3.73 \(\int \frac{x^7 (A+B x^2)}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=104 \[ \frac{a^3 (A b-a B)}{2 b^5 \left (a+b x^2\right )}+\frac{a^2 (3 A b-4 a B) \log \left (a+b x^2\right )}{2 b^5}+\frac{x^4 (A b-2 a B)}{4 b^3}-\frac{a x^2 (2 A b-3 a B)}{2 b^4}+\frac{B x^6}{6 b^2} \]

[Out]

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

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Rubi [A]  time = 0.126427, antiderivative size = 104, 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} \[ \frac{a^3 (A b-a B)}{2 b^5 \left (a+b x^2\right )}+\frac{a^2 (3 A b-4 a B) \log \left (a+b x^2\right )}{2 b^5}+\frac{x^4 (A b-2 a B)}{4 b^3}-\frac{a x^2 (2 A b-3 a B)}{2 b^4}+\frac{B x^6}{6 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^7*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0641138, size = 93, normalized size = 0.89 \[ \frac{\frac{6 a^3 (A b-a B)}{a+b x^2}+6 a^2 (3 A b-4 a B) \log \left (a+b x^2\right )+3 b^2 x^4 (A b-2 a B)+6 a b x^2 (3 a B-2 A b)+2 b^3 B x^6}{12 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

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

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Maple [A]  time = 0.01, size = 122, normalized size = 1.2 \begin{align*}{\frac{B{x}^{6}}{6\,{b}^{2}}}+{\frac{A{x}^{4}}{4\,{b}^{2}}}-{\frac{B{x}^{4}a}{2\,{b}^{3}}}-{\frac{aA{x}^{2}}{{b}^{3}}}+{\frac{3\,B{x}^{2}{a}^{2}}{2\,{b}^{4}}}+{\frac{3\,{a}^{2}\ln \left ( b{x}^{2}+a \right ) A}{2\,{b}^{4}}}-2\,{\frac{{a}^{3}\ln \left ( b{x}^{2}+a \right ) B}{{b}^{5}}}+{\frac{{a}^{3}A}{2\,{b}^{4} \left ( b{x}^{2}+a \right ) }}-{\frac{B{a}^{4}}{2\,{b}^{5} \left ( b{x}^{2}+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(B*x^2+A)/(b*x^2+a)^2,x)

[Out]

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

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Maxima [A]  time = 0.987453, size = 144, normalized size = 1.38 \begin{align*} -\frac{B a^{4} - A a^{3} b}{2 \,{\left (b^{6} x^{2} + a b^{5}\right )}} + \frac{2 \, B b^{2} x^{6} - 3 \,{\left (2 \, B a b - A b^{2}\right )} x^{4} + 6 \,{\left (3 \, B a^{2} - 2 \, A a b\right )} x^{2}}{12 \, b^{4}} - \frac{{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.2057, size = 309, normalized size = 2.97 \begin{align*} \frac{2 \, B b^{4} x^{8} -{\left (4 \, B a b^{3} - 3 \, A b^{4}\right )} x^{6} - 6 \, B a^{4} + 6 \, A a^{3} b + 3 \,{\left (4 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + 6 \,{\left (3 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{2} - 6 \,{\left (4 \, B a^{4} - 3 \, A a^{3} b +{\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{12 \,{\left (b^{6} x^{2} + a b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.848989, size = 102, normalized size = 0.98 \begin{align*} \frac{B x^{6}}{6 b^{2}} - \frac{a^{2} \left (- 3 A b + 4 B a\right ) \log{\left (a + b x^{2} \right )}}{2 b^{5}} - \frac{- A a^{3} b + B a^{4}}{2 a b^{5} + 2 b^{6} x^{2}} - \frac{x^{4} \left (- A b + 2 B a\right )}{4 b^{3}} + \frac{x^{2} \left (- 2 A a b + 3 B a^{2}\right )}{2 b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(B*x**2+A)/(b*x**2+a)**2,x)

[Out]

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

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Giac [A]  time = 1.1271, size = 182, normalized size = 1.75 \begin{align*} -\frac{{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{5}} + \frac{2 \, B b^{4} x^{6} - 6 \, B a b^{3} x^{4} + 3 \, A b^{4} x^{4} + 18 \, B a^{2} b^{2} x^{2} - 12 \, A a b^{3} x^{2}}{12 \, b^{6}} + \frac{4 \, B a^{3} b x^{2} - 3 \, A a^{2} b^{2} x^{2} + 3 \, B a^{4} - 2 \, A a^{3} b}{2 \,{\left (b x^{2} + a\right )} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

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