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

Optimal. Leaf size=60 \[ \frac {a (A b-a B)}{2 b^3 \left (a+b x^2\right )}+\frac {(A b-2 a B) \log \left (a+b x^2\right )}{2 b^3}+\frac {B x^2}{2 b^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {446, 77} \begin {gather*} \frac {a (A b-a B)}{2 b^3 \left (a+b x^2\right )}+\frac {(A b-2 a B) \log \left (a+b x^2\right )}{2 b^3}+\frac {B x^2}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))))

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]]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 50, normalized size = 0.83 \begin {gather*} \frac {\frac {a (A b-a B)}{a+b x^2}+(A b-2 a B) \log \left (a+b x^2\right )+b B x^2}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 81, normalized size = 1.35 \begin {gather*} \frac {B b^{2} x^{4} + B a b x^{2} - B a^{2} + A a b - {\left (2 \, B a^{2} - A a b + {\left (2 \, B a b - A b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.34, size = 91, normalized size = 1.52 \begin {gather*} \frac {\frac {{\left (b x^{2} + a\right )} B}{b^{2}} + \frac {{\left (2 \, B a - A b\right )} \log \left (\frac {{\left | b x^{2} + a \right |}}{{\left (b x^{2} + a\right )}^{2} {\left | b \right |}}\right )}{b^{2}} - \frac {\frac {B a^{2} b}{b x^{2} + a} - \frac {A a b^{2}}{b x^{2} + a}}{b^{3}}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 74, normalized size = 1.23 \begin {gather*} \frac {B \,x^{2}}{2 b^{2}}+\frac {A a}{2 \left (b \,x^{2}+a \right ) b^{2}}+\frac {A \ln \left (b \,x^{2}+a \right )}{2 b^{2}}-\frac {B \,a^{2}}{2 \left (b \,x^{2}+a \right ) b^{3}}-\frac {B a \ln \left (b \,x^{2}+a \right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.03, size = 60, normalized size = 1.00 \begin {gather*} \frac {B x^{2}}{2 \, b^{2}} - \frac {B a^{2} - A a b}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} - \frac {{\left (2 \, B a - A b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 62, normalized size = 1.03 \begin {gather*} \frac {B\,x^2}{2\,b^2}+\frac {\ln \left (b\,x^2+a\right )\,\left (A\,b-2\,B\,a\right )}{2\,b^3}-\frac {B\,a^2-A\,a\,b}{2\,b\,\left (b^3\,x^2+a\,b^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.57, size = 56, normalized size = 0.93 \begin {gather*} \frac {B x^{2}}{2 b^{2}} + \frac {A a b - B a^{2}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {\left (- A b + 2 B a\right ) \log {\left (a + b x^{2} \right )}}{2 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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