frac{3\,\ln \left ( b{x}^{2}+a \right ) Ba}{2\,{b}^{4}}}+{\frac{aA}{{b}^{3} \left ( b{x}^{2}+a \right ) }}-{\frac{3\,{a}^{2}B}{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)^3,x)

[Out]

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

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Maxima [A]  time = 1.0209, size = 127, normalized size = 1.44 \begin{align*} -\frac{5 \, B a^{3} - 3 \, A a^{2} b + 2 \,{\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{2}}{4 \,{\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} + \frac{B x^{2}}{2 \, b^{3}} - \frac{{\left (3 \, B 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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.16418, size = 289, normalized size = 3.28 \begin{align*} \frac{2 \, B b^{3} x^{6} + 4 \, B a b^{2} x^{4} - 5 \, B a^{3} + 3 \, A a^{2} b - 4 \,{\left (B a^{2} b - A a b^{2}\right )} x^{2} - 2 \,{\left ({\left (3 \, B a b^{2} - A b^{3}\right )} x^{4} + 3 \, B a^{3} - A a^{2} b + 2 \,{\left (3 \, B a^{2} b - A a b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{4 \,{\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} 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)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.44463, size = 94, normalized size = 1.07 \begin{align*} \frac{B x^{2}}{2 b^{3}} - \frac{- 3 A a^{2} b + 5 B a^{3} + x^{2} \left (- 4 A a b^{2} + 6 B a^{2} b\right )}{4 a^{2} b^{4} + 8 a b^{5} x^{2} + 4 b^{6} x^{4}} - \frac{\left (- A b + 3 B a\right ) \log{\left (a + b x^{2} \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)**3,x)

[Out]

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

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Giac [A]  time = 1.14964, size = 126, normalized size = 1.43 \begin{align*} \frac{B x^{2}}{2 \, b^{3}} - \frac{{\left (3 \, B a - A b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{4}} + \frac{9 \, B a b^{2} x^{4} - 3 \, A b^{3} x^{4} + 12 \, B a^{2} b x^{2} - 2 \, A a b^{2} x^{2} + 4 \, B a^{3}}{4 \,{\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)^3,x, algorithm="giac")

[Out]

1/2*B*x^2/b^3 - 1/2*(3*B*a - A*b)*log(abs(b*x^2 + a))/b^4 + 1/4*(9*B*a*b^2*x^4 - 3*A*b^3*x^4 + 12*B*a^2*b*x^2
- 2*A*a*b^2*x^2 + 4*B*a^3)/((b*x^2 + a)^2*b^4)